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%Revised Feb. 4, 1999

\redefine\t{\tau}
\redefine\o{\omega}
\redefine\i{\infty}
\define\df{\dfrac}
\define\tf{\tfrac}
\define\s{\sigma}
\define\n{\nolimits_}
\define\Sh{\hbox{\cyr Sh}}
\define\Lim{{\lim_{n \rightarrow \infty}}}
\define\Z{{\Bbb{Z}}}
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  %\define\rom#1{{\rm #1}}
 
\topmatter
\title Ramanujan's Unpublished Manuscript on the Partition and Tau Functions 
 with Proofs and Commentary
\endtitle
\author Bruce C\. Berndt \thanks{The first author thanks the National 
Security Agency for its generous support.}\endthanks and Ken Ono 
\thanks{The second author thanks the National Science Foundation and the 
National Security Agency for their generous support.}\endthanks \endauthor
\address Department of Mathematics, University of Illinois, 1409 West 
Green St., Urbana, IL 61801, USA\endaddress
\email berndt\@math.uiuc.edu \endemail
\address Department of Mathematics, Pennsylvania State University, 
State College, PA 16802, USA\endaddress
\email ono\@math.psu.edu \endemail
\dedicatory Dedicated to our good friend George Andrews on his 60th birthday 
\enddedicatory
\endtopmatter

\rightheadtext{$p(n) \text{ and } \t(n)$}

\document

\centerline{\bf Introduction}

\bigskip

When Ramanujan died in 1920, he left behind an incomplete, unpublished 
manuscript in two parts on the partition function $ p(n) $ and, 
in contemporary terminology, 
Ramanujan's tau-function $ \tau(n)$. The first part, beginning with the Roman
numeral I, is written on 43 pages, 
with the last nine comprising material for insertion in the foregoing part of the  
manuscript.  G\. H\. Hardy extracted a portion of Part I providing 
proofs of Ramanujan's congruences for $ p(n)$ modulo 5, 7, and 11  and 
published it in 1921 \cite{80}, \cite{82, pp\. 232--238}
 under Ramanujan's name. In a footnote, Hardy remarks, ``The manuscript 
contains a large number of further results.  It is very incomplete, and 
will require very careful editing before it can be published in full. I 
have taken from it the three simplest and most striking results, \dots ."  
In 1952, J\. M\. Rushforth \cite{89} published several further 
results, mostly on $ \tau(n), $ from Part I. In 1977, 
R\. A\. Rankin \cite{85} discussed several congruences for 
$ \tau(n) $ found in Part I.  Part II has not been discussed in the
literature. Part I was not made available 
to the public until 1988 when it was photocopied in its original handwritten 
form and published with  Ramanujan's lost notebook \cite{83}. The
existence of Part II was first pointed out by B\. J\. Birch 
\cite{26} in 1975, but, like Part I, it
also was  hidden from the public until 1988, when a handwritten copy
made by G\. N\. Watson was photocopied for \cite{83}.  
Several theorems and proofs in this manuscript had not previously appeared 
before 1988. 

The manuscript arises from the last three years of Ramanujan's life.  It may 
have been written in nursing homes and sanitariums in 1917--1919, when we
know, from letters that Ramanujan wrote to Hardy during this time 
\cite{25, pp\. 192--193}, that Ramanujan was thinking deeply 
about partitions, or, more likely, it may have been written in India during 
the last year 
of his life.  According to Rushforth \cite{89}, the manuscript was
sent to Hardy by ``Ramanujan a few months before the latter's death in 1920."
If this is true, then it probably was enclosed with Ramanujan's last letter
to Hardy, dated January 12, 1920 \cite{25, pp\. 220--223}. 
There is no mention of the manuscript in the extant portion of the letter;
 part of the letter has been lost.   The
manuscript was given by Hardy in 1928 to G\. N\. Watson, who had it in his 
possession until he died in 1965.  At the suggestion of Rankin, Part I  was
sent shortly thereafter to the library of Trinity College, 
Cambridge,  where it still resides. Watson's copy of Part II can be found in
the library of Oxford's Mathematical Institute. We do not know if Ramanujan's
orginal copy of Part II exists. For further historical information, see
 Rankin's two papers \cite{85}, \cite{87}. 

Since many of the proofs in this manuscript had not been published before 
their appearance in handwritten form with the lost notebook 
\cite{83}, since many details were omitted by Ramanujan, since  
mathematicians have established results either proved or asserted in the 
manuscript since it was written, and 
since the manuscript contains many unproved claims, the purpose of this 
paper is to present the manuscript in its entirety, offer some additional 
details, and provide extensive commentary on it.  
Although many of the results in this manuscript have been proven
or explained within a greater context in the works of P\.  Deligne,
J.--P\. Serre, H\. P\. F\. Swinnerton--Dyer, and others, we
 were delighted to find a number
of surprising new  gems. For example, Ramanujan's claims (14.1)--(14.6)
and many of the assertions in both Sections 15 and 16 were unexpected
and entirely new to both authors. Moreover, in proving the claims in Section
14,  the second
author was led, by the ``shape of Ramanujan's claims,''
 to several new general results regarding the distribution
of the partition function modulo every prime $m\geq 5$ \cite{70}.
Part II, beginning with Section 20, is also fascinating, for it contains
Ramanujan's proof, albeit lacking in many details, of his conjectured
congruences for $ p(n) $ modulo arbitrary integral powers of 5. 

Several editorial decisions needed to be made in our presentation of 
the manuscript. 

1) The nine pages of insertions at the end of Part I were interposed 
at their intended positions. 

2) None of Ramanujan's footnotes, such as ``For a direct proof of this see," 
were completed in the manuscript. We have executed their completions, but we
 do not claim that they are what Ramanujan had in mind. 

3) Due to Ramanujan's failure to tag 
certain equalities, the manuscript  contains incomplete references, such 
as ``\dots deduce
 from (\ ) and (\ ) \dots ."  We have added the tags and inserted the equation 
numbers.  Difficulties arose when tags needed to be inserted at places 
between already existing tags with consecutive numbers.  We appended 
letters on such tags; e.g., (6.6a) lies between (6.6) and (6.7). 

4) As with most of his mathematics, Ramanujan provided very few details 
in this manuscript. In Part I, 
Ramanujan indicates, at more than one place, that this is the first of 
two papers that he intends to write on $ p(n) $ and $ \t(n). $ It is 
clear that as Ramanujan wrote the manuscript he continued to discover 
more and more theorems on the subject, and so he more and more frequently 
recorded his results with the promise that he would provide details in his
 next paper.  Thus, details become more sparse as the manuscript progresses,
 so that in the last third of the manuscript there are hardly any details at all. 
However, rather than returning in Part II to the details omitted in Part I,
Ramanujan sketched his proofs of the congruences for $ p(n) $ modulo any
power of 5 or 7.  In Hardy's extraction \cite{80}, 
he considerably amplified Ramanujan's arguments.  Similarly, Rushforth 
\cite{89}  provided many details omitted by Ramanujan.  
In his paper providing proofs of the general congruences modulo $ 5^n $ and $
7^{[n/2]+1}$, Watson \cite{104} had to supply most of the details
omitted by Ramanujan. 
We have followed their leads and have supplied more details for some of 
Ramanujan's arguments.  However, for those parts of the manuscript 
examined by Hardy, Rushforth, and Watson, we have not added details here, as 
readers can find these in the aforementioned papers. So that readers 
remain clear about what was written by Ramanujan, we have placed our 
additions in square brackets. 

5) We have taken the liberty of making minor editorial changes without 
comments.  Such alterations include correcting misprints, adding 
punctuation, and introducing notation.  In particular, Ramanujan generally 
wrote infinite series in expanded form without resorting to summation signs, 
 which we have  supplied. 

Many unproved claims can be found in the manuscript.  Since Ramanujan's 
death, some have been proved by others, often without realizing that 
Ramanujan had originally found them. Some claims are false, and others 
had not been proved. Because of the desire to make minimal additions 
within Ramanujan's manuscript, we have deferred discussions of most of
Ramanujan's unproved claims to the end of this paper, where many 
references to the literature are cited. 


\bigskip
   
\bigskip


\centerline{ \bf PROPERTIES OF $ p(n) $ AND $ \t(n)$}

\centerline{\bf DEFINED BY THE FUNCTIONS}

\centerline{\bf $ \sum_{n=0}^{\infty}p(n)q^n = (q;q)_{\infty}^{-1},$}

\centerline{$ \sum_{n=1}^{\infty}\t(n)q^n = q(q;q)_{\infty}^{24}$}

\bigskip
\bigskip
\centerline{S\. RAMANUJAN} 
 

\bigskip
\bigskip

\centerline{\bf I}

\medskip

\noindent{\bf 0.} I have shown elsewhere by very simple arguments that
$$ \align p(5n-1) \equiv & 0 \pmod5, \\
p(7n-2) \equiv & 0 \pmod7. 
\endalign
$$
In the case of $ \t(n) $ such simple arguments give the following results.

\medskip

\centerline{\bf Modulus 2}

\medskip

It is easy to see that the coefficients of $ q^n $ in the expansion of 
$$ q(q;q)_{\i}^{24} \qquad \text{and} \qquad q(q^8;q^8)_{\i}^{3} $$
are both odd or both even, [where here and in the sequel
$$ (a;q)_{\i} = \prod_{n=0}^{\i}(1-aq^n), $$
where $ |q| < 1.$]  But [by Jacobi's identity \cite{48, 
p\. 285, Thm\. 357}, \cite{21, p\. 39, Entry 24(ii)}],
$$  q(q^8;q^8)_{\i}^{3} = \sum_{n=0}^{\i}(-1)^n(2n+1)q^{(2n+1)^2}. $$
It follows that $ \t(n) $ is odd or even according as $ n $ is an odd square 
or not.  Thus we see that the number of values of $ n $ not exceeding $ n $ 
for which $ \t(n) $ is odd is only
$$ \left[\df{1+\sqrt{n}}{2}\right]. $$

\medskip

\centerline{\bf Modulus 5}

\medskip

Further let $ J $ be any function of $ q $ with integral coefficients 
but not the same function throughout. It is easy to see that
$$ q(q;q)_{\i}^{24} =  q(q;q)_{\i}^{4}(q^5;q^5)_{\i}^{4} +5J. $$
But the coefficient of $ q^{5n} $ in 
$$ q(q;q)_{\i}^{4} $$
is a multiple of 5.\footnotemark\footnotetext{Recall that $ p(5n+4) \equiv 0 
\pmod5.$}  It follows that
$$ \t(5n) \equiv 0 \pmod5. $$

\medskip

\centerline{\bf Modulus 7}

\medskip

This is the simplest of all cases.  Here we have
$$ q(q;q)_{\i}^{24} =  q(q;q)_{\i}^{3}(q^7;q^7)_{\i}^{3} +7J. $$
But since
$$ q(q;q)_{\i}^{3} = q\sum_{n=0}^{\i}(-1)^n(2n+1)q^{n(n+1)/2}, $$
it is easy to see that the coefficients of $ q^{7n}, q^{7n-1}, q^{7n-2} $ 
and $ q^{7n-4} $ are all multiples of 7. It follows that
$$ \t(7n), \t(7n-1), \t(7n-2), \t(7n-4) \equiv 0 \pmod7. $$

\medskip

\centerline{\bf Modulus 23}

\medskip

We have
$$ q(q;q)_{\i}^{24} =  q(q;q)_{\i}(q^{23};q^{23})_{\i} +23J. $$
But [by Euler's pentagonal number theorem \cite{48, p\. 284, 
Thm\. 353}, \cite{21, Entry 22(iii)}],
$$ q(q;q)_{\i}= \sum(-1)^{\nu}q^{1+\frac{1}{2}\nu(3\nu+1)} $$
where the summation extends over all values of $ \nu $ from $ -\i $ to $ \i. $
Now
$$1+\frac{1}{2}\nu(3\nu+1) = (6\nu+1)^2 - \df{23\nu(3\nu+1)}{2}. $$
The residues of a square number for modulus 23 cannot be
$$ 5, 7, 10, 11, 14, 15, 17, 19, 20, 21, 22. $$
It follows from this that 
$$ \left\{\aligned &\t(23n-1), \t(23n-2),
\t(23n-3), \t(23n-4),\\ &\t(23n+5), \t(23n-6), \t(23n+7), \t(23n-8), \qquad
\equiv 0 \pmod{23},\\ 
&\t(23n-9), \t(23n+10), \t(23n+11).
\endaligned \right. $$

\medskip

\centerline{\bf Modulus 5}

\medskip

\noindent{\bf 1.} Let
$$ \align P :=& 1-24\sum_{n=1}^{\i}\df{nq^n}{1-q^n},  \\
 Q :=& 1+240\sum_{n=1}^{\i}\df{n^3q^n}{1-q^n} \\
\intertext{and}
R :=& 1-504\sum_{n=1}^{\i}\df{n^5q^n}{1-q^n},\endalign
 $$
so that\footnotemark\footnotetext{For an elementary proof, see 
\cite{77, eq\. (44)}.}
$$ Q^3-R^2 = 1728q(q;q)_{\i}^{24}.\tag1.1 $$

Let $ \s_s(n) $ denote the [sum of
the] $s^{\text {th}} $ powers of the divisors of $ n. $  Then it is easy to 
see that
$$ Q = 1+5J; \qquad R =P+5J. \tag1.2 $$
Hence, 
$$ Q^3-R^2 = Q-P^2+5J. \tag1.3 $$
But\footnotemark\footnotetext{See \cite{77, eq\. (36)}.}
$$ Q-P^2 = 288\sum_{n=1}^{\i}n\s_1(n)q^n; \tag1.4 $$
and it is obvious that
$$ (q;q)_{\i}^{24}=\df{(q^{25};q^{25})_{\i}}{(q;q)_{\i}} +5J. \tag1.5 
$$
It follows from (1.1) and  (1.3)--(1.5), that
$$ q\df{(q^{25};q^{25})_{\i}}{(q;q)_{\i}}=\sum_{n=1}^{\i}n\s_1(n)q^n 
+5J. \tag1.6 $$
In other words

$$ (q^{25};q^{25})_{\i}\sum_{n=0}^{\i}p(n)q^{n+1} =\sum_{n=1}^{\i}n\s_1(n)q^n 
+5J. \tag1.7 $$
But the coefficient of $ q^{5n} $ in the right hand side is a multiple of 5.  It 
follows that
$$ p(5n-1)\equiv 0 \pmod5. \tag1.8 $$
It also follows from (1.7) that
$$\align & p(n-1)-p(n-26)-p(n-51)+p(n-126)\\
+&p(n-176)-p(n-301) - \cdots -n\s_1(n) \equiv 
0\pmod5, \endalign $$
where  1, 26, 51, 126, \dots  are numbers of the form $ 
\tfrac{1}{2}(5\nu+1)(15\nu+2) $ and \linebreak
 $ \tfrac{1}{2}(5\nu-1)(15\nu-2). $ 
The number of values of $ n $ not exceeding 200 for which $ p(n) \equiv 0, 1, 2, 
3, 4 \pmod5 $ is 69, 33, 34, 34, 30, respectively; and the least value of 
$ n $ for which $ p(n) \equiv 4 \pmod5 $ is 30.  
These being so it appears that $ p(n) 
\equiv 0 \pmod5 $ for about $ \tfrac{1}{3} $ of the values of $ n $ while $ p(n) 
\equiv 1, 2, 3 \text{ or } 4 \pmod5 $ for 
about $ \tfrac{1}{6} $ of the values of 
$ n $ each. It seems extremely difficult to prove any result in this direction 
concerning $ p(n), $ but the problem is much easier concerning $ \t(n). $

\bigskip

\noindent{\bf 2.} It follows from (1.5) and (1.6) that 
$$ \left\{\aligned\t(n)-n\s_1(n) &\equiv 0 \pmod5, \\
\lambda(n)-n\s_1(n) &\equiv 0 \pmod5, 
\endaligned\right.\tag2.1
$$
where
$$ \sum_{n=1}^{\i}\lambda(n)q^n = q\df{(q^{25};q^{25})_{\i}}{(q;q)_{\i}}, $$
so that $ \lambda(n+1) $ is the number of partitions of $ n $ as the sum of 
integers which are not multiples of 25. 
But if $ n $ be written in the form
$$ 2^{a_2}\cdot 3^{a_3}\cdot 5^{a_5}\cdot 7^{a_7}\cdots, $$
where the $ a$'s are zeroes or positive integers, then
$$ n\s_1(n) = \prod_p \df{p^{a_p}(p^{1+a_p}-1)}{p-1}, \qquad 
p = 2, 3, 5, \dots . \tag2.2 $$
But 
$$ \df{p^{a_p}(p^{1+a_p}-1)}{p-1} \equiv 0 \pmod5 \tag2.3 $$
if
$$ a_p \geq 1, \quad p=5 $$
or
$$ \align a_p&\equiv 1\pmod2, \qquad p\equiv 4 \pmod5, \\
\intertext{or}
a_p&\equiv 3\pmod4, \qquad p\equiv 2 \text{ or } 3 \pmod5, \\
\intertext{or}
a_p&\equiv 4\pmod5, \qquad p\equiv 1 \pmod5, 
\endalign
$$
and for no other values.  Suppose now that
$$ \left\{ \aligned t_n &= 0, \qquad \t(n) \equiv 0 \pmod5, \\
t_n &= 1, \qquad \t(n) \not\equiv 0 \pmod5. \endaligned \right. \tag2.4
$$
Then it follows from (2.3) that
$$ \sum_{n=1}^{\i}\df{t_n}{n^s} = \prod\nolimits_1\prod\nolimits_2
\prod\n3,\tag2.5 $$
where
$$ \prod\n1 = \prod_p\df{1}{1-p^{-2s}}, $$
$ p $ being a prime of the form $ 5k-1 $ and 
$$ \prod\n2 = \prod_p\df{1-p^{-3s}}{(1-p^{-s})(1-p^{-4s})}, $$
$ p $ being a prime of the form $ 5k\pm 2 $ and 
$$ \prod\n3 = \prod_p\df{1-p^{-4s}}{(1-p^{-s})(1-p^{-5s})}, $$
$ p $ being a prime of the form $ 5k+1. $ 

It is easy to prove from (2.5) that
$$ \sum_{k=1}^nt_k = o(n). \tag2.6 $$
It can be shown by transcendental methods that
$$ \sum_{k=1}^nt_k \sim \df{Cn}{(\log n)^{1/4}}, \tag2.7$$
and
$$ \sum_{k=1}^nt_k = C\int_1^n\df{dx}{(\log x)^{1/4}} + 
O\left(\df{n}{(\log n)^r}\right),\tag2.8 $$
where $ C $ is a constant and $ r $ is any positive number. 

The proof of (2.6) is quite elementary and very similar to that for showing that 
$ \pi(x) = o(x),$\footnotemark\footnotetext{See Landau's 
{\it Primzahlen} \cite{60, pp\. 641--669}.}  $ \pi(x) $ being the
number of primes not exceeding $ x. $ The result (2.6) can be stated roughly
in other words that $ \t(n) $ and $ \lambda(n) $ are divisible by 5 for
almost all values of $ n, $ while (2.7) and (2.8) give a lot more information. 

\medskip

\centerline{\bf Modulus 25}

\medskip

\noindent{\bf 3.}  It is easily seen from (1.2) that
$$ \align
Q^3-R^2 =& 2(Q^2-PR)-(Q-P^2)+Q(Q-1)^2-(R-P)^2 \tag3.1\\
=&2(Q^2-PR)-(Q-P^2)+25J.
\endalign
$$
But\footnotemark\footnotetext{See \cite{77, Table II}.}
$$ Q^2-PR = 1008\sum_{n=1}^{\i}n\s_5(n)q^n; \tag3.2 $$
and it is obvious that
$$ (q;q)_{\i}^{24} = \df{(q^5;q^5)_{\i}^5}{(q;q)_{\i}}+25J. \tag3.3 $$
Now remembering that
$$ \s_5(n)-\s_1(n) \equiv 0 \pmod5,\tag3.4 $$
it follows from (1.4) and (3.1)--(3.3) that
$$ 6q\df{(q^5;q^5)_{\i}^5}{(q;q)_{\i}} 
=\sum_{n=1}^{\i}\left\{2n\s_5(n)-n\s_1(n)\right\}q^n+25J.\tag3.5 $$
[By extracting those terms with exponents that are multiples of 5 and by 
employing the congruence $ p(5n-1) \equiv 0 \pmod5$,] we easily deduce that
$$ (q;q)_{\i}^5\sum_{n=1}^{\i}p(5n-1)q^n
=\sum_{n=1}^{\i}\left\{10n\s_5(n)-5n\s_1(n)\right\}q^n+25J, 
$$
and hence [by (3.4)] that
$$ (q^5;q^5)_{\i}\sum_{n=1}^{\i}p(5n-1)q^n
=5\sum_{n=1}^{\i}n\s_1(n)q^n +25J. \tag3.6
$$

Since the coefficient of $ q^{5n} $ is a multiple of 25 it follows that
$$ p(25n-1) \equiv 0\pmod{25}. \tag3.7 $$
It also follows from (3.6) that
$$ \align &p(5n-1)- p(5n-26)- p(5n-51)+ p(5n-126)\\
+& p(5n-176)-\cdots -5n\s_1(n) \equiv 0\pmod{25}, 
\endalign
$$
where 1, 26, 51, 126, \dots are the same as in (1.9). 

\bigskip

\noindent{\bf 4.}
It is easy to see [by Fermat's little theorem] that 
$$ n\s_9(n)-2n\s_5(n)+n\s_1(n) \equiv 0 \pmod{25}. \tag4.1 $$
It follows from this and (3.3) and (3.5) that
$$ \t(n)-n\s_9(n)\equiv 0 \pmod{25}. \tag4.2 
$$
It appears that, if $ k $ be any positive integer, it is possible to find 
two integers $ a $ and $ b $ such that
$$ \t(n)-n^a\s_b(n)\equiv 0 \pmod{5^k}, \tag4.3 $$
if $ n $ is not a multiple of 5. Thus for instance
$$ \t(n)-n^{41}\s_{29}(n) \equiv 0 \pmod{125}, \tag4.4 $$
if $ n $ is not a multiple of 5. I have not yet proved these results.
If $ n $ is a multiple of 5, then
$$ \t(n) -4830\t\left(\frac{n}{5}\right)+ 5^{11}\t\left(\frac{n}{25}\right) =0 $$ 
in virtue of (7.6), $ \t(x) $ being considered as 0 if $ x $ is not an integer. 

It also appears that the coefficient of $ q^n $ in the left hand side of 
(3.5) can be exactly determined in terms of the real divisors of $ n. $ Thus
$$  q\df{(q^5;q^5)_{\i}^5}{(q;q)_{\i}} 
= \sum_{n=1}^{\i}\left(\df{n}{5}\right)\df{q^n}{(1-q^n)^2},  \tag4.5
$$
[where $ \left(\frac{n}{p}\right) $ denotes the Legendre symbol]. 
 The allied function
$$  \df{(q;q)_{\i}^5}{(q^5;q^5)_{\i}} = 1-5\sum_{n=1}^{\i}\left(\df{n}{5}\right)
\df{nq^n}{1-q^n}. \tag4.6 $$
It follows from (4.5) that
$$ (q;q)_{\i}^5\sum_{n=1}^{\i}p(5n-1)q^n
=5\sum_{n=1}^{\i}\left(\df{n}{5}\right)\df{q^n}{(1-q^n)^2} $$
and hence that\footnotemark\footnotetext{For a direct proof of this result 
see \cite{78}.}
$$ \sum_{n=0}^{\i}p(5n+4)q^n = 5\df{(q^5;q^5)_{\i}^5}{(q;q)^6_{\i}}.\tag{4.7} $$

\bigskip

\centerline{\bf Modulus 7}

\bigskip

\noindent{\bf 5.} Since\footnotemark\footnotetext{See \cite{77, 
Table I}.}
$$ Q^2 = 1 + 480\sum_{n=1}^{\i}\df{n^7q^n}{1-q^n}, \tag5.1 $$
it is easy to see that 
$$ Q^2=P+7J; \qquad R = 1+7J; \tag5.2 $$
and so
$$ (Q^3-R^2)^2 = P^3-2PQ+R+7J. \tag5.3 $$
But\footnotemark\footnotetext{See \cite{77, Tables II and III, resp.}.}
$$\left\{\aligned PQ-R =& 720\sum_{n=1}^{\i}n\s_3(n)q^n, \\
P^3-3PQ+2R =& - 1728\sum_{n=1}^{\i}n^2\s_1(n)q^n; \endaligned
\right. \tag5.4
$$
and it is obvious that
$$ (q;q)^{48}_{\i} = \df{(q^{49};q^{49})_{\i}}{(q;q)_{\i}}+7J. \tag5.5 $$
It follows from all these that
$$ q^2\df{(q^{49};q^{49})_{\i}}{(q;q)_{\i}} 
= \sum_{n=1}^{\i}\left\{n^2\s_1(n)-n\s_3(n)\right\}q^n +7J.\tag5.6 $$
In other words
$$(q^{49};q^{49})_{\i}\sum_{n=0}^{\i}p(n)q^{n+2} 
= \sum_{n=1}^{\i}\left\{n^2\s_1(n)-n\s_3(n)\right\}q^n +7J. \tag5.7$$
It follows that
$$ p(7n-2) \equiv 0 \pmod7, \tag5.8 $$
and 
$$ \align &p(n-2)-p(n-51)-p(n-100)+p(n-247) \\
+&p(n-345)-\cdots +n\s_3(n)-n^2\s_1(n) \equiv 0 \pmod7, \tag5.9
\endalign 
$$
where 2, 51, 100, 247, \dots are the numbers of the form 
$ \tfrac{1}{2}(7\nu+1)(21\nu+4) $ and $ \tfrac{1}{2}(7\nu-1)(21\nu-4). $

The number of values of $ n $ not exceeding 200 for which $ p(n)
 \equiv 0, 1, 2, 3, 4, 5, 6 \pmod7 $ is 50, 33, 22, 28, 23, 23, 21, 
respectively, and the least value of $ n $ for which 
$ p(n) \equiv 6 \pmod7 $ is 73. It appears that 
$ p(n) \equiv 0 \pmod7 $ for about $ \tfrac{1}{4} $ of the values of $ n $ while 
$ p(n) \equiv  1, 2, 3, 4, 5, 6 \pmod7 $ for about $ \tfrac{1}{8} $ of the 
values of $ n $ each.

\medskip

\noindent{\bf 6.}
It follows from (5.2) that
$$ Q^3-R^2 = PQ -R +7J. \tag6.1 $$
It is easy to see from this and (5.4) that
$$ \t(n)-n\s_3(n) \equiv 0 \pmod7. \tag6.2 $$
Now if $ n = 2^{a_2}\cdot 3^{a_3}\cdot 5^{a_5}\cdot 7^{a_7}\cdots, $ then
$$ n\s_3(n) = \prod_p p^{a_p}\df{p^{3(1+a_p)}-1}{p^3-1}, 
\qquad p=2, 3, 5, 7, \dots .\tag6.3 $$
But
$$ p^{a_p}\df{p^{3(1+a_p)}-1}{p^3-1} \equiv 0 \pmod7, \tag6.4 $$
if 
$$ a_p \equiv 6 \pmod7, \qquad p \equiv 1, 2, \text{ or } 4 \pmod7, $$
or
$$ a_p \equiv 1 \pmod2, \qquad p \equiv 3, 5, \text{ or } 6 \pmod7, $$
or 
$$ a_p \geq 1, \qquad p=7. $$

Suppose now that
$$ \align
t_n =& 1, \qquad \t(n)\not\equiv 0 \pmod7, \\
t_n =& 0, \qquad \t(n)\equiv 0 \pmod7. 
\endalign
$$ 
Then it follows from (6.4) that
$$ \sum_{n=1}^{\i}\df{t_n}{n^s} = \prod\n1\prod\n2 \tag6.5 $$
where
$$ \prod\n1 = \prod_p \df{1-p^{-6s}}{(1-p^{-s})(1-p^{-7s})}, $$
$ p $ being a prime of the form $ 7k +1, 7k+2, 7k+4, $ and 
$$ \prod\n2 =\prod_p\df{1}{1-p^{-2s}}, $$
$ p $ being a prime of the form $ 7k+3, 7k +5, 7k+6. $ It is easy to prove 
from (6.5)  by quite elementary methods that
$$ \sum_{k=1}^n t_k = o(n).\tag6.6 $$
It can be shown by transcendental methods that 
$$ \sum_{k=1}^n t_k \sim \df{Cn}{(\log n)^{1/2}}; \tag6.6a $$ 
and 
$$ \sum_{k=1}^n t_k = C\int_1^n\df{dx}{(\log x)^{1/2}} 
+ O\left(\df{n}{(\log n)^r}\right), \tag6.7$$
where $ r $ is any positive number and 
$$ \align C =& \df{6^{1/2}}{7^{3/4}} 
  \df{1-2^{-6}}{1-2^{-7}}\df{1-11^{-6}}{1-11^{-7}}
\df{1-23^{-6}}{1-23^{-7}}\df{1-29^{-6}}{1-29^{-7}}\cdots\\
&\times 
\df{1}{ \left\{(1-3^{-2})(1-5^{-2})(1-13^{-2})(1-17^{-2})(1-19^{-2})
\cdots\right\}^{1/2}}, \endalign $$
2, 11, 23, \dots being primes of the form $ 7k +1, 7k +2, $ and 
$ 7k+4 $ while 3, 5, 13, \dots being primes of the form $ 7k+3, 7k+5 $ 
and $ 7k+6. $ Thus we see that $ \t(n) $ is divisible by 7 for almost all 
values of $ n; $ and at the same time the number of values of $ n $ for
 which $ \t(n) $ is divisible by 7 is far more numerous than that for which 
$ \t(n) $ is divisible by 5. 

Now if 
$$ \sum_{n=1}^{\i}\lambda(n)q^n = q^2\df{(q^{49};q^{49})_{\i}}{(q;q)_{\i}}, $$
so that $ \lambda(n+2) $ is the number of partitions of $ n $ as the sum of 
integers which are not multiples of 49, it is clear from (5.6) that
$$ \lambda(n)-n^2\s_1(n)+n\s_3(n)\equiv 0 \pmod7. \tag6.8 $$
But it is easy to show that $ n^2\s_1(n) $ and $ n\s_3(n) $ are divisible 
by 7 for almost all values of $ n. $ It follows that $ \lambda(n) $ is 
divisible by 7 for almost all values of $ n. $ It can even be shown that the 
number of values of $ j $ not exceeding $ n $ for which $ \lambda(j) $ is 
\underbar{not} divisible by $ j $ is 
$$ O\left(\df{n}{(\log n)^{1/6}}\right). \tag6.9$$
The index $ \tfrac{1}{6} $ in (6.9) is easily obtained by considering 
$ n^2\s_1(n) $ and $ n\s_3(n) $ separately; but whether this is the right 
index or not can be known only by considering 
$$ n^2\s_1(n)-n\s_3(n) $$
taken together, which seems rather complicated to deal with. 

\bigskip

\centerline{\bf Modulus 49}

\bigskip
  
\noindent{\bf 7.}
We have
$$ \align
(Q^3-R^2)^2 =& (3P^2Q^2-4PQR -2Q^3 +3R^2) \\
&-2(P^3-2PQ+R)+2P(Q^2-P)^2 -(1+2PQ)(R-1)^2 \\
&+\left\{Q(Q^2-P)-R^2+1\right\}^2 \\
=& (3P^2Q^2-4PQR-2Q^3+3R^2)-2(P^3-2PQ+R)+49J
\endalign
$$
in virtue of (5.2). But
\footnotemark\footnotetext{See \cite{77, eq\. (44),
 Table II, Table III, resp.}.}
$$ \left\{\aligned &Q^3-R^2 =1728\sum_{n=1}^{\i}\t(n)q^n, \\
&3Q^3+2R^2-5PQR =1584\sum_{n=1}^{\i}n\s_9(n)q^n, \\
&5Q^3+4R^2-18PQR+9P^2Q^2 = 8640\sum_{n=1}^{\i}n^2\s_7(n)q^n; \endaligned
\right. \tag7.1
$$
and it is obvious that
$$ (q;q)_{\i}^{48} = \df{(q^7;q^7)_{\i}^7}{(q;q)_{\i}} +49J. \tag7.2$$
Now remembering that 
$$ \left\{\aligned\s_7(n)-\s_1(n) \equiv & 0 \pmod7, \\
\s_9(n)-\s_3(n) \equiv & 0\pmod7, \endaligned \right. \tag7.3
$$
it follows from the above equations and (5.4) that
$$\align & q^2\df{(q^7;q^7)_{\i}^7}{(q;q)_{\i}} = \sum_{n=1}^{\i}\left\{2n
\s_9(n) -4n^2\s_7(n) \right. \\
+&\left. 2n\s_3(n) -2n^2\s_1(n)+2\t(n)\right\}q^n +49J.\tag7.4
\endalign
$$
>From this [and (6.2)] we deduce that
$$ (q;q)_{\i}^7\sum_{n=0}^{\i}p(7n+5)q^{n+1} =
\sum_{n=1}^{\i}\left\{28n\s_3(n)+2\t(7n)\right\}q^n +49J. \tag7.5 $$
I have stated in my previous paper 
that\footnotemark\footnotetext{\cite{77, eq\. (101)}.}
$$ \sum_{n=1}^{\i}\df{\t(n)}{n^s} = \prod_p\df{1}{1-\t(p)p^{-s}+p^{11-2s}}, 
\tag7.6 $$
where $ p $ assumes all prime values. This has since been proved by Mr Mordell.
\footnotemark\footnotetext{On Mr Ramanujan's empirical expansions of 
modular functions, {\it Proc\. Cambridge Philos\. Soc\.} 19 (1919),
117--124. A simpler proof is given in Hardy's lectures \cite{47}.}
Now by actual calculation we find that 
$$ \t(7) \equiv 14 \pmod{49}. $$
It follows from this and (7.6) that
$$ \t(7n)-14\t(n) \equiv 0 \pmod{49}. $$
It is easy to see from this and (7.5) that
$$ (q^7;q^7)_{\i}^7\sum_{n=0}^{\i}p(7n+5)q^{n+1} =
7\sum_{n=1}^{\i}n\s_3(n)q^n +49J. \tag7.7
$$
Now if 
$$ n\equiv 3, 5, 6 \pmod7, $$
then $ n $ must contain an odd power of a prime $ p $ of the form $ 7k+3, 
7k+5 $ or $ 7k+6 $ as a divisor since all perfect squares are of the form 
$ 7k, 7k+1, 7k+2 $ or $ 7k+4; $ and so $ \s_3(n) $ is divisible by $ p^3+1 $ 
which is divisible by 7. Also it is obvious that if $ n $ is a multiple of 7 
then $ n\s_3(n) $ is also divisible by 7. It follows that if 
$$ n \equiv 0, 3, 5, 6 \pmod7, $$
then
$$ n\s_3(n) \equiv 0 \pmod7. $$
It is easy to see from this and (7.7) that
$$ p(49n-2), p(49n-9), p(49n-16), p(49n-30) \equiv 0 \pmod{49}. \tag7.8$$
It also follows from (7.7) that
$$ \align
& p(7n-2)-p(7n-51)-p(7n-100)+p(7n-247) \\
+& p(7n-345) - \cdots - 7n\s_3(n) \equiv 0 \pmod{49}, \endalign $$
where 2, 51, 100, 247, \dots are the same as in (5.9). 

\bigskip

\noindent{\bf 8.} It appears that
$$ q(q;q)_{\i}^3(q^7;q^7)_{\i}^3 +8q^2\df{(q^7;q^7)_{\i}^7}{(q;q)_{\i}} =
\sum_{n=1}^{\i}\left(\df{n}{7}\right)q^n\df{1+q^n}{(1-q^n)^3} \tag8.1
$$
[where $\left(\frac{n}{7}\right)$ denotes the Legendre symbol], while the 
allied function 
$$ 49q (q;q)_{\i}^3(q^7;q^7)_{\i}^3 +8\df{(q;q)_{\i}^7}{(q^7;q^7)_{\i}} = 8
-7\sum_{n=1}^{\i}\left(\df{n}{7}\right)\df{n^2q^n}{1-q^n}. \tag8.2 $$
Now remembering that
$$ (q;q)_{\i}^3 = \sum_{n=0}^{\i}(-1)^n(2n+1)q^{n(n+1)/2} $$
and picking out the terms $ q^7, q^{14}, q^{21}, \dots $ from both sides in 
(8.1) we obtain
$$  -7q (q;q)_{\i}^3(q^7;q^7)_{\i}^3 +8(q;q)_{\i}^7
\sum_{n=1}^{\i}p(7n-2)q^n = 49\sum_{n=1}^{\i}\left(\df{n}{7}\right)
q^n\df{1+q^n}{(1-q^n)^3}, 
$$
the series in the right hand side being the same as that in (8.1). It follows
from this and (8.1) that\footnotemark\footnotetext{For a direct proof 
of this see \S. [Ramanujan evidently intended to give a proof of (8.3) 
elsewhere. In his paper \cite{78}, (8.3) is stated without proof.
See the notes at the end of this paper for references to proofs of (8.3).]}
$$ \sum_{n=0}^{\i}p(7n+5)q^n = 7\df{(q^7;q^7)_{\i}^3}{(q;q)_{\i}^4} +
49q^2\df{(q^7;q^7)_{\i}^7}{(q;q)_{\i}^8}.\tag8.3 $$

It also appears that if
$$ \sum_{n=1}^{\i}\lambda(n)q^n= q(q;q)_{\i}^3(q^7;q^7)_{\i}^3, $$
then 
$$ \sum_{n=1}^{\i}\df{\lambda(n)}{n^s} = \df{1}{1+7^{1-s}}\prod\n1\prod\n2, \tag8.4
$$
where
$$ \prod\n1 = \prod_p\df{1}{1-p^{2-2s}}, $$
$ p $ being a prime of the form $ 7k+3, 7k+5, $ or $ 7k+6, $ and 
$$ \prod\n2 = \prod_p\df{1}{1+(2p-a^2)p^{-s}+p^{2-2s}} $$
$ p $ being a prime of the form $ 7k+1, 7k+2, $ or $ 7k+4 $ and $ a $ and 
$ b $ being integers such that $ 4p = a^2+7b^2. $ Thus $ \lambda(n) $ can 
be completely ascertained. It follows from this and (8.1) and (8.2) that the 
coefficients of $ q^n $ in 
$$ \df{(q;q)_{\i}^7}{(q^7;q^7)_{\i}}, \qquad 
q^2\df{(q^7;q^7)_{\i}^7}{(q;q)_{\i}} $$
can be completely ascertained.

Now it is easy to see that 
$$ 3n^9-2n^3 \equiv 0, 1, \text{ or } -1 \pmod{49}, $$
according as $ n \equiv 0 \pmod7, n \equiv 1, 2, 4 \pmod7, $ or $ n 
\equiv 3, 5, 6 \pmod7. $ Also the coefficient of $ q^n $ in 
$ q(1+q)/(1-q)^3 $ is $ n^2. $ Hence the right side in (8.1) can be written as 
$$ \sum_{n=1}^{\i}\left\{3n^2\s_7(n)-2n^2\s_1(n)\right\}q^n +49J. \tag8.5 $$
It follows from this, (7.3), (7.4) and (8.1) that
$$ \t(n)-3\lambda(n)+n\s_9(n)+n\s_3(n) \equiv 0 \pmod{49}, \tag8.6 $$
where $ \lambda(n) $ is the same as in (8.4).  From the formulae (8.4) and 
(8.6) all the residues  of $ \t(n) $ for modulus 49 can be completely ascertained. 

\bigskip

\centerline{\bf Modulus 11}

\bigskip

\noindent{\bf 9.}
In this case we start with the 
series\footnotemark\footnotetext{See \cite{77, Table I}.}
$$ \left\{\aligned 1-264\sum_{n=1}^{\i}\df{n^9q^n}{1-q^n} =& QR,\\
691+65520\sum_{n=1}^{\i}\df{n^{11}q^n}{1-q^n} =& 441Q^3+250R^2. \endaligned
\right. \tag9.1 $$
It follows that
$$ QR=1+11J; \qquad Q^3-3R^2 =-2P+11J. \tag9.2 $$
It is easy to see from this that
$$ \align (Q^3-R^2)^5=& (Q^3-3R^2)^5-Q(Q^3-3R^2)^3-R(Q^3-3R^2)^2 -5QR +11J\\
=& P^5-3P^3Q-4P^2R-5QR+11J. \endalign
$$
But\footnotemark\footnotetext{See \cite{77, Table III, Table II}.}
$$ \left\{\aligned & P^5-10P^3Q+20P^2R-15PQ^2+4QR =
 -20736\sum_{n=1}^{\i}n^4\s_1(n)q^n, \\
& P^3Q-3P^2R+3PQ^2-QR = 3456\sum_{n=1}^{\i}n^3\s_3(n)q^n, \\
& P^2R-2PQ^2+QR = -1728\sum_{n=1}^{\i}n^2\s_5(n)q^n, \\
& PQ^2 -QR = 720\sum_{n=1}^{\i}n\s_7(n)q^n; \endaligned \right. \tag9.3
$$
and it is obvious that
$$ (q;q)_{\i}^{120} = \df{(q^{121};q^{121})_{\i}}{(q;q)_{\i}}+11J. \tag9.4$$
It is easy to see from all these that
$$ q^5\df{(q^{121};q^{121})_{\i}}{(q;q)_{\i}} = 
\sum_{n=1}^{\i}\left\{-n^4\s_1(n)+3n^3\s_3(n)+3n^2\s_5(n)-5n\s_7(n)\right\}q^n 
+11J. \tag9.5$$
It follows from this that
$$ p(11n-5) \equiv 0\pmod{11}; \tag9.6 $$
and 
$$ \align &p(n-5)-p(n-126)-p(n-247)+p(n-610)
+p(n-852)\\-&\cdots +n^4\s_1(n)-3n^3\s_3(n)-3n^2\s_5(n)+5n\s_7(n) \equiv 0 
\pmod{11},\tag9.7 \endalign $$
where  5, 126, 247, 610, \dots  are numbers of the form 
$ \tfrac{1}{2}(11\nu+2)(33\nu+5) $ and \linebreak
 $ \tfrac{1}{2}(11\nu-2)(33\nu-5). $
It is only to prove the general result (9.7) we require all the details in 
(9.3). But we don't require all these details in order to prove (9.6) and 
the proof can be very much simplified as follows: 
we have\footnotemark\footnotetext{See \cite{77, eq\. (30)}.}
$$ q\df{dP}{dq} = \df{P^2-Q}{12}, \qquad q\df{dQ}{dq} = \df{PQ-R}{3}, \qquad
q\df{dR}{dq} = \df{PR-Q^2}{2}. \tag9.8 $$
Now using (9.2) and (9.8) we can show that
\footnotemark\footnotetext{As mentioned in the beginning, the $J$'s are 
not the same functions.}  
$$ (Q^3-R^2)^5=q\df{dJ}{dq} + 11J.$$
 It follows from this and  (9.4) that
$$ q^5\df{(q^{121};q^{121})_{\i}}{(q;q)_{\i}} =q\df{dJ}{dq}+11J. \tag9.9 $$
Since the coefficient of $ q^{11n} $ in the right hand side is a multiple of 
11 it follows that
$$ p(11n-5)\equiv 0\pmod{11}. $$
The number of values of $ n $ not exceeding 200 for which $ p(n) \equiv 
$ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 $ \pmod{11} $ is 77, 23, 24, 14, 15, 
14, 5, 12, 8, 8, 0, respectively.  Even though these values seem to be very 
irregular it appears from the residues of $ p(n) $ for moduli 5 and 7 and 
also from the next section that $ p(n) \equiv 0 \pmod{11} $ for about 
$ \tf{1}{6} $ of the values of $ n $ while $ p(n) \equiv $ 1, 2, 3, 4, 5, 6, 
7, 8, 9, 10 $ \pmod{11} $ for about $ \tf{1}{12} $ of the values of $ n $ each. 

\bigskip

\noindent{\bf 10.}  Mr H\. B\. C\. Darling observed the remarkable fact 
(before I began to write this paper) that $ p(n) $ is divisible by 11 for 
45 values of $ n $ not exceeding 100.  This can be explained by the formula 
(9.7) and the congruency of
$$ n^4\s_1(n)-3n^3\s_3(n)-3n^2\s_5(n) +5n\s_7(n) \tag10.1 $$
for modulus 11. It can be shown by quite elementary methods that (10.1) 
is divisible by 11 for almost all values of $ n. $ [A proof of this fact is 
sketched in Section 19.] It can even be shown 
that the number of values of $ n $ not exceeding $ n $ for which (10.1) 
is \underbar{not} divisible by 11 is 
$$ O\left(\df{n}{(\log n)^{1/10}}\right) \tag10.2$$
by considering the divisibility of the four terms in (10.1) separately; 
but a better result can be found only by considering all the four terms 
in (10.1) taken together.  The same remarks apply to the function 
$ \lambda(n) $ defined by 
$$ \sum_{n=1}^{\i}\lambda(n)q^n = q^5\df{(q^{121};q^{121})_{\i}}{(q;q)_{\i}};
\tag10.3 $$
so that $ \lambda(n+5) $ is the number of partitions of $ n $ as the sum of 
integers which are not multiples of 121; 
that is to say $ \lambda(n) $ is divisible by 11 for almost all values of 
$ n; $ and the number of values of $ \lambda(n) $ not divisible by 11 is 
of the form (10.2). It appears from (10.3) that the number of values of 
$ n $ for which 
$ p(n) \equiv 0 \pmod{11} $ cannot be so high as 45\% if $ n $ exceeds 120.  
Thus the number of values of $ p $ divisible by 11 is 
$$ \align 45\%, \qquad & 0 < n \leq 40 \\
45\%, \qquad & 40 < n \leq 80 \\
45\%, \qquad & 80 < n \leq 120 \\
35\%, \qquad & 120 < n \leq 160 \\
22\tf{1}{2}\%, \qquad & 160 < n \leq 200.
\endalign
$$
It is also very remarkable that, in the table of the first 200 values of 
$ p(n), $ there is not a single value of $ p(n) $ of the form $ 11k-1. $ 
This is probably due to such a high percentage of the values of $ p(n) $ 
divisible by 11 in the beginning. 

I have not yet investigated completely the residues of $ \t(n) $ for 
modulus 11. But it appears that if 
$$ \sum_{n=1}^{\i}\lambda(n)q^n = q(q;q)_{\i}^2(q^{11};q^{11})_{\i}^2, $$
then
$$ \sum_{n=1}^{\i}\df{\lambda(n)}{n^s} = 
\df{1}{1-11^{-s}}\prod_p\df{1}{1-\lambda(p)p^{-s}+p^{1-2s}}, \tag10.4 $$
$ p $ assuming all prime values except 11, and that $ \lambda(p) $ can be 
determined also. If that is so then the residues of $ \t(n) $ for modulus 
11 can also be ascertained since it is easily seen that 
$$ \t(n)-\lambda(n) \equiv 0 \pmod{11}. \tag10.5 $$

Again it is easy to show by using (7.6) [and the values $ \t(2) = -24, 
\t(3) = 252, \t(5) = 4830, \t(7) = -16744,$ and $ \t(11) = 534612,$ which 
can be found in a table in Ramanujan's paper \cite{77}, 
\cite{82, p\. 153}] that
$$ \align \sum_{n=1}^{\i}\df{\t(n)}{n^s} =& \df{1}{1+2^{1-s}+2^{1-2s}}
\df{1}{(1-3^{3-s})^2}\df{1}{(1-5^{2-s})(1-5^{4-s})}\\
&\times \df{1}{(1+7^{2-s})(1-7^{4-s})}
\df{1}{1-11^{-s}}\cdots + 11j, \tag10.5a
\endalign
$$
where $ j $ is a Dirichlet series of the form
$$ \sum\df{a_n}{n^s}, $$
$ a_n $ being an integer.

>From this we can deduce a number of results such as
$$ \t(2^{4\lambda-1}n) \equiv 0 \pmod{11} \tag10.61 $$
if $ n $ is an odd integer; 
$$ \t(3^{11\lambda-1}n) \equiv 0 \pmod{11} \tag10.62 $$
if $ n $ is not a multiple of 3;
$$ \t(5^{5\lambda-1}n) \equiv 0 \pmod{11} \tag10.63 $$
if $ n $ is not a multiple of 5;
$$ \t(7^{10\lambda-1}n) \equiv 0 \pmod{11} \tag10.64 $$
if $ n $ is not a multiple of 7;
$$ \t(11^{\lambda}n)-\t(n) \equiv 0 \pmod{11} \tag10.7 $$
and so on.  [The five congruences above can be established by expanding 
the appropriate factors in (10.5a) in geometric series.  For example, consider
$$ \align \df{1}{1+2^{1-s}+2^{1-2s}} =& -\df{i}{2^{1-s}+1-i} 
+\df{i}{2^{1-s}+1+i} \\
=& -\df{i}{1-i}\sum_{n=0}^{\i}\left(\df{2^{1-s}}{i-1}\right)^n
+\df{i}{1+i}\sum_{n=0}^{\i}\left(\df{2^{1-s}}{-i-1}\right)^n\\
=&i\sum_{n=0}^{\i}2^{n(1-s)}e^{-3\pi i(n+1)/4}-i
\sum_{n=0}^{\i}2^{n(1-s)}e^{3\pi i(n+1)/4}. \endalign
$$
Since $ \sin\{3\pi(n+1)/4\} = 0 $ if and only if $ n \equiv -1 \pmod4, $ the 
assertion (10.61) follows from (10.5a).]

Even though (10.61)--(10.64) are very analogous to one another 
further equations are not necessarily quite similar to these; sometimes there 
are more than one equation and sometimes there are equations of the form
$$ \t(19n) \equiv 0 \pmod{11} \tag10.8 $$
if $ n $ is not a multiple of 19, and 
$$ \t(29n) \equiv 0 \pmod{11} \tag10.9 $$
if $ n $ is not a multiple of 29. 

It is very likely that the primes 19, 29, \dots occurring in equations like (10.8) 
and (10.9) are such that the sum of their reciprocals is a
\underbar{divergent} 
series. If this assertion is true then $ \t(n) $ is divisible by 11 
for almost all values of $ n $ which is easily seen from (10.2). 


\bigskip

\centerline{\bf Moduli 2 and 3}

\bigskip

\noindent{\bf 11.}  [It will be convenient to introduce Ramanujan's 
theta-functions $ \varphi(q) $ and $ \psi(q), $ defined by
$$ \varphi(q) := \sum_{n=-\i}^{\i}q^{n^2} = \df{(-q;-q)_{\i}}{(q;-q)_{\i}}
\tag11.1a $$
and 
$$ \psi(q) := \sum_{n=0}^{\i}q^{n(n+1)/2} =
\df{(q^2;q^2)_{\i}}{(q;q^2)_{\i}}, 
\tag11.1b $$
where the product representations are easy consequences of Jacobi's triple 
product identity.]

Before we proceed to consider higher moduli we shall 
see what the analogous formulae are in the cases of moduli 2 and 3.  It 
is easy to see that [by (11.1b)]
$$ \df{(q^4;q^4)_{\i}}{(q;q)_{\i}} = \df{(q^2;q^2)_{\i}}{(q;q^2)_{\i}} +2J 
 = \psi(q) +2J. \tag11.1 $$
It follows that
$$ p(n)-p(n-4)-p(n-8)+p(n-20)+p(n-28) -\cdots \tag11.2 $$
is odd or even according as $ n $ is a triangular number or not, 4, 8, 20,
 \dots being numbers of the form $ 2\nu(3\nu+1) $ and $ 2\nu(3\nu-1). $

$ p(n) $ is odd for 110 values of $ n $ not exceeding 200 and even for 90 
values of $ n $ in the same range. Thus $ p(n) $ seems to be odd for more
 values of $ n $ than those for which $ p(n) $ is even.

If 
$$ \sum_{n=0}^{\i}\lambda(n)q^n = \df{(q^4;q^4)_{\i}}{(q;q)_{\i}} $$
so that $ \lambda(n) $ is the number of partitions of $ n $ as the sum of 
integers which are not multiples of 4 then [by (11.1) and (11.1b)] $ 
\lambda(n) $ is odd or even 
according as $ n $ is a triangular number or not. 

Again we have
$$ \df{(q^9;q^9)_{\i}}{(q;q)_{\i}} = \df{(q^3;q^3)_{\i}^3}{(q;q)_{\i}}+3J. 
\tag11.3 $$
But it can be shown \cite{23} that
$$ q\df{(q^9;q^9)_{\i}^3}{(q^3;q^3)_{\i}} = 
\sum_{n=1}^{\i}\chi_0(n)\df{q^n}{1+q^n+q^{2n}} \tag11.4 $$
[where $ \chi_0(n) $ is the principal character modulo 3]. But the right
 hand side in (11.4) is of the form
$$ \sum_{n=1}^{\i}\chi_0(n)\df{q^n}{(1-q^n)^2}+3J; $$
and the coefficient of $ q^{3n+1} $ in the above series is $ \s_1(3n+1). $ 
It follows from this and (11.3) and (11.4) that 
$$ \df{(q^9;q^9)_{\i}}{(q;q)_{\i}} = \sum_{n=0}^{\i}\s_1(3n+1)q^n +3J. \tag11.5 $$
>From this we easily deduce that
$$ \align &p(n)-p(n-9)-p(n-18)+p(n-45)+p(n-63)\\
-&p(n-108)-\cdots-\s_1(3n+1)\equiv 0\pmod3, \tag11.6\endalign
$$
where 9, 18, 45, \dots are numbers of the form $ \tf{9}{2}\nu(3\nu+1) $
and $ \tf{9}{2}\nu(3\nu-1). $

The number of values of $ n $ not exceeding 200 for which 
$ p(n) \equiv 0, 1, 2 \pmod3 $ is 66, 68, 66 respectively. Thus it appears 
that $ p(n) \equiv 0, 1, 2 \pmod3 $ for about $ \tf{1}{3} $ of the number of 
values of $ n $ each. 

It follows from (11.5) that if 
$$ \sum_{n=0}^{\i}\lambda(n)q^n = \df{(q^9;q^9)_{\i}}{(q;q)_{\i}} $$
so that $ \lambda(n) $ is the number of partitions of $ n $ as the sum of 
integers which are not multiples of 9, then
$$ \lambda(n)-\s_1(3n+1) \equiv 0 \pmod3. $$

Again the left hand side of (11.4) is of the form
$$ q(q;q)_{\i}^{24} +3J \tag11.7 $$
while the right hand side of (11.4) is of the form
$$\sum_{n=1}^{\i}\df{n^2q^n}{(1-q^n)^2} +3J. $$
It follows that 
$$ \t(n)-n\s_1(n)\equiv 0 \pmod3.\tag11.8 $$

Suppose now that
$$ \left\{\aligned t_n =0, \qquad&\lambda(n) \equiv 0 \pmod3, \\
t_n =1,\qquad &\lambda(n) \not\equiv 0 \pmod3, \endaligned \right. $$
and that
$$ \left\{\aligned T_n =0,\qquad &\t(n) \equiv 0 \pmod3, \\
T_n =1, \qquad &\t(n) \not\equiv 0 \pmod3. \endaligned \right. $$
Then we can easily deduce from (11.7), (11.8), and (2.2) that
$$ \sum_{n=0}^{\i}\df{t_n}{(3n+1)^s} =\sum_{n=0}^{\i}\df{T_n}{n^s} =
 \prod\n1\prod\n2 $$
where
$$ \prod\n1 = \prod_p\df{1}{1-p^{-2s}} $$
$ p $ assuming prime values of the form $ 3k-1 $ and 
$$ \prod\n2 = \prod_p\df{1+p^{-s}}{1-p^{-3s}} $$
$ p $ assuming prime values of the form $ 3k+1.$ We easily deduce from this that 
$$ \left\{\aligned \sum_{k=1}^nt_k = o(n), \\
\sum_{k=1}^nT_k = o(n). \endaligned \right. $$
In other words $ \lambda(n) $ and $ \t(n) $ are divisible by 3 for almost 
all values of $ n. $ We can show by transcendental methods that 
$$ \left\{\aligned \sum_{k=1}^nt_k = C\int_1^n\df{dx}{(\log x)^{1/2}} 
+O\left(\df{n}{(\log n)^r}\right), \\
\sum_{k=1}^nT_k = \df{C}{3}\int_1^n\df{dx}{(\log x)^{1/2}} 
+O\left(\df{n}{(\log n)^r}\right) \endaligned \right.\tag11.8a $$
where $ r $ is any positive number and 
$$  C = \df{2^{1/2}}{3^{1/4}}   
\df{1-7^{-2}}{1-7^{-3}}\df{1-13^{-2}}{1-13^{-3}}\df{1-19^{-2}}{1-19^{-3}}\cdots 
 \df{1}{\left\{(1-2^{-2})(1-5^{-2})(1-11^{-2})\cdots\right\}^{1/2}}
$$
in both cases, 2, 5, 11, \dots being primes of the form $ 3k-1 $ and 7, 13,
19, \dots being primes of the form $ 3k+1. $






\bigskip

\centerline{\bf Further properties of $ \t(n)$}

\bigskip

\noindent{\bf 12.}  
It is easy to see [from (11.1b)] that
$$  (q;q)_{\i}^{24} = \df{(q^2;q^2)_{\i}^8}{(q;q^2)_{\i}^8} + 32J 
= \psi^8(q) +32J. $$
But \cite{21, p\. 139, Ex\. (ii)}
$$ q\psi^8(q) = 
\sum_{n=1}^{\i}\df{n^3q^n}{1-q^{2n}}, $$
and 
$$ \sum_{n=1}^{\i}n^4q^n = \df{q}{1-q^2}+16J, $$
and
$$\sum_{n=1}^{\i}n^8q^n = \df{q}{1-q^2}+32J, $$
[since 
$$ \sum_{n=1}^{\i}n^4q^n  \equiv 1\cdot q+0\cdot q^2 +1\cdot q^3 + 0\cdot q^4
+ \cdots = \df{q}{1-q^2} \pmod{16},$$
as $ n^4 \equiv 0,1 \pmod{16}$, according as $ n $ is even or odd, and
$$ \sum_{n=1}^{\i}n^8q^n \equiv 1\cdot q+0\cdot q^2 +1\cdot q^3 + 0\cdot q^4
+ \cdots = \df{q}{1-q^2} \pmod{32},
 $$
as $ n^8 \equiv 0,1 \pmod{32}$, according as $ n $ is even or odd.] 
It is easy to see from all these that
$$ \left\{\aligned \t(n)-n^3\s_1(n) \equiv & 0 \pmod{16}; \\
\t(n) -n^3\s_5(n) \equiv & 0 \pmod{32}. \endaligned \right.\tag12.1
$$
Again we have
$$ (q;q)_{\i}^{24} = \df{(q^3;q^3)_{\i}^9}{(q;q)_{\i}^3} + 27J. $$
But it can be shown that \cite{22, p\. 143, Thm\. 8.7}
$$ q\df{(q^3;q^3)_{\i}^9}{(q;q)_{\i}^3} =
 \sum_{n=1}^{\i}\df{n^2q^n}{1+q^n+q^{2n}}. \tag12.2$$
Now it is easy to see that
$$ \sum_{n=1}^{\i}n^3q^n = \df{q}{1+q+q^2} +9J $$
and
$$ \sum_{n=1}^{\i}n^9q^n = \df{q}{1+q+q^2} +27J, $$
[since 
$$\sum_{n=1}^{\i}n^3q^n \equiv 1\cdot q -1\cdot q^2+0\cdot q^3 +1\cdot q^4
-1\cdot q^5+0\cdot q^6 +\cdots =\df{q-q^2}{1-q^3}= \df{q}{1+q+q^2} \pmod9,$$
as $ n^3 \equiv 0,1,-1 \pmod9, $ according as $ n \equiv 0,1,-1 \pmod3, $ and
$$
\sum_{n=1}^{\i}n^9q^n  \equiv 1\cdot q -1\cdot q^2+0\cdot q^3 +1\cdot q^4
-1\cdot q^5+0\cdot q^6 +\cdots = \df{q}{1+q+q^2} \pmod{27},$$
as $ n^9 \equiv 0,1,-1 \pmod{27}, $ according as $ n \equiv 0,1,-1 \pmod3$.]
It follows that
$$ \left\{\aligned \t(n) -n^2\s_1(n) \equiv & 0 \pmod9, \\
\t(n)-n^2\s_7(n) \equiv & 0 \pmod{27}. \endaligned \right. \tag12.3
$$
It is easy to deduce from (2.1), (4.2),  (12.1) and (12.3) that
$$ \left\{\aligned \t(n) -n\s_1(n) \equiv & 0 \pmod{30}, \\
\t(n) -n^2\s_1(n) \equiv & 0 \pmod{36}, \\
\t(n) -n^3\s_1(n) \equiv & 0 \pmod{48}, \\
\t(n) -n^5\s_1(n) \equiv & 0 \pmod{120}, \endaligned \right.\tag12.4
$$
$$ \left\{\aligned \t(n) -n\s_3(n) \equiv & 0 \pmod{42}, \\
\t(n) -n^2\s_3(n) \equiv & 0 \pmod{60}, \\
\t(n) -n^4\s_3(n) \equiv & 0 \pmod{168},\endaligned \right.\tag12.5
$$
$$ \left\{\aligned \t(n) -n^3\s_5(n) \equiv & 0 \pmod{288}, \\
\t(n) -n^2\s_7(n) \equiv & 0 \pmod{540}, \\
\t(n) -n\s_9(n) \equiv & 0 \pmod{1050}.\endaligned \right.\tag12.6
$$

Again it easily follows from the second equation in (9.1) that
$$ \t(n)-\s_{11}(n) \equiv 0 \pmod{691}. \tag12.7 $$

It is easy to deduce from this that $ \t(n) $ is divisible by 691 for 
almost all values of $ n, $ and by transcendental methods that the number of
 values of $ n $ not exceeding $ n $ for which $ \t(n) $ is \underbar{not} 
divisible by 691 is of the form
$$ C\int_1^n\df{dx}{(\log x)^{1/690}}+ O\left(\df{n}{(\log n)^r}\right)\tag12.7a  $$
where $ C $ is a constant and $ r $ is any positive number.

It is easy to prove that 
$$ q(-q;-q)_{\i}^{24} = q(q;q)_{\i}^{24} + 48q^2(q^2;q^2)_{\i}^{24}
+2^{12} q^4(q^4;q^4)_{\i}^{24}.\tag12.7b $$
[To prove (12.7b), set, after Ramanujan, 
$$ f(-q) := (q;q)_{\i}. $$
Thus, (12.7b) can be written in the equivalent formulation
$$ qf^{24}(q) =qf^{24}(-q) +48q^2f^{24}(-q^2)+2^{12}q^4f(-q^4). \tag12.7c $$
To prove (12.7c), we use the catalogue of evaluations for $ f $ found 
in Entry 12 of Chapter 17 in Ramanujan's second notebook 
\cite{21, p\. 124}, in particular, 
$$ \align
f(q) =& \sqrt{z}2^{-1/6}\left\{x(1-x)/q\right\}^{1/24}, \qquad
f(-q) = \sqrt{z}2^{-1/6}(1-x)^{1/6}(x/q)^{1/24},\tag12.7c \\
f(-q^2) =& \sqrt{z}2^{-1/3}\left\{x(1-x)/q\right\}^{1/12}, \qquad
f(-q^4) = \sqrt{z}2^{-2/3}(1-x)^{1/24}(x/q)^{1/6}, 
\endalign
$$
where $ x = k^2, $ with $ k $ being the modulus, and $ z = (2/\pi)K, $ with
 $ K $ being the complete elliptic integral of the first kind. 
Using these evaluations in (12.7c), we easily verify its truth.]
>From this it is easy to deduce that 
$$ \t(2n)+24\t(n)+2^{11}\t(\tfrac{1}{2}n) =0 \tag12.8 $$
where $ n $ is any integer and $ \t(x) = 0 $ if $ x $ is not an integer. 

[Recall that $ \varphi $ and $ \psi $ are defined in (11.1a) and (11.1b), 
respectively.] Again it is  easy to prove that
$$ q\psi^8(q)\varphi^{16}(-q) = qf^{24}(-q). $$
[To prove this identity, use (12.7c) and the evaluations 
\cite{21, Entry 11(i), p\. 123; Entry 10(ii), p\. 122}
$$ \left. \psi(q) = \sqrt{\tfrac{1}{2}z}(x/q)^{1/8} \qquad \text{and} \qquad 
\varphi(-q) = \sqrt{z}(1-x)^{1/4}. \right]\tag12.8a $$
But [by the binomial theorem], 
$$
\varphi^{16}(-q)
= -4\varphi^{4}(-q)+16\varphi^{2}(-q) -11 +256J. 
$$
Hence
$$ \align qf^{24}(-q) =& 4\left\{1-\varphi^{4}(-q) 
\right\}q\psi^8(q)-16\left\{1-\varphi^{2}(-q)\right\}q\psi^8(q)+q\psi^8(q) +256J \\
=& 4\left\{1-\varphi^{4}(-q)\right\}q\psi^4(q^2)
-16\left\{1-\varphi^{2}(-q)\right\}q\psi^4(q^2)
 +q\psi^8(q) +256J.
\endalign
$$
But
$$ \align q\psi^8(q)=
&\sum_{n=0}^{\i}\df{n^3q^n}{1-q^{2n}}, \tag12.8b\\
q\psi^4(q^2) 
=&\sum_{n=0}^{\i}\df{(2n+1)q^{2n+1}}{1-q^{4n+2}}, \tag12.8c \\
q\psi^4(q^2)
\varphi^{4}(-q) =&
\sum_{n=1}^{\infty}(-1)^{n-1}\df{n^3q^n}{1-q^{2n}},\tag12.8d \\
q\psi^4(q^2)
\varphi^{2}(-q) 
=&\sum_{n=1}^{\i}(-1)^{n-1}\df{n^2q^n}{1+q^{2n}} \\
=&\sum_{n=0}^{\i}(-1)^{n}\df{(2n+1)^2q^{2n+1}}{1-q^{4n+2}}
-\sum_{n=1}^{\i}\df{(2n)^2q^{2n}}{1+q^{4n}} +16J. \tag12.8e
\endalign
$$
[The identities (12.8b) and (12.8c) are, respectively, Examples (ii) and 
(iii) in Section 17 of Chapter 17 in Ramanujan's second notebook 
\cite{21, p\. 139}.

 By Entry 11(iii) in Chapter 17 of Ramanujan's second notebook 
\cite{21, p\. 123},
$$ \psi(q^2) = \tfrac{1}{2}\sqrt{z}(x/q)^{1/4}. \tag12.8f $$
It follows from (12.8a) and (12.8f) that
$$ q\psi(q^2)\varphi^4(-q) = \tfrac{1}{16}z^4x(1-x). \tag12.8g $$
On the other hand by Entry 14(ii), (ix) in Chapter 17 of the second 
notebook \cite{21, p\. 130}, 
$$ \align
\sum_{n=1}^{\infty}(-1)^{n-1}\df{n^3q^n}{1-q^{2n}} =&
\sum_{n=1}^{\infty}(-1)^{n-1}n^3\left(\df{q^n}{1+q^{n}}
+\df{q^{2n}}{1-q^{2n}}\right) \\
=& \df{1}{16}\left(1+\sum_{n=1}^{\infty}(-1)^{n-1}\df{n^3q^n}{1+q^{n}}-1
+16\sum_{n=1}^{\infty}(-1)^{n-1}\df{n^3q^{2n}}{1-q^{2n}}\right)\\
=&\tfrac{1}{16}z^4x(1-x). \tag12.8h
\endalign
$$
The equality (12.8d) is now a trivial consequence of (12.8g) and (12.8h). 

To prove (12.8e),  first observe, by (12.8a) and (12.8f), that
$$ q\psi^4(q^2)\varphi^{2}(-q) = \tfrac{1}{16}z^3x\sqrt{1-x}. \tag12.8i $$
Next, 
$$\align \sum_{n=1}^{\i}(-1)^{n-1}\df{n^2q^n}{1+q^{2n}} =& 
-\sum_{n=1}^{\i}\df{4n^2q^{2n}}{1+q^{4n}}+
\sum_{n=1}^{\i}\df{(2n+1)^2q^{2n+1}}{1+q^{4n+2}}\\
=&-8\sum_{n=1}^{\i}\df{n^2q^{2n}}{1+q^{4n}} +
\sum_{n=1}^{\i}\df{n^2q^{n}}{1+q^{2n}}\\
=&-8\sum_{n=1}^{\i}\df{n^2q^{2n}}{1+q^{4n}} + \tfrac{1}{16}z^3x, \tag12.8j
\endalign
$$
by Entry 17(ii) in Chapter 17 of Ramanujan's second notebook 
\cite{21, p\. 138}.  To evaluate the sum on the far right side 
of (12.8j), we apply the process of duplication \cite{21, p\. 125} 
to Entry 17(ii) cited above.  Accordingly,
$$ \align
-8\sum_{n=1}^{\i}\df{n^2q^{2n}}{1+q^{4n}}=& - 
\frac{1}{2}\left(\frac{1}{2}z(1+\sqrt{1-x})\right)^3\left(\df{1-\sqrt{1-x}}
{1+\sqrt{1-x}}\right)^2 \\
=&-\tfrac{1}{16}z^3x(1-\sqrt{1-x}),\tag12.8k
\endalign
$$
after simplification. Putting (12.8k) into (12.8j), we readily find that
$$ \sum_{n=1}^{\i}(-1)^{n-1}\df{n^2q^n}{1+q^{2n}} = 
\tfrac{1}{16}z^3x\sqrt{1-x}. \tag12.8m $$
Combining (12.8i) and (12.8k), we complete the proof of the first part of (12.8e). 

To prove the second part of (12.8e), it clearly suffices to prove that
$$ S := \sum_{n=0}^{\i}\df{(2n+1)^2q^{2n+1}}{1+q^{4n+2}}
\equiv \sum_{n=0}^{\i}(-1)^{n}\df{(2n+1)^2q^{2n+1}}{1-q^{4n+2}} =:T 
\pmod{16} \tag12.8n 
$$
Now, 
$$\align
S =& \sum_{n=0}^{\i}\df{(2n+1)^2q^{2n+1}}{1-q^{4n+2}}-
2 \sum_{n=0}^{\i}\df{(2n+1)^2q^{6n+3}}{1-q^{8n+4}}\\
=& T +2 \sum_{n=0}^{\i}\df{(4n+3)^2q^{4n+3}}{1-q^{8n+6}}-
2 \sum_{n=0}^{\i}\df{(2n+1)^2q^{6n+3}}{1-q^{8n+4}}\\
\equiv & T + 2\sum_{n=0}^{\i}\df{q^{4n+3}}{1-q^{8n+6}} 
-2\sum_{n=0}^{\i}\df{q^{6n+3}}{1-q^{8n+4}} \pmod{16} \\
=& T +2\sum_{n=0}^{\i}\df{q^{6n+3}}{1-q^{8n+4}}\pmod{16}
-2\sum_{n=0}^{\i}\df{q^{6n+3}}{1-q^{8n+4}}\pmod{16}\\
=& T\pmod{16}, 
\endalign
$$
where in the antepenultimate line above we expanded the summands of the 
first series in geometric series and then reversed the order of summation. 
This completes the proof of (12.8n), and hence the proof of the second 
equality of (12.8e).]


It follows from all these that
$$ \align q(q;q)_{\i}^{24} =&-3\sum_{n=0}^{\i}\df{(2n+1)^3q^{2n+1}}{1-q^{4n+2}} + 
5\sum_{n=1}^{\i}\df{(2n)^3q^{2n}}{1-q^{4n}} 
-12\sum_{n=0}^{\i}\df{(2n+1)q^{2n+1}}{1-q^{4n+2}} \\
&+16\sum_{n=0}^{\i}(-1)^n\df{(2n+1)^2q^{2n+1}}{1-q^{4n+2}} 
-16\sum_{n=1}^{\i}\df{(2n)^2q^{2n}}{1+q^{4n}} +256J. 
\endalign
$$
Now equating only the odd powers of $ q $ we obtain
$$ \align\sum_{n=0}^{\i}\t(2n+1)q^{2n+1} =&
-3\sum_{n=0}^{\i}\df{(2n+1)^3q^{2n+1}}{1-q^{4n+2}} 
+16\sum_{n=0}^{\i}(-1)^n\df{(2n+1)^2q^{2n+1}}{1-q^{4n+2}}\\
&-12\sum_{n=0}^{\i}\df{(2n+1)q^{2n+1}}{1-q^{4n+2}}+256J. \endalign $$
But if $ n $ be of the form $ 4k+1 $ then it is easy to see that
$$ n^{11}+3n^3-16n^2+12n \equiv 0 \pmod{256}. $$
Changing $ n $ to $ -n $ in this formula we see that if $ n $ be of the form 
$ 4k-1 $ then
$$ n^{11}+3n^3+16n^2+12n \equiv 0 \pmod{256}. $$
It follows that
$$ \sum_{n=0}^{\i}\t(2n+1)q^{2n+1} =
\sum_{n=0}^{\i}\df{(2n+1)^{11}q^{2n+1}}{1-q^{4n+2}}+256J. $$
In other words, 
$$ \t(n) -\s_{11}(n) \equiv 0 \pmod{256} \tag12.9 $$
for all \underbar{odd} values of $ n, $ while the formula (12.8) combined 
with this enable us to find the residues of $ \t(n) $ for modulus $ 2^{11} $ 
for \underbar{even}  values of $ n. $ Thus
$$ \t(n) +24\s_{11}(n) \equiv 0 \pmod{2048} $$
for all values of $ n. $

It follows from (12.7) and (12.9) that 
$$ \t(n) -\s_{11}(n) \equiv 0 \pmod{176896} $$
for all odd values of $ n. $







\bigskip

\centerline{\bf Modulus 13}

\bigskip

\noindent{\bf 13.} In this case we start with the second series in (9.1) 
and the series
$$ 1-24\sum_{n=1}^{\i}\df{n^{13}q^n}{1-q^n} = Q^2R. \tag13.1 $$
It follows from these that
$$ Q^3-3R^2 = -2+13J; \qquad Q^2R = P+13J. \tag13.2 $$
Hence we have
$$ \align
(Q^3-R^2)^7 =& -2(R^2-1)^7+13J \\
=& -5R^6(3R^2-2)^4 -2R^4(3R^2-2)^3 +6R^4(3R^2-2)^2 \\
&-6R^2(3R^2-2)^2-6R^2(3R^2-2)-2(R^2-1)+13J \\
=& -5P^6-2P^4Q+6P^3R-6P^2Q^2-6PQR -(Q^3-R^2)+13J. \endalign
$$
But\footnotemark\footnotetext{See \cite{77}, where not all these
 equalities are given, but where the same methods can be employed to provide
 proofs.}
$$ \left\{\aligned 
& 5(P^6-15P^4Q+40P^3R-45P^2Q^2+24PQR)\\ 
&-(9Q^3+16R^2)=-248832\sum_{n=1}^{\i}n^5\s_1(n)q^n, \\
& 7(P^4Q-4P^3R+6P^2Q^2-4PQR)+(3Q^3+4R^2) = 41472\sum_{n=1}^{\i}n^4\s_3(n)q^n,\\
& 2(P^3R-3P^2Q^2+3PQR)-(Q^3+R^2) = -5184\sum_{n=1}^{\i}n^3\s_5(n)q^n, \\
& 9(PQ-R)^2+5(Q^3-R^2)=8640\sum_{n=1}^{\i}n^3\s_7(n)q^n, \\
& 5PQR-(3Q^3+2R^2) = -1584\sum_{n=1}^{\i}n\s_9(n)q^n, \\
& Q^3-R^2 = 1728\sum_{n=1}^{\i}\t(n)q^n; \endaligned \right. \tag13.3 
$$
and it is obvious that
$$ (q;q)^{168}_{\infty} = \df{(q^{169};q^{169})_{\i}}{(q;q)_{\i}}+13J. \tag13.4 $$
It is easy to see from all these that
$$\align &q^7\df{(q^{169};q^{169})_{\i}}{(q;q)_{\i}} =
(q^{169};q^{169})_{\i}\sum_{n=0}^{\i}p(n)q^{n+7} \\
=& \sum_{n=1}^{\i}\left\{n^5\s_1(n)-4n^4\s_3(n)-3n^3\s_5(n)+6n^2\s_7(n)
 -3n\s_9(n)+3\t(n)\right\}q^n+13J.\tag13.5\endalign
$$
It is easy to see by actual calculation that $\t(13) \equiv 8 \pmod{13} $ 
in virtue of (7.6) and hence $ \t(13n) -8\t(n) \equiv 0 \pmod{13}. $ It 
follows from this and (13.5) that
$$ \sum_{n=1}^{\i}p(13n-7)q^n\ (q^{13};q^{13})_{\i} = 
11\sum_{n=1}^{\i}\t(n)q^n +13J. \tag13.6 $$
It is not necessary to know all the details above in order to prove (13.6). 
The proof can be very much simplified as follows; using (9.8) and (13.2) we
 can show that
$$ (Q^3-R^2)^7 = q\df{dJ}{dq}+3(Q^3-R^2) +13J. \tag13.7 $$
It follows from this that
$$ q^7\df{(q^{169};q^{169})_{\i}}{(q;q)_{\i}} = 
q\df{dJ}{dq}+3\sum_{n=1}^{\i}\t(n)q^n+13J. \tag13.8 $$
>From this we easily deduce (13.6).

Again picking out the terms $ q^{13}, q^{26}, q^{39}, \dots $ 
in (13.6) we obtain [using the congruence $ \t(13n) \equiv 8\t(n) \pmod{13}$]
$$ \sum_{n=1}^{\i}p(13^2n-7)q^n\ (q;q)_{\i} = 10\sum_{n=1}^{\i}\t(n)q^n+13J. 
\tag13.9 $$

It follows from (13.5) that if 
$$ \sum_{n=1}^{\i}\lambda(n)q^n = q^7\df{(q^{169};q^{169})_{\i}}{(q;q)_{\i}}
 $$ so that $ \lambda(n+7) $ is the number of partitions of $ n $ as the sum
 of integers which are not multiples of 169, then
$$ \align &\lambda(n) - n^5\s_1(n)+4n^4\s_3(n)+3n^3\s_5(n)\\
&-6n^2\s_7(n)+3n\s_9(n)-3\t(n) \equiv 0 \pmod{13}. \endalign
$$
The results analogous to (10.61)--(10.9) in the case of modulus 13 are
$$ \t(5^{12\lambda-1}n) \equiv 0 \pmod{13} $$
if $ n $ is not a multiple of 5; 
$$ \t(7n) \equiv 0 \pmod{13} $$
if $ n $ is not a multiple of 7; 
$$ \t(11n) \equiv 0 \pmod{13} $$
if $ n $ is not a multiple of 11; 
$$ \t(13n)-8\t(n) \equiv 0 \pmod{13} $$
if $ n $ is any integer; 
$$ \t(19^{4\lambda-1}n) \equiv 0 \pmod{13} $$
if $ n $ is not a multiple of 19; 
$$ \t(23^{3\lambda-1}n) \equiv 0 \pmod{13} $$
if $ n $ is not a multiple of 23; 
$$ \t(29^{6\lambda-1}n) \equiv 0 \pmod{13} $$
if $ n $ is not a multiple of 29; and so on. 

\medskip

\noindent{\bf 14.} The formulae (13.6) and (13.9) can be written as
$$ \sum_{n=0}^{\i}p(13n+6)q^n=11(q;q)_{\i}^{11}+13J; \tag14.1 $$
and
$$ \sum_{n=0}^{\i}p(13^2n+162)q^n=23(q;q)_{\i}^{23}+13J. \tag14.2 $$
Since I began to write this paper I have found by a different method that 
if $ \lambda $ be any positive odd integer then
$$ \sum_{n=0}^{\i}p\left(13^{\lambda}n
+\df{11\cdot 13^{\lambda}+1}{24}\right)q^n
=-2^{(5\lambda-3)/2}(q;q)_{\i}^{11}+13J; \tag14.3 $$
and if $ \lambda $ be any positive even integer then
$$ \sum_{n=0}^{\i}p\left(13^{\lambda}n+\df{23\cdot 
13^{\lambda}+1}{24}\right)q^n=-2^{(5\lambda-2)/2}(q;q)_{\i}^{23}+13J. \tag14.4 $$
I shall reserve the discussion of these results to another paper. 

A number of results such as the following can be deduced from (14.3) and 
(14.4).  [Note that
$$ (q;q)_{\i}^{11} = 1 -11q+44q^2-55q^3-110q^4+374q^5-143q^6 + \cdots $$
and
$$ \left.(q;q)_{\i}^{23} = 
1 -23q+230q^2-1265q^3+3795q^4-3519q^5-16445q^6 + \cdots .\right]$$
If $ \lambda $ be any positive odd integer then
$$ \left\{ \aligned &p\left(\df{11\cdot 13^{\lambda}+1}{24}\right)
+2^{(5\lambda-3)/2}, \qquad  p\left(\df{35\cdot 13^{\lambda}+1}{24}\right)
+2^{(5\lambda-1)/2},\\
&p\left(\df{59\cdot 13^{\lambda}+1}{24}\right)-2^{(5\lambda+3)/2},\qquad 
p\left(\df{83\cdot 13^{\lambda}+1}{24}\right)-2^{5(\lambda+1)/2},\\
&p\left(\df{107\cdot 13^{\lambda}+1}{24}\right)-2^{(5\lambda+7)/2},\qquad 
p\left(\df{131\cdot 13^{\lambda}+1}{24}\right)-2^{5(\lambda+1)/2},\\
&p\left(\df{155\cdot 13^{\lambda}+1}{24}\right), \qquad  \endaligned \right. 
\tag14.5
$$
and so on are all divisible by 13; and if 
$ \lambda $ be any positive even integer then
$$ \left\{ \aligned &p\left(\df{23\cdot 
13^{\lambda}+1}{24}\right)+2^{(5\lambda-2)/2}, \qquad 
 p\left(\df{47\cdot 13^{\lambda}+1}{24}\right)+2^{(5\lambda+6)/2},\\
&p\left(\df{71\cdot 13^{\lambda}+1}{24}\right)-2^{(5\lambda+2)/2},\qquad 
p\left(\df{95\cdot 13^{\lambda}+1}{24}\right)-2^{(5\lambda+2)/2},\\
&p\left(\df{119\cdot 13^{\lambda}+1}{24}\right)-2^{(5\lambda-2)/2},\qquad 
p\left(\df{143\cdot 13^{\lambda}+1}{24}\right)+2^{(5\lambda+2)/2},\\
&p\left(\df{167\cdot 13^{\lambda}+1}{24}\right), \qquad   \endaligned
\right. 
\tag14.6
$$
and so on are all divisible by 13.
In other words if $n $ is fixed and $ \lambda + n $ is an even integer then
 the residue of 
$$p\left(\df{13^{\lambda}(12n-1)+1}{24}\right) \tag14.7$$
for modulus 13 can be completely ascertained. 

\bigskip

\centerline{\bf General Theory}

\bigskip

\centerline{\bf Modulus $ \varpi$}

\centerline{\bf where $ \varpi $ is a prime greater than 3}

\medskip

\noindent{\bf 15.} We start with the two series
$$ v_{\varpi-1}+(-1)^{(\varpi-1)/2}2(\varpi-1)\delta_{\varpi-1}
\sum_{n=1}^{\i}\df{n^{\varpi-2}q^n}{1-q^n} 
= \sum K^{\prime}_{\ell, m}Q^{\ell}R^m,\tag15.1 $$
where $ K^{\prime}_{\ell, m} $ is a constant integer and the summation 
extends over all positive integral values of $ \ell $ and $ m $ 
(including zero) such that
$$ 4\ell+6m = \varpi-1; $$
and 
$$ v_{\varpi+1}+(-1)^{(\varpi+1)/2}2(\varpi+1)\delta_{\varpi+1}
\sum_{n=1}^{\i}\df{n^{\varpi}q^n}{1-q^n} = \sum K_{\ell, m}Q^{\ell}R^m,\tag15.2$$
where $ K_{\ell,m} $ is a constant integer and the summation extends over 
all positive integral values of $ \ell $ and $ m $ (including zero) such that
$$ 4\ell+6m=\varpi+1. $$
In both the series $ v_s $ and $ \delta_s $ are the numerator and the 
denominator of $ B_s $ in its lowest terms where 
$$ B_2 = \df{1}{6}, \quad B_4 = \df{1}{30}, \quad B_6=\df{1}{42}, 
\quad B_8 = \df{1}{30}, \quad B_{10} = \df{5}{66}, \dots $$
are the Bernoulli numbers. Now by von Staudt's Theorem
$$ \delta_{\varpi-1}\equiv 0 \pmod{\varpi}, $$
and also we have
$$ n^{\varpi}-n \equiv 0 \pmod{\varpi}. $$
And so the left hand side in (15.1) is of the form
$$ c^{\prime}+\varpi J \tag15.3 $$
where $ c^{\prime} $ is a constant integer while that in (15.2) is of the form
$$ k + cP+\varpi J \tag15.31 $$
where $ c $ and $ k $ are constant integers.

It appears that $ k $ can be taken as zero always. This involves the assertion that
$$ 6v_{\varpi+1}+ (-1)^{(\varpi+1)/2}\df{\varpi+1}{2}\delta_{\varpi+1}\equiv
0 \pmod{\varpi}. 
\tag15.4 $$
I have not yet proved this result but in every particular case this can 
actually found to be true.  Thus (15.31) can be replaced by 
$$ cP+\varpi J. \tag15.5 $$
Now using (15.3), (15.5) and (9.8) we can show in particular cases that
$$ (Q^3-R^2)^{(\varpi^2-1)/24} = q\df{dJ}{dq}+(Q^3-R^2) \sum
k_{\ell,m}Q^{\ell}R^m + \varpi J \tag15.6 $$
where $  k_{\ell,m} $ is a constant integer and the summation extends over 
all positive integral values of $ \ell $ and $ m $ (including zero) such that
$$ 4\ell+6m =\varpi -13. $$
But it is obvious that
$$ (q;q)^{\varpi^2-1}_{\infty}
= \df{(q^{\varpi^2};q^{\varpi^2})_{\i}}{(q;q)_{\i}} +\varpi J. \tag15.7 $$
It follows from (15.6) and (15.7) that
$$q^{(\varpi^2-1)/24}\df{(q^{\varpi^2};
q^{\varpi^2})_{\i}}{(q;q)_{\i}}= q\df{dJ}{dq} 
+(Q^3-R^2) \sum k_{\ell,m}Q^{\ell}R^m  +\varpi J \tag15.8 $$
where the remark about the summation in (15.6) applies here also.  From 
this we can always deduce in every particular case that
$$ \align &\sum_{n=1}^{\i}p\left(n\varpi + 
\varpi\left[\df{\varpi}{24}\right]-\df{\varpi^2-1}{24}\right)q^{n+[\varpi/24]}
\ (q^{\varpi^2};q^{\varpi^2})_{\i} \\=& (Q^3-R^2)^{1+[\varpi/24]}
\sum k_{\ell,m}Q^{\ell}R^m  +\varpi J \tag15.9 \endalign $$
where $ k_{\ell,m} $ is a constant integer and the summation extends over all
positive integral values of $ \ell $ and $ m $ (including zero) such that
$$ 4\ell+6m = \varpi-13 $$
and $ [t] $ denotes as usual the greatest integer in $ t. $

Even though all these results are very difficult to prove in general they
 can be easily proved when $ \varpi \leq 23. $

\bigskip

\centerline{\bf Moduli 17, 19 and 23}

\bigskip

\noindent{\bf 16.} In these cases we can easily prove that
$$ \sum_{n=1}^{\i}p(17n-12)q^n \ (q^{17};q^{17})_{\i} 
=7\sum_{n=1}^{\i}\t_2(n)q^n +17J, \tag16.1 $$
where
$$\sum_{n=1}^{\i}\t_2(n)q^n = Qq(q;q)^{24}_{\i}; $$
$$ \sum_{n=1}^{\i}p(19n-15)q^n \ (q^{19};q^{19})_{\i} 
=5\sum_{n=1}^{\i}\t_3(n)q^n +19J, \tag16.2 $$
where
$$\sum_{n=1}^{\i}\t_3(n)q^n = Rq(q;q)^{24}_{\i}; $$
and
$$ \sum_{n=1}^{\i}p(23n-22)q^n \ (q^{23};q^{23})_{\i} =
\sum_{n=1}^{\i}\t_5(n)q^n +23J, \tag16.3 $$
where
$$\sum_{n=1}^{\i}\t_5(n)q^n = QRq(q;q)^{24}_{\i}. $$
I have stated without proof in my previous
 paper\footnotemark\footnotetext{See \cite{77, eq\. (108)}.} that
$$ \left\{\aligned \sum_{n=1}^{\i}\df{\t_2(n)}{n^s} = 
\prod_p\df{1}{1-\t_2(p)p^{-s}+p^{15-2s}}, \\
\sum_{n=1}^{\i}\df{\t_3(n)}{n^s} = 
\prod_p\df{1}{1-\t_3(p)p^{-s}+p^{17-2s}}, \\
\sum_{n=1}^{\i}\df{\t_4(n)}{n^s} = 
\prod_p\df{1}{1-\t_4(p)p^{-s}+p^{19-2s}}, \\
\sum_{n=1}^{\i}\df{\t_5(n)}{n^s} = 
\prod_p\df{1}{1-\t_5(p)p^{-s}+p^{21-2s}}, \\
\sum_{n=1}^{\i}\df{\t_7(n)}{n^s} = 
\prod_p\df{1}{1-\t_7(p)p^{-s}+p^{25-2s}}, \endaligned
\right.\tag16.4
$$
where
$$\sum_{n=1}^{\i}\t_4(n)q^n = Q^2q(q;q)^{24}_{\i} $$
and
$$\sum_{n=1}^{\i}\t_7(n)q^n = Q^2Rq(q;q)^{24}_{\i}, $$
and $ p $ assumes all prime values. All these seem to be capable of 
proof as the case of $ \t(n) $ by Mordell's 
method.\footnotemark\footnotetext{loc\. cit.}

Now using (16.4) we can deduce from (16.1), (16.2) and (16.3) that 
$$ \sum_{n=1}^{\i}p(n17^2-12)q^n \ (q;q)_{\i} 
= c_2\sum_{n=1}^{\i}\t_2(n)q^n +17J, \tag16.5 $$
$$ \sum_{n=1}^{\i}p(n19^2-15)q^n \ (q;q)_{\i} 
= c_3\sum_{n=1}^{\i}\t_3(n)q^n +19J, \tag16.6 $$
and
$$ \sum_{n=1}^{\i}p(n23^2-22)q^n \ (q;q)_{\i} 
= c_5\sum_{n=1}^{\i}\t_5(n)q^n +23J, \tag16.7 $$
where $ c_2, c_3 $ and $ c_5 $ are constants. 

I have found that there are formulae quite analogous to those for modulus 
13 even in these cases. I shall reserve the discussion of these as well as 
those for higher primes to another paper; but I shall consider in the II 
part of this paper the analogous formulae for the smaller primes 5, 7, and 11.

The corresponding formulae for primes greater than 23 are not quite 
analogous.  For instance in the cases of 

\centerline{\underbar{moduli 29 and 31}}

\noindent we have
$$ \sum_{n=1}^{\i}p(29n-6)q^{n+1}\ (q^{29};q^{29})_{\i} 
= 8\sum_{n=1}^{\i}\Omega_2(n)q^n +29J, \tag16.8$$
where
$$\sum_{n=1}^{\i}\Omega_2(n)q^n =Qq^2(q;q)_{\i}^{48}; $$
and
$$ \sum_{n=1}^{\i}p(31n-9)q^{n+1}\ (q^{31};q^{31})_{\i} 
= 10\sum_{n=1}^{\i}\Omega_3(n)q^n +31J, \tag16.9$$
where
$$\sum_{n=1}^{\i}\Omega_3(n)q^n =Rq^2(q;q)_{\i}^{48}. $$

The functions
$$\sum_{n=1}^{\i}\df{\Omega_2(n)}{n^s}, \qquad
\sum_{n=1}^{\i}\df{\Omega_3(n)}{n^s}  $$
are obviously not capable of a single product as in (16.4); but they are,
 as a matter of fact, the differences of two such products. 

\medskip

\noindent{\bf 17.} I have not yet investigated the residues of $ \t(n) $ for
 other moduli besides what was stated before but the case 23 seems to be 
(comparatively) simple. For it appears that if
$$ \sum_{n=1}^{\i}\lambda(n)q^n = q(q;q)_{\i}(q^{23};q^{23})_{\i} $$
so that
$$ \t(n)-\lambda(n)\equiv 0 \pmod{23} \tag17.1 $$
then
$$ \sum_{n=1}^{\i}\df{\lambda(n)}{n^s} = 
\df{1}{1-23^{-s}}\prod\n1\prod\n2\prod\n3, \tag17.2 $$
where
$$ \prod\n1 = \prod_p\df{1}{1-p^{-2s}}, $$
$ p $ assuming all prime values of the 
form\footnotemark\footnotetext{This can be written as $ p^{11}\equiv -1 \pmod{23}.$}
$$ p\equiv 5, 7, 10, 11, 14, 15, 17, 19, 20, 21, 22 \pmod{23}\tag17.3 $$
and
$$ \prod\n2 = \prod_p\df{1}{1+p^{-s}+p^{-2s}}$$
$ p $ assuming all prime values of the 
form\footnotemark\footnotetext{This can be written as $ p^{11}\equiv 1 \pmod{23}.$}
$$ p \equiv 1, 2, 3, 4, 6, 8, 9, 12, 13, 16, 18 \pmod{23} \tag17.4 $$
except of the form $ 23a^2+b^2, $ and 
$$ \prod\n3 = \prod_p\df{1}{(1-p^{-s})^2} $$
$ p $ assuming all primes of the form $ 23a^2+b^2. $
Thus $ \lambda(n) $ can be completely determined and consequently the
residues of $ \t(n) $ for modulus 23 can be completely ascertained.

Suppose now that
$$ \left\{\aligned t_n=0, \qquad & \t(n) \equiv 0 \pmod{23}; \\
t_n =1, \qquad & \t(n) \not\equiv 0 \pmod{23}. \endaligned \right. \tag17.5 $$
Then it is easy to see from ( ) that
$$ \sum_{n=1}^{\i}\df{t_n}{n^s} = \prod\n1\prod\n2\prod\n3, \tag17.6 $$
where
$$ \prod\n1 = \prod_p\df{1}{1-p^{-2s}}, $$
$ p $ assuming all primes of the form (17.3),
$$ \prod\n2 = \prod_p\df{1+p^{-s}}{1-p^{-3s}}, $$
$ p $ assuming all primes of the form (17.4) except those of the form 
$ 23a^2+b^2, $ and 
$$ \prod\n3 = \prod_p\df{1-p^{-22s}}{(1-p^{-s})(1-p^{-23s})} $$
$ p $ assuming all primes of the form $ 23a^2+b^2. $ 

It is easy to prove from (17.6) by quite elementary methods that
$$ \sum_{k=1}^nt_k = o(n); \tag17.7 $$
and by transcendental methods that
$$ \sum_{k=1}^nt_k = C\int_1^n\df{dx}{(\log x)^{1/2}} 
+ O\left(\df{n}{(\log n)^r}\right),\tag17.8 $$
where $ r $ is any positive number and 
$$ \align C =& \df{66^{1/2}}{23^{3/4}}   \df{1-2^{-2}}{1-2^{-3}}
\df{1-3^{-2}}{1-3^{-3}}\df{1-13^{-2}}{1-13^{-3}}\df{1-29^{-2}}{1-29^{-3}}
 \cdots \\ &\times \df{1}{\left\{(1-5^{-2})(1-7^{-2})(1-11^{-2})(1-17^{-2})
\cdots\right\}^{1/2}} \\
& \times \df{1-59^{-22}}{1-59^{-23}}\df{1-101^{-22}}{1-101^{-23}}
\df{1-167^{-22}}{1-167^{-23}}\cdots , \endalign $$
2, 3, 13,\dots being primes of the form (17.4) except those of the form 
$ 23a^2+b^2, $ and 5, 7, 11, 17, \dots being primes of the form (17.3) 
and 59, 101, 167, \dots are those of the form $ 23a^2+b^2. $  Thus we see 
that $ \t(n) $ is almost always divisible by 23. 

We have also shown that among the values of $ \t(n), $ multiples of 3, 7 
and 23 are more or less equally numerous while the multiples of 5 are less 
numerous than these and multiples of 2 are the most numerous. 

Since
$$ \align (1-p^{-s})(1-p^{11-s}) =& (1-p^{-2s})-(p^{11}+1)(p^{-s}-p^{-2s})\\
=&(1-p^{-s})^2-(p^{11}-1)(p^{-s}-p^{-2s}) \endalign $$
it is easy to see from (17.2) and (12.7) that if the prime divisors of $ n $ 
are of the form (17.3) or of the form 
$ 23a^2+b^2$\footnotemark\footnotetext{Some may be of one form
 and some may be of the other form.} then
$$ \t(n) -\s_{11}(n) \equiv 0 \pmod{15893}, \tag17.9$$
15893 being $ 23\cdot 691. $ If, in addition to the restrictions on the 
values of $ n $ in (17.9), we impose the restriction that $ n $ is odd also 
then if follows from (12.9) that
$$ \t(n) -\s_{11}(n) \equiv 0 \pmod{4068608}, $$
4068608 being $ 23\cdot 256 \cdot 691. $

\bigskip

\centerline{\bf Modulus 121}

\bigskip

\noindent{\bf 18.}
The case of modulus $ \varpi^2 $ seems to be much more complicated than the 
case of modulus $ \varpi $ even though the method is practically the same as 
may be seen from the case of modulus 49. I shall now consider the case of 
modulus 121. 

It is easy to show by using (9.2) that 
$$ \align (Q^3-R^2)^5 =& P(Q^3-3R^2)(3P^3-PQ+4R) +4QR(4P^3Q-3P^2R+2QR) \\
& -26P^5+23P^3Q+16P^2R-22PQ^2 +9QR +121J. \tag18.1 \endalign $$
>From this we can deduce that
$$ \align q^5\df{(q^{11};q^{11})_{\i}^{11}}{(q;q)_{\i}} =&
\sum_{n=1}^{\i}\left[n^4\left\{a_1\s_1(n)+b_1\s_{11}(n)\right\}
+n^3\left\{a_2\s_3(n)+b_2\s_{13}(n)\right\}\right.\\
&+n^2\left\{a_3\s_5(n)+b_3\s_{15}(n)\right\}
+n\left\{a_4\s_7(n)+b_4\s_{17}(n)\right\}\tag18.2\\
&\left. +c_1n^2\t_2(n)+c_2n\t_3(n)+c_3\t_4(n)\right]q^n +121J \endalign  $$
where the $a$'s, $b$'s and $c$'s are constant integers and $\t_2(n), \t_3(n) 
$ and $ \t_4(n) $ are the same as in (16.4).  But it is easy to show that
$$ \left\{\aligned &\t_2(n)-n\s_3(n), \\
&\t_3(n)-n\s_5(n),\qquad  \equiv 0  \pmod{11}. \\
&\t_4(n)-n\s_7(n), 
\endaligned\right. \tag18.3$$
It is easy to see from (16.4) that 
$$ \t_4(11n)-\t_4(11)\t_4(n) \equiv 0 \pmod{121}, \tag18.4 $$
and by actual calculation we find that
$$ \t_4(11) \equiv 0 \pmod{11}. \tag18.5 $$
It is also obvious that
$$ \s_{17}(n)-\s_7(n) \equiv 0 \pmod{11}. \tag18.6 $$
Now remembering (18.3)--(18.6) and picking out the terms $ q^{11}, q^{22},
 q^{33}, \dots $ in ( ) we obtain
$$ \sum_{n=1}^{\i}p(11n-5)q^n\ (q^{11};q^{11})_{\i} 
= 11 \sum_{n=1}^{\i}n\s_7(n)q^n + 121J. \tag18.7 $$
It follows from this that 
$$ p(121n-5)\equiv 0 \pmod{121}, \tag18.8 $$
and 
$$\align & p(11n-5)-p(11n-126)-p(11n-247)\\+&p(11n-610)+\cdots-11n\s_7(n) 
\equiv 0 \pmod{121}.  \tag18.9\endalign $$

\medskip

\noindent{\bf 19.} In concluding the first part of this paper I shall 
consider the numbers which are the divisors of $ \t(n) $ for almost all 
values of $ n. $

Suppose that $ \varpi_1, \varpi_2, \varpi_3, 
\dots $ are an infinity of primes such that
$$ \sum_{n=1}^{\i}\df{1}{\varpi_n} \tag19.1 $$
is a \underbar{divergent} series and also suppose that $ a_2, a_3, a_5, a_7,
 \dots $ assume some or all of the positive integers (including zero) but that 
$ a_{\varpi_1}, a_{\varpi_2}, a_{\varpi_3}, 
\dots $ \underbar{never} assume the value 
unity. Then it is easy to show that the number of numbers of the form
$$ 2^{a_2}\cdot3^{a_3}\cdot5^{a_5}\cdot7^{a_7}\cdots \tag19.2 $$
not exceeding $ n $ is of the form
$$ o(n). \tag19.3 $$
In particular if $ a_{\varpi} $ \underbar{never} assumes the value unity for all
 prime values of $ \varpi $ of the form
$$ \varpi\equiv c \pmod{k}, \tag19.4 $$
where $ c $ and $ k $ are any two integers which are prime to each other, 
then the number of numbers of the form (19.2) is of the form 
$$ o(n) \tag19.5 $$
and more accurately is of the form
$$ O\left(\df{n}{(\log n)^{1/(k-1)}}\right) \tag19.6 $$
where $ k $ is the same as in (19.4). 

Thus for example if $ s $ be an odd positive integer, the number of values 
of $ n $ not exceeding $ n $ for which $ \s_s(n) $ is \underbar{not} 
divisible by $ k, $ where $ k $ is any positive integer, is of the form 
$$ o(n) \tag19.7 $$
and more accurately is of the form 
$$ O\left(\df{n}{(\log n)^{1/(k-1)}}\right). \tag19.8 $$
For if $ n $ be written in the form
$$ 2^{a_2}\cdot3^{a_3}\cdot5^{a_5}\cdot7^{a_7}\cdots  $$
then we have
$$ \s_s(n) = \prod_p\df{p^{s(1+a_p)}-1}{p^s-1}, \qquad p = 2, 3, 5, 7, 11,
 \dots .$$
Since $ s $ is odd, $ \s_s(n) $ is divisible by $ k $ at any rate when 
$ a_p = 1 $ for all values of $ p $ of the form 
$$ p\equiv -1 \pmod{k} $$
and hence the results stated follow. Thus we see that, if $ s $ is odd, 
$ \s_s(n) $ is divisible by any given integer for almost all values of $ n. $

It follows from all these and the formulae in Sections 4, 8, 12, and 17,  that
$$ \t(n) \equiv 0 \pmod{ 2^{5}\cdot3^{3}\cdot5^{2}\cdot7^{2}\cdot 23
 \cdot 691} \tag19.9 $$
for almost all values of $ n. $

It appears that $ \t(n) $ is almost always divisible by any power of 2, 
3, and 5. It also appears from Section 9 that there are reasons to suppose that 
$ \t(n) $ is almost always divisible by 11 also. But I have no evidence at 
present to say anything about the other powers of 7 and other primes one way
 or the other. 

Among the values of $ \t(n) $ multiples of 2, 3, 5, 7 and 23 are very 
numerous from the beginning but multiples of 691 begin at a very late 
stage. For instance $ \t(n) $ is divisible by 23 for 132 values of 
$ n $ not exceeding 200 while the first value of $ n $ for which 
$ \t(n) $ is divisible by 691 is 1381 and this is the only such value 
of $ n $ among the first 5000 values. 

\bigskip

\centerline{\bf II}

\medskip

\centerline{\bf Moduli 5 and 25}

\medskip

\noindent{\bf 20.} In this second part we shall use $ J_1, J_2, J_3 $ and $
G_1, G_2, G_3 $ to denote functions of $ q $ with integral powers of $ q $
as well as integral coefficients. These are the same functions in the same
section unlike $ J.$ We shall also use $ J $ in the same sense as in the
first part. 

We start with Euler's identities
$$ (q;q)_{\i} = \sum_{n=-\i}^{\i}(-1)^nq^{n(3n-1)/2} \tag20.1 $$
and Jacobi's identity
$$ (q;q)_{\i}^3 =\sum_{n=0}^{\i}(-1)^n(2n+1)q^{n(n+1)/2}. \tag20.11 $$

It is easy to see from (20.1) that
$$ \df{(q^{1/5};q^{1/5})_{\i}}{(q^5;q^5)_{\i}} = J_1-q^{1/5}+q^{2/5}J_2.
\tag20.2 $$
Now cubing both sides we obtain
$$ \align \df{\sum_{n=0}^{\i}(-1)^n(2n+1)q^{n(n+1)/10}}
{\sum_{n=0}^{\i}(-1)^n(2n+1)q^{5n(n+1)/2}}=& (J_1^3-3J_2^2q) 
- q^{1/5}(3J_1^2-J_2^3q) +3J_1q^{2/5}(1+J_1J_2) \\ &- q^{3/5}(1+6J_1J_2)
+3J_2q^{4/5}(1+J_1J_2). \tag20.3 \endalign $$
But it is easy to see that
$$
\df{\sum_{n=0}^{\i}(-1)^n(2n+1)q^{n(n+1)/10}}
{\sum_{n=0}^{\i}(-1)^n(2n+1)q^{5n(n+1)/2}} = G_1+q^{1/5}G_2+5q^{3/5}.
\tag20.31 $$
Hence
$$ J_1(1+J_1J_2)=0, \qquad 1+6J_1J_2 = -5, \qquad J_2(1+J_1J_2) =0. \tag20.4 $$
These three equations give one and the same relation between $ J_1 $ and $
J_2, $ viz\. 
$$ J_1J_2 = -1. $$
Using this we obtain
$$ \df{(q^5;q^5)_{\i}}{(q^{1/5};q^{1/5})_{\i}} =
\df{1}{J_1-q^{1/5}+q^{2/5}J_2} \tag20.5 $$
$$= \df{(J_1^4+3J_2q)+q^{1/5}(J_1^3+2J_2^2q) +q^{2/5}(2J_1^2+J_2^3q)+
q^{3/5}(3J_1+J_2^4q)+5q^{4/5}}{J_1^5-11q+q^2J_2^5} $$
by rationalizing the denominator $ J_1-q^{1/5}+q^{2/5}J_2. $ It follows from
(20.5) that
$$ \sum_{n=0}^{\i}p(5n+4)q^n\ (q^5;q^5)_{\i} = \df{5}{J_1^5-11q+q^2J_2^5}.
\tag20.6 $$
But we see from (20.2) that 
$$ \df{(\o q^{1/5};\o q^{1/5})_{\i}}{(q^5;q^5)_{\i}} = J_1 -\o q^{1/5}
+\o^2q^{2/5}J_2, \tag20.21 $$
where $ \o^5 =1. $ Now writing the five values of $ \o $ in (20.21) and
multiplying them together we obtain
$$ \df{(q;q)_{\i}^6}{(q^5;q^5)_{\i}^6} = J_1^5-11q+q^2J_2^5. \tag20.7 $$
It follows from this and (20.6) that 
$$ \sum_{n=0}^{\i}p(5n+4)q^n = 5\df{(q^5;q^5)_{\i}^5}{(q;q)_{\i}^6}. \tag20.8 $$
It follows that
$$ p(5n-1) \equiv 0 \pmod5. \tag20.81 $$

Again the right hand side in (20.8) is of the form
$$ 5\df{(q^5;q^5)_{\i}^4}{(q;q)_{\i}} +25J. $$
It follows from this and (20.81) that the coefficients of $ q^4, q^9,
q^{14}, \dots $ in this are all multiples of 25 and consequently the
coefficient of $ q^{5n-1} $ in the left hand side of (20.8) is a multiple of
25. In other words
$$ p(25n-1) \equiv 0 \pmod{25}. \tag20.82 $$
It follows also from (20.8) that
$$ \sum_{n=0}^{\i}p(5n+4)q^n = 5(q;q)_{\i}^{19}+125J. $$

\medskip

\centerline{\bf Modulus 125}

\medskip

\noindent{\bf 21.} Changing $ q $ to $ q^{1/5} $ in (20.8) and arguing as
before, using (20.5) and (20.7) we find that
$$ \align  \sum_{n=0}^{\i}p(25n+24)q^n =& 5^2\cdot63
\df{(q^5;q^5)_{\i}^6}{(q;q)_{\i}^7}
+5^5\cdot52q\df{(q^5;q^5)_{\i}^{12}}{(q;q)_{\i}^{13}}
+5^7\cdot63q^2\df{(q^5;q^5)_{\i}^{18}}{(q;q)_{\i}^{19}}\\
&+5^{10}\cdot6q^3\df{(q^5;q^5)_{\i}^{24}}{(q;q)_{\i}^{25}}
+5^{12}q^4\df{(q^5;q^5)_{\i}^{30}}{(q;q)_{\i}^{31}}.\tag21.1
\endalign
$$
Now
$$\df{(q^5;q^5)_{\i}^6}{(q;q)_{\i}^7} =\sum_{n=0}^{\i}(-1)^n(2n+1)q^{n(n+1)/2}
(q^5;q^5)_{\i}^4 +5J \quad \text{etc.}\tag21.2 $$
and the coefficients of $ q^{5n-1},q^{5n-2}, q^{5n-3} $ in 
$ \sum_{n=0}^{\i}(-1)^n(2n+1)q^{n(n+1)/2} $ are easily seen to be zero or
multiples of 5. It follows that the coefficients of $ q^{5n-1},q^{5n-2},
 q^{5n-3} $ in the left hand side of (21.1) are multiples of 125. In
other words 
$$ \cases p(125n-1) \\
p(125n-26) \equiv 0 \pmod{125}  \\
p(125n-51). 
\endcases \tag21.3
$$
It is also easy to see from (21.1) that
$$   \sum_{n=0}^{\i}p(25n+24)q^n = 75(q;q)_{\i}^{23} +125J. \tag21.4 $$
The right hand side in (21.4) can be written in the form
$$ 75\df{(q;q)_{\i}^{48}}{(q^{25};q^{25})_{\i}} + 125J. \tag21.5 $$
But it is easy to show that
$$ (Q^3-R^2)^2 = -2 \sum_{n=1}^{\i}(n^3-n)\sigma_1(n)q^n+5J. \tag21.6 $$
[To prove (21.6), we need Ramanujan's formula \cite{77, Table
III}, \cite{82, p\. 142}
$$ 6912\sum_{n=1}^{\infty}n^3\sigma_1(n)q^n = 6P^2Q-8PR+3Q^2-P^4. $$
Using this formula together with (1.4) and (1.2), we can readily prove that
$$ 2\sum_{n=1}^{\i}(n^3-n)\sigma_1(n)q^n = -1+2P^2-P^4 + 5J. $$
On the other hand, from (1.2) and (1.3),
$$ (Q^3-R^2)^2 = 1-2P^2+P^4 +5J. $$
The last two equalities yield (21.6).]
It follows that
$$   \sum_{n=0}^{\i}p(25n+24)q^{n+2}\ (q^{25};q^{25})_{\i} 
= 25\sum_{n=1}^{\i}(n^3-n)\sigma_1(n)q^n+125J. \tag21.7 $$
In other words
$$ \align &p(25n-26)-p(25n-651)-p(25n-1276)\\&
+p(25n-3151) +\cdots -25(n^3-n)\sigma_1(n) \equiv 0
\pmod{125}. \tag21.8\endalign  $$

$ p(199) $ is the coefficient of $ q^7 $ in (21.2). 
$$ \align p(199) =& 5^2\cdot63\cdot12195+5^2\cdot52\cdot60541 + 5^7\cdot63\cdot66862
+ 5^{10}\cdot6\cdot29575+5^{12}\cdot 6448\\ =&  3646072432125. \endalign $$
\medskip

\centerline{\bf Moduli $ 5^4, 5^5, \dots $}

\medskip

\noindent{\bf 22.} Changing again $ q $ to $ q^{1/5} $ in (21.1) and arguing
as before using (20.5) and (20.7) we can show that
$$ \sum_{n=0}^{\i}p(125n+99)q^n = \sum_{r=1}^{25}a_r\df{(q^5;q^5)_{\i}^{6r-1}}
{(q;q)_{\i}^{6r}}, \tag22.1 $$
where the $ a$'s are positive integers such that $ a_1 = p(99) = 5^3\cdot
1353839 $ and $ a_2, a_3, a_4, \dots $ contain higher powers of 5 than $ a_1
$ as factors.  It is easy to see from this that
$$ \sum_{n=0}^{\i}p(125n+99)q^n =4\cdot 5^3(q;q)_{\i}^{19}+5^4J. \tag22.2$$
In this way arguing as before, we can show that if $ \lambda $ be any
positive odd integer, then
$$ \sum_{n=0}^{\i}p\left(\df{19\cdot5^{\lambda}+1}{24}+5^{\lambda}n\right)q^n
=\sum_{\nu=1}^{5^{\lambda -1}}a_{\nu}\df{(q^5;q^5)_{\i}^{6\nu-1}}
{(q;q)_{\i}^{6\nu}}, \tag22.3 $$
where the $a$'s are positive integers such that $ a_2, a_3, a_4, \dots $
contain higher powers of 5 than $ a_1 $ as factors; and if $ \lambda $ be a
positive even integer then
$$ \sum_{n=0}^{\i}p\left(\df{23\cdot5^{\lambda}+1}{24}+5^{\lambda}n\right)q^n
=\sum_{\nu=1}^{5^{\lambda -1}}a_{\nu}\df{(q^5;q^5)_{\i}^{6\nu}}
{(q;q)_{\i}^{6\nu+1}}, \tag22.4 $$
where the $a$'s have the same properties as before. We deduce from (22.3)
and (22.4) that if $ \lambda $ is a positive odd integer then
$$ \sum_{n=0}^{\i}p\left(\df{19\cdot5^{\lambda}+1}{24}+5^{\lambda}n\right)q^n
= c_{\lambda}\cdot 5^{\lambda}(q;q)_{\i}^{19}+5^{\lambda+1}J, \tag22.5$$
and if $ \lambda $ is a positive even integer then 
$$ \sum_{n=0}^{\i}p\left(\df{23\cdot5^{\lambda}+1}{24}+5^{\lambda}n\right)q^n
= c_{\lambda}\cdot 5^{\lambda}(q;q)_{\i}^{23}+5^{\lambda+1}J, \tag22.6 $$
where $ c_{\lambda} $ in both cases is a constant.

We easily deduce from these that if $ \lambda $ is an odd integer greater
than 1, then
$$ \cases p\left(5^{\lambda}n-\df{5^{\lambda-1}-1}{24}\right) \\
p\left(5^{\lambda}n-\df{5^{\lambda+1}-1}{24}\right)\text{\footnotemark}
 \quad \equiv 0 \pmod{5^{\lambda}} \\
p\left(5^{\lambda}n-\df{49\cdot 5^{\lambda-1}-1}{24}\right), \endcases\tag22.7 $$
and if $ \lambda $ is a positive even integer, then
$$ p\left(5^{\lambda}n-\df{5^{\lambda}-1}{24}\right) \equiv 0
\pmod{5^{\lambda}}. \tag22.8 $$
\footnotetext{$\lambda$ may also be 1 in this formula.}
\medskip

\noindent{\bf 23.} We have seen that we can take $ c_1 = 1, c_2 = -2, c_3 =
4 $ in (22.5) and (22.6).  It appears from Section 22 that $ c_{\lambda} $ may
probably be some simple function such as $ (-2)^{\lambda}$. If we calculate
a few more values of $ c_{\lambda}, $ we can definitely know what it is.
Then we can make use of the formulae (22.5) and (22.6) to determine
completely the residues of 
$$ p\left(5^{\lambda}n-\df{5^{\lambda+1}-1}{24}\right) $$
for odd values of $ \lambda $ and those of
$$ p\left(5^{\lambda}n-\df{5^{\lambda}-1}{24}\right) $$
for even values of $ \lambda $ for modulus $ 5^{\lambda+1}.$ [To determine these
residues, we need the expansions
$$ \align (q;q)_{\i}^{19}=& 1- 19q+152q^2-627q^3+1140q^4+988q^5-9063q^6 \\
&+14212q^7+7410q^8-44270q^9+22781q^{10}+38114q^{11} \\
&+36176q^{12}-137256q^{13}-154850q^{14}+480605q^{15}+\cdots 
\endalign
$$
and
$$ \align (q;q)_{\i}^{23}=& 1- 23q+230q^2-1265q^3+3795q^4-3519q^5-16445q^6 \\
&+64285q^7-64515q^8-120175q^9+354706q^{10}-123763q^{11} \\
&-407560q^{12}-48530q^{13}+817190q^{14}+1464341q^{15}+\cdots
\endalign
$$
in, respectively, (22.5) and (22.6).] 
 Thus for
instance if follows immediately from (22.5) and (22.6) that if $ \lambda $
is an odd integer then 
$$ \align
& p\left(5^{\lambda}-\df{5^{\lambda+1}-1}{24}\right)-5^{\lambda}c_{\lambda}, \qquad
\qquad p\left(2\cdot
5^{\lambda}-\df{5^{\lambda+1}-1}{24}\right)-5^{\lambda}c_{\lambda},\\
& p\left(3\cdot 5^{\lambda}-\df{5^{\lambda+1}-1}{24}\right)-2\cdot
5^{\lambda}c_{\lambda}, \qquad
 p\left(4\cdot 5^{\lambda}-\df{5^{\lambda+1}-1}{24}\right)+ 2\cdot
5^{\lambda}c_{\lambda},\\
& p\left(5\cdot 5^{\lambda}-\df{5^{\lambda+1}-1}{24}\right),\qquad\qquad\qquad
 p\left(6\cdot 5^{\lambda}-\df{5^{\lambda+1}-1}{24}\right)-2\cdot
5^{\lambda}c_{\lambda},\\
& p\left(7\cdot 5^{\lambda}-\df{5^{\lambda+1}-1}{24}\right)-2\cdot
5^{\lambda}c_{\lambda},\qquad
 p\left(8\cdot 5^{\lambda}-\df{5^{\lambda+1}-1}{24}\right)-2\cdot
5^{\lambda}c_{\lambda},\\
& p\left(9\cdot 5^{\lambda}-\df{5^{\lambda+1}-1}{24}\right),\qquad\qquad\qquad
 p\left(10\cdot 5^{\lambda}-\df{5^{\lambda+1}-1}{24}\right),\\
& p\left(11\cdot 5^{\lambda}-\df{5^{\lambda+1}-1}{24}\right)-
5^{\lambda}c_{\lambda},\qquad
 p\left(12\cdot 5^{\lambda}-\df{5^{\lambda+1}-1}{24}\right)+
5^{\lambda}c_{\lambda},\\
& p\left(13\cdot
5^{\lambda}-\df{5^{\lambda+1}-1}{24}\right)-5^{\lambda}c_{\lambda},\qquad
 p\left(14\cdot 5^{\lambda}-\df{5^{\lambda+1}-1}{24}\right)+
5^{\lambda}c_{\lambda},\\
& p\left(15\cdot 5^{\lambda}-\df{5^{\lambda+1}-1}{24}\right),\qquad\qquad\qquad
 p\left(16\cdot 5^{\lambda}-\df{5^{\lambda+1}-1}{24}\right),
\endalign
$$
and so on are all multiples of $ 5 ^{\lambda+1}$; and if $ \lambda $ is an
even integer, then
$$ \align
& p\left(5^{\lambda}-\df{5^{\lambda}-1}{24}\right) -5^{\lambda}c_{\lambda},\qquad
 p\left(2\cdot 5^{\lambda}-\df{5^{\lambda}-1}{24}\right)-2\cdot
5^{\lambda}c_{\lambda},\\
& p\left(3\cdot 5^{\lambda}-\df{5^{\lambda}-1}{24}\right),\qquad\qquad
 p\left(4\cdot 5^{\lambda}-\df{5^{\lambda}-1}{24}\right),\\ 
& p\left(5\cdot 5^{\lambda}-\df{5^{\lambda}-1}{24}\right),\qquad\qquad
 p\left(6\cdot 5^{\lambda}-\df{5^{\lambda}-1}{24}\right)
-5^{\lambda}c_{\lambda},\\
& p\left(7\cdot 5^{\lambda}-\df{5^{\lambda}-1}{24}\right),\qquad\qquad
 p\left(8\cdot 5^{\lambda}-\df{5^{\lambda}-1}{24}\right),\\
& p\left(9\cdot 5^{\lambda}-\df{5^{\lambda}-1}{24}\right),\qquad\qquad
 p\left(10\cdot 5^{\lambda}-\df{5^{\lambda}-1}{24}\right),\\
& p\left(11\cdot 5^{\lambda}-\df{5^{\lambda}-1}{24}\right)
-5^{\lambda}c_{\lambda},\qquad
 p\left(12\cdot 5^{\lambda}-\df{5^{\lambda}-1}{24}\right) -2\cdot
5^{\lambda}c_{\lambda},\\
& p\left(13\cdot 5^{\lambda}-\df{5^{\lambda}-1}{24}\right),\qquad\qquad
 p\left(14\cdot 5^{\lambda}-\df{5^{\lambda}-1}{24}\right),\\
& p\left(15\cdot 5^{\lambda}-\df{5^{\lambda}-1}{24}\right),\qquad\qquad
 p\left(16\cdot 5^{\lambda}-\df{5^{\lambda}-1}{24}\right)
-5^{\lambda}c_{\lambda},
\endalign
$$
and so on are all multiples of $ 5^{\lambda +1}.$

\medskip

\centerline{\bf Moduli 7 and 49}

\medskip

\noindent{\bf 24.} It is easy to see from (20.1) that
$$ \df{(q^{1/7};q^{1/7})_{\i}}{(q^{7};q^{7})_{\i}} = J_1 +q^{1/7}J_2
-q^{2/7} +q^{5/7}J_3. \tag24.1$$
Now cubing both sides we obtain
$$ \align
&\df{\sum_{n=0}^{\i}(-1)^n(2n+1)q^{n(n+1)/14}}
{\sum_{n=0}^{\i}(-1)^n(2n+1)q^{7n(n+1)/2}}\\ =& (J_1^3+3J_2^2J_3q-6J_1J_3q) 
+q^{1/7}(3J_1^2J_2-6J_2J_3q+J_3^2q^2) 
+3q^{2/7}(J_1J_2^2-J_1^2+J_3q)\\ &+ q^{3/7}(J_2^3-6J_1J_2+3J_1J_3^2q) 
+3q^{4/7}(J_1-J_2^2+J_2J_3^2q)\\&+ 3q^{5/7}(J_2+J_1^2J_3-J_3^2q) 
+q^{6/7}(6J_1J_2J_3-1). 
\endalign
$$
But it is easy to see that
$$ 
\df{\sum_{n=0}^{\i}(-1)^n(2n+1)q^{n(n+1)/14}}
{\sum_{n=0}^{\i}(-1)^n(2n+1)q^{7n(n+1)/2}} = G_1+q^{1/7}G_2+q^{3/7}G_3
-7q^{6/7}. 
$$
Hence
$$ \cases J_1J_2^2-J_1^2+J_3q &= 0, \\
J_1-J_2^2+J_2J_3^2q &= 0, \\
J_2+J_1^2J_3-J_3^2q &= 0, \\
6J_1J_2J_3-1        &= -7. \endcases \tag24.2
$$
All these four equations give the two independent relations
$$ J_1J_2J_3 = -1, \qquad \df{J_1^2}{J_3}+\df{J_2}{J_3^2} =q.\tag24.2a $$
Now write (24.1) in the form
$$ \df{(\o q^{1/7};\o q^{1/7})_{\i}}{(q^{7};q^{7})_{\i}} = J_1 +\o
q^{1/7}J_2 -\o^2 q^{2/7} +\o^5 q^{5/7}J_3, \tag24.3 $$
where $ \o^7=1.$ Again writing the seven values of $ \o $ in (24.3) and
multiplying them together and using (24.2a) we can show that
$$ \align
J_1^7+J_2^7q+J_3^7q^5 =& \df{(q;q)_{\i}^8}{(q^7;q^7)_{\i}^8} +
14q\df{(q;q)_{\i}^4}{(q^7;q^7)_{\i}^4}+57q^3, \tag{24.4}\\
J_1^3J_2+J_2^3J_3q+J_3^3J_1q^2 
=& - \df{(q;q)_{\i}^4}{(q^7;q^7)_{\i}^4} -8q,\tag{24.5} \\
J_1^2J_2^3+J_3^2J_1^3q+J_2^2J_3^3q^2 =&- \df{(q;q)_{\i}^4}{(q^7;q^7)_{\i}^4}
-5q. \tag{24.6}
\endalign
$$
Again taking the reciprocals of both sides in (24.1) and rationalizing the
denominator by using as in Section 20, we can show that
$$
\sum_{n=0}^{\i}p(7n+5)q^n = 7\df{(q^7;q^7)_{\i}^3}{(q;q)_{\i}^4} +
49q\df{(q^7;q^7)_{\i}^7}{(q;q)_{\i}^8}. $$

\medskip






$$ 7^2\cdot 2546, 7^4\cdot 48 \cdot 934, 7^5\cdot 1418989, 7^8\cdot 335400. $$
$$ \{p(47)q^3+\cdots \}(q^{49};q^{49})_{\i} 
=7\sum_{n=1}^{\i}\left\{22n^4\sigma_0(n)-21n^2\sigma_1(n)-\tau(n)\right\}q^n
+7^3J. $$





\bigskip

%\centerline{\bf A page from Ramanujan's Lost Notebook}

%\medskip

%$$ \align \t_2(n) -\s_{15}(n) &\equiv 0 \pmod{3617} \\
%\t_2(n) -n\s_{13}(n) &\equiv 0 \pmod{16170} \\
%\t_2(n) -2n\s_{9}(n) +n^2\s_3(n) &\equiv 0 \pmod{600} \\
%\t_3(n) -\s_{17}(n) &\equiv 0 \pmod{43867} \\
%\t_3(n) -n\s_{15}(n) &\equiv 0 \pmod{6006} \\
%\t_3(n) -n^2\s_{1}(n) &\equiv 0 \pmod{540} \\
%\t_3(n) -6n^2\s_{9}(n)+5n\s_3(n) &\equiv 0 \pmod{150} \\
%\t_3(n) +n\s_{9}(n) +n\s_3(n)-3\t(n) &\equiv 0 \pmod{588} \\
%\t_4(n) -\s_{19}(n) &\equiv 0 \pmod{174611}
%\endalign $$

%\bigskip

\centerline{\bf COMMENTARY}

\bigskip


0. The designation, Section 0, for the first batch of Ramanujan's insertions
is due to the present authors. 

K\. G\. Ramanathan \cite{75} also observed that $ \t(n) $ is 
even unless $ n $ is an odd square. 

The congruences $ \t(7n-r) \equiv 0 \pmod7, r = 0,1, 2, 4, $ were evidently 
first proved by J\. R\. Wilton \cite{107}.  G\. H\. Hardy, in his 
book {\it Ramanujan} \cite{47, pp\. 165--166} also gives a proof, 
as does Ramanathan \cite{76}. 

The congruences $ \t(23n-r) \equiv 0 \pmod{23}, $ where $ r $ is a quadratic 
residue modulo 23, were also first established by Wilton \cite{107}. 

\medskip

1. Without the insertions, the beginning of the paper actually begins with the 
definitions of the Eisenstein series $ P, Q, $ and $ R, $ which are denoted 
by $ L, M, $ and $ N, $ respectively, in Ramanujan's notebooks 
\cite{81}. Since the remainder of  this section was extracted for 
\cite{80} with additional details supplied by Hardy, we have 
not added  more details here.  However, it seems appropriate here to provide 
an introduction to congruences for the partition function  in arithmetic 
progressions, since a large portion of the manuscript focuses on this topic.

In this manuscript Ramanujan proves his well known congruences
for $p(n)$, namely,
$$\align
p(5n+4)&\equiv 0\pmod 5,\\
p(7n+5)&\equiv 0\pmod 7,\\
p(11n+6)&\equiv 0\pmod{11}.
\endalign
$$
These congruences are the first cases of the infinite families,
$$
\align
    p(5^kn+\delta_{5,k})&\equiv 0 \pmod{5^k},\tag{C1.1}\\
    p(7^kn+\delta_{7,k})&\equiv 0 \pmod{7^{ [k/2]+1}},\\
    p(11^kn+\delta_{11,k})&\equiv 0 \pmod{11^k}, \\
\endalign
$$
where $\delta_{p,k}:=1/24 \pmod{p^k}.$
The literature on these congruences is extensive, and there are now many
proofs and approaches to them, e.g\., \cite{3},  \cite{5}, 
\cite{30}, \cite{38}, \cite{39}, 
\cite{40}, \cite{41},
 \cite{42},  \cite{49}, \cite{50},
\cite{51},  \cite{53}, 
\cite{64}, \cite{68}, \cite{73},
   and \cite{104}.

These congruences are indeed surprising for they appear
to be examples of a very rare and isolated phenomenon.
In fact, Ramanujan \cite{79}, \cite{82, p\. 230}
remarked that
 ``It appears that there are no equally simple 
properties for any
moduli involving primes other than these three."



In view of Ramanujan's claim, it is natural to ask about the frequency
of congruences for $p(n)$ and the possibility of finding new ones.
In this direction, the second author has made some progress 
\cite{67}, \cite{69} towards quantifying the rarity of 
such congruences,
and A\. O\. L\. Atkin and  J\. N\. 
O'Brien \cite{6}, \cite{7} have found other 
congruences for
$p(n)$. For instance, Atkin has proved that
$$
  p(17303n+237)\equiv 0\pmod{13}.
$$
It is reasonable to conclude that such congruences
are quite rare, but not so rare that one cannot find
infinitely many such congruences.






\medskip

2. The congruence $ \t(n) \equiv n\s(n) \pmod5 $ was established by Wilton
\cite{106}, and is also proved in Hardy's 
book \cite{47, pp\. 166--167}.  This congruence was generalized by 
R\. P\. Bambah and S\. Chowla \cite{15}, who proved that, if $ n $
 is not a multiple of 5, then
$$ \t(n) \equiv 5n^2\s_7(n)-4n\s_9(n) \pmod{5^3}. $$

The asymptotic formula (2.7) can be proved by using the method devised by 
E\. Landau in his book \cite{60, Sect\. 183} to determine an 
asymptotic formula for the number of integers $ \leq x $ that can be 
represented as a sum of two squares. Alternatively, one can appeal to 
a general Tauberian theorem, such as that proved by H\. Delange 
\cite{35}. However, as first pointed out by G\. K\. Stanley
\cite{98}, the claim (2.8) is false.  Indeed, 
by using the ideas of Landau \cite{60, Sects\. 176--183}, 
one can establish an asymptotic formula of the shape
$$ \sum_{n\leq x}t_n = C\df{x}{(\log x)^{1/4}}
\left(1+\sum_{n=1}^{r-1}\df{c_n}{(\log x)^n} + 
O\left(\df{1}{(\log x)^r}\right)\right), $$
for certain constants $ c_n, 1 \leq n \leq r-1. $  However, generally, 
these constants are not equal to those which would be obtained by 
successive integrations by parts in (2.8). Ramanujan made a similar error 
in his first letter to Hardy \cite{82, p\. xxiv}, 
\cite{25, p\. 24} when he claimed that the number of integers 
$ \leq x $ that can be represented as a sum of two squares is asymptotic to 
a constant times a similar integral. 
See either the sections of Landau's book cited above or Hardy's book
 \cite{47, pp\. 60--63}. In Sections 6, 11, and 17, Ramanujan
records similar asymptotic formulas, and, in contrast to the asymptotic
formula in this section, calculates the leading coefficients in each case.  
  R\. A\. Rankin \cite{87} has verified that the leading terms,
including the coefficients, are correct in each of the instances cited by
Ramanujan. 


\medskip

4. The congruence (4.2) was first proved in print by Wilton
\cite{106} and later by Bambah \cite{10}.

 Rankin \cite{85, p\. 5} has pointed out that Ramanujan's 
conjecture (4.3) is false for $ k \geq 4. $ Observe that 443 is prime and
 that its powers are congruent to $ \pm 1, \pm 443 \pmod{5^4}. $ From 
Watson's \cite{105} table of values for $ \tau(n), $ 
$ \tau(443) \equiv -58 \pmod{5^4}. $ Hence, no integers $ a $ and $ b $ 
exist for which (4.3) holds with $ n = 443 $ and $ k \geq 4. $

However, the congruence (4.4) is true; a congruence equivalent to (4.4) 
was first proved by J.--P\. Serre \cite{91}, \cite{101}. 

The equality below (4.4) is a special instance of the relation
$$ \tau(p^{n+1}) = \tau(p)\tau(p^n)-p^{11}\tau(p^{n-1}), \qquad n > 1,\tag{C4.1} $$
where $ p $ is a prime, which along with (7.6), were first proved by 
L\. J\. Mordell \cite{62}, after Ramanujan had made these 
conjectures in his paper \cite{77, Sect\. 18}, 
\cite{82, p\. 153}. 

Proofs of either of the famous equalities (4.5) or 
(4.6) (or both) have been given by, in chronological order, Ramanujan 
\cite{78}, \cite{82, pp\. 210--213}, H\. B\. C\. Darling 
\cite{34}, L\. J\. Mordell \cite{62}, H\. 
Rademacher and H\. S\. Zuckerman \cite{73}, 
\cite{72, pp\. 186--202}, S\. D\. Chowla \cite{31}, 
 D\. Kruswijk \cite{58}, 
W\. N\. Bailey \cite{8}, \cite{9},
J\. M\. Dobbie \cite{37}, N\. J\. Fine \cite{39}, 
 S\. Raghavan \cite{74},  H\. H\. Chan \cite{30}, and
M\. D\. Hirschhorn \cite{49}, \cite{50}. These 
proofs are quite varied.  Some authors use $q$--series; some, such as 
Rademacher, Zuckerman, and Raghavan, use the the theory of modular forms; 
Chan's proof uses a variant of one of Ramanujan's trigonometric series 
identities in  \cite{77}. 

As indicated by Ramanujan, (4.6) is a companion to (4.5).  Bailey, Chan, 
Darling, Mordell, and Raghavan in the aforementioned papers have also given 
proofs of (4.6). In contrast to (4.5), equality (4.6) can be found in 
Ramanujan's notebooks \cite{21, p\. 257, Entry 9(i)}. 

\medskip

5. Since this section was also extracted by Hardy for \cite{80}, 
we have not added details here. 

\medskip

6. The congruence (6.2) was established by Ramanathan \cite{76}, 
Gupta \cite{46}, and Bambah \cite{11}. 

The comments made in Section 2 about Ramanujan's asymptotic formulas have 
analogues here.  Although the asymptotic formula (6.6a) is correct, 
Ramanujan's stronger claim (6.7) is false, since the constants obtained by 
integrating by parts in (6.7) do not generally match those obtained in a
 proper asymptotic expansion of $ \sum_{k=1}^nt_k. $

\medskip

7. The content of this section can be found with more detail in Rushforth's
 paper \cite{89}.  (In the second equality of (7.1), 
Rushforth \cite{89, eq\. (7.3)} wrote $ -2R^2 $ for $  2R^2, 
$ and in (7.4) \cite{89, penultimate equality on p\. 407} 
Rushforth wrote $ -2n\s_3(n) $ for $ 2n\s_3(n)$.)

\medskip

8. Ramanujan's proof of (8.1) can be found in his paper \cite{78}, 
\cite{82}, while other proofs of (8.1) have been given by 
Mordell \cite{62}, 
Rademacher and Zuckerman \cite{73}, N\. J\. Fine
\cite{39},  O\. Kolberg \cite{55},
Raghavan \cite{74}, and Chan \cite{30}.  Further 
identities akin to (8.1) and (8.2) have been established by Rademacher 
\cite{71}, \cite{72, pp\. 252--279}. These authors then 
continue to prove (8.3). 

Equality (8.4) is true, and its truth is equivalent to
the assertion that
$\eta^3(z)\eta^3(7z)$ is a Hecke eigenform with complex multiplication
in $S_3(\Gamma_0(7),\chi_{-7})$, where \linebreak $S_k(\Gamma_0(N),\chi)$  denotes
the complex vector space of  cusp forms of weight $ k $ with respect
to the congruence subgroup $\Gamma_0(N)$ with Nebentypus character
$\chi$ \cite{88}. (The notation $S_k(\Gamma_0(N))$, with 
$\chi$ absent, simply
means that the character $\chi$ is trivial.) Here the character $\chi_{-7}$ denotes
the usual Kronecker character for the field $\Q(\sqrt{-7})$.
That this form is an eigenform follows immediately from
the fact that this space is one dimensional \cite{33}. 
To deduce (8.4) in a more
elementary fashion, first notice that Jacobi's identity
$$
  q\prod_{n=1}^{\infty}(1-q^{8n})^3=\eta^3(8z)=
  \sum_{n=0}^{\infty}(-1)^n(2n+1)q^{(2n+1)^2}
$$
implies that
$$
  \eta^3(8z)\eta^3(56z)=\sum_{x,y\geq 0}(-1)^{x+y}(2x+1)(2y+1)
  q^{(2x+1)^2+7(2y+1)^2}.
$$
Proving Ramanujan's claim now follows  after a straightforward
computation.

The claims regarding the Euler product expansions 
$\Pi_1$ and $\Pi_2$ follow easily from the theory of complex
multiplication.


In regard to the congruence (8.6), we remark that O\. Kolberg 
\cite{57} proved the beautiful congruence
$$ \t(n) \equiv n\s_9(n) \pmod{49}, \qquad \text{if } 
\left(\frac{n}{7}\right)=-1. $$

\medskip

9. This proof is given in more detail in \cite{80}. 

\medskip

10. Equality (10.4) is true,
and its truth is equivalent to the assertion that
$\eta^2(z)\eta^2(11z)$ is an eigenform of the Hecke operators
acting on $S_2(\Gamma_0(11))$. That this form is an eigenform
follows immediately from the fact that this space is one dimensional.

Some of Ramanujan's congruences for $\tau(n)$
are immediate consequences of its multiplicative properties.
For instance, Ramanujan \cite{77}, \cite{82, p\. 153,
 eq\. (103)} conjectured and Mordell \cite{62} proved that, 
  if $m$ and $n$ are relatively prime integers, then
$$\tau(mn)=\tau(m)\tau(n). \tag{C10.1}
$$
For example, the congruences (10.8) and (10.9) follow
easily since $\tau(19)=10661420$ $\equiv 0\pmod{11}$ 
and $\tau(29)=128406630  \equiv 0\pmod{11}$. Other congruences follow from
(C4.1) or from (C4.1) and (C10.1) together.



In 1969, P\. Deligne \cite{36} proved Serre's conjecture 
\cite{92} 
 on the existence of $\ell$-adic Galois representations
        $\rho_\ell$ attached to modular forms on $\Gamma_{0}(N)$.
Then, in $1972$, Swinnerton--Dyer \cite{99} determined
 the possible images of
$\tilde{\rho}_{\ell}$, the reduction $\mod \ell$ of $\rho_{\ell}$, and showed
that `small' images imply certain congruences for the coefficients
of modular forms. 

The existence of these representations and their study has been
at the forefront of arithmetic geometry ever since Serre
formulated his original conjectures. Every congruence
for $\tau(n)$ involving divisor functions, and the congruence
$$
  \tau(n)\equiv 0\pmod{23} 
$$
for $\fracwithdelims(){n}{23}=-1,$ follows from this theory.
For more details,  readers should consult \cite{36}, 
\cite{92}, \cite{99}, \cite{100}, 
\cite{101}, \cite{102}.



\medskip

11. Ramanujan's speculation that $ p(n) $ is odd more often than it is even 
is not substantiated by more extensive calculations. Indeed, it is a long 
outstanding conjecture that asymptotically $ p(n) $ is equally often even 
and odd. In Sections 1, 5, 9, and 11, based on a table of values for $ p(n), 
1 \leq n \leq 200, $ computed by P\. A\. MacMahon, Ramanujan offers 
conjectures on the distribution of $ p(n) $ modulo 5, 7, 11, and 3, 
respectively. We examine these conjectures in detail.

If $D(r,M)$ denotes the proportion of integers $n$ for which
$p(n)\equiv r\pmod M$ (assuming that such densities exist),
Ramanujan conjectured (in Sections 11, 11, 1, 5, and 9, respectively) that
$$
D(0,2) < D(1,2),
$$
$$
D(i,3)=\tfrac{1}{3}, \ \ \ \ \ {\text {\rm for}}\ 0\leq i \leq 2, 
$$ 
$$
D(i,5)=\cases \frac{1}{3}, \ \ \ \ \ &{\text {\rm if}}\ i=0,\\
\frac{1}{6}, \ \ \ \ \ &{\text {\rm if}}\ 1\leq i \leq 4, 
\endcases 
$$
$$
D(i,7)=\cases \frac{1}{4}, \ \ \ \ \ &{\text {\rm if}}\ i=0,\\
\frac{1}{8}, \ \ \ \ \ &{\text {\rm if}}\ 1\leq i \leq 6, 
\endcases 
$$
$$ 
D(i,11)=\cases \frac{1}{6}, \ \ \ \ \ &{\text {\rm if}}\ i=0,\\
\frac{1}{12}, \ \ \ \ \ &{\text {\rm if}}\ 1\leq i \leq 10. 
\endcases
$$

>From elementary considerations, we show that Ramanujan's conjectures for 
$ D(i,M), M = 5, 7, 11,$ are unreasonable. Remove the values $ n = 5k+4, 
7k+5, 11k+6, $ from consideration when $ M = 5, 7, 11, $ respectively. 
Assuming that the remaining values of $ p(n) $ are distributed randomly 
among the $ M $ residue classes in each of these three cases, we would expect that
$$ D(i,M) = \cases \frac{2M-1}{M^2}, \qquad \text{if } i=0, \\
\frac{M-1}{M^2}, \qquad \text{if } 1 \leq i\leq M. 
\endcases
$$
In particular, we expect that $ D(0,5) =\tfrac{9}{25} $ and $  D(i,5) = 
\tfrac{4}{25}, 1 \leq i\leq 4, $ in contrast to Ramanujan's conjectures. 
Similar discrepancies exist for $ M = 7, 11. $

Let $\delta(r,M)$ denote the proportion of integers $n\leq 100000$
for which $p(n)\equiv r\pmod M$. Here are some values of $\delta(r,M)$
for $M\in \{2, 3, 5, 7, 11, 13\}$.  
$$
\matrix
\underline{r} & \underline{\delta(r,2)} & \underline{\delta(r,3)}
& \underline{\delta(r,5)} & \underline{\delta(r,7)} & \underline{\delta
(r,11)} & \underline{\delta(r,13)}\\
0 & 0.498 & 0.333 & 0.362 & 0.272 & 0.174 & 0.080 \\ 
1 & 0.502 & 0.332 & 0.158 & 0.121 & 0.083 & 0.078 \\ 
2 &  * &    0.334 & 0.161 & 0.122 & 0.083 & 0.076  \\
3 &  * &        *   & 0.160 & 0.122 & 0.082 & 0.077\\
4 &  * &        *   & 0.158 & 0.122 & 0.084 & 0.077\\
5 &  * &        *   &    *   & 0.120 & 0.083 & 0.076\\
6 &  * &        *   &    *   & 0.120 & 0.083 & 0.075\\
7 &  * &        *   &    *   &   *    & 0.081 & 0.077\\
8 &  * &        *   &    *   &   *    & 0.082 & 0.076\\
9 &  * &        *   &    *   &   *    & 0.081 & 0.078\\
10 & * &        *   &    *   &   *    & 0.082 & 0.075\\
11 & * &        *   &    *   &   *    & *     & 0.076\\
12 & * &        *   &    *   &   *    & *      & 0.077\\
\endmatrix
$$

As this data suggests, if the densities $\delta(r,M)$ are well defined,
then Ramanujan's conjectures are mostly incorrect. 
The data suggests that he may be correct when $M=3$, but not for
any other values. 
At present, very little is known about the
densities $\delta(r,M)$ apart from lower bounds for
$\delta(0,M)$ for those $M$ possessing congruences of the
sort discussed in the commentary for Section 1. 
At present, by the work of S. Ahlgren \cite{1},
\cite{2},  
 and J.--L\. Nicolas, I\. Z\.  Ruzsa, A\. S\'ark\"ozy,
 and J.--P\. Serre \cite{66},
it is known that
$$\align
&\# \{ n\leq X \ : \ p(n)\equiv 0\pmod 2 \} \gg \sqrt{X}, \\
&\# \{ n\leq X \ : \ p(n)\equiv 1\pmod 2 \} \gg\ \frac{\sqrt{X}}{\log X},\\
&\# \{ n\leq X \ : \ p(n)\not \equiv 0\pmod M \} \gg \frac{\sqrt{X}}{\log X}.
\endalign
$$
On the other hand, the second author \cite{70} 
has shown that if $M\geq 5$ is prime, then
$$
  \# \{ n \leq X \ : \ p(n)\equiv 0\pmod M \} \gg_{M} X.
$$  

In a similar direction, M. Newman \cite{65} conjectured 
that every positive integer $M$ has the property that
each residue class $m\pmod M$ has infinitely many
integers $n$ for which $p(n)\equiv m\pmod M$.
 A. O. L. Atkin \cite{6}, O. Kolberg \cite{56}, 
and  M. Newman  \cite{65}  have verified
this conjecture for each $M\in \{2, 5, 7, 13\}$.
Because of the validity of (14.3) and (14.4), Ramanujan had also proved 
this conjecture when $M=13$. Motivated by Ramanujan's work, the second 
author \cite{70} has proved Newman's conjecture for 
every prime $ M < 1000$, with the exception of $ M=3.$  
He also has found a simple 
criterion for verifying Newman's conjecture for any prime $ M \geq 5.$


The equality (11.4) can be found in a fragment published with Ramanujan's 
lost notebook \cite{83, p\. 354, eq\. (1.42)}. A proof may be
 found in Berndt's paper \cite{23, Entry 21}. 

The congruence (11.8) has been proved several times in the literature.  Most 
frequently, it is given in the equivalent formulation
$$ \t(n) \equiv \cases \s(n) \pmod3, \qquad &\text{if }(3,n)=1, \\
0 \pmod3, \qquad &\text{if } 3|n. \endcases $$
For proofs, see papers by D\. P\. Banerji \cite{19}, Bambah and 
Chowla \cite{12}, Gupta \cite{45}, and Bambah, Chowla,
 Gupta, and Lahiri \cite{18}.  Bambah and Chowla 
\cite{15} proved the generalization
$$ \t(n) \equiv (n^2+k)\s_7(n) \pmod{3^4}, \qquad (3,n) =1, $$
where $ k = 0, $ if $ n \equiv 1 \pmod3, $ and $ k = 9, $ if $ n \equiv 2 \pmod3.$

The asymptotic formulas in (11.8a) need to be corrected in the same manner 
that the asymptotic formulas in Sections 2 and 6 needed to be recast. 

\medskip

12. The sums $ \sum_{n=1}^{\i}n^aq^n, $ where $ a $ is a positive integer, 
can be explicitly evaluated in terms of Eulerian polynomials 
\cite{20, p\. 113, Entry 4}. 

Bambah, Chowla, and Gupta \cite{17} and Bambah, Chowla, Gupta, 
and Lahiri \cite{18} proved the congruence
$$ \t(n) \equiv \s(n) \pmod8, \qquad \text{if $ n $ is odd}, $$
which, in fact, is implied by the first congruence in (12.1). 

We have been unable to find the identity (12.2) in the literature prior
 to the work of Ramanujan.  On page 257 in his second notebook 
\cite{81}, Ramanujan actually offers a general formula for
$$ S_{2r} := \sum_{n=1}^{\i}\df{n^{2r}q^n}{1+q^n+q^{2n}}, $$
which was first proved by Berndt, S\. Bhargava, and F\. G\. Garvan 
\cite{24}, \cite{22, p\. 143}.  The values of $ S_{2r}, 
1 \leq r \leq 4, $ are explicitly given by Ramanujan. The formula for 
$ S_2 $ is given  without proof in an equivalent form in a paper by J\. M\.
 and P\. B\. Borwein \cite{28}, and this equivalent formula is
 proved in \cite{29} by the Borweins and Garvan. A particularly
 simple proof of (12.2), based on an identity of N\. J\. Fine, has been 
given by S\. H\. Son \cite{97, Lemma 2.6}. 

A proof of the first congruence in (12.3) was given by Bambah and Chowla 
\cite{14}. The second congruence in (12.3) was established in 
another paper by the same authors \cite{15}. 

 Bambah and Chowla \cite{14} proved the second congruence in (12.4). 

The first proof in print of (12.7) was evidently given by J\. R. Wilton
\cite{106}. Later proofs were found by G\. N\. Watson 
\cite{103} and  D\. H\. Lehmer \cite{61}. 

As with corresponding results in Sections 2, 6, and 11, the asymptotic 
formula (12.7a) needs to be corrected. 

Bambah and Chowla \cite{16} gave the first published proof of (12.9). 

The congruence below (12.9) is false, in general.  For example, it is false
for $ n = 1, 3, 4, 5.$
\medskip

13. In Sections 2, 6,  10,  11, and 13
Ramanujan considers the $t$-regular partition functions
$\lambda(n)$ whose generating function are given by
$$
\sum_{n=0}^{\infty}\lambda(n)q^n=
\sum_{n=0}^{\infty}b_t(n)q^n:=\prod_{n=1}^{\infty}\frac{(1-q^{tn})}{
(1-q^n)}.
$$
The dependence of $\lambda$ on $t$ is always clear from the context.
For instance, in Section 2, he considers the case
where $t=25$. In this case he shows that $\lambda(n)$
is almost always a multiple of 5.
A recent paper by B. Gordon and the second author \cite{43} makes 
considerable progress in describing this phenomenon for all $t$.
Let $p_1^{a_1}p_{2}^{a_2}\cdots p_m^{a_{m}}$ be the prime
factorization of $t$.
By \cite{43, Thm\. 1}, if $p_i$ is a prime for which
$p_i^{a_i}\geq \sqrt{t}$, then, for every positive integer $k$,
almost every integer $n$ has the property that $b_t(n)$ is a multiple
of $p_i^k$. This theorem immediately implies all of Ramanujan's
claims of this sort for the functions $\lambda(n)$.

Equality (13.6) has been proved by H\. S\. Zuckerman 
\cite{108} and W\. H\. Simons \cite{96}. 

\medskip

14.  The claims (14.1)--(14-6) are among the most fascinating
results in the unpublished manuscript. For example,
these results indicate that
$$\sum_{n=0}^{\infty}p(13n+6)q^{24n+11}
\equiv 11\eta^{11}(24z)\pmod{13}.
$$
M\. Newman \cite{64} has proved some of these claims. However, 
the second author \cite{70} has shown that this phenomenon also holds
with respect to other moduli. In particular, 
if $m\geq 5$ is prime and $k$ is a positive integer, then
$$
\sum_{n=0}^{\infty}p\left (\frac{m^kn+1}{24}\right )q^n
$$
is the reduction modulo $m$ of a holomorphic cusp form of weight
 $\frac{m^2-m-1}{2}$. This implies that results
like (14.1)--(14.6) exist for every prime $m\geq 5$, not just
$m=13$. Moreover, using the theory of Hecke
operators of half-integral weight, the Shimura correspondence, and the 
theory of Galois representations, the second author \cite{70} 
has proved that for every prime
$m\geq 5$ that there are integers $0\leq b < a$ for which
$$
  p(an+b)\equiv 0\pmod m
$$
for every non-negative integer $n$. 

%The equalities (14.3) and (14.4) may be proved
%by showing that the Atkin operator $U_{13}$ when applied
%iteratively to the power series $\sum_{n=0}^{\infty}p(13n+6)q^{24n+11}$
%forms a cycle of power series $\pmod{13}$ of length 24.
%The proof of this result and its natural generalizations will
%appear in a paper by the second author \cite{70}. 

\medskip

15. 
 In Section 15 Ramanujan gives a brief description of the method
he employs to obtain generating functions of the type
$$
  \sum p(\varpi n + b_{\varpi})q^n \pmod \varpi,
$$
where $\varpi > 3$ is prime.
Let $B_k$ denote the $k$th Bernoulli  number in contemporary notation.
 Note that Ramanujan's convention for Bernoulli numbers is different from 
the contemporary one in which the Bernoulli numbers $ B_k $ are defined by
$$ \df{x}{e^x-1} = \sum_{k=0}^{\i}\df{B_k}{k!}x^k,  \qquad |x| <
2\pi. \tag{C15.1}$$
For every positive even integer $k$, define
positive coprime integers $v_k$ and $\delta_k$ by
$$
  \frac{v_k}{\delta_k}:=|B_k|.
$$

If $k\geq 4$ is an even integer,  let
$$E_{2k}(z):=1-\frac{4k}{B_{2k}}\sum_{n=1}^{\infty}\sigma_{2k-1}(n)q^n
$$
denote the normalized  Eisenstein series of weight $ 2k $ with respect
to the full modular group $\Gamma_0(1)$, where $ B_k $ is defined by
(C15.1). 

Now assume that $\varpi > 3$ is prime.
It is easy to check that the left hand side of (15.1)
is $v_{\varpi-1}\cdot E_{\varpi-1}(z)$ and that
the left hand side of (15.2) represents
$v_{\varpi+1}\cdot E_{\varpi+1}(z)$. That both of these
$q$-series have integer coefficients is obvious, and that
they have the desired representation as sums of $Q^{\ell}R^m$
is well known \cite{86, p\. 199, Thm\. 6.1.3}. 
In fact, every holomorphic modular form
with respect to the full modular group has such a representation.

As Ramanujan claims, (15.3) and (15.31) are easy deductions
from Fermat's Little Theorem and the von Staudt and Claussen
theorem. However, the claim that $k=0$ in (15.31) is not entirely
clear. In fact, this is one of the questions in the manuscript
that Ramanujan admits still requires proof. 

\proclaim{Proposition} Ramanujan's claim that
$k\equiv 0\pmod \varpi$ in \rom{(15.31)} is true.
\endproclaim
\demo{Proof}
A simple calculation verifies Ramanujan's assertion that the
truth of (15.4) implies that $k$ is indeed 0, or more precisely
$0\pmod \varpi$. In particular, it suffices to prove that
$$
  12v_{\varpi+1}+(-1)^{\frac{\varpi+1}{2}}\delta_{\varpi+1}\equiv 0
\pmod \varpi. \tag{C15.2}
$$
Using the well known Voronoi congruences \cite{52, p\. 237, Prop. 15.2.3}, 
we find, for every integer $a$ coprime to $\varpi$, that
$$
  (a^2-1)v_{\varpi+1}\equiv a\cdot (-1)^{\frac{\varpi-1}{2}}\delta_{\varpi+1}
\sum_{j=1}^{\varpi-1}j \fracwithdelims[]{ja}{\varpi} \pmod \varpi,
$$
since the sign of $B_{2k}$
is $(-1)^{k+1}$ for every positive integer $k$.
Therefore we find that
$$
\left ( \frac{a^2-1}{a}\right ) v_{\varpi+1}+(-1)^{\frac{\varpi+1}{2}}
\delta_{\varpi+1}\sum_{j=1}^{\varpi-1}j\fracwithdelims[]{ja}{\varpi}
\equiv 0\pmod \varpi. \tag{C15.3}
$$
In view of (C15.2) and (C15.3), it suffices to prove that, for each
 integer $a$ coprime
to $\varpi$,
$$
\sum_{j=1}^{\varpi-1}j\fracwithdelims[]{ja}{\varpi}\equiv \frac{a^2-1}{12a}
\pmod \varpi. \tag{C15.4}
$$

We now prove (C15.4) by examining Dedekind sums. If $k$ is a positive integer,
and $h$ is coprime to $k$, then the Dedekind sum $s(h,k)$ is defined by
$$
  s(h,k):=\sum_{j=1}^{k-1}\frac{j}{k}\left ( \frac{hj}{k}-
\fracwithdelims[]{hj}{k}-\frac{1}{2}\right ).
$$
It is easy to verify that
$$
 12 a \varpi s(a,\varpi)=\frac{12a^2}{\varpi}\cdot
\frac{\varpi(\varpi-1)(2\varpi-1)}{6}-12a\sum_{j=1}^{\varpi-1}j
\fracwithdelims[]{ja}{\varpi}-6a\cdot \frac{\varpi(\varpi-1)}{2}.
$$
However, by \cite{4, p\. 64, Th. 3.8},  it is known that
$$
  12a\varpi s(a,\varpi)\equiv a^2+1\pmod \varpi,
$$
and so we find that
$$
  a^2+1\equiv 2a^2-12a\sum_{j=1}^{\varpi-1}j\fracwithdelims[]{ja}{\varpi}
\pmod \varpi.
$$
This is (C15.4), and this
completes the proof of Ramanujan's claim.
\enddemo 

The remainder of Section 15 is straightforward and follows from Ramanujan's
collection of formulas involving the operator $q\frac{d}{dq}$.

\medskip

16. The equalities (16.3) and (16.7) are proved in Rushforth's paper 
\cite{89}.  As with the key results in Section 14, the claims
(16.1)--(16.3) and (16.5)--(16.9) follow from the work of Ono
\cite{70}. 

In (16.4), Ramanujan claims that the Dirichlet series
$$\align
&\sum_{n=1}^{\infty}\frac{\tau_2(n)}{n^s},
\ \ \ \sum_{n=1}^{\infty}\frac{\tau_3(n)}{n^s},\\
&\sum_{n=1}^{\infty}\frac{\tau_4(n)}{n^s},
\ \ \ \sum_{n=1}^{\infty}\frac{\tau_5(n)}{n^s},
\ \ \ \sum_{n=1}^{\infty}\frac{\tau_7(n)}{n^s}
\endalign
$$
have Euler products. This is easily verified since
all corresponding modular forms are eigenforms
of Hecke operators, for they each lie in a one dimensional
space of cusp forms \cite{33}.

At the end of Section 16, Ramanujan claims that the two Dirichlet series
$$
\sum_{n=1}^{\infty}\frac{\Omega_2(n)}{n^s}\ \ \ {\text {\rm and}}\ \ \ 
\sum_{n=1}^{\infty}\frac{\Omega_3(n)}{n^s}
$$
are both differences of two series with Euler products.
In terms of classical modular forms, 
$\sum_{n=1}^{\infty}\Omega_2(n)q^n\in S_{28}(\Gamma_0(1))$ and
$\sum_{n=1}^{\infty}\Omega_3(n)q^n\in S_{30}(\Gamma_0(1))$.
Both of these two spaces are two dimensional \cite{33}, and one can
easily check that
$$\align
&S_{28}(\Gamma_0(1))=\C Q\Delta^2 \oplus \C Q^4\Delta ,\\
&S_{30}(\Gamma_0(1))=\C R\Delta^2 \oplus \C R^3\Delta,
\endalign
$$
where $ \Delta  :=\Delta(q) = q(q;q)^{24}_{\i}. $
It is easy to show that the space $S_{28}(\Gamma_0(1))$
is spanned by the eigenforms
$$\align
  &f_1:=(-5076+108\sqrt{18209})Q\Delta^2+Q^4\Delta, \\ 
 &f_2:=(-5076-108\sqrt{18209})Q\Delta^2+Q^4\Delta.
\endalign
$$
(For calculations of this sort, see N\. Koblitz's text \cite{54,
 p\. 173, Prop\. 51}.)
Since $ \sum_{n=1}^{\infty}a_1(n)q^n$ 
and $ \sum_{n=1}^{\infty}a_2(n)q^n$ are eigenforms, the 
two Dirichlet series
$$\sum_{n=1}^{\infty}\frac{a_1(n)}{n^s}\ \ {\text {\rm and}}\ \ 
\sum_{n=1}^{\infty}\frac{a_2(n)}{n^s}
$$
have Euler products as in (16.4) with weight 28.
The ``difference" to which  Ramanujan alludes is the 
identity
$$
    Q\Delta^2=\sum_{n=1}^{\infty}\Omega_2(n)q^n=
\frac{1}{216\sqrt{18209}}(f_1-f_2).
$$

Similarly, one can easily verify that the space $S_{30}(\Gamma_0(1))$
is spanned by the eigenforms
$$\align
&g_1:=(5856+2208\sqrt{-83})R\Delta^2+R^3\Delta,\\
&g_2:=(5856-2208\sqrt{-83})R\Delta^2+R^3\Delta.
\endalign
$$
Since $ \sum_{n=1}^{\infty}b_1(n)q^n$ and $ \sum_{n=1}^{\infty}
b_2(n)q^n$ are eigenforms, it easily follows that
the two Dirichlet series
$$
\sum_{n=1}^{\infty}\frac{b_1(n)}{n^s} \ \ {\text {\rm and}}\ \ 
\sum_{n=1}^{\infty}\frac{b_2(n)}{n^s}
$$
have Euler products as in (16.4) with weight 30. 
The difference to which Ramanujan  alludes is the 
identity
$$
  R\Delta^2=\sum_{n=1}^{\infty}\Omega_3(n)q^n=\frac{1}{4416\sqrt{-83}}
(g_1-g_2).
$$



\medskip

17. The claim (17.2) is equivalent to the
assertion that $\eta(z)\eta(23z)$ is an eigenform with complex
multiplication 
in the space $S_1(\Gamma_0(23),\chi_{-23})$
\cite{44}, \cite{88}, \cite{33}. Here
$\chi_{-23}$ is the usual Kronecker character for the
quadratic field $\Q(\sqrt{-23})$.
Although the claims regarding the Euler products $\Pi_1, \Pi_2$ and
$\Pi_3$ follow easily from the theory of complex multiplication,
one can more easily obtain them from Euler's
pentagonal number theorem,
$$
  (q;q)_{\i}=\sum_{n=-\infty}^{\infty}(-1)^n
q^{n(3n+1)/2}.
$$
Here one would use an argument similar to that briefly outlined in the
commentary for Section 8. 
In analogy with our comments in Sections 2 and 6, the claim (17.8) is 
false, but the leading term in the asymptotic expansion is indeed 
$ Cx/\sqrt{\log x}. $

\medskip

18. The proof of (18.1) is quite difficult, but it is given in Rushforth's 
paper \cite{89}.  Ramanujan has omitted many details in his 
assertion (18.2).  For the remainder of the proof of Ramanujan's congruence 
modulo $ 11^2 $ to be completed, it is necessary to explicitly determine 
the constants $ a_4, b_4, $ and $ c_3 $ in (18.2). Rushforth does not prove 
(18.2) but proceeds by a different route to (18.7). The third congruence in 
(18.3) is proved in Rushforth's paper \cite{89}.    The equation 
to which Ramanujan refers before (18.7) is not given in the manuscript, but 
would arise from (18.2) by using (18.3)--(18.6). 

\medskip

19. 
This manuscript contains many results on the divisibility of
$\tau(n)$. 
In several sections Ramanujan
concludes that $\tau(n)$ is a multiple of a given integer $M$
for almost all $n$. In other words,  for such $M$,
$$
  \lim_{X\rightarrow \infty}\frac{\# \{ 1\leq n \leq X \ : \ \tau(n)\equiv
0\pmod M\}}{X}=1.
$$ 
Specifically, Ramanujan finds in (19.9) that $\tau(n)$
is a multiple of $2^5\cdot 3^3 \cdot 5^2 \cdot 7^2 \cdot 23
\cdot 691$ for almost all $n$. Various authors have proved versions of (19.9)
with varying exponents 
on the six primes.  It was first proved by Chowla \cite{32} that, 
in fact, the conclusion still holds if the powers of 2, 3, 5, 7, 23, and 691 
are replaced by any set of six positive integral powers. 


Ramanujan obtains his results by employing the congruences
for $\tau(n)$ with modulus $M\in \{
2^5, 3^3,
5^2, 7^2, 23, 691\}$. In each case, he finds that
a positive density of primes $p$
has the property that $\tau(p)\equiv 0\pmod M$.
A Tauberian argument based on the multiplicativity of
$\tau(n)$ then leads to  his conclusion \cite{94, Sect\. 2}. 

Results of this type depend upon the divisibility of divisor functions.
Improving on Watson's theorem \cite{103}, Rankin
\cite{84}  found an asymptotic formula for the number of positive
integers $ \leq x $ for which $ \sigma_{\nu}(n) $ is not divisible by the
prime number $ k. $  These results were generalized by E\. J\. Scourfield
\cite{90}. 


J.--P\. Serre \cite{93}, \cite{94}, \cite{95}
has obtained a substantial generalization of
Ramanujan's claims for all  modular forms of integral weight 
with respect to  congruence subgroups of the
full modular group. In particular, if $\sum_{n=1}^{\infty}
a(n)q^n$ ($q:=e^{2\pi i z}$) is the Fourier expansion
of a modular form of integral weight with integral coefficients,
then for every positive integer $M$,
$a(n)\equiv 0\pmod M$ for almost all $n$.
M\. R\. Murty and V\. K.\ Murty \cite{63} have obtained
an interesting improvement on the original formulation
of Serre's result.

Serre's theorem is based on the existence of $\ell$-adic
Galois representations
associated to modular forms (see the comments on Section 10).
In addition to providing an arithmetic and group theoretic description of
congruences for Fourier coefficients $a(n)$ of the
types found by Ramanujan for $\tau(n)$,
their mere existence
implies, by the Chebotarev Density
Theorem, that a positive proportion of primes $p$ have the 
property that $a(p)\equiv 0\pmod M$.



Bambah and Chowla \cite{13} state without proofs several 
congruences for $ \t(n). $ Lahiri \cite{59} gives an enormous 
number of congruences involving $ \t(n). $ Van der Blij's  beautiful paper
 \cite{27}, giving congruences and other properties of $ \t(n), $
 is particularly recommended. Except for those employing the theory of
$\ell$-adic Galois representations, almost all the authors giving proofs of 
congruences for $ \t(n) $ whom we have cited use ideas similar to those
 employed by Ramanujan in this manuscript. 

\medskip

 20.--23. These sections contain Ramanujan's proof of his
congruence for $ p(n) $ modulo any positive integral power of 5, with
(22.5)--(22.8) being the principal congruences. Observe that (22.7) and
(22.8) include (C1.1). As mentioned earlier, the ideas here were expanded
into a more detailed proof given in 1938 by Watson \cite{104},
who does not mention Part II of Ramanujan's unpublished manuscript in his
paper, although according to Rushforth \cite{89}, Watson received
a copy from Hardy in 1928.

The details in Section 20 are reasonably ample, but beginning with Section
21, the details are sparse.  In particular, (21.1) is difficult to prove.
The proof given by Watson may follow along somewhat different lines
than those indicated by Ramanujan. Readers can likely follow the details for
the remainder of Section 21.  We have added some details for (21.6), which is
not used in Watson's work.  The heart of Ramanujan's proof lies in
(22.1)--(22.6), for which Ramanujan provides no details.  These are developed
in Watson's paper \cite{104}. 
\medskip

 24. Clearly, Ramanujan intended to follow the same lines of
attack for powers of 7 as he did for powers of 5 in Sections 20--23.
If he had completed his argument, he would have undoubtedly seen that his
original conjecture modulo powers of 7 needed to be corrected. Most likely,
his declining health prevented him from working out the remaining details,
which were completed by Watson \cite{104}. 
To verify the equations (24.4)--(24.6), it suffices to notice that all
three equations are essentially claims about the presentation
of modular functions with respect to $\Gamma_0(7)$.  In each
case, one may multiply both sides of the claimed identity by
$(q^7;q^7)^8$. After doing so, one needs to compare, up to a
shifted power of $q$, the Fourier expansions of two cusp forms
of weight 4.  One can then easily deduce these claims from the results
in \cite{33}. 


At the end of Part II are two detatched fragments. They actually appear at
the end of Section 21 in Watson's copy of the manuscript, but it seems to us
that they are better placed at the end of the section pertaining to the
moduli 7 and 49. 

\medskip

We are grateful to Paul Bateman, Michael Hirschhorn, and Robert A\. Rankin
 for helpful comments. 


\Refs

\magnification=\magstephalf

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\by S\. Ramanujan
\paper On certain arithmetical functions
\jour Trans\. Cambridge Philos\. Soc\.
\vol 22
\yr 1916
\pages 159--184
\endref

\ref
\no 78
\by S\. Ramanujan
\paper Some properties of $ p(n), $ the number of partitions of $ n$
\jour Proc\. Cambridge Philos\. Soc\. 
\vol 19
\yr 1919
\pages 207--210
\endref

\ref
\no 79
\by S\. Ramanujan
\paper Congruence properties of partitions
\jour Proc\. London Math\. Soc\. 
\vol 18
\yr 1920
\pages xix
\endref

\ref
\no 80
\by S\. Ramanujan
\paper Congruence properties of partitions
\jour Math\. Z\.
\vol 9
\yr 1921
\pages 147--153
\endref


\ref
\no 81
\by S\. Ramanujan
\book Notebooks (2 volumes)
\publ Tata Institute of Fundamental Research
\publaddr Bombay
\yr 1957
\endref

\ref
\no 82
\by S\. Ramanujan
\book Collected Papers
\publ Chelsea
\publaddr New York
\yr 1962
\endref

\ref
\no 83
\by S\. Ramanujan
\book The Lost Notebook and Other Unpublished Papers
\publ Narosa
\publaddr New Delhi
\yr 1988
\endref

\ref
\no 84
\by R\. A\. Rankin
\paper The divisibility of divisor functions
\jour Glasgow Math\. J\.
\vol 5
\yr 1961
\pages 35--40
\endref

\ref
\no 85
\paper Ramanujan's unpublished work on congruences
\by R\. A\. Rankin
\inbook Modular Functions of One Variable V 
\bookinfo Lecture Notes in Math\., No\. 601
\publ Springer--Verlag
\publaddr Berlin
\yr 1977
\pages 3--15
\endref

\ref
\no 86
\book Modular Forms and Functions
\by R\. A\. Rankin
\publ Cambridge University Press
\publaddr Cambridge
\yr 1977
\endref

\ref
\no 87
\paper Ramanujan's manuscripts and notebooks
\by R\. A\. Rankin
\jour Bull\. London Math\. Soc\. 
\vol 14
\yr 1982
\pages 81--97
\endref

\ref
\no 88
\paper Galois representations attached to eigenforms with Nebentypus
\by K\. Ribet
\inbook Modular Functions of One Variable V
\bookinfo Lecture Notes in Math\., No\. 601
\publ Springer--Verlag
\publaddr Berlin
\yr 1977
\pages 17--51
\endref


\ref
\no 89
\by J\. M\. Rushforth
\paper  Congruence properties of the partition function and associated functions
\jour Proc\. Cambridge Philos\. Soc\.
\vol 48
\yr 1952
\pages 402--413
\endref

\ref
\no 90
\by E\. J\. Scourfield
\paper On the divisibility of $ \sigma_{\nu}(n) $
\jour Acta Arith\. 
\yr 1964
\vol 10
\pages 245--285
\endref

\ref
\no 91
\by J.--P\. Serre
\paper Une interpr\'etation des congruences relatives \`a la fonction
 $ \tau $ de Ramanujan
\inbook S\'eminaire Delange--Pisot--Poitou: 1967/68, Th\'eorie des Nombres,
 Fasc\. 1, Exp\. 14
\yr 1969, 17 pp.
\publ Secr\'etariat Math\'ematique
\publaddr Paris
\endref

\ref
\no 92
\by J.--P\. Serre
\paper Congruences et formes modulaires [d'apr\`es H\. P\. F\. Swinnerton--Dyer]
\inbook S\'eminaire Bourbaki, 24e ann\'ee (1971/1972), Exp\. No\. 416
\bookinfo Lecture Notes in Math\., No\. 317
\publ Springer--Verlag
\publaddr Berlin
\yr 1973
\pages 319--338
\endref

\ref
\no 93
\by J.--P\. Serre
\paper Divisibilit\'e des coefficients des formes modulaires de poids entier
\jour C\. R\. Acad\. Sci\. (Paris), S\'er A
\vol 279
\yr 1974
\pages 670--682
\endref


\ref
\no 94
\by J.--P\. Serre
\paper Divisibilit\'e de certaines fonctions arithm\'etiques
\jour L'Enseign\. Math\.
\vol 22
\yr 1976
\pages 227--260
\endref


\ref
\no 95
\by J.--P\. Serre
\paper Quelques applications du th\'eor\'eme de densit\'e de Chebotarev
\jour Publ\. Math\. I\. H\. E\. S\. 
\vol 54
\yr 1981
\pages 123--201
\endref

\ref
\no 96
\by W\. H\. Simons
\paper Congruences involving the partition function $ p(n)$
\jour Bull\. Amer\. Math\. Soc\.
\vol 50
\yr 1944
\pages 883--892
\endref

\ref
\no 97
\by S\. H\. Son
\paper Some integrals of theta functions in Ramanujan's lost notebook
\inbook Number Theory, Fifth Conference of the Canadian Number Theory Association
\eds R\. Gupta and K\. S\. Williams
\publ American Mathematical Society
\publaddr Providence, RI
\yr 1998
\pages 329--339
\endref

\ref
\no 98
\by G\. K\. Stanley
\paper Two assertions made by Ramanujan
\jour J\. London Math\. Soc\.
\vol 3
\yr 1928
\pages 232--237
\endref

\ref
\no 99
\by H\. P\. F\. Swinnerton--Dyer
\paper On $\ell$-adic representations and congruences for coefficients of
 modular forms 
\inbook Modular Functions of One Variable III
\bookinfo Lecture Notes in Math\., No\. 350
\publ Springer--Verlag
\yr 1973
\publaddr Berlin
\pages 1--55
\endref 

\ref
\no 100
\by H\. P\. F\. Swinnerton--Dyer
\paper Correction to: ``On $\ell$-adic representations and congruences 
for coefficients of  modular forms" 
\inbook Modular Functions of One Variable IV
\bookinfo Lecture Notes in Math\., No\. 476
\publ Springer--Verlag
\yr 1975
\publaddr Berlin
\page 149
\endref

\ref
\no 101
\by H\. P\. F\. Swinnerton--Dyer
\paper On $\ell$--adic representations and congruences for coefficients of
 modular forms (II)
\inbook Modular Functions of One Variable V
\bookinfo Lecture Notes in Math\., No\. 601
\publ Springer--Verlag
\yr 1977
\publaddr Berlin
\pages 63--90
\endref

\ref
\no 102
\by H\. P\. F\. Swinnerton--Dyer
\paper Congruence properties of $ \tau(n)$
\inbook Ramanujan Revisited
\eds G\. E\. Andrews, R\. A\. Askey, B\. C\. Berndt, K\. G\. Ramanathan, 
and R\. A\. Rankin
\publ Academic Press
\publaddr Boston
\yr 1988
\pages 289--311
\endref


\ref
\no 103
\by G\. N\. Watson
\paper \"Uber Ramanujansche Kongruenzeigenschaften der Zerf\"allungsanzahlen, I
\jour Math\. Z\.
\vol 39
\yr 1935
\pages 712--731
\endref

\ref
\no 104
\by G\. N\. Watson
\paper Ramanujans Vermutung \"uber Zerf\"allungsanzahlen
\jour J\. Reine Angew\. Math\.
\vol 179
\yr 1938
\pages 97--128
\endref

\ref
\no 105
\by G\. N\. Watson
\paper A table of Ramanujan's function $ \tau(n)$
\jour Proc\. London Math\. Soc\.
\vol 51
\yr 1948
\pages 1--13
\endref

\ref
\no 106
\by J\. R\. Wilton
\paper On Ramanujan's arithmetical function $ \sigma_{r,s}(n)$
\jour Proc\. Cambridge Philos\. Soc\.
\vol 25
\yr 1929
\pages 255--264
\endref

\ref
\no 107
\by J\. R\. Wilton
\paper Congruence properties of Ramanujan's function $ \t(n)$
\jour Proc\. London Math\. Soc\.
\vol 31
\yr 1930
\pages 1--10
\endref

\ref
\no 108
\by H\. S\. Zuckerman
\paper Identities analogous to Ramanujan's identities involving the
 partition function
\jour Duke Math\. J\.
\yr 1939
\vol 5
\pages 88--110
\endref

\endRefs



\enddocument
