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\topmatter
\title Parity of the partition function 
\endtitle
\rightheadtext{Parity  of the partition function}
\author Ken Ono\endauthor
\address Department of Mathematics, The University of Illinois,
Urbana, Illinois 61801 \endaddress
\email ono\@symcom.math.uiuc.edu \endemail
%\curraddr School of Mathematics, Institute for Advanced Study,
%Princeton, New Jersey 08540 \endcurraddr
\keywords Parity conjecture, partitions, modular forms\endkeywords
%  Math Subject Classifications 
\subjclass Primary 05A17; Secondary  11P83 \endsubjclass
\thanks The author is supported by NSF grant DMS-9508976. \endthanks
\communicatedby{Don Zagier}
\date February 28, 1995\inRevisedForm May 3, 1995 \enddate
\cvol{1}
\cvolno{1}
\cyear{1995}
\cvolyear{1995}
\abstract
Let $p(n)$ denote the number of partitions of a non-negative integer
$n$. A well-known conjecture asserts that every arithmetic progression 
contains infinitely many integers $M$ for which $p(M)$ is odd, as well
as infinitely many integers $N$ for which $p(N)$ is even (see Subbarao [22]).
From the works of various authors, this conjecture has been verified for
every arithmetic progression with
 modulus $t$ when $t=1,2,3,4,5,10,12,16,$ and $40.$
Here we announce  that there indeed are infinitely many  
integers $N$ in every arithmetic progression for which $p(N)$ is even; 
and that there are infinitely many integers $M$ 
in every arithmetic progression for which $p(M)$ is odd
so long as there is at least one such $M$.
In fact if there is such an $M$, then the smallest such
$M\leq  10^{10}t^7$.
Using these results and a fair bit
of machine computation, we have verified the conjecture for every arithmetic
progression with modulus $t\leq 100,000$. 
\endabstract
%\thanks The author is supported in part by a grant from the National
%Science Foundation.
%\endthanks
\endtopmatter

\document
\advance\pageno by 34
\noindent

\head 1. Introduction \endhead

\noindent
A partition of a non-negative integer $n$ is a non-increasing sequence
of positive integers whose sum is $n$. Euler's generating
  function for $p(n)$, the
 number of partitions of an integer $n,$ is:
$$
   \sum_{n=0}^{\infty} p(n)q^n= \prod_{n=1}^{\infty}\frac{1}{1-q^n}.
\tag{1}
$$
Ramanujan discovered various surprising congruences for $p(n)$
when $n$ is in certain special arithmetic progressions; for
example:
$$
   p(5n+4)\equiv 0 \mod 5,
$$
$$
   p(7n+5)\equiv 0 \mod 7,
$$
and
$$
   p(11n+6)\equiv 0 \mod 11.
$$
There are now many proofs of these congruences (and their generalizations)
in the literature (see [1, 2, 3, 4, 5, 6, 7, 11, 12, 23 ] for instance).
Surprisingly there do not seem to be any such congruences modulo
2 or 3. In fact the parity of $p(n)$ seems to be quite random, and it 
is believed that the partition function is `equally often' even and odd;
that is, that $p(n)$ is even for  $\sim \frac{1}{2}x$ positive
integers $n\leq x$ (see Parkin and Shanks [19]). 


In [22] Subbarao made the following conjecture on the parity of
$p(n)$, for those integers $n$ belonging to any given arithmetic progression:


\proclaim{Conjecture} For any arithmetic progression $r \pmod t$, there are
infinitely many integers 
$M\equiv r \pmod t$ for which $p(M)$ is odd,
and there are infinitely many integers 
$N\equiv r \pmod t$ for which $p(N)$ is even.
\endproclaim 

From the works of Garvan, Kolberg, Hirschhorn, Stanton, and Subbarao
(see [6, 9, 10, 13, 16, 22],),
this conjecture has been proved for every arithmetic
progression with modulus $t$ when
$t=1,2,3,4,5,10,12,16$ and $40$.

Using very different methods, we go some way towards a proof
of the conjecture.
Using the theory of modular forms, we
announce:
\proclaim{Main Theorem 1}
For any arithmetic progression $r \pmod t$, there are
infinitely many integers 
$N\equiv r \pmod t$ for which $p(N)$ is even.
\endproclaim

\proclaim{Main Theorem 2}
For any arithmetic progression $r \pmod t$, there are
infinitely many integers 
$M\equiv r \pmod t$ for which $p(M)$ is odd,
provided there is one such $M$. Furthermore, if there does exist an 
$M\equiv r \pmod t$ for which $p(M)$ is odd,
then the smallest such $M$ is less than $C_{r,t}$, where 
$$ C_{r,t}:=\frac{2^{23}A\cdot 3^7t^6}{d^2}\prod_{p\mid
6t}\left (1-\frac{1}{p^2} \right )-A,$$
with $d:={\text {\rm gcd}}(24r-1,t)$ and $A>\frac{t}{24}$
is a power of $2.$
\endproclaim

From the two theorems we obtain an algorithm to determine the truth
of our parity conjecture for any given arithmetic progression
$r \pmod t$:\ Compute $p(N) \pmod 2$ for $N=r, r+t, r+2t, \dots$
for all such $N$ up to $C_{r,t}$.
As soon as we find one odd number
we have verified the conjecture. If all these numbers are even
then we have proved that the conjecture is false.

Ken Burrell (Universal Analytics, Inc.)  ran an
 efficient version of this algorithm on a
CRAY C-90, giving the following result:

\proclaim{Main Corollary} For all $0\leq r < t \leq 10^5$, 
there are infinitely many integers 
$M\equiv r \pmod t$ for which $p(M)$ is odd.
\endproclaim

\head 2. The main ideas \endhead


\noindent
First we briefly recall essential preliminaries concerning modular
forms. For more on the theory of modular forms see [15].



Let $A=\left ( \matrix a & b\\ c& d\\ \endmatrix \right )\in SL_{2}(\Z)$
act on $\frak{H},$ the upper half of the complex plane, by the linear
fractional transformation $Az=\frac{az+b}{cz+d}.$ If $N$ is a positive
integer, then we define the following
{\it congruence subgroups} of $SL_{2}(\Z)$ of {\it level} $N$:
$$
   \Gamma_{0}(N):= \left \{ \left ( \matrix a & b \\
					    c & d\\ \endmatrix \right
) | \  ad-bc=1, \ c\equiv 0 \mod N \right \}.
$$
and
$$
    \Gamma_{1}(N):=\left \{ \left ( \matrix a & b \\
     c & d\\ \endmatrix \right ) \mid \ ad-bc=1, \ a\equiv d \equiv 1
\mod N, \ {\text {\rm and}}\ c\equiv 0 \mod N \right \}.
$$


A meromorphic function $f(z)$ on $\frak{H}$ is called a {\it modular
function} with positive integer 	weight $k$ with respect 
to congruence subgroup $\Gamma$  if
$$
    f\left ( \frac{az+b}{cz+d} \right ) = (cz+d)^k f(z)
$$
for all $z\in \frak{H}$ and all $\left ( \matrix a & b \\ c & d \endmatrix
\right ) \in \Gamma.$  If $f(z)$ is holomorphic on $\frak{H}$ and
at the cusps of $\Gamma$ (i.e. the rationals), then $f(z)$ is known
as a {\it modular form} of weight $k$ with respect to $\Gamma.$ If $f(z)$
vanishes at the cusps of $\Gamma,$ then $f(z)$ is known as a {\it cusp
form}.   

We denote the finite dimensional space of modular forms (resp. cusp forms)
 of weight
$k$ with respect to $\Gamma_{1}(N)$ by $M_{k}(N)$ (resp. $S_{k}(N)$).
In the variable $q=e^{2\pi i z},$ a holomorphic modular form $f(z)
\in M_{k}(N)$
 admits a Fourier
expansion of the form
$$
     f(z)=\sum_{n=0}^{\infty} a(n)q^n.
$$




Of particular interest are certain modular forms in $M_{k}(N)$ with
nice modular transformation properties with respect to $\Gamma_{0}(N).$
If $\chi$ is a Dirichlet character $\mod N,$ then we
 say that a form $f(z)\in M_{k}(N)$ (resp. $S_{k}(N)$ ) is
modular form of weight $k$ with Nebentypus character $\chi$ if
$$
    f\left ( \frac{az+b}{cz+d} \right ) = \chi(d) (cz+d)^kf(z)
$$
for all $z\in \frak{H}$ and all $\left ( \matrix a & b \\ c & d \endmatrix
\right ) \in \Gamma_{0}(N).$ The space of such modular forms
 (resp. cusp forms) is denoted by  
$M_{k}(N,\chi)$ (resp. $S_{k}(N,\chi)$).






The Dedekind eta-function is the principal modular form of interest 
in this paper; it is defined  by the infinite product
$$
   \eta(z):=q^{\frac{1}{24}}\prod_{n=1}^{\infty}(1-q^n). 
$$
A function $f(z)$ is called an {\it eta-product}
if it is expressible as a finite product of the
form
$$
    f(z)=\prod_{\delta \mid N}\eta^{r_{\delta}}(\delta z)
$$
where $N$ and each $r_{\delta}$ is an integer.
Probably the most famous of all eta-products is 
Ramanujan's $\Delta-$function, defined by 
$\Delta(z):=\eta^{24}(z)=q\prod_{n=1}^{\infty}(1-q^n)^{24}$.
This is the unique normalized weight $12$ cusp form on $SL_{2}(\Z)$.
More generally, Gordon, Hughes, and Newman (see [8,17,18]) 
examined the general {\it modular} properties of eta-products.


 We construct
 modular forms that are eta-products  whose Fourier expansions modulo $2$
are determined by the values of $p(n)$ modulo 2.
\proclaim{Proposition 1}
For a given positive integer $\ t$, let 
$A > \frac{t}{24}$ be a power of $2.$  Define $f_{t,A}(z)$ by
$$
    f_{t,A}(z):=\frac{\eta(24z)}{\eta(48z)}\Delta^{A}(24tz)=
    \sum_{n\geq 1} a_{t}(n)q^{24n-1}.
$$
Then $f_{t,A}(z)$ is a cusp form in 
$S_{12A}(1152t,\fracwithdelims(){2}{d})$.
Moreover the Fourier expansion of $f_{t,A}(z) \mod 2$ can be factored
as:
$$ f_{t,A}(z)=\sum_{n=0}^{\infty}a_{t}(n)q^{24n-1}\equiv \left(
   \sum_{n=0}^{\infty}p(n)q^{24n-1}\right)\left(\sum_{n=0}^{\infty}q^{
24At(2n+1)^2} \right)\mod 2. \tag{2} $$
\endproclaim
 \demo{Proof}
Using the well known properties of the Dedekind eta-function,
it is relatively
straightforward to deduce that $f_{t,A}(z)$ is a modular form
of weight $12A.$ 
It is also straightforward
to deduce that $f_{t,A}(z)$ is a cusp form.

The essential feature of the cusp form $f_{t,A}(z)$ is the convenient
fact that $f_{t,A}(z)$ is {\it essentially} the product of the
generating function for $p(n)$ and a theta function $\mod 2.$ 

Since $\dsize 
\frac{1}{1-X^n}=\frac{1+X^n}{1-X^{2n}}\equiv \frac{1-X^n}{1-X^{2n}}
\mod 2,$ it follows that
$$
   \sum_{n=0}^{\infty}p(n)q^n\equiv \prod_{n=1}^{\infty}
   \frac{1-q^n}{1-q^{2n}} \mod 2.
$$
In terms of the eta-functions, we find that
$$
   \frac{\eta(24z)}{\eta(48z)}=\frac{1}{q}\prod_{n=1}^{\infty}
   \frac{1-q^{24n}}{1-q^{48n}}\equiv \sum_{n=0}^{\infty}
   p(n)q^{24n-1} \mod 2.  \tag{3}
$$

The following infinite product identity was proved by Jacobi:
$$
  \frac{\eta^2(16z)}{\eta(8z)}=q\prod_{n=1}^{\infty}\frac{(1-q^{16n})^2}
  {(1-q^{8n})}=\sum_{n=0}^{\infty}q^{(2n+1)^2}.
$$
Therefore since $(1-X)^2\equiv (1-X^2) \mod 2,$ we find that
$$
  \Delta(z)=q\!\prod_{n=1}^{\infty}\!(1-q^n)^{24}=
  q\prod_{n=1}^{\infty}\!\frac{(1-q^n)^{32}}{(1-q^n)^8}\equiv
  q\prod_{n=1}^{\infty}\!\frac{(1-q^{16n})^2}{(1-q^{8n})}\equiv \sum_{n=0}
  ^{\infty}q^{(2n+1)^2} \!\!\mod 2. \tag{4}
$$

The factorization of $f_{t,A}(z)$ now follows easily from (3) and (4).
$\hfill \square $
\enddemo

\noindent
 Serre [20] proved the following remarkable theorem 
regarding
the divisibility of Fourier coefficients of holomorphic integer
weight modular forms.
\proclaim{Theorem } {\bf (Serre)} Let $f(z)$ be a
holomorphic  modular form of positive integer
 weight $k$ on some congruence
subgroup of $SL_{2}(\Z)$ with Fourier expansion
$$
    f(z)=\sum_{n=0}^{\infty} a(n)q^{n}
$$
where $a(n)$ are algebraic integers in some number field. If
$m$ is a positive integer, then there exists a positive constant $\alpha$
such that the set of integers $n \leq x$ for which $a(n)$ is not divisible
by $m$ has cardinality $\ll \frac{x}{\log^{\alpha}x}.$
\endproclaim

\noindent
With this theorem we obtain
\proclaim{Main Theorem 1}
For any arithmetic progression $r \pmod t$, there are
infinitely many integers 
$N\equiv r \pmod t$ for which $p(N)$ is even.
\endproclaim

\demo{Proof}
Comparing coefficients in $(2)$, it is easy to deduce that
$$
   a_t(Atk^2+n)\equiv \sum_{ i\geq 1, \ i\  {\text {\rm odd}}}
   p(At(k^2-i^2)+n) \mod 2. \tag{5}
$$

Now suppose every $ N\geq n_0$ for which $N\equiv r\pmod t$
has the property that
 $p(N)$ is odd.  If
 $k\equiv 1\mod 4,$ then every integer $n\equiv r\pmod t$ in the interval
$[Atk^2+n_0, At(k+2)^2+r-t]$ has the property that
$a_t(n)$ is odd since there are $\dsize \frac{k+1}{2}$ many odd summands
in $(5).$
 After combining all such intervals,
we find 
 a set of positive integers with positive density
for which $a_t(n)\not \equiv 0\mod 2.$ This would contradict Serre's theorem.
$\hfill \square$    
\enddemo 

\noindent
Now we need to establish that there are infinitely 
many $M\equiv r\pmod t $ where
$p(M)$ is odd  provided that there is at least one $M.$ 
 To do this we first deduce a technical lemma about the reduction 
modulo $m$ of the Fourier expansions of holomorphic modular forms.
Main Theorem 2  follows as a consequence,
 for if there were only finitely
many $M\equiv r\pmod t$ for which $p(M)$ is odd, then the reduction
$\mod 2$ of the relevant modular form contradicts the lemma.

For a given positive integer $m$ and formal power series 
$f:=\sum_{n\in \Z}a(n)q^n$ with algebraic integer coefficients, we define 
${\text {\rm Ord}}_{m}(f)$ to be the smallest integer $n$ for which 
$a(n)$ is not divisible by $m$. A special case of a theorem of
Sturm [21] allows us to computationally determine whether $m$ divides
$a(n)$ for every integer $n$ 
(that is, to determine whether ${\text {\rm Ord}}_{m}(f)=\infty$).

If
 $f(z)=\sum_{n=0}^{\infty} a(n)q^n\in M_{k}(N)$ for some positive
integer $N$
with algebraic integer Fourier
coefficients from a fixed number field and $m$ is a positive integer,
then Sturm proved that if
$$
 {\text {\rm    Ord}}_{m}(f) > \frac{k}{12}N^2 \prod_{p\mid N}(1-\frac{1}{p^2}),
$$
then ${\text {\rm Ord}}_{m}(f) = \infty$ (i.e. $a(n)\equiv 0 \mod m$
\ for all $n$).
Now we prove the essential lemma about the reduction of a  holomorphic
modular form $\mod m.$
\noindent
\proclaim{Lemma 1} Let $f(z)=\sum_{n=0}^{\infty} a(n)q^n$
 where the coefficients
$a(n)$ are algebraic integers in some number field. Let $s$ and $w$
be positive integers and $b_{1}, b_{2}, \dots b_{s}$  distinct
non-zero integers.
If $m$ is a positive integer and 
$$
    f(z)\equiv 
   \sum_{1 \leq i \leq s} \sum_{ \ n=0}^{\infty}a_{i}(n)
q^{wn^2+b_{i}} \mod m
$$
where  $a_{i}(n)\not \equiv 0 \mod m$ for infinitely many $n\geq 0,$
 then $f(z)$ is not in $M_{k}(N)$ for any
pair of positive integers $k,$ and $N.$
\endproclaim
\demo{Proof}
Suppose that $f\in M_k(N)$ for some $k$ and $N$. If $p\equiv 1\pmod N$ is prime, then the image of $f$ under the Hecke operator $T_p$ satisfies 
   $$\align f(z)|T_p & =\sum_{n\geq 0}(a(pn)+p^{k-1}a(n/p))\,q^n\\
              &\equiv \sum\Sb i,n\\ p|wn^2+b_i\endSb a_i(n)\,q^{(wn^2+b_i)/p}
              +p^{k-1}\,\sum_{i,n} a_i(n)\,q^{(wn^2+b_i)p}\qquad\pmod m\,, \tag 6\endalign$$
and $f|T_p$ again belongs to $M_k(N)$.  We claim that $\,\text{Ord}_m(f|T_p)<\infty$ 
for every sufficiently large prime $p$ and $\,\text{Ord}_m(f|T_p)>C$ for almost every $p$, 
for any given constant $C$.  Taking $C=\frac{kN^2}{12}\prod\limits_{l|N}(1-\l^{-2})$ gives 
a contradiction to Sturm's theorem.

To see that $\,\text{Ord}_m(f|T_p)<\infty$, we observe that only finitely many of the
infinitely many exponents on the right-hand side of (6) can coincide, so that the expression
cannot vanish identically modulo $m$.  Indeed, if $(wn^2+b_i)p^{\pm1}=(wl^2+b_j)p^{\pm1}$
for some $n\ne l$, then $w(n+l)(n-l)=b_j-b_i$ implies that both $n$ and $l$ are bounded,
while if $(wn^2+b_i)p=(wl^2+b_j)p^{-1}$ then $w(pn+l)(pn-l)=b_j-p^2b_i$ gives the same
conclusion if no $b_j$ is divisible by $p^2$.  

For the reverse direction, we observe that if $C\frac{t}{24}$ is a power of $2.$
\endproclaim

\demo{ Proof}
Recall from Proposition 1 that
$$
   f_{t,A}(z):=\frac{\eta(24z)}{\eta(48z)}\Delta^{A}(24tz)=\sum_{n\geq 1}
a_{t}(n)q^{24n-1} \in S_{12A}(1152t,\chi).
$$
and
$$
   f_{t,A}(z)\equiv \sum_{n=0}^{\infty}p(n)q^{24n-1}\sum_{n=0}^{\infty}
   q^{24 A t(2n+1)^2} \mod 2.
$$
Let $d:={\text {\rm gcd}}(24r-1, t).$
Therefore by Lemma 2 we define  
$$    f_{24r-1, 24t}(z)=
\sum_{n\equiv 24r-1 \mod
24t} a_{t}(n)q^{n} \in S_{12A}\left ( \frac{2^{13}\cdot
3^4t^3}{d} \right ),
$$
which when reduced $\mod 2$ is:
$$
    f_{24r-1,24t}(z)\equiv \sum_{n\equiv r \mod t}p(n)q^{24n-1}
        \sum_{n=0}^{\infty}q^{24 A t(2n+1)^2} \mod 2. 
$$
Note that the arithmetic progression $r\mod t$ corresponds to the
arithmetic progression $24r-1 \mod 24t.$
If $p(M)$ is odd for at least one $M\equiv r\pmod t$ but only finitely many,
then the $\mod 2$ factorization above contradicts Lemma 1.
 This proves that if $p(M)$ is
odd for at least one $M\equiv r\pmod t,$ then $p(M)$ is odd for infinitely many
such $M.$

The computation of the constant $C_{r,t}$ follows easily
from the bound in
 Sturm's theorem.
\enddemo \hfill $\square$

\noindent
This theorem then proves that if $0\leq r < t,$ and if $p(M)$ is
ever odd for an $M\equiv r\pmod t$ (hence infinitely often),
then  the first odd value must occur
where $M < C_{r,t}.$
It is easy to verify that $C_{r,t} < 10^{10}t^7$ since
$\frac{t}{12} > 2^j$ when we choose the minimal $j$ such that
$2^j > \frac{t}{24}.$
As a consequence of the two main
theorems we find that the conjecture holds for an arithmetic
progression $r \pmod t$ provided there is at least one $N\equiv r\mod
t$ for which $p(N)$ is odd.
By computing $p(n) \mod 2$ for all $n\leq 5,000,000$ we found that
every arithmetic progression with modulus $t\leq 100,000$ contains an
integer $M$ for which $p(M)$ is odd. Therefore we obtain:

\noindent
\proclaim{Main Corollary } 
For all $0\leq r < t \leq 10^5,$ there are infinitely many integers $M
\equiv r\pmod t $
for which $p(M)$ is odd.
\endproclaim
\noindent
\head 3. Acknowledgements \endhead
\noindent
I am indebted  Ken T. Burrell
( Universal Analytics, Inc.) 
whose computations are the content of the Main Corollary. The computations
were completed on a CRAY C-90 at the San Diego Supercomputing Center.
I am indebted to Andrew Granville for his  advice
regarding this research. I also thank the referee whose comments
improved the paper in many ways. In particular, the proof of Lemma 1
is an improved version of the original proof as suggested by the
referee.


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\enddocument