ct.
Abstra
ELECTRONIC RESEARCH ANNOUNCEMENTS
OF THE AMERICAN MATHEMATICAL SOCIETY
Volume 1, Issue 1, 1995
PARITY OF THE PARTITION FUNCTION
KEN ONO
(Communicated by Don Zagier)
Letp(n) denote the number of partitions of a non-negative integer
n. A well-known conjecture asserts that every arithmetic progression contains
in nitely many integers M for which p(M) is odd, as well as in nitely many
integersN for whichp(N) is even (see Subbarao [22]). From the works of var-
ious authors, this conjecture has been veri ed for every arithmetic progression
with modulust whent =1;2;3;4;5;10; 12; 16; and 40: Here we announce that
there indeed are in nitely many integers N in every arithmetic progression for
which p(N) is even; and that there are in nitely many integers M in every
arithmetic progression for which p(M) is odd so long as there is at least one
10 7such M. In fact if there is such an M, then the smallest such M 10 t .
Using these results and a fair bit of machine computation, we have veri ed the
conjecture for every arithmetic progression with modulus t 100; 000.
1. Introduction
A partition of a non-negative integern is a non-increasing sequence of positive inte-
gers whose sum isn. Euler’s generating function forp(n), the number of partitions
of an integer n; is:
1 1
X Y 1n(1) p(n)q = :
n1−q
n=0 n=1
Ramanujan discovered various surprising congruences forp(n)whennis in certain
special arithmetic progressions; for example:
p(5n+4) 0mod5;
p(7n+5) 0mod7;
and
p(11n+6) 0mod1:
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
Received by the editors February 28, 1995, and, in revised form, May 3, 1995.
1991 Mathematics Subject Classi cation . Primary 05A17; Secondary 11P83.
Key words and phrases. Parity conjecture, partitions, modular forms.
The author is supported by NSF grant DMS-9508976.
c 1995 American Mathematical Society
35
36 KEN ONO
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
1often’ even and odd; that is, thatp(n)isevenfor xpositive integersn x (see2
Parkin and Shanks [19]).
In [22] Subbarao made the following conjecture on the parity of p(n), for those
integersn belonging to any given arithmetic progression:
Conjecture. For any arithmetic progression r (mod t), there are in nitely many
integers M r (mod t) for which p(M) is odd, and there are in nitely many
integers N r (mod t) for which p(N) is even.
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 whent=1;2;3;4;5;10; 12; 16 and 40.
Using very di erent methods, we go some way towards a proof of the conjecture.
Using the theory of modular forms, we announce:
Main Theorem 1. For any arithmetic progression r (mod t), there are in nitely
many integers N r (mod t) for which p(N) is even.
Main Theorem 2. For any arithmetic progression r (mod t), there are in nitely
many integers M r (mod t) for which p(M) is odd, provided there is one such
M. Furthermore, if there does exist anM r (mod t) for whichp(M) is odd, then
the smallest such M is less than C ,wherer;t
23 7 6
Y2 A 3 t 1
C := 1− −A;r;t 2 2d p
pj6t
twith d := gcd(24r− 1;t) andA> is a power of 2:
24
From the two theorems we obtain an algorithm to determine the truth of our
parity conjecture for any given arithmetic progression r (mod t): Compute p(N)
(mod 2) for N =r;r +t;r+2t;::: for all such N up to C . As soon as we ndr;t
one odd number we have veri ed the conjecture. If all these numbers are even then
we have proved that the conjecture is false.
Ken Burrell (Universal Analytics, Inc.) ran an e cient version of this algorithm
on a CRAY C-90, giving the following result:
5Main Corollary. For al l 0 r be a power of 2: De ne24
f (z) byt;A
X (24z) A 24n−1f (z):= (24tz)= a(n)q :t;A t
(48z)
n 1
2Then f (z) isacuspforminS (1152t; ). Moreover the Fourier expansiont;A 12A d
of f (z)mod2 can be factored as:t;A
! !
1 1 1
X X X
224n−1 24n−1 24At(2n+1)(2) f (z)= a (n)q p(n)q q mod 2:t;A t
n=0 n=0 n=0
Proof. Using the well known properties of the Dedekind eta-function, it is relatively
straightforward to deduce that f (z) is a modular form of weight 12A: It is alsot;A
straightforward to d that f (z) is a cusp form.t;A
The essential feature of the cusp formf (z) is the convenient fact thatf (z)t;A t;A
is essentially the product of the generating function for p(n) and a theta function
mod 2:
n n1 1+X 1−X
Since = mod 2; it follows that
n 2n 2n1−X 1−X 1−X
1 1 n
X Y 1−qnp(n)q mod 2:
2n1−q
n=0 n=1
In terms of the eta-functions, we nd that
1 124n
Y X(24z) 1 1−q 24n−1(3) = p(n)q mod 2:
48n(48z) q 1−q
n=1 n=0
The following in nite product identity was proved by Jacobi:
1 12 16n 2
Y X(16z) (1−q ) 2(2n+1)=q = q :
8n(8z) (1−q )
n=1 n=0
2 2Therefore since (1−X) (1−X )mod2; we nd that
(4)
1 1 1 1n 32 16n 2
Y Y Y X(1−q ) (1−q ) 2n 24 (2n+1) ( z)=q (1−q ) =q q q mod 2:
n 8 8n(1−q ) (1−q )
n=1 n=1 n=1 n=0
The factorization off (z) now follows easily from (3) and (4).t;A
Serre [20] proved the following remarkable theorem regarding the divisibility of
Fourier coe cients of holomorphic integer weight modular forms.
Theorem. (Serre) Let f (z) be a holomorphic modular form of positive integer
weight k on some congruence subgroup of SL (Z) with Fourier expansion2
1
X
nf(z)= a(n)q
n=0
where a(n) are algebraic integers in some number eld. If m is a positive integer,
then there exists a positive constant such that the set of integers n x for which
xa(n) is not divisible by m has cardinality :log x
With this theorem we obtain
PARITY OF THE PARTITION FUNCTION 39
Main Theorem 1. For any arithmetic progression r (mod t), there are in nitely
many integers N r (mod t) for which p(N) is even.
Proof. Comparing coec ients in (2), it is easy to deduce that
X
2 2 2
(5) a (Atk +n) p(At(k −i )+n)mod2:t
i 1;iodd
Now suppose every N n for which N r (mod t) has the property that0
p(N) is odd. If k 1mod4;then every integer n r (mod t)intheinterval
2 2[Atk +n ;At(k+2) +r−t] has the property that a (n) is odd since there are0 t
k +1
many odd summands in (5): After combining all such intervals, we nd a set
2
of positive integers with positive density for which a (n)6 0mod2: This wouldt
contradict Serre’s theorem.
Now we need to establish that there are in nitely many M r (mod t)wherep(M)
is odd provided that there is at least oneM: To do this we rst 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
nitely many M r (mod t)forwhichp(M) is odd, then the reduction mod 2
of the relevant modular form contradicts the lemma.
P
nFor a given positive integer m and formal power series f := a(n)q withn2Z
algebraic integer coe cients, we de ne Ord (f) to be the smallest integer n form
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 Ord (f)=1 ).m
P
1 nIf f(z)= a(n)q 2 M (N) for some positive integer N with algebraickn =0
integer Fourier coe cients from a xed number eld and m is a positive integer,
then Sturm proved that if
Yk 12Ord (f)> N (1− );m 212 p
pjN
then Ord (f)=1(i.e. a(n)