on the distribution of parity in the partition function

On the Distribution of Parityin the Partition Function

By Thomas R. Parkin and Daniel Shanks

1. Introduction. Let p(n) be the number of (unrestricted) partitions of n, and

define p(0) = 1. Then p(n) is generated by

(D E p(n)xn = "ft —"T, = 11 + ¿ (-l)V<3n-1)/2 + xn(3n+v/2]n=0 n=l (1 — X ) l n=l

There is little known about p(n) modulo 2; in particular, there are no known criteria

for the parity of p(n) comparable in simplicity with Ramanujan's famous sufficient

condition for divisibility by 5 :

(2) 5 | p(5fc + 4) .

Kolbcrg [1] proved, but by contradiction and without identifying the arguments

n, that i nitely many p(n) are even, and infinitely many are odd. His proof is

almost as simple as Euclid's proof that there are infinitely many primes, but like

that proof it offers only very little more in the way of exact information concern-

ing questions of distribution.

From Gupta's tables [2], [3] we find the following cumulative distribution into

odds and evens for 0 ^ n :£ 499.

n S 99 n < 199 n ¿ 299 n < 399 n g 499

Odds 58 111 171 222 277

Evens 42 89 129 17S 223

In the absence of any known reason to the contrary, and because of the rather un-

smooth recursion for p(n) implied by (1), it would be natural to guess that the

evens and odds are equinumerous, i.e., that the ratio of their counts has the limit

1 as the upper bound for n —» °°. But the early preponderance of the odds, as just

tabulated, would make us hesitate to conjecture that this is true. Nonetheless, it

seemed to us not unlikely that this early preponderance might wash out as later

returns came in (from upstate, so to speak). But it dees seem unlikely that a theo-

retical proof of this could be attained with known techniques.

We have therefore examined the question empirically with a computer, and

have put an even stronger question. Consider the number m = 1.74264258- • -,

which when written in binary :

(3) m = 1.10111110000111011101

has its fcth bit to the right of the binary point 0 or 1 according as p(fc) is even or

odd. (m stands for Major MacMahon.) We now ask if m is normal with respect to

the base 2. If so, this not only implies the previously supposed equinumerosity, but

Received February 13, 1967.

466

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DISTRIBUTION OF PARITY IN THE PARTITION FUNCTION 467

also implies that all possible pairs, 00, 01, 10, and 11, have an asymptotic density

of I, etc.Here, however, we must note that the corresponding proposition modulo 5 is

definitely false. Thus, if

(4) r = 1.12302102021210112002 • • •

is a number written in quinary with its fcth place = p(k) modulo 5, we know from

(2) that r is not normal. In fact, r is not even simply normal since it is further known

that more than 20% of the p(n) are divisible by 5. For, in addition to (2), Morris

Newman shows, in the following paper [4], that

(5) 5|p(5-194fc+15147)

also, and still other independent linear functions also have this property. (A. O. L.

Atkin has obtained more general results; these will appear in [5].)

One of our reasons for stressing this failure modulo 5 is because of the character

of our main problem. Suppose, for instance, that our empirical investigation shows

that parity does appear to be equinumerous, and even normal. Then one might well

remark: "So what? Isn't that what one expects?" But the failure modulo 5 puts

the problem in a more interesting light.

We have determined the parity of p(n) up to n = 2,039,999. In what follows

we will indicate our method, our results, and some related investigations.

2. Notation and Nomenclature. Let a„ be the nth bit of m in (3):

(6) an = p(n) (mod 2) .

Let the finite sequence

amam+iam+2 • • • am+k—i

be called the mth fc-tuple. Thus 1101 is the 0th 4-tuple and 11111 is the 3rd 5-tuple.There are 2* possible types of fc-tuple, and let us designate these 2* types by the

integer, which, when written in binary, is the fc-tuple itself. Thus 1101 is the 13th

type of 4-tuple and 11111 is the 31st type of 5-tuple. Let

2»be the number of t type fc-tuples that appear to the left of, but not including, the

nth fc-tuple. (We find it convenient, because of (11) below, to count the 0th fc-tuple

here, and therefore to omit the nth, so that the argument n in 2~Ak) in) means that

n fc-tuples have been counted.) Thus, from (3),

(2) (2) (2) (2)

E (10) = 2 , £ (10) = 1 , £ (10) =2 £ (10) = 50 12 3

and referring to our previous table,

o) (i)

2Z (500) = 223, £ (500) = 277 .0 1

Then equinumerosity means


468 THOMAS R. PARKIN AND DANIEL SHANKS

(1) (1)

(7) I(n)~2W~ñ"lo 1 ¿

while the stronger normality means that

(W(8) £ in) ~ 2~kn

t

as n —-> =o for all £ and all fc.

Note that if one has counted the fc-tuples 2~Ak) in), one can obtain the counts of

j-tuples with j < fc simply by addition. Thus

(8) (8) (7)

Z in) + Z in) = Z (n)

and generally

«o (*) (*-«(9) Z in) + Z («) = Z (n)

2 i 2 <+1 Í

for all fc and all t.

To test normality we have counted the 256 types of 8-tuples out to n = 2 • 106,

and we deduced from these the counts, successively, of 7-tuples, 6-tuples, etc.

3. Computing the Parity Individually or En Masse. That the first two terms of

equation (1) are equal is fairly obvious. For the simplest proof of the equality of

the second and third terms, see [6]. Together, these equations imply Euler's recur-

rence : For n 2: 1,

QQ) Pin) = Pin - 1) + pin - 2) - p(n - 5) - p(n - 7)

+ ••• + (-Di+1p(n - e,-)

where e,; = \ i(3i =F 1), and where the series breaks off just before n — e¿ becomes

negative. One may thus compute the an en masse by recurrence using (10) modulo

2. For n large about § (6n)1/2 terms are needed to compute an if the previous a„_e¿

are already known.

But MacMahon [7] found the more efficient recurrences :

ctin = an + a„_7 + a„_9 + ■ • • + an-a¡ with a,- = i(8i Tl)

, . ain+i = an + a„_5 + a«-n + • • ■ + an-a¡ with ß, = i (Si =F 3) , , „,

atn+s = an + a„_3 + an-i3 + • • • + an-yi with y{ = i (Si =F 5)

a4n+6 = an + an-i + an_i5 + • • • + ans¡ with 5,- = i(Si =F 7)

(Note that 4n + 2 = 4(n — 1) + 6, but the formulas are neater as given.) We

will give a proof of (11) presently. For now, let us note the savings possible.

(1) The number of terms for an (not ain) with n large is now ~J (2n)1/2 so that

the use of (11) requires only V 3/8 = 0.2165 as much arithmetic as the use of (10).

(2) To compute an out to n = N we now need to save the an only to n = [2V/4],

so that only 0.25 as much storage is necessary.

Aside from this more efficient computation en masse, there also arises the pos-

sibility of iterating (11), and thus of computing an individual an with no mass



storage whatsoever, since each application of (11) reduces the arguments by a

factor of 4. We will discuss this possibility briefly later.

4. MacMahon's Congruences. In [7] MacMahon gave a proof of (11) based upon

self-conjugate partitions, and in [8] he used (11) to compute the parities out to

n = 1000. Subsequently, independently, and in effect, but not explicitly, G. N.

Watson [9] reproved (11) using theta functions. Still later, H. Gupta [10] gave still

another proof, this time using Ramanujan's tau function.

Perhaps the most direct proof, since it involves knowledge of none of these

special concepts or functions, is this : Since

-. i n i 2/1 i i n i in1 + xn + x*" + • • • -a 1 - xn + xin-= —j^ (mod 2)

1 — X 1 + X

we have

(1 - x)(l - x2)(1 - x3) ■■■ (1 - x)(1 + x2)(1 - x3)(1 + X*) ■ ■

d _ A) (l - A) ■ ■(1 - x)(l -x)(l -x4)(l -*")

(mod 2)

Thusco -. co -• co + In

n r^ - n r-^-t • n r^î3 (™d2)-i=l 1 — X "=1 1 — X «=1 1 — X

Since the product on the right equals Y^ñ-o xn(-n+lV2 (see [11] for the shortest proof)

we have

(12) Z Pin)xn = Z Pin)xin Z xMn+1)/2 (mod 2) ,in V~* n(n+l)/2__ pyrijx = ¿^ p\ri)C

n=0

and comparing like powers of x, congruences (11) follow quickly.

It may be of interest to indicate the quite extraneous considerations that led us

to this problem. One of us was in the process of reviewing [12] The Groups of Order

2" (n ^ 6), by Marshall Hall, Jr. and James K. Senior, Macmillan, New York,

1964. The abelian groups there are designated as belonging to a family Ti, and the

number of such groups of order 2" is, of course, pin). It may be noted, see pages

103-104, that the lattice diagrams of these groups suggest that they fall into dual

pairs. The question of whether p(n) is even or odd is therefore the question of

whether there are an even or odd number of lattices which are self-dual.

This leads one to consider self-conjugate partitions and thus to rediscover (11)

with (essentially) MacMahon's proof. But the proof above is somewhat simpler.

Naturally, after having "discovered" the efficient congruences (11), one is eager to

exploit them.

5. Normality. We show in Table 1 the value of \ m = .676- • • in octal to 3200

places. In this one can read an for 0 ^ n ^ 9599. We have placed in the UMT

file of this journal the complete 213-page value of \ m out to n = 2,039,999. In

Tables 2 and 4 we list the counts of the 8-tuples J2f> (n) for t = 0(1)255 and

n = 106 and 2 • 106, respectively. For example,



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DISTRIBUTION OF PARITY IN THE PARTITION" FUNCTION" 471

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(8)

E do6)(8)

3952 and £ (2-106) = 7916

These tables are read first across, and then down, for increasing t.

From Tables 2 and 4 we compute the counts of fc-tuples for k = 7, 6, • • •, 1

at n = 106 and 2-106, respectively. This is done by use of the recursion (9), and

the results are listed in Tables 3 and 5 in the obvious way. Thus

Z do6)(6) (5)

7900, 2~2 (106) = 15848, £ (2-106) = 62655

The initial impression of this data is that no type of fc-tuple is favored over

other types, that the various types are equidistributed, and that the data here is

consistent with the hypothesis of normality. We have attempted no elaborate sta-

tistical tests of this, but we did examine Good's psi-square serial test [13], [14] to a

limited extent. Let

(13)2*-i / (« \

^2 = 2V1Z Z(n)-2~*n(=0 \ t '

Good showed that if the bits of a binary number are random, then ipk2 has an

expectation 2k — 1. We list these \pk2 for fc = 1(1)6 and n = 106, 2 • 106 together with

their expectation in Table 6.

Table 6

n 106 n = 2-106 Expect.

123456

0.7961.6317.737

23.10644.32987.733

0.5061.1922.6629.429

21.77056.850

137

153163

Now note: We are testing here for randomness, but we are really interested in

normality. The former implies the latter, but what of the converse? The data in

Table 6 is consistent with randomness, and therefore also with normality. At n

= 2-106 (but not at n = 106) the distribution is even "too good." It seems to us

conceivable (but admittedly, we are now going somewhat beyond our competence)

that real numbers may exist with the \pk2 consistently too small. While such be-

havior would not be random, it could still imply normality—in fact, the smaller

the \pk2 are, the better.

6. Equinumerous Evens and Odds. Turning now to fc = 1 in greater detail—

and the question whether even and odd partition numbers are equinumerous—we

list in Table 7 the number of odds, X.Í1' in), and the ratio of odds to evens

Zi" W/ES1' in) forn = 50,000(50,000)2 • 106.Since these steps An = 50,000 are large and therefore do not allow a completely

accurate picture of the variations in the ratio function, we supplement Table 7 with



the description in Table 8. This lists 11 regions, A through K, within each of which

the ratio remains continually greater than 1, or continually less than 1. Thus, the

early preponderance of the odds, that we already noted, continues throughout region

A until n = 6672. Between these regions there are many small oscillations of the

ratio function around the value 1. For example, between regions G and H, the

difference:

odds — evens

varies between +56 and —65, and the ratio equals 1 for 176 different values of n

(including, as in Table 7, n = 400,000).

Table 7

n-10-

5101520253035404550r,rt60657075SO859095

100

Odds

25016502007504199766

124703149758175105200000225123250016274917299972324951349834374718399531424656449744474475499554

Ratio

1.001281.008031.001090.995330.995260.99678

.00120

.00000

.00109

.000120.999400.999810.999700.999050.998500.997660.998380.998860.997790.99822

n-10"

105110115120125130135140145150155160165170175180185190195200

Odds

524597549632574646599770624669649700674581699672724763749745774859799757824694849627874724899622924804949733974570999497

Ratio

0.998470.998660.998770.999230.998940.999080.998760.999060.999350.999320.999640.999390.999260.999120.999370.999160.999580.999440.999110.99899

Table 8

Region Limits Ratio Extreme \pi(n) At n

A15CDEFGIIIJK

1-667116287-4878149185-151211

162951-332867333373-363347363769-375013376961-395293406565-494241538051-601509637169-645423646475-2040000+

>1<1>1<1>1<1>1>1<1>1<1

+ 1.-1

+2.-1.

996*662882684

+0.553-0.158

+0.204+0.692-0.499

+0.154-1.165

1230*2101778823

241706347684367246386259434150569769641119812968

* Only n > 1000 examined here.



Consistent with the definition (13) is the designation \¡/i(n) for the normalized

difference:

(14)odds - evens Zi" (w) - Zo" O)

= *i(n)An An

As in the previous section, our main interest here is not so much in the distribution

of \pi(n) as in its extreme values, and in Table 8 we list the extreme value it takes

on in each interval. For instance, in region B, at n = 21017 there are 10629 evens

and 10388 odds for an extreme value

t/-! (21017) = -1.662.

In regions E through J parity is very much equidistributed. The worst normalized

difference occurs in region C at n = 78823, with 39816 odds and only 39007 evens.

(On Table 7, this n lies between the first two entries, and has a ratio = 1.02074.)

It is reasonable to conjecture that

(15) ti(n) = 0(A)

for any positive e. If this is true, then we have not merely that the ratio

we also know its rate of convergence :

(16) Iratio - 1| < an'l/2+t

for some a, and any e.

1, but

7. Runs. The data in Section 5 was extended only to 8-tuples. To go beyond

would require massive amounts of data, but the following special cases are of some

interest. How often should one expect say, 15, and only 15 consecutive odd partition

numbers? Since this presumes that the partition numbers immediately prior to such

a sequence and immediately subsequent are both even, we are in fact asking for the

count of 17-tuples of type 2(216 — 1) = 65534. As above, the expectation to n

= 2 • 106 is

(17)

Z (2-106)65534

2_17(2-106) = 15.26.

Actually, there are 16 such runs of exactly 15 successive odds—the first run begin-

ning with p(108417), and the sixteenth beginning with p(1936252).

In Table 9 we indicate the number of runs ïï 15 out to n = 2 • 106. There are

no runs here greater than 20. All of this data seems to be as expected.

Table 9

k Even Runs Odd Runs Expectation

151617181920

Total

1075221

27

1645400

29

15.37.63.81.91.00.5

30.1



Curio-collectors may wish to know that the 20 partition numbers

pin) , 1517214 Sná 1517233

are all even, while

pin), 617995 í»í 618012

constitutes the first sequence of exactly 18 odd partition numbers.

8. Remarks on the Presumed Normality. The last three sections, taken together,

do make a good empirical case for normality (modulo 2). We are indebted to Dr.

A. O. L. Atkin for a reason why the modulus 2 and also the modulus 3 would be

expected to be special for the partition numbers. All known congruence relations

for these numbers can be deduced from the so-called modular forms. Entering here

in a fundamental way is the linear function

24m - 1 ,

and while this can be divisible by any prime greater than 3, 2 and 3 are clearly

special. Therefore, Atkin would also expect normality (modulo 3). We have not

examined this.

Of course, such considerations are merely suggestive, and, so far, have not led

to a proof of normality for either modulus, 2 or 3.

Another aspect of the distinction here between the apparent normality (modulo

2) and the distinct nonnormality (modulo 5), as exemplified in (2) and (5), is that

one is reminded of the numbers of Wolfgang Schmidt. As is known, he showed [15],

[16] that there exist real numbers x normal to one base r without being normal to

another s. Perhaps we should clarify the difference between the phenomena pres-

ently under investigation and Schmidt's phenomena. Given any sequence of integers,

ain), we could construct two different real numbers as in our equations (3) and (4),

and they may, as apparently is the case here, be normal to one base while not to

another. On the other hand, a Schmidt number x gives rise to two different integer

sequences:

a(n) = [xrn] and b(n) = [xsn].

Finally, we wish to draw the main inference. Some time ago, Professor Freeman

Dyson wrote one of us, "Atkin and I were never able to do anything with modulo 2

[for the partition function]." But if the parity is normal, and this is what our

investigation strongly suggests, it appears to be a valid inference that "nothing"

can be done—"nothing" surely as simple as the congruence (2), or even as profound

as the congruence (5). There remains the problem of proving the presumed normality,

but no doubt that will be very difficult. Rather more promising is the weaker prob-

lem of showing that every fc-tuple occurs, that is :

(*)Z in) > 0 (every t, fc)t

for a sufficiently large n. Happily, this implies the (only seemingly stronger) result:

(«Z in) -> » (all t, fc) .



9. Iterated Computation of the Parity ; An Unsolved Problem. As we indicated

at the end of Section 3, by iterating equations (11) one can determine individual

parities independently of any stored table of an except for

ao = 1, a2 = 0 .

This leads to an unsolved problem of interest. Let us introduce an abbreviated no-

tation; instead of

(1200 = Ö50 + sZ43 + CMi + a20 + 0,16

we write

200 = 50, 43, 41, 20, 16 .

The algorithm is standardized by use of the three rules:

(a) Replace the largest term on the right by its equivalent in (11).

(b) Whenever two repetitions of an argument appear on the right, cancel them

both (since their sum is even in any case).

(c) Repeat until 0 or 2 or 0, 2 is all that remains on the right. Example:

For 200 one has the sequence :

50, 48, A, 20, 16, 11, 10, 10, 7, 10, 5, 5, 4, 2, 1, 0, 1,1, 0.

Here we have italicized each term replaced by its equivalent, and used boldface for

each pair eliminated by cancelling. Thus p(200) = p(2) = even.

In the computation for 200 we listed 19 terms, and cancelled A pairs. We define

tin) and ein)

to be these two functions. Thus

¿(200) = 19 , c(200) = 4 .

Let us compute these functions for n = 100, 200, 300, 400, 500, 600. To do the

algorithm efficiently, it is best not to use (11) directly, but, after having decided

whether the current term to be replaced is of the form

4n, 4n + 1 , 4n + 3 , or 4n + 6 ,

respectively, we write down n, and then subtract according to the differences:

7, 2, 21, 4, 35, 6, 49, 8, etc.,5, 6, 15, 12, 25, 18, 35, 24, etc.,

3, 10, 9, 20, 15, 30, 21, 40, etc., or

1, 14, 3, 28, 5, 42, 7, 56, etc.

Here is a brief Table 10.

Table 10

100200300400500600

tin)

111930385850

ein)

240

111017



We raise the questions whether

(17) tin) = Oin)1

(18) c(n) = 0(n)l

Clearly, t(n) will generally increase with n, but "luck" plays a part; for 400 and

600 there is much cancellation of large terms, while for 500 there is relatively little.

The real point of our query is the question whether the parity of an individual

p(n) can be determined in 0(n) operations. If one computed such an individual

parity by our previous, en masse, table building, technique the computation would

require

f 0(Vn)dn = 0(n3'2)

operations. We do not know whether (17) is true.

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on the distribution of parity in the partition function

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