on the distribution of parity in the partition function
TRANSCRIPT
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
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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
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DISTRIBUTION OF PARITY IN THE PARTITION FUNCTION 469
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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470 THOMAS R. PARKIN AND DANIEL SHANKS
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DISTRIBUTION OF PARITY IN THE PARTITION" FUNCTION" 471
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472 THOMAS R. PARKIN AND DANIEL SHANKS
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DISTRIBUTION OF PARITY IN THE PARTITION FUNCTION 473
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474 THOMAS R. PARKIN AND DANIEL SHANKS
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DISTRIBUTION OF PARITY IN THE PARTITION FUNCTION 475
(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
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476 THOMAS R. PARKIN AND DANIEL SHANKS
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.
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DISTRIBUTION OF PARITY IN THE PARTITION FUNCTION 477
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
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478 THOMAS R. PARKIN AND DANIEL SHANKS
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) .
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DISTRIBUTION OF PARITY IN THE PARTITION FUNCTION 479
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
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480 THOMAS R. PARKIN AND DANIEL SHANKS
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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