a formula for the partition function that ``counts

A formula for the partition function that “counts”

Drew Sills

Georgia Southern University

April 27, 2017

Drew Sills A formula for the partition function that “counts”

Joint work with

Yuriy Choliy

Department of Chemistry and Chemical BiologyRutgers University


A partition of an integer n is a representation of n as a sum ofpositive integers where order of summands (parts) does notmatter.

43+12+22+1+11+1+1+1,

so, we have five partions of 4.


Let p(n) denote the number of partitions of n.

p(0) = 1p(1) = 1p(2) = 2p(3) = 3p(4) = 5p(5) = 7p(6) = 11p(7) = 15p(8) = 22p(9) = 30p(10) = 42p(100) = 190,569,192p(200) = 3,972,999,029,388p(500) = 2,300,165,032,574,323,995,027



p(0) = 1

p(1) = 1p(2) = 2p(3) = 3p(4) = 5p(5) = 7p(6) = 11p(7) = 15p(8) = 22p(9) = 30p(10) = 42p(100) = 190,569,192p(200) = 3,972,999,029,388p(500) = 2,300,165,032,574,323,995,027



p(0) = 1p(1) = 1

p(2) = 2p(3) = 3p(4) = 5p(5) = 7p(6) = 11p(7) = 15p(8) = 22p(9) = 30p(10) = 42p(100) = 190,569,192p(200) = 3,972,999,029,388p(500) = 2,300,165,032,574,323,995,027



p(0) = 1p(1) = 1p(2) = 2

p(3) = 3p(4) = 5p(5) = 7p(6) = 11p(7) = 15p(8) = 22p(9) = 30p(10) = 42p(100) = 190,569,192p(200) = 3,972,999,029,388p(500) = 2,300,165,032,574,323,995,027



p(0) = 1p(1) = 1p(2) = 2p(3) = 3

p(4) = 5p(5) = 7p(6) = 11p(7) = 15p(8) = 22p(9) = 30p(10) = 42p(100) = 190,569,192p(200) = 3,972,999,029,388p(500) = 2,300,165,032,574,323,995,027



p(0) = 1p(1) = 1p(2) = 2p(3) = 3p(4) = 5

p(5) = 7p(6) = 11p(7) = 15p(8) = 22p(9) = 30p(10) = 42p(100) = 190,569,192p(200) = 3,972,999,029,388p(500) = 2,300,165,032,574,323,995,027



p(0) = 1p(1) = 1p(2) = 2p(3) = 3p(4) = 5p(5) = 7p(6) = 11p(7) = 15p(8) = 22p(9) = 30p(10) = 42

p(100) = 190,569,192p(200) = 3,972,999,029,388p(500) = 2,300,165,032,574,323,995,027



p(0) = 1p(1) = 1p(2) = 2p(3) = 3p(4) = 5p(5) = 7p(6) = 11p(7) = 15p(8) = 22p(9) = 30p(10) = 42p(100) = 190,569,192p(200) = 3,972,999,029,388p(500) = 2,300,165,032,574,323,995,027


Let pN(n) denote the number of partitions of n into parts lessthan or equal to N.

3 + 3 = 3 + 2 + 1 = 3 + 1 + 1 + 1 = 2 + 2 + 2= 2 + 2 + 1 + 1 = 2 + 1 + 1 + 1 = 1 + 1 + 1 + 1 + 1 + 1

So, p3(6) = 7


Generating Functions

The generating function for the sequence {an}∞n=0 is

GF [{an}∞n=0; x ] :=∞∑

n=0

anxn.

GF [{1}∞n=0; x ] =1

1− x

GF[{

1n!

}∞n=0

; x]

= ex


L. Euler (1707-1783)


(1

1− x

)(1

1− x2

)(1

1− x3

)

= (1 + x + x2 + x3 + x4 + · · · )× (1 + x2 + x4 + x6 + x8 + · · · )× (1 + x3 + x6 + x9 + x12 + · · · )


(1

1− x1

)(1

1− x2

)(1

1− x3

)= (1 + x1 + x1+1 + x1+1+1 + x1+1+1+1 + · · · )× (1 + x2 + x2+2 + x2+2+2 + x2+2+2+2 + · · · )× (1 + x3 + x3+3 + x3+3+3 + x3+3+3+3 + · · · )


(1

1− x1

)(1

1− x2

)(1

1− x3

)= (1 + x1 + x1+1 + x1+1+1 + x1+1+1+1 + · · · )× (1 + x2 + x2+2 + x2+2+2 + x2+2+2+2 + · · · )× (1 + x3 + x3+3 + x3+3+3 + x3+3+3+3 + · · · )

=1 + x1 + (x2 + x1+1) + (x3 + x2+1 + x1+1+1) + · · ·


(1

1− x1

)(1

1− x2

)(1

1− x3

)= (1 + x1 + x1+1 + x1+1+1 + x1+1+1+1 + · · · )× (1 + x2 + x2+2 + x2+2+2 + x2+2+2+2 + · · · )× (1 + x3 + x3+3 + x3+3+3 + x3+3+3+3 + · · · )

=1 + x1 + (x2 + x1+1) + (x3 + x2+1 + x1+1+1) + · · ·

=∞∑

n=0

p3(n)xn = GF [p3(n); x ].


GF [pN(n); x ] =∞∑

n=0

pN(n)xn =N∏

j=1

11− x j


Euler’s generating function for p(n)

GF [p(n); x ] =∞∑

n=0

p(n)xn =∞∏

j=1

11− x j .


Euler’s pentagonal number theorem

(1−x)(1−x2)(1−x3) · · · = 1−x−x2+x5+x7−x12−x15+x22+x26−· · · .

∞∏j=1

(1− x j) =∞∑

k=−∞(−1)kxk(3k−1)/2.

∞∑k=−∞

(−1)kxk(3k−1)/2∞∏

j=1

11− x j = 1.

∞∑k=−∞

(−1)kxk(3k−1)/2∞∑

n=0

p(n)xn = 1.


Euler’s partition recurrence

p(n) = p(n−1) + p(n−2)−p(n−5)−p(n−7) + p(n−12) + · · ·


G. H. Hardy (1877–1947)S. Ramanujan (1887–1920)


For |z| < 1, let f (z) :=∑n≥0

p(n)zn =∏j≥0

11− z j

Fix n. Note that f (z)/zn+1 has a pole of order n + 1 atz = 0, and essential singularities along |z| = 1.Residue of f (z)/zn+1 at z=0 is p(n).

p(n) =1

2πi

∫C

f (z)

zn+1 dz where C is any positively oriented,

simple closed contour enclosing the origin and inside theunit circle.


For |z| < 1, let f (z) :=∑n≥0

p(n)zn =∏j≥0

11− z j

Fix n. Note that f (z)/zn+1 has a pole of order n + 1 atz = 0, and essential singularities along |z| = 1.

Residue of f (z)/zn+1 at z=0 is p(n).

p(n) =1

2πi

∫C

f (z)




For |z| < 1, let f (z) :=∑n≥0

p(n)zn =∏j≥0

11− z j


p(n) =1

2πi

∫C

f (z)




For |z| < 1, let f (z) :=∑n≥0

p(n)zn =∏j≥0

11− z j


p(n) =1

2πi

∫C

f (z)

zn+1 dz

where C is any positively oriented,



For |z| < 1, let f (z) :=∑n≥0

p(n)zn =∏j≥0

11− z j


p(n) =1

2πi

∫C

f (z)




Hardy & Ramanujan’s first asymptotic formula for p(n)

p(n) ∼ 14n√

3eπ√

2n/3 as n→∞

(where f (n) ∼ g(n) means limn→∞ f (n)/g(n) = 1.)

At n = 200, RHS gives about 4,100,251,432,188p(200) = 3,972,999,029,388.



p(n) ∼ 14n√

3eπ√

2n/3 as n→∞


At n = 200, RHS gives about 4,100,251,432,188

p(200) = 3,972,999,029,388.



p(n) ∼ 14n√

3eπ√

2n/3 as n→∞


At n = 200, RHS gives about 4,100,251,432,188p(200) = 3,972,999,029,388.


P. A. MacMahon (1854–1929)


Hardy & Ramanujan’s “exact” asymptotic formula forp(n)

p(n) =1

2π√

2

α√

n∑k=1

√k∑

05h<k(h,k)=1

ωh,ke−2πihn/k ddn

eπk

√23 (n−

124 )√

n − 124

+ O(n−1/4),

with α an arbitrary constant and ωh,k a certain complex 24k throot of unity.

f (n) = O(g(n)) means that |f (n)/g(n)| is bounded for largeenough n.


Hardy & Ramanujan’s “exact” asymptotic formula forp(n)

p(n) =1

2π√

2

α√

n∑k=1

√k∑

05h<k(h,k)=1

ωh,ke−2πihn/k ddn

eπk

√23 (n−

124 )√

n − 124

+ O(n−1/4),

with α an arbitrary constant and ωh,k a certain complex 24k throot of unity.f (n) = O(g(n)) means that |f (n)/g(n)| is bounded for largeenough n.


Hans Rademacher (1892–1969)


Rademacher’s convergent series for p(n)

p(n) =1

π√

2

∞∑k=1

√k

×∑

05h<k(h,k)=1

e−2πinh/kωh,kddn

sinh(πk

√23

(n − 1

24

))√

n − 124

.


Calculating p(1) with Rademacher’s series

k = 1 term: 1.133558447k = 2 term: −0.130590021k = 3 term: −0.0297786023k = 4 term: 0.0157484033k = 5 term: 0k = 6 term: 0.0143908980


Ken Ono and Jan Bruinier


Bruinier–Ono formula for p(n)

p(n) =1

24n − 1Tr(n),

whereTr(n) :=

∑Q∈Qn

P(αQ).

Discriminant −24n + 1 = b2 − 4ac positive definite integralbinary quadratic forms Q(x , y) = ax2 + bxy + cy2 with theproperty 6 | a. The group Γ0(6) acts on such forms, and let Qnbe any set of representatives of those equivalence classes witha > 0 and b ≡ 1 (mod 12). For each Q(x , y), we let αQ be theCM point in the upper half of the complex plane, for whichQ(αQ,1) = 0. Also,

P(z) := −(

12πi

ddz

+1

2π=(z)

)F (z)


where

F (z) :=E2(z)− 2E2(2z)− 3E2(3z) + 6E2(6z)

2η(z)2η(2z)2η(3z)2η(6z)2

and

η(z) := q1/24∞∏

j=1

(1− qj)

andE2(z) := 1− 24

∑n≥1

∑d |n

dqn

with the convention that q := e2πiz .


To show p(1) = 1,

we have

Q1 = {Q1,Q2,Q3}= {6x2 + xy + y2,12x2 + 13xy + 4y2,18x2 + 25xy + 9y2}.

The CM points are

αQ1 =−1 +

√−23

12, αQ2 =

−13 +√−23

24, αQ3 =

−25 +√−23

36.

P(αQ1) ≈ 13.965486281P(αQ2) ≈ 4.517256859− 3.097890591iP(αQ3) ≈ 4.517256859 + 3.097890591i


To show p(1) = 1, we have

Q1 = {Q1,Q2,Q3}= {6x2 + xy + y2,12x2 + 13xy + 4y2,18x2 + 25xy + 9y2}.

The CM points are

αQ1 =−1 +

√−23

12, αQ2 =

−13 +√−23

24, αQ3 =

−25 +√−23

36.

P(αQ1) ≈ 13.965486281P(αQ2) ≈ 4.517256859− 3.097890591iP(αQ3) ≈ 4.517256859 + 3.097890591i


Ferrers graph

5 + 4 + 3 + 3 + 1 + 1 + 1 + 1

• • • • •• • • •• • •• • •••••


Ferrers graph with Durfee Square

5 + 4 + 3 + 3 + 1 + 1 + 1 + 1

• • • • •• • • •• • •• • •••••


Let D(n, k) denote the number of partitions of nwith Durfee square of order k .

p(n) =

b√

nc∑k=0

D(n, k).


D(n, k) =

Uk∑mk=0

Uk−1∑mk−1=0

· · ·U2∑

m2=0

(1 + n − k2 −

k∑h=2

hmh

)k∏

i=2

(mi+1),

where

Uj := Uj(n, k) = bn − k2 −

∑kh=j+1 hmh

jc.


How about a simple approximation formula?

Consider the partial fraction decomposition of the generatingfunction of D(n, k):

x(1− x)2 =

1(1− x)2 −

11− x

x4

(1− x)2(1− x2)2 =1/4

(1− x)4 −3/4

(1− x)3 +11/16

(1− x)2 −1/8

1− x

+1/16

(1 + x)2 −1/8

1 + x

etc., expand RHS as a series and extract coefficient of xn.


How about a simple approximation formula?Consider the partial fraction decomposition of the generatingfunction of D(n, k):

x(1− x)2 =

1(1− x)2 −

11− x

x4

(1− x)2(1− x2)2 =1/4

(1− x)4 −3/4

(1− x)3 +11/16

(1− x)2 −1/8

1− x

+1/16

(1 + x)2 −1/8

1 + x

etc., expand RHS as a series and extract coefficient of xn.


D(n,1) = n,

D(n,2) =(n − 1)(2n2 − 4n − 3)

48+ (−1)n n − 1

16,

D(n,3) =(n − 3)(6n4 − 72n3 + 184n2 + 192n − 235)

25920− (−1)n n − 3

64

+(ωn + ω−n)(n − 3) +

(n3

)81

(where ω := e2πi/3 and(n

3

)is the Legendre symbol)


D̃(n,1) = n,

D̃(n,2) =(n − 1)(2n2 − 4n − 3)

48+ (−1)n n−1

16 ,

D̃(n,3) =(n − 3)(6n4 − 72n3 + 194n2 + 192n − 235)

25920− (−1)n n−3

64

+(ωn+ω−n)(n−3)+(n

3)81


Approximate p(n) by

pD(n) :=

b√

nc∑k=1

D̃(n, k)

and by the k = 1 term of the Hardy–Ramanujan–Rademacherformula:

pR(n) :=

cosh(π√

23

(n − 1

24

))2√

3(n − 1

24

) −sinh

(π√

23

(n − 1

24

))2π√

2(n − 1

24

)3/2 .


n p(n) pD(n)− p(n) pR(n)− p(n) pR(n)− pD(n)5 7 0.25 0.26210 0.01210

10 42 −0.37905 −0.37221 0.0068415 176 0.39120 0.56047 0.1692720 627 −1.24394 −1.24232 0.0016225 1958 2.10036 2.09834 −0.0020230 5604 −3.72589 −3.72044 0.0054540 37,338 −7.39250 −7.39081 0.0017050 204,226 −14.9227 −14.9235 −0.0008060 966,467 −33.6090 −33.6385 −0.0294675 8,118,264 79.2210 79.2222 0.00129

100 1.9 × 108 −347.2173 −347.2167 0.00069150 4.1 × 1010 −4253.1144 −4253.1138 0.00058200 4.0 × 1012 −36202.1049 −36202.1042 0.00062300 9.2 × 1015 −1442614.889 −1442614.887 0.00168500 2.3 × 1021 −560997650.0056 −560997650.0066 −0.00093


pR(n)− pD(n) for 1 ≤ n ≤ 1600.

Can we find a tight bound and an explanation for the small sizeof |pD(n)− pR(n)|?


Thank you for listening!


a formula for the partition function that ``counts

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