The Combinatorics of -Series and Partitions, Center for

George Andrews’ 75th Birthday / 1

“The Combinatorics of q-Series and Partitions”,Center for Combinatorics, Nankai Univ., Aug 2-4, 2013

Partition Analysis and

Partition Congruences

Peter Paule(joint work with Cristian-Silviu Radu)

Johannes Kepler University LinzResearch Institute for Symbolic Computation (RISC)

George Andrews’ 75th Birthday / Preamble 2

Preamble


From: Doron Zeilberger<[email protected]>

Date: Wed, July 31, 2013 4:25 pm

To: <[email protected]>

Subject: AndrewsFest Dec. 2-4, 2013

... 2. If appropriate, could you read (at the

banquet or any other suitable time), the following

brief greeting.

‘‘I was looking forward to celebrating with you

12-8=4 months early, our great guru’s turning

three-quarters-of-a-century young. Due to unforeseen

circumstances, I could not make it in body, but I

will be with you in spirit. I told George that I

would give the talk that I intended for this

celebration at my own famous Experimental Mathematics

seminar, first thing in the Fall semester. But in

order to make it more timely, I will deliver it one

day after George’s actual birthday, i.e. on 4+1=5

Dec. 2013).’’


From: Doron Zeilberger<[email protected]>

Date: Wed, July 31, 2013 4:25 pm

To: <[email protected]>

Subject: AndrewsFest Dec. 2-4, 2013

... 2. If appropriate, could you read (at the

banquet or any other suitable time), the following

brief greeting.

‘‘I was looking forward to celebrating with you

12-8=4 months early, our great guru’s turning

three-quarters-of-a-century young. Due to unforeseen

circumstances, I could not make it in body, but I

will be with you in spirit. I told George that I

would give the talk that I intended for this

celebration at my own famous Experimental Mathematics

seminar, first thing in the Fall semester. But in

order to make it more timely, I will deliver it one

day after George’s actual birthday, i.e. on 4+1=5

Dec. 2013).’’


FYI, here is the title and abstract of the talk that

I would have given at Nankai, and hope to give at my

seminar on Dec. 5, 2013. It will be posted soon on

our seminar’s webpage.

Speaker: Doron Zeilberger

Place and Time: Hill 705, Dec. 5, 2013, 5:00-5:48pm

Title: George Eyre Andrews (b. Dec. 4, 1938):

A Reluctant REVOLUTIONARY

Abstract:


FYI, here is the title and abstract of the talk that

I would have given at Nankai, and hope to give at my

seminar on Dec. 5, 2013. It will be posted soon on

our seminar’s webpage.

Speaker: Doron Zeilberger

Place and Time: Hill 705, Dec. 5, 2013, 5:00-5:48pm

Title: George Eyre Andrews (b. Dec. 4, 1938):

A Reluctant REVOLUTIONARY

Abstract:


Abstract: One of the greatest combinatorialists and

number theorists of our time is George Andrews, who

made partitions an active and hot mathematical area,

and helped make Ramanujan a household name. But all

this dwarfs compared with his PIONEERING use of

Symbolic Computation, starting with very creative use

of the early computer algebra system SCRATCHPAD, and

continuing to this day with the still-going-strong

saga, joint with the RISC-Linz gang, on implementing

and beautifully applying MacMahon’s Omega calculus.

Alas, George, being a tie-and-jacket wearing

traditionalist, does not like revolutions, and hence

even he does not realize the full impact of his

mathematical legacy, that is FAR larger than the sum

of its many impressive parts.

George Andrews’ 75th Birthday / Partition Analysis 6

Partition Analysis


From The Selected Works of George E Andrews:

“The no. of partitions of N of the form N = b1 + · · ·+ bnsatisfying

bnn≥ bn−1n− 1

≥ · · · ≥ b22≥ b1

1≥ 0

equals the no. of partitions of N into odd parts each ≤ 2N − 1.

This problem cried out for MacMahon’s Partition Analysis, . . .

Given that Partition Analysis is an algorithm for producingpartition generating functions, I was able to convince Peter Pauleand Axel Riese to join an effort to automate this algorithm.”


From The Selected Works of George E Andrews:

“The no. of partitions of N of the form N = b1 + · · ·+ bnsatisfying

bnn≥ bn−1n− 1

≥ · · · ≥ b22≥ b1

1≥ 0

equals the no. of partitions of N into odd parts each ≤ 2N − 1.

This problem cried out for MacMahon’s Partition Analysis, . . .

Given that Partition Analysis is an algorithm for producingpartition generating functions, I was able to convince Peter Pauleand Axel Riese to join an effort to automate this algorithm.”


A bit of Personal History:

From: George E Andrews<[email protected]>

Date: Mon, 8 Apr 1996 16:13:11 -0400

To: [email protected]

Subject: Some thoughts on a joint project

Dear Peter,

Over a month ago I promised you a description of

a joint proposal that you and I might pursue. I have

now written up enough of what I know to describe a

reasonable undertaking.


A bit of Personal History:


Date: Mon, 8 Apr 1996 16:13:11 -0400


Subject: Some thoughts on a joint project

Dear Peter,

Over a month ago I promised you a description of

a joint proposal that you and I might pursue. I have

now written up enough of what I know to describe a

reasonable undertaking.


With this letter I am also e-mailing two papers,

one by Bousquet-Melou & Erikkson, and one by me. The

first is a beautiful combinatorial proof of an

absolutely lovely generalization of Euler’s Theorem,

which they call the Lecture Hall Partition Theorem.

My paper provides a second proof of this theorem, but

that is not its main point, which is the

resuscitation of MacMahon’s Partition Analysis ...


... MacMahon developed his Partition Analysis to

prove his conjectures on plane partitions. On page

187 of Combinatory Analysis, Vol.2, MacMahon admits

defeat in using Partition Analysis to prove his plane

partition conjectures. This admission effectively

killed the subject of Partition Analysis as

originally conceived.


To his credit, MacMahon developed his theory of

lattice permutations which eventually proved his

conjectures. Richard Stanley, in his Ph.D. thesis,

beautifully extended this aspect of MacMahon’s work.

However all this left the original Partition Analysis

completely moribund. The main point is that MacMahon

never proved a single deep theorem using his

Partition Analysis, although one can sense its power

in some elementary examples (e.g. p.183 or p.430 in

Combinatory Analysis, Vol.2).

Now I believe that Partition Analysis merits

another chance.


To his credit, MacMahon developed his theory of

lattice permutations which eventually proved his

conjectures. Richard Stanley, in his Ph.D. thesis,

beautifully extended this aspect of MacMahon’s work.

However all this left the original Partition Analysis

completely moribund. The main point is that MacMahon

never proved a single deep theorem using his

Partition Analysis, although one can sense its power

in some elementary examples (e.g. p.183 or p.430 in

Combinatory Analysis, Vol.2).

Now I believe that Partition Analysis merits

another chance.


First of all, it can be used to prove the beautiful

theorem of Bousquet-Melou and Eriksson. Second, I

believe (and I am only part way through this at the

moment, so I may have to revise this suggestion

later) that I can use Partition Analysis to prove

that the generating function for all plane partitions

is ∏n≥1

1

(1− qn)n,

as well as the other MacMahon theorem on plane

partitions. If and when I get time to verify and

write down my ideas, you will receive them first.

This effectively retrieves MacMahon’s original

failure.


Third, and most important, if one could turn

partition analysis into an effective automated tool,

it seems to me that there are countless conceivable

applications ...

I would be very interested in your thoughts on

these ideas. They are not fully formed and may be

subject to obstacles that I have not currently

envisioned.

Best wishes,

George


My reply:

From: [email protected]

Date: Tue, 9 Apr 1996

To: George E Andrews<[email protected]>

Subject: Re: Some thoughts on a joint project

Dear George,

Best thanks for your very interesting

proposal-sketch and the papers. Of course I will

make more detailed comments on that matter (which at

the first glance looks extremely appealing), but this

will be a little bit later ...

Best wishes,

Peter


My reply:

From: [email protected]

Date: Tue, 9 Apr 1996

To: George E Andrews<[email protected]>


Dear George,

Best thanks for your very interesting

proposal-sketch and the papers. Of course I will

make more detailed comments on that matter (which at

the first glance looks extremely appealing), but this

will be a little bit later ...

Best wishes,

Peter


Andrews’ reaction:


Date: Wed, 10 Apr 1996 13:00:23 -0400



Dear Peter,

There is, of course, no rush on the material I

sent. Indeed my own thoughts about it are still

somewhat uncertain. I have made further progress

since I wrote to you; ... It might well form the

beginning for joint work if I can manage part of my

1997-1998 sabbatical in Austria ...

Best wishes,

George


Andrews’ reaction:


Date: Wed, 10 Apr 1996 13:00:23 -0400



Dear Peter,

There is, of course, no rush on the material I

sent. Indeed my own thoughts about it are still

somewhat uncertain. I have made further progress

since I wrote to you; ... It might well form the

beginning for joint work if I can manage part of my

1997-1998 sabbatical in Austria ...

Best wishes,

George


Example (PA and the Omega package)

Find a suitable closed form of

L(x1, x2, x3) :=∑

b1,b2,b3∈N s.t. 2b3 − 3b2 ≥ 0, b2 − 2b1 ≥ 0

xb11 xb22 x

b33

= Ω=

∑b1,b2,b3≥0

λ2b3−3b21 λb2−2b12 xb11 xb22 x

b33 .

In[1]:= << Omega2.m

Omega Package by Axel Riese (in cooperation with George E. Andrewsand Peter Paule) - c©RISC, JKU Linz - V 2.47




L(x1, x2, x3) :=∑

b1,b2,b3∈N s.t. 2b3 − 3b2 ≥ 0, b2 − 2b1 ≥ 0

xb11 xb22 x

b33

= Ω=

∑b1,b2,b3≥0

λ2b3−3b21 λb2−2b12 xb11 xb22 x

b33 .

In[1]:= << Omega2.m





L(x1, x2, x3) :=∑

b1,b2,b3∈N s.t. 2b3 − 3b2 ≥ 0, b2 − 2b1 ≥ 0

xb11 xb22 x

b33

= Ω=

∑b1,b2,b3≥0

λ2b3−3b21 λb2−2b12 xb11 xb22 x

b33 .

In[1]:= << Omega2.m



In[2]:= LCrude = OSum[ x1b1 x2b2 x3b3,

2 b3 - 3 b2 ≥ 0, b2 - 2 b1 ≥ 0 , b1 ≥ 0, λ]

Out[2]= Ω≥

λ1, λ2

1(1− x1

λ22

)(1−λ2 x2

λ31

)(1−λ21 x3)

In[3]:= L=OR[LCrude]

Out[3]= 1+x2 x32

(1−x3)(1−x22 x33)(1−x1 x22 x33)

In[4]:= L /. x1->q, x2->q, x3->q

Out[4]= 1+q3

(1−q)(1−q5)(1−q6)



2 b3 - 3 b2 ≥ 0, b2 - 2 b1 ≥ 0 , b1 ≥ 0, λ]

Out[2]= Ω≥

λ1, λ2

1(1− x1

λ22

)(1−λ2 x2

λ31

)(1−λ21 x3)


Out[3]= 1+x2 x32

(1−x3)(1−x22 x33)(1−x1 x22 x33)

In[4]:= L /. x1->q, x2->q, x3->q

Out[4]= 1+q3

(1−q)(1−q5)(1−q6)



2 b3 - 3 b2 ≥ 0, b2 - 2 b1 ≥ 0 , b1 ≥ 0, λ]

Out[2]= Ω≥

λ1, λ2

1(1− x1

λ22

)(1−λ2 x2

λ31

)(1−λ21 x3)


Out[3]= 1+x2 x32

(1−x3)(1−x22 x33)(1−x1 x22 x33)

In[4]:= L /. x1->q, x2->q, x3->q

Out[4]= 1+q3

(1−q)(1−q5)(1−q6)



2 b3 - 3 b2 ≥ 0, b2 - 2 b1 ≥ 0 , b1 ≥ 0, λ]

Out[2]= Ω≥

λ1, λ2

1(1− x1

λ22

)(1−λ2 x2

λ31

)(1−λ21 x3)


Out[3]= 1+x2 x32

(1−x3)(1−x22 x33)(1−x1 x22 x33)

In[4]:= L /. x1->q, x2->q, x3->q

Out[4]= 1+q3

(1−q)(1−q5)(1−q6)


Remark: q-Engel Expansions

In[1]:= << Engel.m

Package Engel written by Burkhard Zimmermann (in cooperation withG.E. Andrews, A. Knopfmacher, and P. Paule) Revised in 2008, 2009,2011 by Ralf Hemmecke. Contains qNormal.m, written by Axel Riese. -c©RISC, JKU Linz - V 1.11

In[2]:= A = 1/qfac[q, q2];

In[3]:= Engel[A, q, 15]

Out[3]= 1 + q1−q + q3

(1−q)(1−q2) + q6

(1−q)(1−q2)(1−q3) +O(q)10

This means, the program guesses Euler’s

∞∑k=0

q(k+12 )

(q; q)k=

∞∏n=0

1

1− q2n+1.



In[1]:= << Engel.m


In[2]:= A = 1/qfac[q, q2];

In[3]:= Engel[A, q, 15]

Out[3]= 1 + q1−q + q3

(1−q)(1−q2) + q6

(1−q)(1−q2)(1−q3) +O(q)10


∞∑k=0

q(k+12 )

(q; q)k=

∞∏n=0

1

1− q2n+1.



In[1]:= << Engel.m


In[2]:= A = 1/qfac[q, q2];

In[3]:= Engel[A, q, 15]

Out[3]= 1 + q1−q + q3

(1−q)(1−q2) + q6

(1−q)(1−q2)(1−q3) +O(q)10


∞∑k=0

q(k+12 )

(q; q)k=

∞∏n=0

1

1− q2n+1.

George Andrews’ 75th Birthday / Combinatorial Construction of Modular Forms 19

Combinatorial Construction of Modular Forms


Omega and Mathematical Discovery

ca1

AAA

U ca2

-

ca3AAAAAA

-

U

ca4

-

ca5AAAAAA

-

U

ca6

. . . . . . . . . .

ca7 . . . . . . . . . . ca2k−1

AAAAAA

-

U

ca2k−2

-

ca2k+1

AAAU

ca2k

c a2k+2

A k-elongated partition diamond of length 1

ca1

AAA

U ca2

. . . . . . . .ca3

. . . . . . . . ca2k

AAA

ca2k+1

U ca2k+2 cAAA

U ca2k+3

. . . . . . . .ca2k+4

. . . . . . . . ca4k+1

AAA

ca4k+2

U ca4k+3 . . . . . . cAAA

U c

. . . . . . . .c

. . . . . . . . ca(2k+1)n−1

AAA

ca(2k+1)n

U ca(2k+1)n+1

A k-elongated partition diamond of length n

1


Omega and Mathematical Discovery

ca1

AAA

U ca2

-

ca3AAAAAA

-

U

ca4

-

ca5AAAAAA

-

U

ca6

. . . . . . . . . .

ca7 . . . . . . . . . . ca2k−1

AAAAAA

-

U

ca2k−2

-

ca2k+1

AAAU

ca2k

c a2k+2

A k-elongated partition diamond of length 1

ca1

AAA

U ca2

. . . . . . . .ca3

. . . . . . . . ca2k

AAA

ca2k+1

U ca2k+2 cAAA

U ca2k+3

. . . . . . . .ca2k+4

. . . . . . . . ca4k+1

AAA

ca4k+2

U ca4k+3 . . . . . . cAAA

U c

. . . . . . . .c

. . . . . . . . ca(2k+1)n−1

AAA

ca(2k+1)n

U ca(2k+1)n+1

A k-elongated partition diamond of length n

1


Generating function:

hn,k(q) =

∏n−1j=0 (1 + q(2k+1)j+2)(1 + q(2k+1)j+4) · · · (1 + q(2k+1)j+2k)∏(2k+1)n+1

j=1 (1− qj)

This generating function is beautiful indeed, but George Andrewswanted more!


Date: Jan 2005

To: [email protected],[email protected]

Subject: I had a brainstorm


Generating function:

hn,k(q) =

∏n−1j=0 (1 + q(2k+1)j+2)(1 + q(2k+1)j+4) · · · (1 + q(2k+1)j+2k)∏(2k+1)n+1

j=1 (1− qj)

This generating function is beautiful indeed, but George Andrewswanted more!


Date: Jan 2005

To: [email protected],[email protected]

Subject: I had a brainstorm


Dear Axel and Peter,

I have had a BRAINSTORM! (I can almost see the

smile on Peter’s face ...) To begin with let me draw

your attention to Thm.2 in PAXI and Cor.2.1 in PAVIII

... However, THIS IS JUST THE BEGINNING.

Namely, I always had a slight disappointment in

both the earlier results because the generating

function ... was not a modular form. However, I now

know how to produce the appropriate directed graph to

provide ... a generating function that is not only a

modular form but, in fact, a product of instances of

Dedekind’s eta-function.


Recall the generating function:

hn,k(q) =

∏n−1j=0 (1 + q(2k+1)j+2)(1 + q(2k+1)j+4) · · · (1 + q(2k+1)j+2k)∏(2k+1)n+1

j=1 (1− qj)

The result of Andrews’ brainstorm: delete the source:

h∗n,k(q) =

∏n−1j=0 (1 + q(2k+1)j+1)(1 + q(2k+1)j+3) · · · (1 + q(2k+1)j+2k−1)∏(2k+1)n

j=1 (1− qj)

and glue the diamonds together:


Recall the generating function:

hn,k(q) =

∏n−1j=0 (1 + q(2k+1)j+2)(1 + q(2k+1)j+4) · · · (1 + q(2k+1)j+2k)∏(2k+1)n+1

j=1 (1− qj)

The result of Andrews’ brainstorm: delete the source:

h∗n,k(q) =

∏n−1j=0 (1 + q(2k+1)j+1)(1 + q(2k+1)j+3) · · · (1 + q(2k+1)j+2k−1)∏(2k+1)n

j=1 (1− qj)

and glue the diamonds together:


∞∑n=0

∆k(n)qn:=hn,k(q)h∗n,k(q)

=

∞∏n=1

(1− q2n)(1− q(2k+1)n)

(1− qn)3(1− q(4k+2)n)

= q(k+1)/12 η(2τ)η((2k + 1)τ)

η(τ)3η((4k + 2)τ)

cb(2k+1)n+1

AAAc

b(2k+1)n−1

. . . . . . . .cb(2k+1)n

. . . . . . . . cAAA

c

c

K

K

. . . cAAA

c. . . . . . . .

c. . . . . . . .K

cb7AAAAAA

cb6

K

cb5

Kc

b4

AAAAAA

cb3

cb2

b2k+2

b2k+1

b2k

ca1

AAA

U ca2

. . . . . . . .ca3

. . . . . . . . ca2k

AAA

ca2k+1

U ca2k+2 . . . cAAA

U c

. . . . . . . .c

. . . . . . . . ca(2k+1)n−1

AAA

ca(2k+1)n

U ca(2k+1)n+1

A broken k-diamond of length 2n

1


∞∑n=0

∆k(n)qn:=hn,k(q)h∗n,k(q)

=

∞∏n=1

(1− q2n)(1− q(2k+1)n)

(1− qn)3(1− q(4k+2)n)

= q(k+1)/12 η(2τ)η((2k + 1)τ)

η(τ)3η((4k + 2)τ)

cb(2k+1)n+1

AAAc

b(2k+1)n−1

. . . . . . . .cb(2k+1)n

. . . . . . . . cAAA

c

c

K

K

. . . cAAA

c. . . . . . . .

c. . . . . . . .K

cb7AAAAAA

cb6

K

cb5

Kc

b4

AAAAAA

cb3

cb2

b2k+2

b2k+1

b2k

ca1

AAA

U ca2

. . . . . . . .ca3

. . . . . . . . ca2k

AAA

ca2k+1

U ca2k+2 . . . cAAA

U c

. . . . . . . .c

. . . . . . . . ca(2k+1)n−1

AAA

ca(2k+1)n

U ca(2k+1)n+1

A broken k-diamond of length 2n

1


Numerous Congruences for ∆k(n)

Example [Sellers & Radu, 2012]

∞∑n=0

∆2(3n+ 1)qn ≡ 2q

∞∏n=1

(1− q10n)4

(1− q5n)2(mod 3)

This implies for all n ≥ 0:

∆2(15n+ 1) ≡ 0 (mod 3),

∆2(27n+ 16) ≡ 0 (mod 3),

∆2(147n+ 58) ≡ 0 (mod 3),

etc.

Note. More people became interested in such congruences, e.g.,S.H. Chan, W.Y.C. Chen, S. Cui, A.R.B. Fan, S.S. Fu, N. Gu,M.D. Hirschhorn, M. Jameson, E. Mortenson, X. Xiong, R.T. Yu.




∞∑n=0

∆2(3n+ 1)qn ≡ 2q

∞∏n=1

(1− q10n)4

(1− q5n)2(mod 3)


∆2(15n+ 1) ≡ 0 (mod 3),

∆2(27n+ 16) ≡ 0 (mod 3),

∆2(147n+ 58) ≡ 0 (mod 3),

etc.





∞∑n=0

∆2(3n+ 1)qn ≡ 2q

∞∏n=1

(1− q10n)4

(1− q5n)2(mod 3)


∆2(15n+ 1) ≡ 0 (mod 3),

∆2(27n+ 16) ≡ 0 (mod 3),

∆2(147n+ 58) ≡ 0 (mod 3),

etc.


George Andrews’ 75th Birthday / Partition Congruences 26

Partition Congruences


Radu’s Algorithmic Approach [Ramanujan J., 2009]

I Input:∞∑n=0

a(n)qn:=∏δ|M

∞∏n=1

(1− qδn

)rδ;

Question: Given positive integers m, t, p, is it true that

a(mn+ t) ≡ 0 (mod p), n ≥ 0 ?

I Output: TRUE or FALSE, plus a proof certificate.



I Input:∞∑n=0

a(n)qn:=∏δ|M

∞∏n=1

(1− qδn

)rδ;

Example:

∞∑n=0

a(n)qn:=

∞∏n=1

(1− qn)−1 =

∞∑n=0

p(n)qn.

Note. p(n) is the no. of partitions of n: 3 = 2 + 1 = 1 + 1 + 1.


a(mn+ t) ≡ 0 (mod p), n ≥ 0 ?




I Input:∞∑n=0

a(n)qn:=∏δ|M

∞∏n=1

(1− qδn

)rδ;


a(mn+ t) ≡ 0 (mod p), n ≥ 0 ?




I Input:∞∑n=0

a(n)qn:=∏δ|M

∞∏n=1

(1− qδn

)rδ;


a(mn+ t) ≡ 0 (mod p), n ≥ 0 ?



Application 1

I Input:

∞∑n=0

a(n)qn:=∏δ|M

∞∏n=1

(1− qδn

)rδ=

∞∏n=1

(1− q3n)−1(1− qn)−3

and

∞∑n=0

b(n)qn:=

∞∏n=1

(1− qn)14

(1− q7n)2

∞∑n=0

a(n)qn.

Ono [1996] proved: For t = 22, 40, 49 it true that

a(63n+ t) ≡ b(63n+ t) ≡ 0 (mod 7), n ≥ 0.


Application 1

I Input:

∞∑n=0

a(n)qn:=∏δ|M

∞∏n=1

(1− qδn

)rδ=

∞∏n=1

(1− q3n)−1(1− qn)−3

and

∞∑n=0

b(n)qn:=

∞∏n=1

(1− qn)14

(1− q7n)2

∞∑n=0

a(n)qn.

Ono [1996] proved: For t = 22, 40, 49 it true that

a(63n+ t) ≡ b(63n+ t) ≡ 0 (mod 7), n ≥ 0.


I Ono’s Proof: transforms the a(63n+ t) problem into theb(63n+ t) problem, and needs to compute

148,147 coefficients.

I Radu’s Proof: directly applicable to both, the a(63n+ t)and the b(63n+ t) problem; needs to compute

17, resp. 183, coefficients.

Remark. Radu’s algorithm generalizes a method by Rademacherbased on Sturm’s criterion.














Application 2

I Input:

∞∑n=0

a(n)qn:=∏δ|M

∞∏n=1

(1− qδn

)rδ=

∞∏n=1

(1− qn)−3.

From Mathew Boylan’s list [Acta Arithm. 2004]: It istrue that

a(11n+ 7) ≡ 0 (mod 11), n ≥ 0

anda(17n+ 15) ≡ 0 (mod 17), n ≥ 0.


Application 2

I Input:

∞∑n=0

a(n)qn:=∏δ|M

∞∏n=1

(1− qδn

)rδ=

∞∏n=1

(1− qn)−3.

From Mathew Boylan’s list [Acta Arithm. 2004]: It istrue that

a(11n+ 7) ≡ 0 (mod 11), n ≥ 0

anda(17n+ 15) ≡ 0 (mod 17), n ≥ 0.


I Radu’s Algorithm: is directly applicable to bothcongruences; it needs to compute


Remark 1. Radu’s algorithm can prove all entries from Boylan’slist.


I Radu’s Algorithm: is directly applicable to bothcongruences; it needs to compute


Remark 1. Radu’s algorithm can prove all entries from Boylan’slist.


Remark 2 by Zeilberger [from a recent article with Edinah Gnang].“Thanks to the impressive algorithm of Silviu Radu every suchcongruence (and even more general ones), is effectively (and fairlyefficiently!) decidable.” [. . . ] As impressive as Radu’s algorithm is,it is not elementary. It uses the ‘fancy’, and intimidating, theory ofmodular functions, that being analytic, is not quite legitimate toour finitismic and discrete philosophy of mathematics. Hence it isstill interesting (at least to us!) to find elementary,Hirschhorn-style proofs.”


Spin-offs from Algorithmic Insight

EXAMPLE 1: The Subbarao Conjecture (1966)

I Ono [1996]: Given integers 0 ≤ B < A:

p(An+B) ≡ 0 (mod 2) for infinitely many n.

I Radu [2012]: Given integers 0 ≤ B < A:

p(An+B) 6≡ 0 (mod 2) for infinitely many n.

Remark. Radu’s algorithmic insight enabled him to invoke a resultby Deligne and Rapoport [1979]. The same applies to his proof[2012] of a conjecture of Ahlgren and Ono [2002].


















EXAMPLE 2: The Andrews-Sellers Family Conjecture (1994)

I Andrews’ generalized Frobenius partitions [1984]:

∞∑m=0

cφ2(m)qm:=

∞∏n=1

1− q4n−2

(1− q2n−1)4(1− q4n).

I Sellers conjectured [1994]: For all α ≥ 1,

cφ2(5αn+ λα) ≡ 0 (mod 5α), n ≥ 0.

where 12λα ≡ 1 (mod 5α).

I Proof by P. and Radu [2012]: symbolic computation ledto new subspaces of modular functions which where essentialfor the proof.




∞∑m=0

cφ2(m)qm:=

∞∏n=1

1− q4n−2

(1− q2n−1)4(1− q4n).


cφ2(5αn+ λα) ≡ 0 (mod 5α), n ≥ 0.






∞∑m=0

cφ2(m)qm:=

∞∏n=1

1− q4n−2

(1− q2n−1)4(1− q4n).


cφ2(5αn+ λα) ≡ 0 (mod 5α), n ≥ 0.



George Andrews’ 75th Birthday / EXAMPLE 3: Ramanujan’s Congruences 35

EXAMPLE 3: Ramanujan’s Congruences


Ramanujan’s Congruences

p(5n+ 4) ≡ 0 (mod 5),

p(7n+ 5) ≡ 0 (mod 7),

p(11n+ 6) ≡ 0 (mod 11)


Ramanujan’s Congruences

p(5n+ 4) ≡ 0 (mod 5),

p(52n+ 24) ≡ 0 (mod 52),

p(53n+ 99) ≡ 0 (mod 53),

etc.


Ramanujan’s Conjecture (1919)

For ` ∈ 5, 7, 11 and α ∈ 1, 2, 3, . . . :

p(`αn+ µα,`) ≡ 0 (mod `α),

where µα,` ∈ 0, . . . , `α − 1 is uniquely defined by

24µα,` ≡ 1 (mod `α).


A Bit of History

I Ramanujan [1917-1919]: proof sketches for ` = 5 and ` = 7;see Berndt and Ono [1999].

I Watson [1938]: ` = 5 and ` = 7 (corrected version)

I Atkin [1967]: ` = 11

I Gordon [1983]:

p(11αn+ µα,11) ≡ 0 (mod 11α+ε),

where ε is a fixed non-positive integer to be determined.Following Gordon’s proof one can easily determine ε = 0.


A Bit of History



I Atkin [1967]: ` = 11

I Gordon [1983]:

p(11αn+ µα,11) ≡ 0 (mod 11α+ε),



A Bit of History



I Atkin [1967]: ` = 11

I Gordon [1983]:

p(11αn+ µα,11) ≡ 0 (mod 11α+ε),



A Bit of History



I Atkin [1967]: ` = 11

I Gordon [1983]:

p(11αn+ µα,11) ≡ 0 (mod 11α+ε),



Our proof for ` = 11 and ` = 5, 7:

I In his generalization of Watson’s method, Gordon needs tocompute the structure constants of a certain algebra.Although this is not needed in the case ` = 5, 7.

I For ` = 11 we succeeded to do without structure constants,(i.e., we stay with the module structure), so the ingredientshere are the same as in the original proof of Watson.

I This way we obtain a unified framework for all three cases` = 5, 7, 11.

I The price to pay: our approach is more expensivecomputationally.



















George Andrews’ 75th Birthday / The Modular Function Setting 40

The Modular Function Setting


Basic Notions

The level ` congruence subgroup

Γ0(`) :=

(a

c

b

d

)∈ SL2(Z) : c ≡ 0 (mod `)

acts on H∗ := H ∪Q ∪ i∞ (extended upper half plane) via(

a

c

b

d

)(τ) 7→ aτ + b

cτ + d.

This induces an action on functions f : H∗ → C ∪ i∞ by

(f |γ)(τ) := f(aτ + b

cτ + d

)where γ =

(a

c

b

d

).


Basic Notions

The level ` congruence subgroup

Γ0(`) :=

(a

c

b

d

)∈ SL2(Z) : c ≡ 0 (mod `)

acts on H∗ := H ∪Q ∪ i∞ (extended upper half plane) via(

a

c

b

d

)(τ) 7→ aτ + b

cτ + d.

This induces an action on functions f : H∗ → C ∪ i∞ by

(f |γ)(τ) := f(aτ + b

cτ + d

)where γ =

(a

c

b

d

).


Compact Riemann Surfaces X0(`)

X0(`) := orbit space of the action of Γ0(`) on H∗

= [τ ] : τ ∈ H∗

Note.aτ + b

cτ + d−→ a

cwhen τ −→ i∞.

anda

0= i∞.

Set of cusps: For any prime `,

[x] ∈ X0(`) : x ∈ Q ∪ i∞ = [0], [i∞].




= [τ ] : τ ∈ H∗

Note.aτ + b

cτ + d−→ a


anda

0= i∞.


[x] ∈ X0(`) : x ∈ Q ∪ i∞ = [0], [i∞].




= [τ ] : τ ∈ H∗

Note.aτ + b

cτ + d−→ a


anda

0= i∞.


[x] ∈ X0(`) : x ∈ Q ∪ i∞ = [0], [i∞].


Modular Functions (MF) on Compact Riemann Surfaces X0(`)

f : H∗ → C ∪ i∞ induces a MF f∗ : X0(`)→ C ∪ i∞ if

I f is holomorphic on H;

I f is constant on the orbits [τ ], τ ∈ H∗;

I f is meromorphic at points in Q ∪ i∞.


Modular Functions (MF) on Compact Riemann Surfaces X0(`)

f : H∗ → C ∪ i∞ induces a MF f∗ : X0(`)→ C ∪ i∞ if

I f is holomorphic on H;

I f is constant on the orbits [τ ], τ ∈ H∗;

I f is meromorphic at points in Q ∪ i∞.


Fourier expansions of f∗ at cusps [a/c] ∈ X0(`)

For τ ∈ neighborhood(i∞):

f∗([

aτ + b

cτ + d

])) = (f |γ)(τ) =

∑n≥ord[a/c](f∗)

a[a/c](n)qn[a/c].

where γ =(a bc d

)∈ SL2(Z), q[a/c] =

(e2πiτ

) gcd(c2,`)` .

Note. If [a/c] = [a/0] = [i∞] then q[a/c] = q.


Fourier expansions of f∗ at cusps [a/c] ∈ X0(`)

For τ ∈ neighborhood(i∞):

f∗([

aτ + b

cτ + d

])) = (f |γ)(τ) =

∑n≥ord[a/c](f∗)

a[a/c](n)qn[a/c].

where γ =(a bc d

)∈ SL2(Z), q[a/c] =

(e2πiτ

) gcd(c2,`)` .

Note. If [a/c] = [a/0] = [i∞] then q[a/c] = q.


Back to Ramanujan’s congruences

L2β−1,` := q

∞∏n=1

(1− q`n)

∞∑n=0

p(`2β−1n+ µ2β−1,`)qn,

and

L2β,` := q

∞∏n=1

(1− qn)

∞∑n=0

p(`2βn+ µ2β,`)qn.

We consider these series as Fourier expansions at [i∞] of modularfunctions f∗ : X0(`)→ C ∪ i∞ from a suitable subring R(`).


The Rings R(5), R(7), and R(11)

Recall η(τ) = q124

∞∏n=1

(1− qn) (q = e2πiτ ).

From the literature it is well known that we can take

R(5) = Z[z5], R(7) = Z[z7],

andR(11) = 〈J0, J1, . . . , J4〉Z[z11]

where

z5 :=η6(5τ)

η6(τ), z7 :=

η4(7τ)

η4(τ), z11 :=

η12(11τ)

η12(τ),

and with the Jk such that ord[∞](J∗k ) = k as in Atkin (1967).


The Rings R(5), R(7), and R(11)

Recall η(τ) = q124

∞∏n=1

(1− qn) (q = e2πiτ ).

From the literature it is well known that we can take

R(5) = Z[z5], R(7) = Z[z7],

andR(11) = 〈J0, J1, . . . , J4〉Z[z11]

where

z5 :=η6(5τ)

η6(τ), z7 :=

η4(7τ)

η4(τ), z11 :=

η12(11τ)

η12(τ),

and with the Jk such that ord[∞](J∗k ) = k as in Atkin (1967).


The U -operator

U`(f∗):=

∑n

a(`n)qn

where f∗([τ ]) =∑

n a(n)qn is the Fourier expansion of f∗ at i∞.

Definition. For prime `,

U(1)` (f∗) := U`

(η(`2τ)

η(τ)f∗)

and U(2)` (f∗) := U`(f

∗).

Lemma.f∗ ∈ R(`) ⇒ U

(i)` (f∗) ∈ R(`).


The U -operator

U`(f∗):=

∑n

a(`n)qn




U(1)` (f∗) := U`

(η(`2τ)

η(τ)f∗)

and U(2)` (f∗) := U`(f

∗).


(i)` (f∗) ∈ R(`).


The U -operator

U`(f∗):=

∑n

a(`n)qn




U(1)` (f∗) := U`

(η(`2τ)

η(τ)f∗)

and U(2)` (f∗) := U`(f

∗).


(i)` (f∗) ∈ R(`).


Recursive representation of Lα,` in R(`)

Lemma. For ` ∈ 5, 7, 11,

Lα,` ∈ R(`);

the Lα,` satisfy a recursion in R(`) as follows:

L0,` = 1,

L2β−1,` = U(1)` (L2β−2,`) and L2β,` = U

(2)` (L2β−1,`).

This recursion is used in the induction proof of our “MainTheorem”:


Main Theorem

For suitably defined subsets Xs,` of R(`) we have:

Theorem. For ` ∈ 5, 7, 11 and all positive integers β thereexist fβ,` ∈ X1,` if β is odd, and fβ,` ∈ X2,` if β is even, suchthat

Lβ,` = `βfβ,` for ` = 5, 11,

andLβ,` = `d

β+12efβ,` for ` = 7.

Note. Ramanujan’s congruences are obtained as an immediatecorollary.


Main Tool: The Fundamental Lemma

Fundamental Lemma. Let ` ∈ 5, 7, 11. For any w : H→ Cand j ∈ Z:

U`(wzj` ) = −

d`−1∑i=0

a(`)i (z`)U`(wz

j+i−d`` ).

Setting w to the generators of the module R(`), we obtain the

U(i)` actions on all the module elements — provided

we know the 2d`n` (2 · 5 · 1, 2 · 7 · 1, 2 · 55 · 5) initial actions.

This action then serves to express Lβ,` in terms of the generatorswhich reveal the divisibilty properties of these elements.


The R(`)-subsets Xs,` (` = 11: Atkin)

Set

A` :=12

gcd(`− 1, 12)

`

`+ 1and z` :=

(η`η

) 24gcd(`−1,12)

.

Definition. For ` ∈ 5, 7, 11 and s = 1, 2:

Xs,` :=

n`−1∑i=0

Ji,`

∞∑j=0

ai(j)`dA``(`j+ξ

(s,`)i )ezj` : a0(0) = 0 and

with ai(j) ∈ Z nonzero for only finitely many j

The integer exponents ξ(s,`)i are defined as follows:


The Exponents ξ(s,`)i

Definition. For ` ∈ 5, 7, 11 and s = 1, 2 define

ξ(s,`) : 0, . . . , n` − 1 → Z, i 7→ ξ(s,`)i

by

(ξ(1,11)0 , . . . , ξ

(1,11)4 ) := (−5,−1, 1, 2, 6),

(ξ(2,11)0 , . . . , ξ

(2,11)4 ) := (−4, 0, 2, 3, 7);

andξ(1,7)0 := −7 and ξ

(2,7)0 := −10;

andξ(1,5)0 := −6 and ξ

(2,5)0 := −5.

George Andrews’ 75th Birthday / Proof of the Main Theorem 53

Proof of the Main Theorem


The Fundamental Polynomial for any prime ` ≥ 5

The main ingredient in our induction proof is

Theorem. For any prime ` ≥ 5 there exists a monic irreduciblepolynomial F`(X,Y ) ∈ Q[Y ][X] of degree d` := `(`−1)

gcd(`−1,12) in

X (and Y ) such that

F`(z`(τ), z`(`τ)) = 0.

Recall that

z`(τ) =(η(`τ)

η(τ)

) 24gcd(`−1,12)

.


Proving the Existence of the Fundamental Polynomial

Proof.The proof uses a variant of a classical fact from Riemann surfaces;namely, that on a compact Riemann surface two meromorphicfunctions are connected by an algebraic relation.


The Fundamental Polynomial for ` ∈ 5, 7, 11

Theorem. For ` ∈ 5, 7, 11 the fundamental polynomial

F`(X,Y ) = Xd` +

d`−1∑i=0

a(`)i (Y )Xi

has coefficient polynomials a(`)i (Y ) ∈ Q[Y ] of the form

a(`)i (Y ) =

d∑j=d d`−i

`e

s`(i, j)`dA``(`j+i−d`)eY j

with s`(i, j) ∈ Z (determined using computer algebra).


Corollary: The Fundamental Lemma


U`(wzj` ) = −

d`−1∑i=0

a(`)i (z`)U`(wz

j+i−d`` ).

Proof.Applying U` to wzj−d`` F`(z`(τ), z`(`τ)) = 0 gives the desiredresult.


Corollary: The Fundamental Lemma


U`(wzj` ) = −

d`−1∑i=0

a(`)i (z`)U`(wz

j+i−d`` ).

Proof.Applying U` to wzj−d`` F`(z`(τ), z`(`τ)) = 0 gives the desiredresult.

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