Combinatorics, Modular Forms, and Discrete Geometry

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Combinatorics, Modular Forms, and Discrete Geometry / 1 Geometric and Enumerative Combinatorics, IMA University of Minnesota, Nov 10 14, 2014 Combinatorics, Modular

1 Combinatorics, Modular Forms, and Discrete Geometry / 1 Geometric and Enumerative Combinatorics, IMA University of Minnesota, Nov 10 14, 2014 Combinatorics, Modular Forms, and Discrete Geometry Peter Paule (joint work with: G.E. Andrews, S. Radu) Johannes Kepler University Linz Research Institute for Symbolic Computation (RISC)

2 Combinatorics, Modular Forms, and Discrete Geometry / Partition Analysis 2 Partition Analysis

3 Combinatorics, Modular Forms, and Discrete Geometry / Partition Analysis 3 The no. of partitions of N of the form N = b b n satisfying b n n b n 1 n 1 b 2 2 b equals the no. of partitions of N into odd parts each 2n 1. This problem cried out for MacMahon s Partition Analysis,...

Combinatorics, Modular Forms, and Discrete Geometry / Partition Analysis 3 The no. of partitions of N of the form N = b 1 + + b n satisfying b n n b n 1 n 1 b 2 2 b 1 1 0 equals the no.

4 Combinatorics, Modular Forms, and Discrete Geometry / Partition Analysis 3 The no. of partitions of N of the form N = b b n satisfying b n n b n 1 n 1 b 2 2 b 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 producing partition generating functions, I was able to convince Peter Paule and Axel Riese to join an effort to automate this algorithm.

5 Combinatorics, Modular Forms, and Discrete Geometry / Partition Analysis 7 How Zeilberger tells the story of partition analysis (and more):

6 Combinatorics, Modular Forms, and Discrete Geometry / Partition Analysis 8 Example (PA and the Omega package) Find a suitable closed form of L(x 1, x 2, x 3 ) := b 1,b 2,b 3 N s.t. 2b 3 3b 2 0, b 2 2b 1 0 x b 1 1 x b 2 2 x b 3 3

7 Combinatorics, Modular Forms, and Discrete Geometry / Partition Analysis 8 Example (PA and the Omega package) Find a suitable closed form of L(x 1, x 2, x 3 ) := b 1,b 2,b 3 N s.t. 2b 3 3b 2 0, b 2 2b 1 0 x b 1 1 x b 2 2 x b 3 3 = Ω b 1,b 2,b 3 0 λ2b 3 3b 2 1 λ b 2 2b 1 2 x b 1 1 x b 2 2 x b 3 3

8 Combinatorics, Modular Forms, and Discrete Geometry / Partition Analysis 8 Example (PA and the Omega package) Find a suitable closed form of L(x 1, x 2, x 3 ) := b 1,b 2,b 3 N s.t. 2b 3 3b 2 0, b 2 2b 1 0 x b 1 1 x b 2 2 x b 3 3 = Ω b 1,b 2,b 3 0 λ2b 3 3b 2 1 λ b 2 2b 1 2 x b 1 1 x b 2 2 x b 3 3 = Ω 1 1 x 1 λ λ 2x 2 λ λ 2 1 x 3

9 Combinatorics, Modular Forms, and Discrete Geometry / Partition Analysis 8 Example (PA and the Omega package) Find a suitable closed form of L(x 1, x 2, x 3 ) := b 1,b 2,b 3 N s.t. 2b 3 3b 2 0, b 2 2b 1 0 x b 1 1 x b 2 2 x b 3 3 = Ω b 1,b 2,b 3 0 λ2b 3 3b 2 1 λ b 2 2b 1 2 x b 1 1 x b 2 2 x b 3 3 In[1]:= << Omega2.m Omega Package by Axel Riese (in cooperation with George E. Andrews and Peter Paule) - c RISC, JKU Linz - V 2.47

10 Combinatorics, Modular Forms, and Discrete Geometry / Partition Analysis 9 In[2]:= LCrude = OSum[ x1 b1 x2 b2 x3 b3, {2 b3-3 b2 0, b2-2 b1 0, b1 0}, λ]

11 Combinatorics, Modular Forms, and Discrete Geometry / Partition Analysis 9 In[2]:= LCrude = OSum[ x1 b1 x2 b2 x3 b3, {2 b3-3 b2 0, b2-2 b1 0, b1 0}, λ] Out[2]= Ω λ 1, λ 2 ( )( 1 ) 1 x1 λ 2 1 λ 2 x2 2 λ 3 (1 λ 2 x3) 1 1 In[3]:= L=OR[LCrude]

12 Combinatorics, Modular Forms, and Discrete Geometry / Partition Analysis 9 In[2]:= LCrude = OSum[ x1 b1 x2 b2 x3 b3, {2 b3-3 b2 0, b2-2 b1 0, b1 0}, λ] Out[2]= Ω λ 1, λ 2 ( )( 1 ) 1 x1 λ 2 1 λ 2 x2 2 λ 3 (1 λ 2 x3) 1 1 In[3]:= L=OR[LCrude] Out[3]= 1+x2 x3 2 (1 x3)(1 x2 2 x3 3 )(1 x1 x2 2 x3 3 ) In[4]:= L /. {x1->q, x2->q, x3->q}

13 Combinatorics, Modular Forms, and Discrete Geometry / Partition Analysis 9 In[2]:= LCrude = OSum[ x1 b1 x2 b2 x3 b3, {2 b3-3 b2 0, b2-2 b1 0, b1 0}, λ] Out[2]= Ω λ 1, λ 2 ( )( 1 ) 1 x1 λ 2 1 λ 2 x2 2 λ 3 (1 λ 2 x3) 1 1 In[3]:= L=OR[LCrude] Out[3]= 1+x2 x3 2 (1 x3)(1 x2 2 x3 3 )(1 x1 x2 2 x3 3 ) In[4]:= L /. {x1->q, x2->q, x3->q} Out[4]= 1+q 3 (1 q)(1 q 5 )(1 q 6 )

14 Combinatorics, Modular Forms, and Discrete Geometry / Partition Analysis 10 GENERAL THEME: linear Diophantine constraints Find b 1,... b n N such that c 1,1 c 1,n c 2,1 c 2,n..... c m,1 c m,n c 1 1 c 2 b.. b n c m

15 Combinatorics, Modular Forms, and Discrete Geometry / Partition Analysis 10 GENERAL THEME: linear Diophantine constraints Find b 1,... b n N such that c 1,1 c 1,n c 2,1 c 2,n..... c m,1 c m,n c 1 1 c 2 b. =. b n c m

16 Combinatorics, Modular Forms, and Discrete Geometry / Partition Analysis 10 GENERAL THEME: linear Diophantine constraints Find b 1,... b n N such that c 1,1 c 1,n c 2,1 c 2,n..... c m,1 c m,n c 1 1 c 2 b. =. b n c m New algorithm by F. Breuer & Z. Zafeirakopolous [poster: A Linear Diophantine System Solver, Lind Hall 400, 4 pm] Polyhedral Omega is a new algorithm for solving linear Diophantine systems (LDS), i.e., for computing a multivariate rational function representation of the set of all non-negative integer solutions to a system of linear equations and inequalities. Polyhedral Omega combines methods from partition analysis with methods from polyhedral geometry. In particular, we combine MacMahon s iterative approach based on the Omega operator and explicit formulas for its evaluation with geometric tools such as Brion decompositions and Barvinok s short rational function representations. In this way, we connect two recent branches of research that have so far remained separate, unified by the concept of symbolic cones which we introduce.

17 Combinatorics, Modular Forms, and Discrete Geometry / Omega and Mathematical Discovery 11 Omega and Mathematical Discovery

18 Combinatorics, Modular Forms, and Discrete Geometry / Omega and Mathematical Discovery 12 a 1 a 3 a 5 a 7 a 2k 1 a 2k a 2k a 2 a 4 a 6 a 2k a 2k 2 A k-elongated partition diamond of length 1 a 1 a a 2k+2 a 4k a 2 a 2k+1 a 2k+4 a 2k a 2k+3 a 4k+2 a 4k+1 a (2k+1)n a (2k+1)n 1 a (2k+1)n+1 A k-elongated partition diamond of length n

19 Combinatorics, Modular Forms, and Discrete Geometry / Omega and Mathematical Discovery 13 Generating function for k-elongated diamonds of length n: h n,k (q) = n 1 j=0 (1 + q(2k+1)j+2 )(1 + q (2k+1)j+4 ) (1 + q (2k+1)j+2k ) (2k+1)n+1 (1 q j ) Andrews great idea: delete the source:

20 Combinatorics, Modular Forms, and Discrete Geometry / Omega and Mathematical Discovery 13 Generating function for k-elongated diamonds of length n: h n,k (q) = n 1 j=0 (1 + q(2k+1)j+2 )(1 + q (2k+1)j+4 ) (1 + q (2k+1)j+2k ) (2k+1)n+1 (1 q j ) Andrews great idea: delete the source: h n,k (q) = n 1 j=0 (1 + q(2k+1)j+1 )(1 + q (2k+1)j+3 ) (1 + q (2k+1)j+2k 1 ) (2k+1)n (1 q j ) and glue the diamonds together:

21 Combinatorics, Modular Forms, and Discrete Geometry / Omega and Mathematical Discovery 13 Generating function for k-elongated diamonds of length n: h n,k (q) = n 1 j=0 (1 + q(2k+1)j+2 )(1 + q (2k+1)j+4 ) (1 + q (2k+1)j+2k ) (2k+1)n+1 (1 q j ) Andrews great idea: delete the source: h n,k (q) = n 1 j=0 (1 + q(2k+1)j+1 )(1 + q (2k+1)j+3 ) (1 + q (2k+1)j+2k 1 ) (2k+1)n (1 q j ) and glue the diamonds together: b (2k+1)n+1 b (2k+1)n b 2k+1 b 7 b b 2k b 6 b (2k+1)n 1 b 2k b 5 b 4 a 1 b 2 a a 2k+2... a (2k+1)n a 2 a 2k+1 a 2k a (2k+1)n a (2k+1)n 1 A broken k-diamond of length 2n

22 Combinatorics, Modular Forms, and Discrete Geometry / Omega and Mathematical Discovery 14 b (2k+1)n+1 k (m)q m := lim h n,k(q) h n n,k (q) = (1 + qj ) (1 qj ) 2 (1 + q(2k+1)j ) = (1 + qj )(1 q j ) (1 qj ) 3 (1 + q(2k+1)j ) (1 q 2j )(1 q (2k+1)j ) = (1 q j ) 3 (1 q (4k+2)j ) m=0 b (2k+1)n b 2k+1 b 7 b b 2k b 6 b (2k+1)n 1 b 2k b 5 b 4 a 1 b 2 a a 2k+2... a (2k+1)n a 2 A broken k-diamond of length 2n a 2k+1 a 2k a (2k+1)n a (2k+1)n 1

23 Combinatorics, Modular Forms, and Discrete Geometry / Omega and Mathematical Discovery 15 Consequently, k (m)q m = lim h n,k(q) h n n,k (q) m=0 = with η the Dedekind eta function: η(τ) := q 1 24 (1 q 2j )(1 q (2k+1)j ) (1 q j ) 3 (1 q (4k+2)j ) (k+1)/12 η(2τ)η((2k + 1)τ) = q η(τ) 3 η((4k + 2)τ) (1 q n ) (q = e 2πiτ ) n=1

24 Combinatorics, Modular Forms, and Discrete Geometry / Omega and Mathematical Discovery 15 Consequently, k (m)q m = lim h n,k(q) h n n,k (q) m=0 = with η the Dedekind eta function: η(τ) := q 1 24 (1 q 2j )(1 q (2k+1)j ) (1 q j ) 3 (1 q (4k+2)j ) (k+1)/12 η(2τ)η((2k + 1)τ) = q η(τ) 3 η((4k + 2)τ) (1 q n ) (q = e 2πiτ ) n=1 NOTE. η 24 is a modular form of weight 12 for SL 2 (Z), because of ( ) aτ + b η = ɛ(a, b, c, d) i(cτ + d)η(τ) cτ + d where a d b c = 1 and c > 0.

25 Combinatorics, Modular Forms, and Discrete Geometry / Omega and Mathematical Discovery 16 Recall η ( ) aτ + b = ɛ(a, b, c, d) i(cτ + d)η(τ). cτ + d Hence, for τ H (upper half complex plane): η(τ + 1) 24 = η ( ) 1 τ τ + 1 = ɛ(1, 1, 0, 1) 24 i(0 τ + 1) 24 η(τ) 24 = η(τ) 24 The Fourier series expansion ( q-series expansion, q = e 2πiτ ) is WHY η-quotients? η(τ) = q (1 q n ) 24. n=1

26 Combinatorics, Modular Forms, and Discrete Geometry / Omega and Mathematical Discovery 17

27 Combinatorics, Modular Forms, and Discrete Geometry / Omega and Mathematical Discovery 18 Congruences for k (n) Theorem [Andrews & P, Partition Analysis XI]. For all n N, 1 (2n + 1) 0 (mod 3). Proof.

28 Combinatorics, Modular Forms, and Discrete Geometry / Omega and Mathematical Discovery 18 Congruences for k (n) Theorem [Andrews & P, Partition Analysis XI]. For all n N, 1 (2n + 1) 0 (mod 3). Proof. Because of (1 q j ) 3 1 q 3j (mod 3), 1 (m)q m (1 q 2j )(1 q 3j ) = (1 q j ) 3 (1 q 6j ) m=0 (1 q 2j )(1 q 3j ) (1 q 3j )(1 q 6j ) (mod 3). Hence the coefficients of odd powers of q have to be zero.

29 Combinatorics, Modular Forms, and Discrete Geometry / Omega and Mathematical Discovery 19 Recall: Theorem. For all n N, 1 (2n + 1) 0 (mod 3). Algorithmic Proof [Radu 2014]:

30 Combinatorics, Modular Forms, and Discrete Geometry / Omega and Mathematical Discovery 19 Recall: Theorem. For all n N, 1 (2n + 1) 0 (mod 3). Algorithmic Proof [Radu 2014]: 1 (2n + 1)q n (1 q 2j ) 2 (1 q 6j ) 2 = 3 (1 q j ) 6 n=0

31 Combinatorics, Modular Forms, and Discrete Geometry / Omega and Mathematical Discovery 19 Recall: Theorem. For all n N, 1 (2n + 1) 0 (mod 3). Algorithmic Proof [Radu 2014]: 1 (2n + 1)q n (1 q 2j ) 2 (1 q 6j ) 2 = 3 (1 q j ) 6 n=0 NOTE. Human proof [Hirschhorn & Sellers, 2007]

32 Combinatorics, Modular Forms, and Discrete Geometry / Omega and Mathematical Discovery 20 Some conjectures [Andrews & P, PA XI]: For all n N, 2 (10n + 2) 0 (mod 2) and 2 (25n + 14) 0 (mod 5). S.H. Chan [2008] proved this and also

33 Combinatorics, Modular Forms, and Discrete Geometry / Omega and Mathematical Discovery 20 Some conjectures [Andrews & P, PA XI]: For all n N, 2 (10n + 2) 0 (mod 2) and 2 (25n + 14) 0 (mod 5). S.H. Chan [2008] proved this and also 2 (10n + 6) 0 (mod 2) and 2 (25n + 24) 0 (mod 5).

34 Combinatorics, Modular Forms, and Discrete Geometry / Omega and Mathematical Discovery 20 Some conjectures [Andrews & P, PA XI]: For all n N, 2 (10n + 2) 0 (mod 2) and 2 (25n + 14) 0 (mod 5). S.H. Chan [2008] proved this and also 2 (10n + 6) 0 (mod 2) and 2 (25n + 24) 0 (mod 5). NOTE. First proof of the 10-case [Hirschhorn & Sellers, 2007]

35 Combinatorics, Modular Forms, and Discrete Geometry / Omega and Mathematical Discovery 21 Recall [S.H. Chan, 2008]: 2 (25n + 14) 2 (25n + 24) 0 (mod 5). Algorithmic Proof [Radu 2014].

36 Combinatorics, Modular Forms, and Discrete Geometry / Omega and Mathematical Discovery 21 Recall [S.H. Chan, 2008]: 2 (25n + 14) 2 (25n + 24) 0 (mod 5). Algorithmic Proof [Radu 2014]. Human preprocessing: since (1 q j ) 5 1 q 5j (mod 5), 2 (n) d(n) (mod 5), where and 2 (n)q n (1 q 2j )(1 q 5j ) = (1 q j ) 3 (1 q 10j ) m=0 m=0 d(n)q n (1 q 2j )(1 q j ) 2 := (1 q 10j )

37 Combinatorics, Modular Forms, and Discrete Geometry / Omega and Mathematical Discovery 22 Radu s program Ramanujan-Kolberg delivers: q 3 2 ( η(2τ)η(5τ) 10 ) ( ) η(τ) 6 η(10τ) 20 d(25n + 14)q n d(25n + 24)q n m=0 m=0 = 25(2t t t t + 400) where t = η(τ)3 η(5τ) η(2τ)η(10τ) 3 M(Γ 0(10)). NOTE 1.

38 Combinatorics, Modular Forms, and Discrete Geometry / Omega and Mathematical Discovery 22 Radu s program Ramanujan-Kolberg delivers: q 3 2 ( η(2τ)η(5τ) 10 ) ( ) η(τ) 6 η(10τ) 20 d(25n + 14)q n d(25n + 24)q n m=0 m=0 = 25(2t t t t + 400) where t = η(τ)3 η(5τ) η(2τ)η(10τ) 3 M(Γ 0(10)). NOTE 1.The program computes a similar identity also for 2 (n) instead of d(n), but the output is much bigger.

39 Combinatorics, Modular Forms, and Discrete Geometry / Omega and Mathematical Discovery 22 Radu s program Ramanujan-Kolberg delivers: q 3 2 ( η(2τ)η(5τ) 10 ) ( ) η(τ) 6 η(10τ) 20 d(25n + 14)q n d(25n + 24)q n m=0 m=0 = 25(2t t t t + 400) where t = η(τ)3 η(5τ) η(2τ)η(10τ) 3 M(Γ 0(10)). NOTE 1.The program computes a similar identity also for 2 (n) instead of d(n), but the output is much bigger. NOTE 2. There are numerous other congruences for broken diamonds and generalizations.

40 Combinatorics, Modular Forms, and Discrete Geometry / Radu s Ramanujan-Kolberg Package 24 Radu s Ramanujan-Kolberg Package

41 Combinatorics, Modular Forms, and Discrete Geometry / Radu s Ramanujan-Kolberg Package 25 Back to Euler (= limit of Lecture Hall) Define Q(n)q n 1 := 1 q 2j 1 : n=0 Radu s Ramanujan-Kolberg package delivers (computing over E (14)): Q(7n + 3)q n Q(7n + 4)q n Q(7n + 6)q n n=0 = 8 q 5 NOTE. This implies: n=0 n=0 (1 q 2j ) 5 (1 q 14j ) 16 (1 q j ) 13 (1 q 7j ) 8 ( 16 E E E 1 E 4 )

42 Combinatorics, Modular Forms, and Discrete Geometry / Radu s Ramanujan-Kolberg Package 25 Back to Euler (= limit of Lecture Hall) Define Q(n)q n 1 := 1 q 2j 1 : n=0 Radu s Ramanujan-Kolberg package delivers (computing over E (14)): Q(7n + 3)q n Q(7n + 4)q n Q(7n + 6)q n n=0 = 8 q 5 NOTE. This implies: n=0 n=0 (1 q 2j ) 5 (1 q 14j ) 16 (1 q j ) 13 (1 q 7j ) 8 ( 16 E E E 1 E 4 ) Q(7n + 3) Q(7n + 4) Q(7n + 6) 0 (mod 2).

43 Combinatorics, Modular Forms, and Discrete Geometry / Radu s Ramanujan-Kolberg Package 26 Radu s Ramanujan-Kolberg package also delivers: and Q(7n)q n Q(7n + 1)q n Q(7n + 5)q n n=0 = q 6 n=0 n=0 (1 q 2j ) 5 (1 q 14j ) 16 (1 q j ) 13 (1 q 7j ) 8 (3 E E E 1 ) Q(7n + 2)q n n=0 = q 3 (1 q 14j ) 8 (1 q j ) 3 (1 q 2j )(1 q 7j ) 4 (8 E 1 + E 4 8)

44 Combinatorics, Modular Forms, and Discrete Geometry / Radu s Ramanujan-Kolberg Package 27 STEP 1. Find generators of the multiplicative monoid E (14): Solving a problem for nonnegative integers with linear Diophantine constraints, we obtain the generators E 1 = ( η(2τ) η(τ) ) 1 ( η(7τ) η(τ) ) 7 ( ) η(14τ) 7 E (14), η(τ) and ( ) η(2τ) 8 ( ) η(7τ) 4 ( ) η(14τ) 8 E 2 = E (14), η(τ) η(τ) η(τ) ( ) η(2τ) 5 ( ) η(7τ) 5 ( ) η(14τ) 13 E 3 = E (14), η(τ) η(τ) η(τ)

45 Combinatorics, Modular Forms, and Discrete Geometry / Radu s Ramanujan-Kolberg Package 28 and ( ) η(2τ) 1 ( ) η(7τ) 3 ( ) η(14τ) 7 E 4 = E (14), η(τ) η(τ) η(τ) ( ) η(2τ) 5 ( ) η(7τ) 7 ( ) η(14τ) 11 E 5 = E (14), η(τ) η(τ) η(τ) E 6 = ( η(2τ) η(τ) ) 2 ( η(7τ) η(τ) ) 6 ( ) η(14τ) 10 E (14). η(τ) Summary: STEP 1 computes generators E 1,..., E 6 of the multiplicative monoid E (14) consisting of eta quotients which are modular functions with poles only at infinity.

46 Combinatorics, Modular Forms, and Discrete Geometry / Radu s Ramanujan-Kolberg Package 29 A crucial FINITE representation GOAL: We want to represent our object as an element in the infinite dimensional vectorspace E (14) Q = {c 1 e c k e k : c i Q, e j E (14)} = Q[E 1,..., E 6 ]. STEP 2. Represent E (14) as a Q[E 1 ]-module which is freely generated by 1 and E 4 ; i.e., E (14) Q = 1, E 4 Q[E1 ].

47 Combinatorics, Modular Forms, and Discrete Geometry / Radu s Ramanujan-Kolberg Package 30 NOTE 1. Ramanujan [1919] proved for p(n)q n 1 := 1 q j : n=0 p(5n + 4)q n = 5 n=0 (1 q 5j ) 5 (1 q j ) 6 and p(7n + 5)q n n=0 = 7 (1 q 7j ) 3 (1 q j ) q (1 q 7j ) 7 (1 q j ) 8.

48 Combinatorics, Modular Forms, and Discrete Geometry / Radu s Ramanujan-Kolberg Package 31 NOTE 2. An alternative formulation in terms of z 5 :=q (1 q 5j ) 6 (1 q j ) 6 = ( η(5τ) η(τ) ) 6 and z 7 :=q (1 q 7j ) 4 (1 q j ) 4 = ( ) η(7τ) 4 : η(τ) q (1 q 5j ) p(5n + 4)q n = 5 z 5 n=0 and q (1 q 7j ) p(7n + 5)q n = 7 z q z7. 2 n=0

49 Combinatorics, Modular Forms, and Discrete Geometry / Radu s Ramanujan-Kolberg Package 32 NOTE 3. p(5n + 4)q n = 5 n=0 (1 q 5j ) 5 (1 q j ) 6 It would be difficult to find more beautiful formulae than the Rogers-Ramanujan identities... ; but here Ramanujan must take second place to Prof. Rogers; and, if I had to select one formula from all Ramanujan s work, I would agree with Major MacMahon in selecting... [G.H. Hardy]

50 Combinatorics, Modular Forms, and Discrete Geometry / Radu s Ramanujan-Kolberg Package 33 NOTE. The Ramanujan-Kolberg package computes in E (22): p(11n + 6)q n = q 14 (1 q 22j ) 22 (1 q j ) 10 (1 q 2j ) 2 (1 q 11j ) 11 with n=0 (1078t t t t z 1 (187t t t 9581) +z 2 (11t t t 6754)

51 Combinatorics, Modular Forms, and Discrete Geometry / Radu s Ramanujan-Kolberg Package 33 NOTE. The Ramanujan-Kolberg package computes in E (22): p(11n + 6)q n = q 14 (1 q 22j ) 22 (1 q j ) 10 (1 q 2j ) 2 (1 q 11j ) 11 with n=0 (1078t t t t z 1 (187t t t 9581) +z 2 (11t t t 6754) t:= 3 88 w w w 3, z 1 := 5 88 w w w 3 3, z 2 := 1 44 w w w 3, where

52 Combinatorics, Modular Forms, and Discrete Geometry / Radu s Ramanujan-Kolberg Package 33 NOTE. The Ramanujan-Kolberg package computes in E (22): p(11n + 6)q n = q 14 (1 q 22j ) 22 (1 q j ) 10 (1 q 2j ) 2 (1 q 11j ) 11 with n=0 (1078t t t t z 1 (187t t t 9581) +z 2 (11t t t 6754) t:= 3 88 w w w 3, z 1 := 5 88 w w w 3 3, z 2 := 1 44 w w w 3, where w 1 :=[ 3, 3, 7], w 2 :=[8, 4, 8], w 3 :=[1, 11, 11] E (22) and

53 Combinatorics, Modular Forms, and Discrete Geometry / Radu s Ramanujan-Kolberg Package 33 NOTE. The Ramanujan-Kolberg package computes in E (22): p(11n + 6)q n = q 14 (1 q 22j ) 22 (1 q j ) 10 (1 q 2j ) 2 (1 q 11j ) 11 with n=0 (1078t t t t z 1 (187t t t 9581) +z 2 (11t t t 6754) t:= 3 88 w w w 3, z 1 := 5 88 w w w 3 3, z 2 := 1 44 w w w 3, where w 1 :=[ 3, 3, 7], w 2 :=[8, 4, 8], w 3 :=[1, 11, 11] E (22) and [r 2, r 11, r 22 ]:= δ 22 ( ) η(δτ rδ E (22). η(τ)

54 Combinatorics, Modular Forms, and Discrete Geometry / Radu s Ramanujan-Kolberg Package 34 Reference Cristian-Silviu Radu: An Algorithmic Approach to Ramanujan-Kolberg Identities. Journal of Symbolic Computation, 2014.

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