Combinatorics, Modular Functions, and Computer Algebra

Combinatorics, Modular Functions, and Computer Algebra / 1 Combinatorics and Computer Algebra (CoCoA 2015) Colorado State Univ., Fort Collins, July 19-25, 2015 Combinatorics, Modular Functions, and Computer Algebra Peter Paule (joint work with: G.E. Andrews, S. Radu) Johannes Kepler University Linz Research Institute for Symbolic Computation (RISC)

Combinatorics, Modular Functions, and Computer Algebra / Numbers 2 Numbers

Combinatorics, Modular Functions, and Computer Algebra / Numbers 3 1 1 2 3 5 7 11 15 22 30 42 56 77 101 135 176 231 297 385 490 627 792 1002 1255 1575 1958 2436 3010 3718 4565 5604 6842 8349 10143 12310 14883 17977 21637 26015 31185 37338 44583 53174 63261 75175 89134 105558 124754 147273 173525 204226 239943 281589 329931 386155 451276 526823 614154 715220 831820 966467 1121505 1300156 1505499 1741630 2012558 2323520 2679689 3087735 3554345 4087968 4697205 5392783 6185689 7089500 8118264 9289091 10619863 12132164 13848650

Combinatorics, Modular Functions, and Computer Algebra / Numbers 4 Example: p(4) = 5: 4, 3 + 1, 2 + 2, 2 + 1 + 1, 1 + 1 + 1 + 1.

Combinatorics, Modular Functions, and Computer Algebra / Numbers 4 Example: p(4) = 5: 4, 3 + 1, 2 + 2, 2 + 1 + 1, 1 + 1 + 1 + 1. EULER. The generating function of the partition numbers is E(q) := p(n)q n = n=0 n=1 1 1 q n = (1 + q 1 + q 1+1 + q 1+1+1 +... ) (1 + q 2 + q 2+2 + q 2+2+2 +... ) (1 + q 3 + q 3+3 + q 3+3+3 +... ) etc. = + q 1+1+1+2+2+3+ +

Combinatorics, Modular Functions, and Computer Algebra / Numbers 5 Back to p(n): ANY PATTERNS? 3,313, 3,325, and 3,362 of the first 10,000 entries are congruent respectively to 0, 1, and 2 modulo 3. [Ahlgren & Ono] BUT: 3,611 (many more than the expected one-fifth) of the first 10,000 values of p(n) are divisible by 5. [Ahlgren & Ono]

Combinatorics, Modular Functions, and Computer Algebra / Numbers 6 1 1 2 3 5 7 11 15 22 30 42 56 77 101 135 176 231 297 385 490 627 792 1002 1255 1575 1958 2436 3010 3718 4565 5604 6842 8349 10143 12310 14883 17977 21637 26015 31185 37338 44583 53174 63261 75175 89134 105558 124754 147273 173525 204226 239943 281589 329931 386155 451276 526823 614154 715220 831820 966467 1121505 1300156 1505499 1741630 2012558 2323520 2679689 3087735 3554345 4087968 4697205 5392783 6185689 7089500 8118264 9289091 10619863 12132164 13848650

Combinatorics, Modular Functions, and Computer Algebra / Numbers 7 Ramanujan s famous congruences p(5n + 4) 0 (mod 5), p(7n + 5) 0 (mod 7), p(11n + 6) 0 (mod 11)

Combinatorics, Modular Functions, and Computer Algebra / Numbers 8 Proof. The first congruence is implied by p(5n + 4)q n = 5 n=0 j=1 (1 q 5j ) 5 (1 q j ) 6

Combinatorics, Modular Functions, and Computer Algebra / Numbers 8 Proof. The first congruence is implied by p(5n + 4)q n = 5 n=0 j=1 (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]

Combinatorics, Modular Functions, and Computer Algebra / Numbers 9 Ramanujan [1919] found also an identity for the 7-case: p(7n + 5)q n n=0 = 7 j=1 (1 q 7j ) 3 (1 q j ) 4 + 49q j=1 (1 q 7j ) 7 (1 q j ) 8. What about the case p(11n + 6)?

Combinatorics, Modular Functions, and Computer Algebra / Modular Functions 10 Modular Functions

Combinatorics, Modular Functions, and Computer Algebra / Modular Functions 11 Recall E(q) := (1 q n ). n=1 We move from formal power (resp. Laurent) series to complex functions by defining where here and in the following q := q(τ) := exp(2πiτ)(= e 2πiτ ), τ H := {z C : Im(z) > 0}. MODULAR SYMMETRY.

Combinatorics, Modular Functions, and Computer Algebra / Modular Functions 12 We also need subgroups where N N \ {0}: {( ) } a b Γ 0 (N) := SL c d 2 (Z) : N c.

Combinatorics, Modular Functions, and Computer Algebra / Modular Functions 12 We also need subgroups where N N \ {0}: {( ) } a b Γ 0 (N) := SL c d 2 (Z) : N c. Def. A holomorphic function f(τ) is a modular function for Γ 0 (N), in short: f M(N), if ( ) ( ) aτ + b a b f = f(τ) for all Γ cτ + d c d 0 (N) and

Combinatorics, Modular Functions, and Computer Algebra / Modular Functions 13 Examples. Eta quotients like q E(q5 ) 6 E(q) 6 = j=1 (1 q 5j ) 6 (1 q j ) 6 M(5), q E(q7 ) 4 E(q) 4 = j=1 (1 q 7j ) 4 (1 q j ) 4 M(7); functions like q E(q 5 ) p(5n + 4)q n M(5), n=0 or the celebrated modular j-function (Klein invariant), j(τ) = 1 q +744+196884q+21493760q2 +864299970q 3 + M(1). NOTE. Γ 0 (1) = SL 2 (Z).

Combinatorics, Modular Functions, and Computer Algebra / Modular Functions 13 Zero recognition of functions in M(N)? The action ( ) a b Γ 0 (N) H H, τ := aτ + b c d cτ + d induces the action Γ 0 (N) HoloFus(H) HoloFus(H), (f γ)(τ) := f(γτ). The modular functions in M(N) HoloFus(H) are those which are constant on the orbits of the first action. It is crucial to extend this action as follows:

Combinatorics, Modular Functions, and Computer Algebra / Modular Functions 14 The set of orbits X 0 (N) := {[τ] : τ H Q { }} can be turned into a COMPACT Riemann surface.

Combinatorics, Modular Functions, and Computer Algebra / Modular Functions 14 The set of orbits X 0 (N) := {[τ] : τ H Q { }} can be turned into a COMPACT Riemann surface. Each (holomorphic) f M(N) has a meromorphic extension f : X 0 (N) C {0}. ZERO RECOGNITION.

Combinatorics, Modular Functions, and Computer Algebra / Modular Functions 15 To prove (and to find!) Ramanujan s identities automatically, we can restrict to consider M (N) := {f M(N) : f has a pole at [ ]}. Our goal is to present a given f M (N) in the form E(q a 1 ) α 1 E(q a 2 ) α2 E(q u1 ) µ1 E(q u2 ) µ2 f = c 1 E(q b 1 ) β 1E(q b 2) β 2 + + c m E(q v 1 ) ν 1E(q v 2) ν 2 with coefficients c j C and the Eta quotients in M (N).

Combinatorics, Modular Functions, and Computer Algebra / Modular Functions 16 NOTE 1. This monoid is finitely generated, E (N) = f 1,..., f n = {f l 1 1... f n ln : l i N}; the generators f 1,..., f n E (N) for fixed N can be determined algorithmically (e.g., with partition analysis). NOTE 2. In view of E(q a 1 ) α 1 E(q a 2 ) α2 E(q u1 ) µ1 E(q u2 ) µ2 f = c 1 E(q b 1 ) β 1E(q b 2) β 2 + + c m E(q v 1 ) ν 1E(q v 2) ν 2 our task is to constructively decide subalgebra membership f? C[f 1,..., f n ] M (N).

Combinatorics, Modular Functions, and Computer Algebra / Monoids & Subalgebras 17 Monoids & Subalgebras

Combinatorics, Modular Functions, and Computer Algebra / Monoids & Subalgebras 18 McNuggets Partitions GIVEN TASK: buy exactly 43 nuggets.

Combinatorics, Modular Functions, and Computer Algebra / Monoids & Subalgebras 18 McNuggets Partitions GIVEN TASK: buy exactly 43 nuggets. IMPOSSIBLE!

Combinatorics, Modular Functions, and Computer Algebra / Monoids & Subalgebras 18 McNuggets Partitions GIVEN TASK: buy exactly 43 nuggets. IMPOSSIBLE! See: How to order 43 Chicken McNuggets - Numberphile www.youtube.com/watch?v=vntsugys038

Combinatorics, Modular Functions, and Computer Algebra / Monoids & Subalgebras 19 Generators of (additive) submonoids of N M:= 6, 9, 20 N = {0, 1,... } [M] i := {x M : x i (mod 6)}

Combinatorics, Modular Functions, and Computer Algebra / Monoids & Subalgebras 19 Generators of (additive) submonoids of N M:= 6, 9, 20 N = {0, 1,... } [M] i := {x M : x i (mod 6)} [M] 0 = {0, 6, 12,... }, [M] 1 = { 1, 7, 13, 19, 25, 31, 37, 43, 49, 54,... }, [M] 2 = { 2, 8, 14, 20, 26,... }, [M] 3 = { 3, 9, 15,... }, [M] 4 = { 4, 10, 16, 22, 28, 34, 40, 46,... }, [M] 5 = { 5, 11, 17, 23, 29, 35,... }.

Combinatorics, Modular Functions, and Computer Algebra / Monoids & Subalgebras 20 QUESTION. What happens if we add more generators? M + := 4, 6, 9, 20 N = {0, 1,... } [M + ] i := {x M : x i (mod 6)}

Combinatorics, Modular Functions, and Computer Algebra / Monoids & Subalgebras 20 QUESTION. What happens if we add more generators? M + := 4, 6, 9, 20 N = {0, 1,... } [M + ] i := {x M : x i (mod 6)} [M + ] 0 = {0, 6, 12,... }, [M + ] 1 = { 1, 7, 13, 19,... }, [M + ] 2 = { 2, 8, 14,... }, [M + ] 3 = { 3, 9, 15,... }, [M + ] 4 = {4, 10,... }, [M + ] 5 = { 5, 11, 17, 23,... }. NOTE. In contrast to the Frobenius number F (6, 9, 20) = 43, now F (4, 6, 9, 20) = 11 and 43 = 2 4 + 6 + 9 + 1 20 = 4 + 2 6 + 3 9 = [four more]. QUESTION. How to find presentations in terms of generators?

Combinatorics, Modular Functions, and Computer Algebra / Monoids & Subalgebras 21 (a) Greedy Algorithm STRATEGY: Successive subtraction; begin with the largest generator, first taken a maximal number of times; iterate. EXAMPLE: x = 62 and M = 6, 9, 20. 62 3 20 = 2 62 2 20 = 22, 22 2 9 = 4 22 1 9 = 13, 13 2 6 = 1 22 3 6 = 4 62 1 20 = 42, 42 4 9 = 6, 6 1 6 = 0.

Combinatorics, Modular Functions, and Computer Algebra / Monoids & Subalgebras 22 SOLUTION. 62 = 1 6 + 4 9 + 1 20.

Combinatorics, Modular Functions, and Computer Algebra / Monoids & Subalgebras 22 SOLUTION. 62 = 1 6 + 4 9 + 1 20. Also, and 62 = 4 6 + 2 9 + 1 20 62 = 7 6 + 0 9 + 1 20.

Combinatorics, Modular Functions, and Computer Algebra / Monoids & Subalgebras 22 SOLUTION. 62 = 1 6 + 4 9 + 1 20. Also, and 62 = 4 6 + 2 9 + 1 20 62 = 7 6 + 0 9 + 1 20. PREVIEW. The Omega package computes In[2]:=OEqR[ OEqSum[x a y b z c, {6a + 9b + 20c = 62}, λ ]] Out[2]= ( x 7 + x 4 y 2 + xy 4) z.

Combinatorics, Modular Functions, and Computer Algebra / Monoids & Subalgebras 23 (b) Algorithm based on the generators u 1,..., u 5 STRATEGY: Find a representation x = u i + l 6.

Combinatorics, Modular Functions, and Computer Algebra / Monoids & Subalgebras 23 (b) Algorithm based on the generators u 1,..., u 5 STRATEGY: Find a representation x = u i + l 6. EXAMPLE: M = 6, u 1, u 2, u 3, u 4, u 5 = 6, 43, 20, 9, 40, 29. and x = 62. 62 2 (mod 6); hence we recall that [M] 2 = { 2, 8, 14, 20, 26,... }; consequently, 62 = 20 + 7 6 = u 2 + 7 6.

Combinatorics, Modular Functions, and Computer Algebra / Monoids & Subalgebras 24 Application: C-subalgebras of C[z] GIVEN: f 1, f 2, f 3 C[z] with t :=f 1 = z 6 + a z 5 +..., FIND: g 1,..., g k C[z] such that f 2 = z 9 + b z 8 +..., f 3 = z 20 + c z 19 +... ; C[f 1, f 2, f 3 ] = g 1,..., g k C[t] = C[t] + C[t] g 1 + + C[t] g k.

Combinatorics, Modular Functions, and Computer Algebra / Monoids & Subalgebras 25 DECISION PROCEDURE: C-subalgebra membership GIVEN: f and f 1,..., f n from C[z]; FIND: polynomials p 0, p 1,..., p k C[z] such that f = p 0 (t) + p 1 (t) g 1 + + p k (t) g k 1, g 1,..., g k C[t] = C[f 1,..., f n ] APPLICATION. With this procedure the right hand sides of Ramanujan identities can be found automatically: just apply it to given f M (N) and f 1,..., f n E (N). In this context, z = 1 q because of possible poles sitting at [ ]; the f j can be treated as formal Laurent series like f 1 = a 6 q 6 + a 5 q 5 +, f 2 = b 9 q 9 + b 8 q 8 +, f 3 = c 20 q 20 + c 19 +, etc. q19

Combinatorics, Modular Functions, and Computer Algebra / Monoids & Subalgebras 26 Example. Radu s Ramanujan-Kolberg package in STEP 1 computes that: 1 E(q)8 q E(q 7 ) 7 p(7n + 5)q n M (7). n=0

Combinatorics, Modular Functions, and Computer Algebra / Monoids & Subalgebras 26 Example. Radu s Ramanujan-Kolberg package in STEP 1 computes that: 1 E(q)8 q E(q 7 ) 7 p(7n + 5)q n M (7). n=0 In STEP 2 Radu s package determines that the monoid is generated by one element (recall f 1 = t): E 1 E(q)4 (7) = f 1 = q E(q 7 ) 4. In STEP 3 the package determines the module presentation C[f 1,..., f n ] = C[f 1 ] = 1, g 1,..., g k C[t] = 1 C[f1 ].

Combinatorics, Modular Functions, and Computer Algebra / Monoids & Subalgebras 27 QUESTION. How to compute the u 1,..., u 5 from M = 6, 9, 20? ANSWER. To find u 1 = 49, we ask Omega to compute In[3]:=OEqR[ OEqSum[x a y b z c, { 6a + 9b + 20c = 1}, λ ]] Out[3]= x 8 yz 2 (1 x 3 y 2 )(1 x 10 z 3 ).

Combinatorics, Modular Functions, and Computer Algebra / Monoids & Subalgebras 28 Out[3] means that a,b,c 0 s.t. 6a+9b+20c=1 x a y b z c = x 8 yz 2 (1 x 3 y 2 ) (1 x 10 z 3 ) ; in other words, 8 3 10 1 + N 2 + N 0 2 0 3 is the solution set of 6a + 9b + 20c = 1. For example, 1 9 + 2 20 = 1 + 8 6 = 49 = u 1. Analogously one computes u 2,..., u 5.

Combinatorics, Modular Functions, and Computer Algebra / Partition Analysis 29 Partition Analysis

Combinatorics, Modular Functions, and Computer Algebra / Partition Analysis 30 PROBLEM [Polya]. Consider triangles with sides of integer length such that 1 l m n; TASK: express the total number of such triangles in terms of n.

Combinatorics, Modular Functions, and Computer Algebra / Partition Analysis 31 PROBLEM [folklore]. Consider triangles with sides of integer length such that 1 a b c; TASK: find the total number of all such triangles with perimeter n.

Combinatorics, Modular Functions, and Computer Algebra / Partition Analysis 32 SOLUTION IDEA. Determine the generating function T (x, y, z; q) := x a y b z c q a+b+c. c b a 1, s.t. a+b>c Then all Polya triangles are given by the coefficient of z n in T (x, y, z; 1), and the total number of such triangles as the coefficient of z n in T (1, 1, z; 1).

Combinatorics, Modular Functions, and Computer Algebra / Partition Analysis 33 SOLUTION IDEA. Determine the generating function T (x, y, z; q) := x a y b z c q a+b+c. c b a 1, s.t. a+b>c Then all triangles with given perimeter n are given by the coefficient of q n in T (x, y, z; q), and the total number of such triangles as the coefficient of q n in T (1, 1, 1; q).

Combinatorics, Modular Functions, and Computer Algebra / Partition Analysis 34 HOW TO FIND A SUITABLE REPRESENTATION OF T (x, y, z; q) := x a y b z c q a+b+c? c b a 1, s.t. a+b>c Pick up the Omega package, freely available at www.risc.jku.at/research/combinat/software and do the following:

Combinatorics, Modular Functions, and Computer Algebra / Partition Analysis 35

Combinatorics, Modular Functions, and Computer Algebra / Partition Analysis 35 WHAT STANDS BEHIND THIS?

Combinatorics, Modular Functions, and Computer Algebra / Partition Analysis 35 WHAT STANDS BEHIND THIS? MacMahon s Partition Analysis

Combinatorics, Modular Functions, and Computer Algebra / Partition Analysis 36 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. 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.

Combinatorics, Modular Functions, and Computer Algebra / Partition Analysis 37 How Zeilberger tells the story of partition analysis (and more):

Combinatorics, Modular Functions, and Computer Algebra / Partition Analysis 38 MacMahon s Omega Calculus: T (x, y, z; q) := = Ω c b a 1, s.t. a+b>c a,b,c 1 x a y b z c q a+b+c l b a 1 l c b 2 l a+b 1 c 3 x a y b z c q a+b+c

Combinatorics, Modular Functions, and Computer Algebra / Partition Analysis 39 The second step is the elimination of the slack variables:

Combinatorics, Modular Functions, and Computer Algebra / Partition Analysis 39 The second step is the elimination of the slack variables: NOTE. To this end, MacMahon used elimination rules like Ω l k (1 A l) ( 1 B l ) = A k (1 A)(1 A B), k 0.

Combinatorics, Modular Functions, and Computer Algebra / Partition Analysis 40 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 [arxiv:1501.07773] The new algorithm Polyhedral Omega solves linear Diophantine systems by combining methods from partition analysis with methods from polyhedral geometry as Brion decompositions and Barvinok s short rational function representations.

Combinatorics, Modular Functions, and Computer Algebra / Omega Discovery 41 Omega Discovery

Combinatorics, Modular Functions, and Computer Algebra / Omega Discovery 42 MacMahon s plane partitions, e.g., of n = 3: 3, 2 + 1, 2 + 1, 1 + 1 + 1, 1 + 1 + 1, 1 + 1 + 1.

Combinatorics, Modular Functions, and Computer Algebra / Omega Discovery 42 MacMahon s plane partitions, e.g., of n = 3: 3, 2 + 1, 2 + 1, 1 + 1 + 1, 1 + 1 + 1, 1 + 1 + 1 More generally, one can consider digraphs (resp. posets) like a 2 a 5 a 8 P = a 1 a 4 a 7 a 10 a 3 a 6 a 9.

Combinatorics, Modular Functions, and Computer Algebra / Omega Discovery 43 a 1 a 3 a 5 a 7 a 2k 1 a 2k+1.......... a 2k+2.......... a 2 a 4 a 6 a 2k a 2k 2 A k-elongated partition diamond of length 1 a 1 a 3........................ a 2k+2 a 4k+3.............................. 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

Combinatorics, Modular Functions, and Computer Algebra / Omega Discovery 44 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 j=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 j=1 (1 q j ) and glue the diamonds together: b (2k+1)n+1 b (2k+1)n b 2k+1 b 7 b 3................... b 2k+2................ b 6 b (2k+1)n 1 b 2k b 5 b 4 a 1 b 2 a 3................ a 2k+2... a (2k+1)n+1................ a 2 a 2k+1 a 2k a (2k+1)n a (2k+1)n 1 A broken k-diamond of length 2n

Combinatorics, Modular Functions, and Computer Algebra / Omega Discovery 45 Consequently, m=0 k (m)q m := lim n h n,k(q) h n,k (q) = j=1 (1 q 2j )(1 q (2k+1)j ) (1 q j ) 3 (1 q (4k+2)j ) = E(q2 )E(q 2k+1 ) E(q) 3 E(q 4k+2 )

Combinatorics, Modular Functions, and Computer Algebra / Omega Discovery 45 Consequently, m=0 k (m)q m := lim n h n,k(q) h n,k (q) = and think of Ramanujan! j=1 (1 q 2j )(1 q (2k+1)j ) (1 q j ) 3 (1 q (4k+2)j ) = E(q2 )E(q 2k+1 ) E(q) 3 E(q 4k+2 ) b (2k+1)n+1 b (2k+1)n b 2k+1 b 7 b 3................... b 2k+2................ b 6 b (2k+1)n 1 b 2k b 5 b 4 a 1 b 2 a 3................ a 2k+2... a (2k+1)n+1................ a 2 a 2k+1 a 2k a (2k+1)n a (2k+1)n 1 A broken k-diamond of length 2n

Combinatorics, Modular Functions, and Computer Algebra / Omega Discovery 46

Combinatorics, Modular Functions, and Computer Algebra / Omega Discovery 47 Theorem. For all n N, 1 (2n + 1) 0 (mod 3).

Combinatorics, Modular Functions, and Computer Algebra / Omega Discovery 47 Theorem. 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 j=1 j=1 (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.

Combinatorics, Modular Functions, and Computer Algebra / Omega Discovery 48 Recall: Theorem. For all n N, 1 (2n + 1) 0 (mod 3). Algorithmic Proof [Radu 2014]:

Combinatorics, Modular Functions, and Computer Algebra / Omega Discovery 48 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 j=1

Combinatorics, Modular Functions, and Computer Algebra / Omega Discovery 49 NOTE. There is much more: many new families of combinatorial objects, q-series identities, and computer proofs and findings. Let us return to Ramanujan s congruence p(11n + 6) 0 (mod 11) :

Combinatorics, Modular Functions, and Computer Algebra / Omega Discovery 50 Radu s Ramanujan-Kolberg package computes in M (22): E(q) 10 E(q 2 ) 2 E(q 11 ) 11 E(q 22 ) 22 p(11n + 6)q n with n=0 = q 14( 1078t 4 + 13893t 3 + 31647t 2 + 11209t 21967 +z 1 (187t 3 + 5390t 2 + 594t 9581) +z 2 (11t 3 + 2761t 2 + 5368t 6754) ) t:= 3 88 w 1 + 1 11 w 2 1 8 w 3, z 1 := 5 88 w 1 + 2 11 w 2 1 8 w 3 3, z 2 := 1 44 w 1 3 11 w 2 + 5 4 w 3, where

Combinatorics, Modular Functions, and Computer Algebra / Omega Discovery 51 References G.E. Andrews and P. Paule: MacMahon s Partition Analysis XI: Broken diamonds and modular forms. Acta Arithmetica 126 (2007), 281-294. S. Radu: An Algorithmic Approach to Ramanujan-Kolberg Identities, Journal for Symbolic Computation 68 (2014), 1-33. P. Paule and S. Radu: Partition Analysis, Modular Functions, and Computer Algebra. To appear in Recent Trends in Combinatorics, IMA Volume in Mathematics and its Applications, Springer, 2015. (Available at www.dk-compmath.jku.at/publications.)

Combinatorics, Modular Functions, and Computer Algebra / RISC Software & Appls. 52 RISC Software & Appls.

Combinatorics, Modular Functions, and Computer Algebra / RISC Software & Appls. 53 Software of the RISC Algorithmic Combinatorics Group Feel free to pick up packages! www.risc.jku.at/research/combinat/software

Combinatorics, Modular Functions, and Computer Algebra / RISC Software & Appls. 54

Combinatorics, Modular Functions, and Computer Algebra / RISC Software & Appls. 55

Combinatorics, Modular Functions, and Computer Algebra / RISC Software & Appls. 56

Combinatorics, Modular Functions, and Computer Algebra / RISC Software & Appls. 57 Symbolic Summation in QFT JKU Collaboration with DESY (Berlin Zeuthen) (Deutsches Elektronen Synchrotron) Project leader: Partners: Carsten Schneider (RISC) Johannes Blümlein (DESY) Peter Paule (RISC)

Combinatorics, Modular Functions, and Computer Algebra / RISC Software & Appls. 58 Evaluation of Feynman diagrams Behavior of particles Φ(N, ɛ, x)dx Feynman integrals Evaluations required for the LHC experiment at CERN DESY processable by physicists simple sum expressions RISC (symbolic summation) f(n, ɛ, k) multi-sums

Combinatorics, Modular Functions, and Computer Algebra / RISC Software & Appls. 59 Challenges of the project About 1000 difficult Feynman diagrams have been treated so far Resources (some took 50 days of calculation time) About a million multi-sums have been simplified (most were double and triple sums) 9 full time employed researchers at RISC/DESY: J. Ablinger, A. Behring, J. Blümlein, A. Hasselhuhn, A. de Freitas, C. Raab, M. Round, C. Schneider, F. Wissbrock 4 up-to-date mainframe DESY computers at RISC + exploiting DESY s computer farms New computer algebra/special functions technologies (new/tuned algorithms, efficient implementations,...)

Combinatorics, Modular Functions, and Computer Algebra / RISC Software & Appls. 60 = F 3 (N)ε 3 + F 2 (N)ε 2 + F 1 (N)ε 1 + F 0 (N) Simplify N 3 j=0 j k=0 k (j+1 k+1)( k l)( N 1 [ j+n 3 l+n q 3 l+n q s 3 ( 1) j+k l+n q 3 l=0 q=0 s=1 r=0 j+2 )( j+n 3 q )( l+n q 3 s ) ( l+n q s 3 r )r!( l+n q r s 3)!(s 1)! ( l+n q 2)!( j+n 1)(N q r s 2)(q+s+1) 4S 1 ( j + N 1) 4S 1 ( j + N 2) 2S 1 (k) (S 1 ( l + N q 2) + S 1 ( l + N q r s 3) 2S 1 (r + s)) ] + 2S 1 (s 1) 2S 1 (r + s) + 3 further 6 fold sums

Combinatorics, Modular Functions, and Computer Algebra / RISC Software & Appls. 61 F 0 (N) = (using Sigma.m, EvaluateMultiSums.m and J. Ablinger s HarmonicSums.m package) 7 12 S 1(N) 4 + (17N + 5)S 1(N) 3 + ( 35N 2 2N 5 3N(N + 1) 2N 2 (N + 1) 2 + 13S 2(N) + 5( 1)N ) S1 2 2N 2 (N) 2 + ( 4(13N + 5) N 2 (N + 1) 2 + ( 4( 1)N (2N + 1) 13 ) S2 (N) + ( 29 N(N + 1) N 3 ) ( 1)N S 3 (N) + ( 2 + 2( 1) N ) S 2,1 (N) 28S 2,1 (N) + 20( 1)N ) S1 N 2 (N) + ( 3 (N + 1) 4 + ) ( 1)N S 2 (N) 2 2( 1) N S 2 (N) 2 ( 2(3N 5) + S 3 (N) N(N + 1) + ( 26 + 4( 1) N ) S 1 (N) + 4( 1)N N + 1 ) S2 2N 2 (N) + S 2 (N) ( 10S 1 (N) 2 + ( 8( 1)N (2N + 1) N(N + 1) + ( ( 1)N (5 3N) 2N 2 5 (N + 1) 4(3N 1) ) + S1 (N) + 8( 1)N (3N + 1) N(N + 1) N(N + 1) 2 + ( 22 + 6( 1) N ) 16 ) S 2 (N) N(N + 1) + ( ( 1)N (9N + 5) 29 ) S3 (N) + ( 19 N(N + 1) 3N 2 ) 2( 1)N S 4 (N) + ( 6 + 5( 1) N ) S 4 (N) + ( 2( 1)N (9N + 5) 2 N(N + 1) N 8( 1)N (2N + 1) + 4(9N + 1) N(N + 1) + 32S 2,1,1 (N) + ) S2,1 (N) + ( 20 + 2( 1) N ) S 2, 2 (N) + ( 17 + 13( 1) N ) S 3,1 S 2,1 (N) ( 24 + 4( 1) N ) S 3,1 (N) + ( 3 5( 1) N ) S 2,1, ) ( 3 2 S 1(N) 2 3S 1(N) + 3 N 2 ( 1)N S 2 (N) ζ(2) )

Combinatorics, Modular Functions, and Computer Algebra / RISC Software & Appls. 61 F 0 (N) = (using Sigma.m, EvaluateMultiSums.m and J. Ablinger s HarmonicSums.m package) 7 12 S 1(N) 4 + (17N + 5)S 1(N) 3 + ( 35N 2 2N 5 3N(N + 1) 2N 2 (N + 1) 2 + 13S 2(N) + 5( 1)N ) S1 2 2N 2 (N) 2 + ( 4(13N + 5) N 2 (N + 1) 2 + ( 4( 1)N (2N + 1) 13 ) S2 (N) + ( 29 N(N + 1) N 3 ) ( 1)N S 3 (N) + ( 2 + 2( 1) N ) S 2,1 (N) 28S 2,1 (N) + 20( 1)N ) S1 N 2 (N) + ( 3 (N + 1) 4 + ) ( 1)N S 2 (N) 2 2( 1) N S 2 (N) 2 ( 2(3N 5) + S 3 (N) N(N + 1) + ( j 26 + 4( 1) N ) S 1 (N) + 4( 1)N ) 1 N + 1 + ( ( 1)N (5 3N) 2N 2 5 ) i S2 (N + 1) 2N 2 (N) + S 2 (N) ( k 10S 1 (N) 2 + ( 8( 1)N (2N + 1) ( 1) N(N + 1) 4(3N 1) ) N i k=1 1 ζ(2) = j k 2 + S1 + 8( 1)N (3N + 1) j=1 N(N + 1) N(N + 1) 2 + ( 22 + 6( 1) N ) 16k=1) S 2,1,1 (N) = S 2 (N) N(N + 1) + ( ( 1)N (9N + 5) N(N + 1) 29 3N + ( 2( 1)N (9N + 5) 2 N(N + 1) N 8( 1)N (2N + 1) + 4(9N + 1) N(N + 1) + 32S 2,1,1 (N) + i 2 ) i=1 S3 (N) + ( 19 2 ) 2( 1)N S 4 (N) + ( 6 + 5( 1) N ) S 4 (N) ) S2,1 (N) + ( 20 + 2( 1) N ) S 2, 2 (N) + ( 17 + 13( 1) N ) S 3,1 S 2,1 (N) ( 24 + 4( 1) N ) S 3,1 (N) + ( 3 5( 1) N ) S 2,1, ) ( 3 2 S 1(N) 2 3S 1(N) + 3 N 2 ( 1)N S 2 (N) ζ(2)

Combinatorics, Modular Functions, and Computer Algebra / RISC Software & Appls. 62 Symbolic Special Functions Manipulation: Application in Electromagnetic Wave Simulation Collaboration with company CST (Computer Simulation Technology) Partner: RISC: Joachim Schöberl (TU Vienna) Christoph Koutschan (now RICAM) Peter Paule

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Combinatorics, Modular Functions, and Computer Algebra / RISC Software & Appls. 67 The RISC Software Company: the applied branch of RISC Industrial Applications CEO: Wolfgang Freiseisen 50 employees

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Combinatorics, Modular Functions, and Computer Algebra / Appendix: Subalgebra Membership Algorithm 72 Appendix: Subalgebra Membership Algorithm

Combinatorics, Modular Functions, and Computer Algebra / Appendix: Subalgebra Membership Algorithm 73 Application: C-subalgebras of C[z] GIVEN: f 1, f 2, f 3 C[z] with t :=f 1 = z 6 + a z 5 +..., FIND: g 1,..., g k C[z] such that f 2 = z 9 + b z 8 +..., f 3 = z 20 + c z 19 +... ; C[f 1, f 2, f 3 ] = g 1,..., g k C[t] = C[t] + C[t] g 1 + + C[t] g k.

Combinatorics, Modular Functions, and Computer Algebra / Appendix: Subalgebra Membership Algorithm 74 KEY PROCEDURE GIVEN: f 1, f 2, f 3 C[z] with M = deg(f 1 ), deg(f 2 ), deg(f 3 ) = 6, 9, 20 ; select t := f 1 and define m := deg(t) = 6; FIND: g 1,..., g 5 C[f 1, f 2, f 3 ] such that C[f 1, f 2, f 3 ] = C[t, f 2, f 3 ] = C[t, g 1,..., g 5 ] and {deg(g 1 ),..., deg(g 5 )} = {u 1,..., u 5 } {1,..., 5} (mod 6). How to determine the g i?

Combinatorics, Modular Functions, and Computer Algebra / Appendix: Subalgebra Membership Algorithm 75 How to determine the g i? [M] 0 = {0, 6, 12,... }, and t :=f 1 = z 6 + a z 5 +. [M] 1 = { 1, 7, 13, 19, 25, 31, 37, 43, 49, 54,... }, define in view of u 1 = 49 = 9 + 2 20, g 1 :=f 2 f 2 3 = z u 1 +. [M] 2 = { 2, 8, 14, 20, 26,... }, define in view of u 2 = 20, g 2 :=f 3 = z u 2 +. [M] 3 = { 3, 9, 15,... }, define in view of u 3 = 9, g 3 :=f 2 = z u 3 +.

Combinatorics, Modular Functions, and Computer Algebra / Appendix: Subalgebra Membership Algorithm 76 [M] 4 = { 4, 10, 16, 22, 28, 34, 40, 46,... }, define in view of u 4 = 40 = 2 20, g 4 :=f 2 3 = z u 4 +. [M] 5 = { 5, 11, 17, 23, 29, 35,... }, define in view of u 5 = 29 = 9 + 20, g 5 :=f 2 f 3 = z u 5 +. SUMMARY. We determined the g i C[f 1, f 2, f 3 ] as g 1 := f 2 f3 2 = z u 1 +, g 2 := f 3 = z u 2 +, g 3 := f 2 = z u 3 +, g 4 := f3 2 = z u 4 +, g 5 := f 2 f 3 = z u 5 +, such that C[f 1, f 2, f 3 ] = C[t, g 1,... g 5 ] and deg(g i ) i (mod 6).

Combinatorics, Modular Functions, and Computer Algebra / Appendix: Subalgebra Membership Algorithm 77 Example. f = z 62 + C[f 1, f 2, f 3 ]: 62 = 20 + 7 6 = u 2 + 7 6 ; recalling g 2 :=f 3 = z u 2 + one has, f g 2 t 7 = c 1 z 61 +. 61 = 49 + 2 6 = u 1 + 2 6 ; recalling g 1 :=f 2 f3 2 = zu 1 + one has, f g 2 t 7 c 1 g 1 t 2 = c 2 z 60 +. BUT:

Combinatorics, Modular Functions, and Computer Algebra / Appendix: Subalgebra Membership Algorithm 78 Example (contd). f = z 62 + C[f 1, f 2, f 3 ]: For instance, it could be that f g 2 t 7 c 1 g 1 t 2 = z 4 + C[f 1, f 2, f 3 ]. NOTE. BUT 4 / 6, 9, 20 = deg(f 1 ), deg(f 2 ), deg(f 3 ) 4 deg(g) : g C[f 1, f 2, f 3 ]. SOLUTION. We repeat the KEY PROCEDURE {f 1, f 2, f 3 } {g 1, g 2, g 3, g 4, g 5 } by iteratively joining products: {g 1, g 2, g 3, g 4, g 5 } {g i g j } {h 1, h 2, h 3, h 4, h 5 }, {h 1, h 2, h 3, h 4, h 5 } {h i h j } {... }, a.s.o.

Combinatorics, Modular Functions, and Computer Algebra / Appendix: Subalgebra Membership Algorithm 79 DECISION PROCEDURE: C-subalgebra membership GIVEN: f and f 1,..., f n from C[z]; FIND: polynomials p 0, p 1,..., p k C[z] such that f = p 0 (t) + p 1 (t) g 1 + + p k (t) g k 1, g 1,..., g k C[t] = C[f 1,..., f n ] APPLICATION. With this procedure the right hand sides of Ramanujan identities can be found automatically: just apply it to given f M (N) and f 1,..., f n E (N). In this context, z = 1 q because of possible poles sitting at [ ]; the f j can be treated as formal Laurent series like f 1 = a 6 q 6 + a 5 q 5 +, f 2 = b 9 q 9 + b 8 q 8 +, f 3 = c 20 q 20 + c 19 +, etc. q19

Combinatorics, Modular Functions, and Computer Algebra / Appendix: Subalgebra Membership Algorithm 80 Example. Radu s Ramanujan-Kolberg package in STEP 1 computes that: 1 E(q)8 q E(q 7 ) 7 p(7n + 5)q n M (7). n=0 In STEP 2 Radu s package determines that the monoid is generated by one element (recall f 1 = t): E 1 E(q)4 (7) = f 1 = q E(q 7 ) 4. In STEP 3 the package determines the module presentation C[f 1,..., f n ] = C[f 1 ] = 1, g 1,..., g k C[t] = 1 C[f1 ].