Binomial Determinants for Tiling Problems Yield to the Holonomic Ansatz
Hao Du, Christoph Koutschan, Thotsaporn Thanatipanonda, Elaine Wong
Johann Radon Institute for Computational and Applied Mathematics (RICAM) Austrian Academy of Sciences
July 2, 2021 S´eminairede Combinatoire de Lyon `al’ENS + 1 + 1 + 1
µ µ + i + j + s + t − 4 Es,t(n) := det − δi+s,j+t . 16i6n j + t − 1 16j6n
Example: is the determinant of the matrix µ+8 µ+9 µ+10 µ+11 µ+12 µ+9 µ+10 µ+11 µ+12 µ+13 µ+10 µ+11 µ+12 µ+13 µ+14 µ+11 µ+12 µ+13 µ+14 µ+15 µ+12 µ+13 µ+14 µ+15 µ+16
Families of Binomial Determinants Definition: For n ∈ N, for s, t ∈ Z, and for µ an indeterminate, define the following (n × n)-determinants: µ µ + i + j + s + t − 4 Ds,t(n) := det + δi+s,j+t , 16i6n j + t − 1 16j6n
1 / 32 + 1 + 1 + 1
Example: is the determinant of the matrix µ+8 µ+9 µ+10 µ+11 µ+12 µ+9 µ+10 µ+11 µ+12 µ+13 µ+10 µ+11 µ+12 µ+13 µ+14 µ+11 µ+12 µ+13 µ+14 µ+15 µ+12 µ+13 µ+14 µ+15 µ+16
Families of Binomial Determinants Definition: For n ∈ N, for s, t ∈ Z, and for µ an indeterminate, define the following (n × n)-determinants: µ µ + i + j + s + t − 4 Ds,t(n) := det + δi+s,j+t , 16i6n j + t − 1 16j6n µ µ + i + j + s + t − 4 Es,t(n) := det − δi+s,j+t . 16i6n j + t − 1 16j6n
1 / 32 + 1 + 1 + 1
Example: is the determinant of the matrix µ+8 µ+9 µ+10 µ+11 µ+12 µ+9 µ+10 µ+11 µ+12 µ+13 µ+10 µ+11 µ+12 µ+13 µ+14 µ+11 µ+12 µ+13 µ+14 µ+15 µ+12 µ+13 µ+14 µ+15 µ+16
Families of Binomial Determinants Definition: For n ∈ N, for s, t ∈ Z, and for µ an indeterminate, define the following (n × n)-determinants: µ µ + i + j + s + t − 4 Ds,t(n) := det + δi+s,j+t , 16i6n j + t − 1 16j6n µ µ + i + j + s + t − 4 Es,t(n) := det − δi+s,j+t . 16i6n j + t − 1 16j6n
µ History: D0,0(n) was introduced in the work of Andrews in 1979 – 1980 in the context of descending plane partitions:
1 / 32 Families of Binomial Determinants Definition: For n ∈ N, for s, t ∈ Z, and for µ an indeterminate, define the following (n × n)-determinants: µ µ + i + j + s + t − 4 Ds,t(n) := det + δi+s,j+t , 16i6n j + t − 1 16j6n µ µ + i + j + s + t − 4 Es,t(n) := det − δi+s,j+t . 16i6n j + t − 1 16j6n
µ Example: D4,6(5) is the determinant of the matrix µ+8 µ+9 µ+10 µ+11 µ+12 6 7 8 9 10 µ+9 µ+10 µ+11 µ+12 µ+13 6 7 8 9 10 µ+10 + 1 µ+11 µ+12 µ+13 µ+14 6 7 8 9 10 µ+11 µ+12 µ+13 µ+14 µ+15 + 1 6 7 8 9 10 µ+12 µ+13 µ+14 µ+15 µ+16 6 7 8 + 1 9 10 1 / 32 Families of Binomial Determinants Definition: For n ∈ N, for s, t ∈ Z, and for µ an indeterminate, define the following (n × n)-determinants: µ µ + i + j + s + t − 4 Ds,t(n) := det + δi+s,j+t , 16i6n j + t − 1 16j6n µ µ + i + j + s + t − 4 Es,t(n) := det − δi+s,j+t . 16i6n j + t − 1 16j6n
µ+2 Example: D3,5 (5) is the determinant of the matrix µ+8 µ+9 µ+10 µ+11 µ+12 5 6 7 8 9 µ+9 µ+10 µ+11 µ+12 µ+13 5 6 7 8 9 µ+10 + 1 µ+11 µ+12 µ+13 µ+14 5 6 7 8 9 µ+11 µ+12 µ+13 µ+14 µ+15 + 1 5 6 7 8 9 µ+12 µ+13 µ+14 µ+15 µ+16 5 6 7 + 1 8 9 1 / 32 Families of Binomial Determinants Definition: For n ∈ N, for s, t ∈ Z, and for µ an indeterminate, define the following (n × n)-determinants: µ µ + i + j + s + t − 4 Ds,t(n) := det + δi+s,j+t , 16i6n j + t − 1 16j6n µ µ + i + j + s + t − 4 Es,t(n) := det − δi+s,j+t . 16i6n j + t − 1 16j6n
µ Example: D4,6(5) is the determinant of the matrix µ+8 µ+9 µ+10 µ+11 µ+12 6 7 8 9 10 µ+9 µ+10 µ+11 µ+12 µ+13 6 7 8 9 10 µ+10 + 1 µ+11 µ+12 µ+13 µ+14 6 7 8 9 10 µ+11 µ+12 µ+13 µ+14 µ+15 + 1 6 7 8 9 10 µ+12 µ+13 µ+14 µ+15 µ+16 6 7 8 + 1 9 10 1 / 32 + 1 + 1 + 1
Families of Binomial Determinants Definition: For n ∈ N, for s, t ∈ Z, and for µ an indeterminate, define the following (n × n)-determinants: µ µ + i + j + s + t − 4 Ds,t(n) := det + δi+s,j+t , 16i6n j + t − 1 16j6n µ µ + i + j + s + t − 4 Es,t(n) := det − δi+s,j+t . 16i6n j + t − 1 16j6n
µ−1 Example: D5,6 (5) is the determinant of the matrix µ+8 µ+9 µ+10 µ+11 µ+12 6 7 8 9 10 µ+9 + 1 µ+10 µ+11 µ+12 µ+13 6 7 8 9 10 µ+10 µ+11 + 1 µ+12 µ+13 µ+14 6 7 8 9 10 µ+11 µ+12 µ+13 µ+14 µ+15 + 1 6 7 8 9 10 µ+12 µ+13 µ+14 µ+15 µ+16 6 7 8 9 + 1 10 1 / 32 + 1 + 1 + 1
Families of Binomial Determinants Definition: For n ∈ N, for s, t ∈ Z, and for µ an indeterminate, define the following (n × n)-determinants: µ µ + i + j + s + t − 4 Ds,t(n) := det + δi+s,j+t , 16i6n j + t − 1 16j6n µ µ + i + j + s + t − 4 Es,t(n) := det − δi+s,j+t . 16i6n j + t − 1 16j6n
µ−1 Example: E5,6 (5) is the determinant of the matrix µ+8 µ+9 µ+10 µ+11 µ+12 6 7 8 9 10 µ+9 − 1 µ+10 µ+11 µ+12 µ+13 6 7 8 9 10 µ+10 µ+11 − 1 µ+12 µ+13 µ+14 6 7 8 9 10 µ+11 µ+12 µ+13 µ+14 µ+15 − 1 6 7 8 9 10 µ+12 µ+13 µ+14 µ+15 µ+16 6 7 8 9 − 1 10 1 / 32 “Conjecture 37” (Lascoux/Krattenthaler 2005) Let µ be an indeterminate and m, r ∈ Z. If m > r > 1, then
µ m−r 4m+(m−r)(m−r−1)−3r E1,2r−1(2m − 1) = (−1) · 2 2r−3 ! m−1 ! Y Y i!(i + 1)! × i! · (2i)! (2i + 2)! i=0 i=0 2r−2 ! µ 1 Y × (µ − 1) · 2 + r − 2 m−r · (µ + i − 1)2m+2r−2i−1 i=1 r−2 2 ! Y (2m − 2i − 3)! × 2 i=0 (m − i − 2)! (2m + 2i − 1)! (2m + 2i + 1)! m−r−1 b 2 c Y 2 2 × µ + 3i + 3r − 1 − µ − 3m + 3i + 3 2 2 m−r−2i−1 2 m−r−2i i=0
2 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
n odd n even
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t t
n even n odd
s s
3 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
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3 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
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3 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
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3 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
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3 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
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3 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
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3 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
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3 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
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3 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
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3 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
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3 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
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3 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
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3 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
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3 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
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3 / 32 For each path P , let ω(P ) be the product of its edge weights. Let X e(a, b) = ω(P ) and P :a→b e(a1, b1) e(a1, b2) ··· e(a1, bn) e(a2, b1) e(a2, b2) ··· e(a2, bn) M = . . . .. . . . . . e(an, b1) e(an, b2) ··· e(an, bn) Then the determinant of M is the signed sum over all n-tuples P = (P1,...,Pn) of non-intersecting paths from A to B: n X Y det(M) = sign(σ(P )) ω(Pi).
(P1,...,Pn): A→B i=1 where σ denotes a permutation that is applied to B.
Lindstr¨om-Gessel-ViennotLemma Let G be a directed acyclic graph and consider base vertices A = {a1, . . . , an} and destination vertices B = {b1, . . . , bn}.
4 / 32 Then the determinant of M is the signed sum over all n-tuples P = (P1,...,Pn) of non-intersecting paths from A to B: n X Y det(M) = sign(σ(P )) ω(Pi).
(P1,...,Pn): A→B i=1 where σ denotes a permutation that is applied to B.
Lindstr¨om-Gessel-ViennotLemma Let G be a directed acyclic graph and consider base vertices A = {a1, . . . , an} and destination vertices B = {b1, . . . , bn}. For each path P , let ω(P ) be the product of its edge weights. Let X e(a, b) = ω(P ) and P :a→b e(a1, b1) e(a1, b2) ··· e(a1, bn) e(a2, b1) e(a2, b2) ··· e(a2, bn) M = . . . .. . . . . . e(an, b1) e(an, b2) ··· e(an, bn)
4 / 32 Lindstr¨om-Gessel-ViennotLemma Let G be a directed acyclic graph and consider base vertices A = {a1, . . . , an} and destination vertices B = {b1, . . . , bn}. For each path P , let ω(P ) be the product of its edge weights. Let X e(a, b) = ω(P ) and P :a→b e(a1, b1) e(a1, b2) ··· e(a1, bn) e(a2, b1) e(a2, b2) ··· e(a2, bn) M = . . . .. . . . . . e(an, b1) e(an, b2) ··· e(an, bn) Then the determinant of M is the signed sum over all n-tuples P = (P1,...,Pn) of non-intersecting paths from A to B: n X Y det(M) = sign(σ(P )) ω(Pi).
(P1,...,Pn): A→B i=1 where σ denotes a permutation that is applied to B. 4 / 32 t + n − 1
. .
t + 1
t
µ + s − 2 ... µ + s + n − 3
Lindstr¨om-Gessel-ViennotLemma In our context, it implies that the determinant without the Kronecker delta µ + i + j + s + t − 4 det 16i,j6n j + t − 1 counts n-tuples of non-intersecting paths in the lattice N2:
5 / 32 Lindstr¨om-Gessel-ViennotLemma In our context, it implies that the determinant without the Kronecker delta µ + i + j + s + t − 4 det 16i,j6n j + t − 1 counts n-tuples of non-intersecting paths in the lattice N2:
t + n − 1
. .
t + 1
t
µ + s − 2 ... µ + s + n − 3
5 / 32 Lindstr¨om-Gessel-ViennotLemma In our context, it implies that the determinant without the Kronecker delta µ + i + j + s + t − 4 det 16i,j6n j + t − 1 counts n-tuples of non-intersecting paths in the lattice N2:
t + n − 1
. .
t + 1
t
µ + s − 2 ... µ + s + n − 3
5 / 32 Lattice Paths ←→ Rhombus Tilings
6 / 32 Lattice Paths ←→ Rhombus Tilings
6 / 32 Lattice Paths ←→ Rhombus Tilings
6 / 32 Lattice Paths ←→ Rhombus Tilings
6 / 32 By applying this procedure recursively, one obtains
X I = det BI+s−t (s > t), I⊆{1,...,n−s+t}
I where BJ denotes the matrix that is obtained by deleting all rows with indices in I and all columns with indices in J from the matrix µ + i + j + s + t − 4 B = j + t − 1 16i,j6n
Expansion of Kronecker Deltas
··· b1,6 b1,7 + 1 b1,8 ···
··· b b b + 1 ··· Laplace expansion: 2,6 2,7 2,8 = ......
··· b1,6 b1,7 b1,8 ··· ··· b2,6 b2,8 + 1 ··· ··· b b b + 1 ··· 2,6 2,7 2,8 ± ......
7 / 32 I where BJ denotes the matrix that is obtained by deleting all rows with indices in I and all columns with indices in J from the matrix µ + i + j + s + t − 4 B = j + t − 1 16i,j6n
Expansion of Kronecker Deltas
··· b1,6 b1,7 + 1 b1,8 ···
··· b b b + 1 ··· Laplace expansion: 2,6 2,7 2,8 = ......
··· b1,6 b1,7 b1,8 ··· ··· b2,6 b2,8 + 1 ··· ··· b b b + 1 ··· 2,6 2,7 2,8 ± ...... By applying this procedure recursively, one obtains
µ X (s−t)·|I| I Ds,t(n) = (−1) det BI+s−t (s > t), I⊆{1,...,n−s+t}
7 / 32 Expansion of Kronecker Deltas
··· b1,6 b1,7 + 1 b1,8 ···
··· b b b + 1 ··· Laplace expansion: 2,6 2,7 2,8 = ......
··· b1,6 b1,7 b1,8 ··· ··· b2,6 b2,8 + 1 ··· ··· b b b + 1 ··· 2,6 2,7 2,8 ± ...... By applying this procedure recursively, one obtains
µ X (s−t)·|I| I Ds,t(n) = (−1) det BI+s−t (s > t), I⊆{1,...,n−s+t}
I where BJ denotes the matrix that is obtained by deleting all rows with indices in I and all columns with indices in J from the matrix µ + i + j + s + t − 4 B = j + t − 1 16i,j6n 7 / 32 Expansion of Kronecker Deltas
··· b1,6 b1,7 + 1 b1,8 ···
··· b b b + 1 ··· Laplace expansion: 2,6 2,7 2,8 = ......
··· b1,6 b1,7 b1,8 ··· ··· b2,6 b2,8 + 1 ··· ··· b b b + 1 ··· 2,6 2,7 2,8 ± ...... By applying this procedure recursively, one obtains
µ X (s−t+1)·|I| I Es,t(n) = (−1) det BI+s−t (s > t), I⊆{1,...,n−s+t}
I where BJ denotes the matrix that is obtained by deleting all rows with indices in I and all columns with indices in J from the matrix µ + i + j + s + t − 4 B = j + t − 1 16i,j6n 7 / 32 Hexagonal Fusion s = 1 t = 3 n = 6 µ = 5
8 / 32 Hexagonal Fusion s = 1 t = 3 n = 6 µ = 5
8 / 32 Hexagonal Fusion s = 1 t = 3 n = 6 µ = 5
8 / 32 Hexagonal Fusion s = 1 t = 3 n = 6 µ = 5
8 / 32 Hexagonal Fusion s = 1 t = 3 n = 6 µ = 5
8 / 32 Hexagonal Fusion s = 1 t = 3 n = 6 µ = 5
8 / 32 Hexagonal Fusion s = 1 t = 3 n = 6 µ = 5
8 / 32 Combinatorial Interpretation: Holey Hexagon
s = 5 t = 7 n = 8 µ = 8
9 / 32 Combinatorial Interpretation: Holey Hexagon
s = 5 t = 7 n = 8 µ = 8
9 / 32 Combinatorial Interpretation: Holey Hexagon
s = 5 t = 7 n = 8 µ = 8
9 / 32 Proof (by example): 3 6 D2,0(4) E1,0(3)
A Combinatorial Proof Lemma: For n, s ∈ Z such that n > s > 1 and n > 1, µ µ+3 Es,0(n) = Ds−1,0(n − 1), µ µ+3 Ds,0(n) = Es−1,0(n − 1).
10 / 32 A Combinatorial Proof Lemma: For n, s ∈ Z such that n > s > 1 and n > 1, µ µ+3 Es,0(n) = Ds−1,0(n − 1), µ µ+3 Ds,0(n) = Es−1,0(n − 1).
Proof (by example): 3 6 D2,0(4) E1,0(3)
10 / 32 A Combinatorial Proof Lemma: For n, s ∈ Z such that n > s > 1 and n > 1, µ µ+3 Es,0(n) = Ds−1,0(n − 1), µ µ+3 Ds,0(n) = Es−1,0(n − 1).
Proof (by example): 3 6 D2,0(4) E1,0(3)
10 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
n odd n even
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11 / 32 Theorem (Desnanot-Jacobi-Dodgson). Let mi,j i,j∈Z and Ms,t(n) := dets6i
Ms,t(n)Ms+1,t+1(n − 2) =
Ms,t(n − 1)Ms+1,t+1(n − 1) − Ms+1,t(n − 1)Ms,t+1(n − 1).
Schematically:
× = × − ×
Switching Lemma and DJD µ µ µ Lemma: Let As,t(n) be either Ds,t(n) or Es,t(n). For real numbers s, t∈ / {−1, −2,...} with t − s ∈ N and n ∈ Z+,
t−s−1 Y µ + s + i − 1 Aµ (n) = n · Aµ (n). s,t i + s + 1 t,s i=0 n
12 / 32 Schematically:
× = × − ×
Switching Lemma and DJD µ µ µ Lemma: Let As,t(n) be either Ds,t(n) or Es,t(n). For real numbers s, t∈ / {−1, −2,...} with t − s ∈ N and n ∈ Z+,
t−s−1 Y µ + s + i − 1 Aµ (n) = n · Aµ (n). s,t i + s + 1 t,s i=0 n Theorem (Desnanot-Jacobi-Dodgson). Let mi,j i,j∈Z and Ms,t(n) := dets6i
Ms,t(n)Ms+1,t+1(n − 2) =
Ms,t(n − 1)Ms+1,t+1(n − 1) − Ms+1,t(n − 1)Ms,t+1(n − 1).
12 / 32 Switching Lemma and DJD µ µ µ Lemma: Let As,t(n) be either Ds,t(n) or Es,t(n). For real numbers s, t∈ / {−1, −2,...} with t − s ∈ N and n ∈ Z+,
t−s−1 Y µ + s + i − 1 Aµ (n) = n · Aµ (n). s,t i + s + 1 t,s i=0 n Theorem (Desnanot-Jacobi-Dodgson). Let mi,j i,j∈Z and Ms,t(n) := dets6i
Ms,t(n)Ms+1,t+1(n − 2) =
Ms,t(n − 1)Ms+1,t+1(n − 1) − Ms+1,t(n − 1)Ms,t+1(n − 1).
Schematically:
× = × − ×
12 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
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13 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
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13 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
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13 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
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13 / 32 Theorem: For m, r ∈ Z and m > r > 1, m−r µ (−1) (µ − 1) µ + 2r − 1 2m−2 E2r−1,1(2m − 1) = µ (2r − 2)! m + r − 1 m−r+1 2 + r m−r m−r 2 µ 2 Y µ + 2i + 6r − 5 2 + 2i + 3r − 2 × i−1 i . 2 µ 2 i=1 i i 2 + i + 3r − 2 i−1 µ Corollary: Apply switching lemma to obtain E1,2r−1(2m − 1).
Proof of the Lascoux/Krattenthaler Conjecture Lemma: For m, r ∈ Z and m > r > 1, µ D (2m) (m + r − 1)(µ − 1)(µ + 2m + 1)(µ + 2r) 2r,1 = , µ+3 2m(2r − 1)(µ + 2)(µ + 2m + 2r − 1) E2r−1,1(2m − 1) µ E (2m + 1) (m + r)(µ − 1)(µ + 2m + 2)(µ + 2r + 1) 2r+1,1 = . µ+3 2r(2m + 1)(µ + 2)(µ + 2m + 2r + 1) D2r,1 (2m)
14 / 32 µ Corollary: Apply switching lemma to obtain E1,2r−1(2m − 1).
Proof of the Lascoux/Krattenthaler Conjecture Lemma: For m, r ∈ Z and m > r > 1, µ D (2m) (m + r − 1)(µ − 1)(µ + 2m + 1)(µ + 2r) 2r,1 = , µ+3 2m(2r − 1)(µ + 2)(µ + 2m + 2r − 1) E2r−1,1(2m − 1) µ E (2m + 1) (m + r)(µ − 1)(µ + 2m + 2)(µ + 2r + 1) 2r+1,1 = . µ+3 2r(2m + 1)(µ + 2)(µ + 2m + 2r + 1) D2r,1 (2m)
Theorem: For m, r ∈ Z and m > r > 1, m−r µ (−1) (µ − 1) µ + 2r − 1 2m−2 E2r−1,1(2m − 1) = µ (2r − 2)! m + r − 1 m−r+1 2 + r m−r m−r 2 µ 2 Y µ + 2i + 6r − 5 2 + 2i + 3r − 2 × i−1 i . 2 µ 2 i=1 i i 2 + i + 3r − 2 i−1
14 / 32 Proof of the Lascoux/Krattenthaler Conjecture Lemma: For m, r ∈ Z and m > r > 1, µ D (2m) (m + r − 1)(µ − 1)(µ + 2m + 1)(µ + 2r) 2r,1 = , µ+3 2m(2r − 1)(µ + 2)(µ + 2m + 2r − 1) E2r−1,1(2m − 1) µ E (2m + 1) (m + r)(µ − 1)(µ + 2m + 2)(µ + 2r + 1) 2r+1,1 = . µ+3 2r(2m + 1)(µ + 2)(µ + 2m + 2r + 1) D2r,1 (2m)
Theorem: For m, r ∈ Z and m > r > 1, m−r µ (−1) (µ − 1) µ + 2r − 1 2m−2 E2r−1,1(2m − 1) = µ (2r − 2)! m + r − 1 m−r+1 2 + r m−r m−r 2 µ 2 Y µ + 2i + 6r − 5 2 + 2i + 3r − 2 × i−1 i . 2 µ 2 i=1 i i 2 + i + 3r − 2 i−1 µ Corollary: Apply switching lemma to obtain E1,2r−1(2m − 1). 14 / 32 Proof of the Lascoux/Krattenthaler Conjecture Lemma: For n, s ∈ Z and n > s > 1,
µ A (n) (n + s − 2)(µ − 1)(µ + n + 1)(µ + s) s,1 = , µ+3 2n(s − 1)(µ + 2)(µ + n + s − 1) Bs−1,1(n − 1) where (A, B, s, n) is (D,E, 2r, 2m) or (E,D, 2r + 1, 2m + 1).
Theorem: For m, r ∈ Z and m > r > 1, m−r µ (−1) (µ − 1) µ + 2r − 1 2m−2 E2r−1,1(2m − 1) = µ (2r − 2)! m + r − 1 m−r+1 2 + r m−r m−r 2 µ 2 Y µ + 2i + 6r − 5 2 + 2i + 3r − 2 × i−1 i . 2 µ 2 i=1 i i 2 + i + 3r − 2 i−1 µ Corollary: Apply switching lemma to obtain E1,2r−1(2m − 1). 14 / 32 ∗ ∗ ∗ ∗ µ 1 L·D2,1(4) ·R = µ+3 1 E1,1 (3) 1
Matrix Transformations
1 0 0 0 1 0 0 0 0 1 0 0 µ 0 1 0 0 ·D (4) · = 0 0 1 0 2,1 0 0 1 0 0 0 0 1 0 0 0 1
µ+1 µ+2 µ+3 µ+4 1 2 + 1 3 4 µ+2 µ+3 µ+4 + 1 µ+5 1 2 3 4 µ+3 µ+4 µ+5 µ+6 + 1 1 2 3 4 µ+4 µ+5 µ+6 µ+7 1 2 3 4
15 / 32 ∗ ∗ ∗ ∗ µ 1 L·D2,1(4) ·R = µ+3 1 E1,1 (3) 1
Matrix Transformations
1 0 0 0 1 0 0 0 −1 1 0 0 µ 0 1 0 0 ·D (4) · = 0 −1 1 0 2,1 0 0 1 0 0 0 −1 1 0 0 0 1
µ+1 µ+2 µ+3 µ+4 1 2 + 1 3 4 µ+2− µ+1 µ+3− µ+2 − 1 µ+4− µ+3 + 1 µ+5− µ+4 1 1 2 2 3 3 4 4 µ+3− µ+2 µ+4− µ+3 µ+5− µ+4 − 1 µ+6− µ+5 + 1 1 1 2 2 3 3 4 4 µ+4 µ+3 µ+5 µ+4 µ+6 µ+5 µ+7 µ+6 1 − 1 2 − 2 3 − 3 4 − 4 − 1
15 / 32 ∗ ∗ ∗ ∗ µ 1 L·D2,1(4) ·R = µ+3 1 E1,1 (3) 1
Matrix Transformations
1 0 0 0 1 0 0 0 −1 1 0 0 µ 0 1 0 0 ·D (4) · = 0 −1 1 0 2,1 0 0 1 0 0 0 −1 1 0 0 0 1
µ+1 µ+2 µ+3 µ+4 1 2 + 1 3 4 µ+1 µ+2 − 1 µ+3 + 1 µ+4 0 1 2 3 µ+2 µ+3 µ+4 − 1 µ+5 + 1 0 1 2 3 µ+3 µ+4 µ+5 µ+6 0 1 2 3 − 1
15 / 32 ∗ ∗ ∗ ∗ µ 1 L·D2,1(4) ·R = µ+3 1 E1,1 (3) 1
Matrix Transformations
1 0 0 0 1 1 0 0 −1 1 0 0 µ 0 1 0 0 ·D (4) · = 0 −1 1 0 2,1 0 0 1 0 0 0 −1 1 0 0 0 1
µ+1 µ+2 µ+1 µ+3 µ+4 1 2 + 1 + 1 3 4 µ+1 µ+2+ µ+1 − 1 µ+3 + 1 µ+4 0 1 0 2 3 µ+2 µ+3+ µ+2 µ+4 − 1 µ+5 + 1 0 1 0 2 3 µ+3 µ+4 µ+3 µ+5 µ+6 0 1 + 0 2 3 − 1
15 / 32 ∗ ∗ ∗ ∗ µ 1 L·D2,1(4) ·R = µ+3 1 E1,1 (3) 1
Matrix Transformations
1 0 0 0 1 1 0 0 −1 1 0 0 µ 0 1 0 0 ·D (4) · = 0 −1 1 0 2,1 0 0 1 0 0 0 −1 1 0 0 0 1
µ+1 µ+2 µ+2 µ+3 µ+4 1 2 + 1 3 4 µ+1 µ+2+ µ+2 − 1 µ+3 + 1 µ+4 0 1 0 2 3 µ+2 µ+3+ µ+3 µ+4 − 1 µ+5 + 1 0 1 0 2 3 µ+3 µ+4 µ+4 µ+5 µ+6 0 1 + 0 2 3 − 1
15 / 32 ∗ ∗ ∗ ∗ µ 1 L·D2,1(4) ·R = µ+3 1 E1,1 (3) 1
Matrix Transformations
1 0 0 0 1 1 0 0 −1 1 0 0 µ 0 1 0 0 ·D (4) · = 0 −1 1 0 2,1 0 0 1 0 0 0 −1 1 0 0 0 1
µ+1 µ+3 µ+3 µ+4 1 2 3 4 µ+1 µ+3 − 1 µ+3 + 1 µ+4 0 1 2 3 µ+2 µ+4 µ+4 − 1 µ+5 + 1 0 1 2 3 µ+3 µ+5 µ+5 µ+6 0 1 2 3 − 1
15 / 32 ∗ ∗ ∗ ∗ µ 1 L·D2,1(4) ·R = µ+3 1 E1,1 (3) 1
Matrix Transformations
1 0 0 0 1 1 1 0 −1 1 0 0 µ 0 1 1 0 ·D (4) · = 0 −1 1 0 2,1 0 0 1 0 0 0 −1 1 0 0 0 1
µ+1 µ+3 µ+3 µ+3 µ+4 1 2 3 + 2 4 µ+1 µ+3 − 1 µ+3+ µ+3 µ+4 0 1 2 1 3 µ+2 µ+4 µ+4+ µ+4 − 1 µ+5 + 1 0 1 2 1 3 µ+3 µ+5 µ+5 µ+5 µ+6 0 1 2 + 1 3 − 1
15 / 32 ∗ ∗ ∗ ∗ µ 1 L·D2,1(4) ·R = µ+3 1 E1,1 (3) 1
Matrix Transformations
1 0 0 0 1 1 1 0 −1 1 0 0 µ 0 1 1 0 ·D (4) · = 0 −1 1 0 2,1 0 0 1 0 0 0 −1 1 0 0 0 1
µ+1 µ+3 µ+4 µ+4 1 2 3 4 µ+1 µ+3 − 1 µ+4 µ+4 0 1 2 3 µ+2 µ+4 µ+5 − 1 µ+5 + 1 0 1 2 3 µ+3 µ+5 µ+6 µ+6 0 1 2 3 − 1
15 / 32 ∗ ∗ ∗ ∗ µ 1 L·D2,1(4) ·R = µ+3 1 E1,1 (3) 1
Matrix Transformations
1 0 0 0 1 1 1 1 −1 1 0 0 µ 0 1 1 1 ·D (4) · = 0 −1 1 0 2,1 0 0 1 1 0 0 −1 1 0 0 0 1
µ+1 µ+3 µ+4 µ+4 µ+4 1 2 3 4 + 3 µ+1 µ+3 − 1 µ+4 µ+4+ µ+4 0 1 2 3 2 µ+2 µ+4 µ+5 − 1 µ+5+ µ+5 0 1 2 3 2 µ+3 µ+5 µ+6 µ+6 µ+6 0 1 2 3 + 2 − 1
15 / 32 ∗ ∗ ∗ ∗ µ 1 L·D2,1(4) ·R = µ+3 1 E1,1 (3) 1
Matrix Transformations
1 0 0 0 1 1 1 1 −1 1 0 0 µ 0 1 1 1 ·D (4) · = 0 −1 1 0 2,1 0 0 1 1 0 0 −1 1 0 0 0 1
µ+1 µ+3 µ+4 µ+5 1 2 3 4 µ+1 µ+3 − 1 µ+4 µ+5 0 1 2 3 µ+2 µ+4 µ+5 − 1 µ+6 0 1 2 3 µ+3 µ+5 µ+6 µ+7 0 1 2 3 − 1
15 / 32 Matrix Transformations
1 0 0 0 1 1 1 1 −1 1 0 0 µ 0 1 1 1 ·D (4) · = 0 −1 1 0 2,1 0 0 1 1 0 0 −1 1 0 0 0 1
µ+1 µ+3 µ+4 µ+5 1 2 3 4 µ+1 µ+3 − 1 µ+4 µ+5 0 1 2 3 µ+2 µ+4 µ+5 − 1 µ+6 0 1 2 3 µ+3 µ+5 µ+6 µ+7 0 1 2 3 − 1
∗ ∗ ∗ ∗ µ 1 L·D2,1(4) ·R = µ+3 1 E1,1 (3) 1
15 / 32 ! µ = Rs,1(n).
Laplace expansion: a˜1,1 a˜1,2 ··· a˜1,n µ 1 As,1(n) = det µ+3 1 Bs−1,1(n − 1) 1
=a ˜1,1 · Cof1,1(n − 1) + ... +a ˜1,n · Cof1,n(n − 1).
With cn,j := Cof1,j(n − 1)/Cof1,1(n − 1), we obtain µ n As,1(n) X µ+3 = a˜1,j · cn,j Bs−1,1(n − 1) j=1
Proof via the Holonomic Ansatz To show: µ A (n) (n + s − 2)(µ − 1)(µ + n + 1)(µ + s) s,1 = µ+3 2n(s − 1)(µ + 2)(µ + n + s − 1) Bs−1,1(n − 1) | {zµ } =:Rs,1(n)
16 / 32 ! µ = Rs,1(n).
With cn,j := Cof1,j(n − 1)/Cof1,1(n − 1), we obtain µ n As,1(n) X µ+3 = a˜1,j · cn,j Bs−1,1(n − 1) j=1
Proof via the Holonomic Ansatz To show: µ A (n) (n + s − 2)(µ − 1)(µ + n + 1)(µ + s) s,1 = µ+3 2n(s − 1)(µ + 2)(µ + n + s − 1) Bs−1,1(n − 1) | {zµ } =:Rs,1(n) Laplace expansion: a˜1,1 a˜1,2 ··· a˜1,n µ 1 As,1(n) = det µ+3 1 Bs−1,1(n − 1) 1
=a ˜1,1 · Cof1,1(n − 1) + ... +a ˜1,n · Cof1,n(n − 1).
16 / 32 ! µ = Rs,1(n).
Proof via the Holonomic Ansatz To show: µ A (n) (n + s − 2)(µ − 1)(µ + n + 1)(µ + s) s,1 = µ+3 2n(s − 1)(µ + 2)(µ + n + s − 1) Bs−1,1(n − 1) | {zµ } =:Rs,1(n) Laplace expansion: a˜1,1 a˜1,2 ··· a˜1,n µ 1 As,1(n) = det µ+3 1 Bs−1,1(n − 1) 1
=a ˜1,1 · Cof1,1(n − 1) + ... +a ˜1,n · Cof1,n(n − 1).
With cn,j := Cof1,j(n − 1)/Cof1,1(n − 1), we obtain µ n As,1(n) X µ+3 = a˜1,j · cn,j Bs−1,1(n − 1) j=1 16 / 32 Proof via the Holonomic Ansatz To show: µ A (n) (n + s − 2)(µ − 1)(µ + n + 1)(µ + s) s,1 = µ+3 2n(s − 1)(µ + 2)(µ + n + s − 1) Bs−1,1(n − 1) | {zµ } =:Rs,1(n) Laplace expansion: a˜1,1 a˜1,2 ··· a˜1,n µ 1 As,1(n) = det µ+3 1 Bs−1,1(n − 1) 1
=a ˜1,1 · Cof1,1(n − 1) + ... +a ˜1,n · Cof1,n(n − 1).
With cn,j := Cof1,j(n − 1)/Cof1,1(n − 1), we obtain µ n As,1(n) X ! µ µ+3 = a˜1,j · cn,j = Rs,1(n). Bs−1,1(n − 1) j=1 16 / 32 These recurrences can be constructed from precomputed data via ansatz with undetermined coefficients and interpolation.
[1] [1] [1] [1] p0,2 · cn,j+2 + p1,0 · cn+1,j + p0,1 · cn,j+1 + p0,0 · cn,j = 0 [2] [2] [2] [2] p1,1 · cn+1,j+1 + p1,0 · cn+1,j + p0,1 · cn,j+1 + p0,0 · cn,j = 0 [3] [3] [3] [3] p2,0 · cn+2,j + p1,0 · cn+1,j + p0,1 · cn,j+1 + p0,0 · cn,j = 0
[3] p2,0 = −(j − 2n − 4)(j − 2n − 3)(µ + 6n + 5)(µ + 6n + 7)(µ + 6n + 9)(n + r − 1)(n + r)(j + µ + 2n + 3)(j + µ + 2n + 4)(2j4 + 3j3µ − 6j3n + j3 + j2µ2 − 12j2µn − 3j2µ + 12j2n2 − 30j2n − 8j2 − 4jµ2n − 2jµ2 + 24jµn2 − 8jµn − 6jµ + 72jn2 + 12jn − 4j + 8µ2n2 + 4µ2n + 40µn2 + 20µn + 48n2 + 24n)(µ + 2n + 2r)(µ + 2n + 2r + 1)(µ + 2n + 2r + 2)(µ + 2n + 2r + 3)(µ + 4n + 2r + 1)
Proof via the Holonomic Ansatz
Guess: cn,j satisfies a holonomic system of recurrence equations.
17 / 32 [1] [1] [1] [1] p0,2 · cn,j+2 + p1,0 · cn+1,j + p0,1 · cn,j+1 + p0,0 · cn,j = 0 [2] [2] [2] [2] p1,1 · cn+1,j+1 + p1,0 · cn+1,j + p0,1 · cn,j+1 + p0,0 · cn,j = 0 [3] [3] [3] [3] p2,0 · cn+2,j + p1,0 · cn+1,j + p0,1 · cn,j+1 + p0,0 · cn,j = 0
[3] p2,0 = −(j − 2n − 4)(j − 2n − 3)(µ + 6n + 5)(µ + 6n + 7)(µ + 6n + 9)(n + r − 1)(n + r)(j + µ + 2n + 3)(j + µ + 2n + 4)(2j4 + 3j3µ − 6j3n + j3 + j2µ2 − 12j2µn − 3j2µ + 12j2n2 − 30j2n − 8j2 − 4jµ2n − 2jµ2 + 24jµn2 − 8jµn − 6jµ + 72jn2 + 12jn − 4j + 8µ2n2 + 4µ2n + 40µn2 + 20µn + 48n2 + 24n)(µ + 2n + 2r)(µ + 2n + 2r + 1)(µ + 2n + 2r + 2)(µ + 2n + 2r + 3)(µ + 4n + 2r + 1)
Proof via the Holonomic Ansatz
Guess: cn,j satisfies a holonomic system of recurrence equations. These recurrences can be constructed from precomputed data via ansatz with undetermined coefficients and interpolation.
17 / 32 [3] p2,0 = −(j − 2n − 4)(j − 2n − 3)(µ + 6n + 5)(µ + 6n + 7)(µ + 6n + 9)(n + r − 1)(n + r)(j + µ + 2n + 3)(j + µ + 2n + 4)(2j4 + 3j3µ − 6j3n + j3 + j2µ2 − 12j2µn − 3j2µ + 12j2n2 − 30j2n − 8j2 − 4jµ2n − 2jµ2 + 24jµn2 − 8jµn − 6jµ + 72jn2 + 12jn − 4j + 8µ2n2 + 4µ2n + 40µn2 + 20µn + 48n2 + 24n)(µ + 2n + 2r)(µ + 2n + 2r + 1)(µ + 2n + 2r + 2)(µ + 2n + 2r + 3)(µ + 4n + 2r + 1)
Proof via the Holonomic Ansatz
Guess: cn,j satisfies a holonomic system of recurrence equations. These recurrences can be constructed from precomputed data via ansatz with undetermined coefficients and interpolation.
[1] [1] [1] [1] p0,2 · cn,j+2 + p1,0 · cn+1,j + p0,1 · cn,j+1 + p0,0 · cn,j = 0 [2] [2] [2] [2] p1,1 · cn+1,j+1 + p1,0 · cn+1,j + p0,1 · cn,j+1 + p0,0 · cn,j = 0 [3] [3] [3] [3] p2,0 · cn+2,j + p1,0 · cn+1,j + p0,1 · cn,j+1 + p0,0 · cn,j = 0
17 / 32 Proof via the Holonomic Ansatz
Guess: cn,j satisfies a holonomic system of recurrence equations. These recurrences can be constructed from precomputed data via ansatz with undetermined coefficients and interpolation.
[1] [1] [1] [1] p0,2 · cn,j+2 + p1,0 · cn+1,j + p0,1 · cn,j+1 + p0,0 · cn,j = 0 [2] [2] [2] [2] p1,1 · cn+1,j+1 + p1,0 · cn+1,j + p0,1 · cn,j+1 + p0,0 · cn,j = 0 [3] [3] [3] [3] p2,0 · cn+2,j + p1,0 · cn+1,j + p0,1 · cn,j+1 + p0,0 · cn,j = 0
[3] p2,0 = −(j − 2n − 4)(j − 2n − 3)(µ + 6n + 5)(µ + 6n + 7)(µ + 6n + 9)(n + r − 1)(n + r)(j + µ + 2n + 3)(j + µ + 2n + 4)(2j4 + 3j3µ − 6j3n + j3 + j2µ2 − 12j2µn − 3j2µ + 12j2n2 − 30j2n − 8j2 − 4jµ2n − 2jµ2 + 24jµn2 − 8jµn − 6jµ + 72jn2 + 12jn − 4j + 8µ2n2 + 4µ2n + 40µn2 + 20µn + 48n2 + 24n)(µ + 2n + 2r)(µ + 2n + 2r + 1)(µ + 2n + 2r + 2)(µ + 2n + 2r + 3)(µ + 4n + 2r + 1)
17 / 32 c2m,1 = 1, 2m X µ + i + j + 2r − 3 · c − c = 0, (2 i 2m), j − 1 2m,j 2m,i+2r−2 6 6 j=1
Cof1,j (n−1) I Abandon the original definition cn,j := . Cof1,1(n−1) I Use the conjectured holonomic description for cn,j instead.
Cof1,j (n−1) I The first two identities prove c = . n,j Cof1,1(n−1) I The third identity proves the claimed quotient of determinants.
Proof via the Holonomic Ansatz Prove: in the case where (A, B, s, n) is (D,E, 2r, 2m)
2m 2r−1 X µ + j + 2r − 1 X · c − c = Rµ (2m). j 2m,j 2m,j 2r,1 j=1 j=1
18 / 32 c2m,1 = 1, 2m X µ + i + j + 2r − 3 · c − c = 0, (2 i 2m), j − 1 2m,j 2m,i+2r−2 6 6 j=1
I Use the conjectured holonomic description for cn,j instead.
Cof1,j (n−1) I The first two identities prove c = . n,j Cof1,1(n−1) I The third identity proves the claimed quotient of determinants.
Proof via the Holonomic Ansatz Prove: in the case where (A, B, s, n) is (D,E, 2r, 2m)
2m 2r−1 X µ + j + 2r − 1 X · c − c = Rµ (2m). j 2m,j 2m,j 2r,1 j=1 j=1
Cof1,j (n−1) I Abandon the original definition cn,j := . Cof1,1(n−1)
18 / 32 c2m,1 = 1, 2m X µ + i + j + 2r − 3 · c − c = 0, (2 i 2m), j − 1 2m,j 2m,i+2r−2 6 6 j=1
Cof1,j (n−1) I The first two identities prove c = . n,j Cof1,1(n−1) I The third identity proves the claimed quotient of determinants.
Proof via the Holonomic Ansatz Prove: in the case where (A, B, s, n) is (D,E, 2r, 2m)
2m 2r−1 X µ + j + 2r − 1 X · c − c = Rµ (2m). j 2m,j 2m,j 2r,1 j=1 j=1
Cof1,j (n−1) I Abandon the original definition cn,j := . Cof1,1(n−1) I Use the conjectured holonomic description for cn,j instead.
18 / 32 I The third identity proves the claimed quotient of determinants.
Proof via the Holonomic Ansatz Prove: in the case where (A, B, s, n) is (D,E, 2r, 2m)
c2m,1 = 1, 2m X µ + i + j + 2r − 3 · c − c = 0, (2 i 2m), j − 1 2m,j 2m,i+2r−2 6 6 j=1 2m 2r−1 X µ + j + 2r − 1 X · c − c = Rµ (2m). j 2m,j 2m,j 2r,1 j=1 j=1
Cof1,j (n−1) I Abandon the original definition cn,j := . Cof1,1(n−1) I Use the conjectured holonomic description for cn,j instead.
Cof1,j (n−1) I The first two identities prove c = . n,j Cof1,1(n−1)
18 / 32 Proof via the Holonomic Ansatz Prove: in the case where (A, B, s, n) is (D,E, 2r, 2m)
c2m,1 = 1, 2m X µ + i + j + 2r − 3 · c − c = 0, (2 i 2m), j − 1 2m,j 2m,i+2r−2 6 6 j=1 2m 2r−1 X µ + j + 2r − 1 X · c − c = Rµ (2m). j 2m,j 2m,j 2r,1 j=1 j=1
Cof1,j (n−1) I Abandon the original definition cn,j := . Cof1,1(n−1) I Use the conjectured holonomic description for cn,j instead.
Cof1,j (n−1) I The first two identities prove c = . n,j Cof1,1(n−1) I The third identity proves the claimed quotient of determinants.
18 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
n odd n even
s s
t t
n even n odd
s s
19 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
n odd n even
s s
t t
n even n odd
s s
19 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
n odd n even
s s
t t
n even n odd
s s
19 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
n odd n even
s s
t t
n even n odd
s s
19 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
n odd n even
s s
t t
n even n odd
s s
19 / 32 Lemma: For m, r ∈ Z and m > r > 1, 2r(2m − 1)(µ − 3)(µ + 2m + 2r − 2) lim = , ε→0 µ(m + r)(µ + 2m − 3)(µ + 2r − 2)
2m(2r + 1)(µ − 3)(µ + 2m + 2r) lim = . ε→0 µ(m + r + 1)(µ + 2m − 2)(µ + 2r − 1)
Proof of a Conjecture for s = −1
Theorem: For m, r ∈ Z and m > r > 1, µ E−1,2r−1(2m + 1) = m−r 2m ! (−1) (3 − µ)(m + r + 1)m−r Y (µ + i − 3)2r · 22m−2r+1 µ + r − 3 (i) 2 2 m−r+1 i=1 2r m−r 2 µ 2 ! Y µ + 2i + 6r − 3 2 + 2i + 3r − 1 × i i−1 . 2 µ 2 i=1 i i 2 + i + 3r − 1 i−1 20 / 32 Lemma: For m, r ∈ Z and m > r > 1, 2r(2m − 1)(µ − 3)(µ + 2m + 2r − 2) lim = , ε→0 µ(m + r)(µ + 2m − 3)(µ + 2r − 2)
2m(2r + 1)(µ − 3)(µ + 2m + 2r) lim = . ε→0 µ(m + r + 1)(µ + 2m − 2)(µ + 2r − 1)
Proof of a Conjecture for s = −1
µ D−1,2r(2m) µ+3 E−1,2r−1(2m − 1) µ E−1,2r+1(2m + 1) µ+3 D−1,2r(2m)
Theorem: For m, r ∈ Z and m > r > 1, µ E−1,2r−1(2m + 1) = m−r 2m ! (−1) (3 − µ)(m + r + 1)m−r Y (µ + i − 3)2r · 22m−2r+1 µ + r − 3 (i) 2 2 m−r+1 i=1 2r m−r 2 µ 2 ! Y µ + 2i + 6r − 3 2 + 2i + 3r − 1 × i i−1 . 2 µ 2 i=1 i i 2 + i + 3r − 1 i−1 20 / 32 Lemma: For m, r ∈ Z and m > r > 1, 2r(2m − 1)(µ − 3)(µ + 2m + 2r − 2) lim = , ε→0 µ(m + r)(µ + 2m − 3)(µ + 2r − 2)
2m(2r + 1)(µ − 3)(µ + 2m + 2r) lim = . ε→0 µ(m + r + 1)(µ + 2m − 2)(µ + 2r − 1)
Proof of a Conjecture for s = −1
µ D2r,−1(2m) µ+3 E2r−1,−1(2m − 1) µ E2r+1,−1(2m + 1) µ+3 D2r,−1(2m)
Theorem: For m, r ∈ Z and m > r > 1, µ E−1,2r−1(2m + 1) = m−r 2m ! (−1) (3 − µ)(m + r + 1)m−r Y (µ + i − 3)2r · 22m−2r+1 µ + r − 3 (i) 2 2 m−r+1 i=1 2r m−r 2 µ 2 ! Y µ + 2i + 6r − 3 2 + 2i + 3r − 1 × i i−1 . 2 µ 2 i=1 i i 2 + i + 3r − 1 i−1 20 / 32 Lemma: For m, r ∈ Z and m > r > 1, 2r(2m − 1)(µ − 3)(µ + 2m + 2r − 2) lim , ε→0 µ(m + r)(µ + 2m − 3)(µ + 2r − 2)
2m(2r + 1)(µ − 3)(µ + 2m + 2r) lim . ε→0 µ(m + r + 1)(µ + 2m − 2)(µ + 2r − 1)
Proof of a Conjecture for s = −1
µ D (2m) 0 2r,−1 = µ+3 0 E2r−1,−1(2m − 1) µ E (2m + 1) 0 2r+1,−1 = µ+3 0 D2r,−1(2m)
Theorem: For m, r ∈ Z and m > r > 1, µ E−1,2r−1(2m + 1) = m−r 2m ! (−1) (3 − µ)(m + r + 1)m−r Y (µ + i − 3)2r · 22m−2r+1 µ + r − 3 (i) 2 2 m−r+1 i=1 2r m−r 2 µ 2 ! Y µ + 2i + 6r − 3 2 + 2i + 3r − 1 × i i−1 . 2 µ 2 i=1 i i 2 + i + 3r − 1 i−1 20 / 32 Lemma: For m, r ∈ Z and m > r > 1, 2r(2m − 1)(µ − 3)(µ + 2m + 2r − 2) lim , ε→0 µ(m + r)(µ + 2m − 3)(µ + 2r − 2)
2m(2r + 1)(µ − 3)(µ + 2m + 2r) lim . ε→0 µ(m + r + 1)(µ + 2m − 2)(µ + 2r − 1)
Proof of a Conjecture for s = −1
µ D2r+ε,−1+ε(2m) µ+3 = E2r−1+ε,−1+ε(2m − 1) µ E2r+1+ε,−1+ε(2m + 1) µ+3 = D2r+ε,−1+ε(2m)
Theorem: For m, r ∈ Z and m > r > 1, µ E−1,2r−1(2m + 1) = m−r 2m ! (−1) (3 − µ)(m + r + 1)m−r Y (µ + i − 3)2r · 22m−2r+1 µ + r − 3 (i) 2 2 m−r+1 i=1 2r m−r 2 µ 2 ! Y µ + 2i + 6r − 3 2 + 2i + 3r − 1 × i i−1 . 2 µ 2 i=1 i i 2 + i + 3r − 1 i−1 20 / 32 Proof of a Conjecture for s = −1 Lemma: For m, r ∈ Z and m > r > 1, µ D (2m) 2r(2m − 1)(µ − 3)(µ + 2m + 2r − 2) lim 2r+ε,−1+ε = , ε→0 µ+3 µ(m + r)(µ + 2m − 3)(µ + 2r − 2) E2r−1+ε,−1+ε(2m − 1) µ E (2m + 1) 2m(2r + 1)(µ − 3)(µ + 2m + 2r) lim 2r+1+ε,−1+ε = . ε→0 µ+3 µ(m + r + 1)(µ + 2m − 2)(µ + 2r − 1) D2r+ε,−1+ε(2m)
Theorem: For m, r ∈ Z and m > r > 1, µ E−1,2r−1(2m + 1) = m−r 2m ! (−1) (3 − µ)(m + r + 1)m−r Y (µ + i − 3)2r · 22m−2r+1 µ + r − 3 (i) 2 2 m−r+1 i=1 2r m−r 2 µ 2 ! Y µ + 2i + 6r − 3 2 + 2i + 3r − 1 × i i−1 . 2 µ 2 i=1 i i 2 + i + 3r − 1 i−1 20 / 32 n X = a˜ · c + O(ε) i,1 n,i Btr i=2 =
µ row/column adapt 2nd ε = 0 Taylor tr As+ε,−1+ε(n) LAR ˜ ˜ operations column A truncation A
det det det det
linearity of µ = 2 = 2 As+ε,−1+ε(n) A A + O(ε ) A + O(ε ) determinant mod ε2
Laplace n ε · Cofi,1(n − 1) X cn,i := lim a˜i,1 · Cofi,1(n − 1) ε→0 Cof1,1(n − 1) i=2
Proof Layout
µ As+ε,−1+ε(n) µ+3 Bs−1+ε,−1+ε(n − 1)
21 / 32 n X = a˜ · c + O(ε) i,1 n,i Btr i=2 =
row/column adapt 2nd ε = 0 Taylor LAR ˜ ˜tr operations column A truncation A
det det det
= linearity of = A A + O(ε2) A + O(ε2) determinant mod ε2
Laplace n ε · Cofi,1(n − 1) X cn,i := lim a˜i,1 · Cofi,1(n − 1) ε→0 Cof1,1(n − 1) i=2
Proof Layout
µ As+ε,−1+ε(n) µ+3 Bs−1+ε,−1+ε(n − 1)
µ As+ε,−1+ε(n)
det
µ As+ε,−1+ε(n)
21 / 32 n X = a˜ · c + O(ε) i,1 n,i Btr i=2 =
adapt 2nd ε = 0 Taylor ˜ ˜tr column A truncation A
det det det
= linearity of = A A + O(ε2) A + O(ε2) determinant mod ε2
Laplace n ε · Cofi,1(n − 1) X cn,i := lim a˜i,1 · Cofi,1(n − 1) ε→0 Cof1,1(n − 1) i=2
Proof Layout
µ As+ε,−1+ε(n) µ+3 Bs−1+ε,−1+ε(n − 1)
µ row/column As+ε,−1+ε(n) LAR operations
det
µ As+ε,−1+ε(n)
21 / 32 n X = a˜ · c + O(ε) i,1 n,i Btr i=2 =
adapt 2nd ε = 0 Taylor ˜ ˜tr column A truncation A
det det
linearity of = A + O(ε2) A + O(ε2) determinant mod ε2
Laplace n ε · Cofi,1(n − 1) X cn,i := lim a˜i,1 · Cofi,1(n − 1) ε→0 Cof1,1(n − 1) i=2
Proof Layout
µ As+ε,−1+ε(n) µ+3 Bs−1+ε,−1+ε(n − 1)
µ row/column As+ε,−1+ε(n) LAR operations
det det
µ = As+ε,−1+ε(n) A
21 / 32 n X = a˜ · c + O(ε) i,1 n,i Btr i=2 =
ε = 0 Taylor ˜tr truncation A
det det
linearity of = A + O(ε2) A + O(ε2) determinant mod ε2
Laplace n ε · Cofi,1(n − 1) X cn,i := lim a˜i,1 · Cofi,1(n − 1) ε→0 Cof1,1(n − 1) i=2
Proof Layout
µ As+ε,−1+ε(n) µ+3 Bs−1+ε,−1+ε(n − 1)
µ row/column adapt 2nd As+ε,−1+ε(n) LAR ˜ operations column A
det det
µ = As+ε,−1+ε(n) A
21 / 32 n X = a˜ · c + O(ε) i,1 n,i Btr i=2 =
ε = 0 Taylor ˜tr truncation A
det = A + O(ε2) mod ε2
Laplace n ε · Cofi,1(n − 1) X cn,i := lim a˜i,1 · Cofi,1(n − 1) ε→0 Cof1,1(n − 1) i=2
Proof Layout
µ As+ε,−1+ε(n) µ+3 Bs−1+ε,−1+ε(n − 1)
µ row/column adapt 2nd As+ε,−1+ε(n) LAR ˜ operations column A
det det det
linearity of µ = 2 As+ε,−1+ε(n) A A + O(ε ) determinant
21 / 32 n X = a˜ · c + O(ε) i,1 n,i Btr i=2 =
det = A + O(ε2) mod ε2
Laplace n ε · Cofi,1(n − 1) X cn,i := lim a˜i,1 · Cofi,1(n − 1) ε→0 Cof1,1(n − 1) i=2
Proof Layout
µ As+ε,−1+ε(n) µ+3 Bs−1+ε,−1+ε(n − 1)
µ row/column adapt 2nd ε = 0 Taylor tr As+ε,−1+ε(n) LAR ˜ ˜ operations column A truncation A
det det det
linearity of µ = 2 As+ε,−1+ε(n) A A + O(ε ) determinant
21 / 32 n X = a˜ · c + O(ε) i,1 n,i Btr i=2 =
Laplace n ε · Cofi,1(n − 1) X cn,i := lim a˜i,1 · Cofi,1(n − 1) ε→0 Cof1,1(n − 1) i=2
Proof Layout
µ As+ε,−1+ε(n) µ+3 Bs−1+ε,−1+ε(n − 1)
µ row/column adapt 2nd ε = 0 Taylor tr As+ε,−1+ε(n) LAR ˜ ˜ operations column A truncation A
det det det det
linearity of µ = 2 = 2 As+ε,−1+ε(n) A A + O(ε ) A + O(ε ) determinant mod ε2
21 / 32 n X = a˜i,1 · cn,i + O(ε) i=2
Laplace n ε · Cofi,1(n − 1) X cn,i := lim a˜i,1 · Cofi,1(n − 1) ε→0 Cof1,1(n − 1) i=2
Proof Layout
µ As+ε,−1+ε(n) µ+3 Btr Bs−1+ε,−1+ε(n − 1) =
µ row/column adapt 2nd ε = 0 Taylor tr As+ε,−1+ε(n) LAR ˜ ˜ operations column A truncation A
det det det det
linearity of µ = 2 = 2 As+ε,−1+ε(n) A A + O(ε ) A + O(ε ) determinant mod ε2
21 / 32 n X = a˜i,1 · cn,i + O(ε) i=2
ε · Cofi,1(n − 1) cn,i := lim ε→0 Cof1,1(n − 1)
Proof Layout
µ As+ε,−1+ε(n) µ+3 Btr Bs−1+ε,−1+ε(n − 1) =
µ row/column adapt 2nd ε = 0 Taylor tr As+ε,−1+ε(n) LAR ˜ ˜ operations column A truncation A
det det det det
linearity of µ = 2 = 2 As+ε,−1+ε(n) A A + O(ε ) A + O(ε ) determinant mod ε2
Laplace n X a˜i,1 · Cofi,1(n − 1) i=2
21 / 32 n X = a˜i,1 · cn,i + O(ε) i=2
Proof Layout
µ As+ε,−1+ε(n) µ+3 Btr Bs−1+ε,−1+ε(n − 1) =
µ row/column adapt 2nd ε = 0 Taylor tr As+ε,−1+ε(n) LAR ˜ ˜ operations column A truncation A
det det det det
linearity of µ = 2 = 2 As+ε,−1+ε(n) A A + O(ε ) A + O(ε ) determinant mod ε2
Laplace n ε · Cofi,1(n − 1) X cn,i := lim a˜i,1 · Cofi,1(n − 1) ε→0 Cof1,1(n − 1) i=2
21 / 32 Proof Layout µ n As+ε,−1+ε(n) X = a˜ · c + O(ε) µ+3 i,1 n,i Btr Bs−1+ε,−1+ε(n − 1) i=2 =
µ row/column adapt 2nd ε = 0 Taylor tr As+ε,−1+ε(n) LAR ˜ ˜ operations column A truncation A
det det det det
linearity of µ = 2 = 2 As+ε,−1+ε(n) A A + O(ε ) A + O(ε ) determinant mod ε2
Laplace n ε · Cofi,1(n − 1) X cn,i := lim a˜i,1 · Cofi,1(n − 1) ε→0 Cof1,1(n − 1) i=2
21 / 32 µ Ln ·As+ε,−1+ε(n) ·Rn = j µ+s−3+2ε µ+j+s−3+2ε µ+s−3+2ε P −1+ε j−2+ε − −2+ε ± δs,k−2 k=1 (2 j n) 6 6 µ+i+s−5+2ε µ+i+j+s−5+2ε µ+i+s−5+2ε −2+ε j−3+ε − −3+ε ∓ δs,j−i (2 6 i 6 n) (2 6 i,j 6 n) j ε + O(ε2) 1 + O(ε) µ+j+s−3 ± P δ + O(ε) µ+s−2 j−2 s,k−2 k=1 (3 j n) 6 6 −ε + O(ε2) ε + O(ε2) µ+i+j+s−5 ∓ δ + O(ε) (µ+i+s−4)2 µ+i+s−2 j−3 s,j−i (2 6 i 6 n) (2 6 i 6 n) (2 6 i 6 n, 3 6 j 6 n)
µ+3 Bs−1+ε,−1+ε(n − 1) =
Matrix Transformations µ+i+j+s−5+2ε µ j−2+ε ± δs,j−i−1 As+ε,−1+ε(n) = (1 6 i,j 6 n)
22 / 32 j ε + O(ε2) 1 + O(ε) µ+j+s−3 ± P δ + O(ε) µ+s−2 j−2 s,k−2 k=1 (3 j n) 6 6 −ε + O(ε2) ε + O(ε2) µ+i+j+s−5 ∓ δ + O(ε) (µ+i+s−4)2 µ+i+s−2 j−3 s,j−i (2 6 i 6 n) (2 6 i 6 n) (2 6 i 6 n, 3 6 j 6 n)
µ+3 Bs−1+ε,−1+ε(n − 1) =
Matrix Transformations µ+i+j+s−5+2ε µ j−2+ε ± δs,j−i−1 As+ε,−1+ε(n) = (1 6 i,j 6 n) µ Ln ·As+ε,−1+ε(n) ·Rn = j µ+s−3+2ε µ+j+s−3+2ε µ+s−3+2ε P −1+ε j−2+ε − −2+ε ± δs,k−2 k=1 (2 j n) 6 6 µ+i+s−5+2ε µ+i+j+s−5+2ε µ+i+s−5+2ε −2+ε j−3+ε − −3+ε ∓ δs,j−i (2 6 i 6 n) (2 6 i,j 6 n)
22 / 32 µ+3 Bs−1+ε,−1+ε(n − 1) =
Matrix Transformations µ+i+j+s−5+2ε µ j−2+ε ± δs,j−i−1 As+ε,−1+ε(n) = (1 6 i,j 6 n) µ Ln ·As+ε,−1+ε(n) ·Rn = j µ+s−3+2ε µ+j+s−3+2ε µ+s−3+2ε P −1+ε j−2+ε − −2+ε ± δs,k−2 k=1 (2 j n) 6 6 µ+i+s−5+2ε µ+i+j+s−5+2ε µ+i+s−5+2ε −2+ε j−3+ε − −3+ε ∓ δs,j−i (2 6 i 6 n) (2 6 i,j 6 n) j ε + O(ε2) 1 + O(ε) µ+j+s−3 ± P δ + O(ε) µ+s−2 j−2 s,k−2 k=1 (3 j n) 6 6 −ε + O(ε2) ε + O(ε2) µ+i+j+s−5 ∓ δ + O(ε) (µ+i+s−4)2 µ+i+s−2 j−3 s,j−i (2 6 i 6 n) (2 6 i 6 n) (2 6 i 6 n, 3 6 j 6 n)
22 / 32 Matrix Transformations µ+i+j+s−5+2ε µ j−2+ε ± δs,j−i−1 As+ε,−1+ε(n) = (1 6 i,j 6 n) µ Ln ·As+ε,−1+ε(n) ·Rn = j µ+s−3+2ε µ+j+s−3+2ε µ+s−3+2ε P −1+ε j−2+ε − −2+ε ± δs,k−2 k=1 (2 j n) 6 6 µ+i+s−5+2ε µ+i+j+s−5+2ε µ+i+s−5+2ε −2+ε j−3+ε − −3+ε ∓ δs,j−i (2 6 i 6 n) (2 6 i,j 6 n) j ε + O(ε2) 1 + O(ε) µ+j+s−3 ± P δ + O(ε) µ+s−2 j−2 s,k−2 k=1 (3 j n) 6 6 −ε 2 ε 2 µ+i+j+s−5 + O(ε ) + O(ε ) ∓ δs,j−i + O(ε) (µ+i+s−4)2 µ+i+s−2 j−3 (2 6 i 6 n) (2 6 i 6 n) (2 6 i 6 n, 3 6 j 6 n) µ+i+j+s−3+2ε µ+3 j−2+ε ∓ δs,j−i Bs−1+ε,−1+ε(n − 1) = (1 6 i,j 6 n−1) 22 / 32 Matrix Transformations µ+i+j+s−5+2ε µ j−2+ε ± δs,j−i−1 As+ε,−1+ε(n) = (1 6 i,j 6 n) µ Ln ·As+ε,−1+ε(n) ·Rn = j µ+s−3+2ε µ+j+s−3+2ε µ+s−3+2ε P −1+ε j−2+ε − −2+ε ± δs,k−2 k=1 (2 j n) 6 6 µ+i+s−5+2ε µ+i+j+s−5+2ε µ+i+s−5+2ε −2+ε j−3+ε − −3+ε ∓ δs,j−i (2 6 i 6 n) (2 6 i,j 6 n) j ε + O(ε2) 1 + O(ε) µ+j+s−3 ± P δ + O(ε) µ+s−2 j−2 s,k−2 k=1 (3 j n) 6 6 −ε 2 ε 2 µ+i+j+s−5 + O(ε ) + O(ε ) ∓ δs,j−i + O(ε) (µ+i+s−4)2 µ+i+s−2 j−3 (2 6 i 6 n) (2 6 i 6 n) (2 6 i 6 n, 3 6 j 6 n) ε 2 µ+i+j+s−3 µ+3 µ+i+s−1 + O(ε ) j−2 ∓ δs,j−i + O(ε) Bs−1+ε,−1+ε(n−1) = (1 6 i 6 n−1) (1 6 i 6 n−1,2 6 j 6 n−1) 22 / 32 Matrix Transformations µ+i+j+s−5+2ε µ j−2+ε ± δs,j−i−1 As+ε,−1+ε(n) = (1 6 i,j 6 n) µ Ln ·As+ε,−1+ε(n) ·Rn = j µ+s−3+2ε µ+j+s−3+2ε µ+s−3+2ε P −1+ε j−2+ε − −2+ε ± δs,k−2 k=1 (2 j n) 6 6 µ+i+s−5+2ε µ+i+j+s−5+2ε µ+i+s−5+2ε −2+ε j−3+ε − −3+ε ∓ δs,j−i (2 6 i 6 n) (2 6 i,j 6 n) j ε + O(ε2) 1 + O(ε) µ+j+s−3 ± P δ + O(ε) µ+s−2 j−2 s,k−2 k=1 (3 j n) 6 6 −ε 2 ε 2 µ+i+j+s−5 + O(ε ) + O(ε ) ∓ δs,j−i + O(ε) (µ+i+s−4)2 µ+i+s−2 j−3 (2 6 i 6 n) (2 6 i 6 n) (2 6 i 6 n, 3 6 j 6 n) ε 2 µ+i+j+s−5 µ+3 µ+i+s−2 + O(ε ) j−3 ∓ δs,j−i + O(ε) Bs−1+ε,−1+ε(n−1) = (2 6 i 6 n) (2 6 i 6 n,3 6 j 6 n) 22 / 32 n X 1 · c = −1 µ + i + s − 2 n,i i=2 n X µ + i + j + s − 5 · c = ±c (3 j n) j − 3 n,i n,j−s 6 6 i=2
ε · Cofi,1(n − 1) I Abandon the original definition cn,i := lim . ε→0 Cof1,1(n − 1) I Use the conjectured holonomic description for cn,i instead. ε · Cofi,1(n − 1) I The first two identities prove cn,i = lim . ε→0 Cof1,1(n − 1) I The third identity proves the claimed quotient of determinants.
The Holonomic Ansatz
n X −1 2s(n − 1)(µ − 3)(µ + n + s − 2) · c = (µ + i + s − 4) n,i µ(n + s)(µ + n − 3)(µ + s − 2) i=2 2
23 / 32 n X 1 · c = −1 µ + i + s − 2 n,i i=2 n X µ + i + j + s − 5 · c = ±c (3 j n) j − 3 n,i n,j−s 6 6 i=2
I Use the conjectured holonomic description for cn,i instead. ε · Cofi,1(n − 1) I The first two identities prove cn,i = lim . ε→0 Cof1,1(n − 1) I The third identity proves the claimed quotient of determinants.
The Holonomic Ansatz
n X −1 2s(n − 1)(µ − 3)(µ + n + s − 2) · c = (µ + i + s − 4) n,i µ(n + s)(µ + n − 3)(µ + s − 2) i=2 2
ε · Cofi,1(n − 1) I Abandon the original definition cn,i := lim . ε→0 Cof1,1(n − 1)
23 / 32 n X 1 · c = −1 µ + i + s − 2 n,i i=2 n X µ + i + j + s − 5 · c = ±c (3 j n) j − 3 n,i n,j−s 6 6 i=2
ε · Cofi,1(n − 1) I The first two identities prove cn,i = lim . ε→0 Cof1,1(n − 1) I The third identity proves the claimed quotient of determinants.
The Holonomic Ansatz
n X −1 2s(n − 1)(µ − 3)(µ + n + s − 2) · c = (µ + i + s − 4) n,i µ(n + s)(µ + n − 3)(µ + s − 2) i=2 2
ε · Cofi,1(n − 1) I Abandon the original definition cn,i := lim . ε→0 Cof1,1(n − 1) I Use the conjectured holonomic description for cn,i instead.
23 / 32 I The third identity proves the claimed quotient of determinants.
The Holonomic Ansatz
n X 1 · c = −1 µ + i + s − 2 n,i i=2 n X µ + i + j + s − 5 · c = ±c (3 j n) j − 3 n,i n,j−s 6 6 i=2 n X −1 2s(n − 1)(µ − 3)(µ + n + s − 2) · c = (µ + i + s − 4) n,i µ(n + s)(µ + n − 3)(µ + s − 2) i=2 2
ε · Cofi,1(n − 1) I Abandon the original definition cn,i := lim . ε→0 Cof1,1(n − 1) I Use the conjectured holonomic description for cn,i instead. ε · Cofi,1(n − 1) I The first two identities prove cn,i = lim . ε→0 Cof1,1(n − 1)
23 / 32 The Holonomic Ansatz
n X 1 · c = −1 µ + i + s − 2 n,i i=2 n X µ + i + j + s − 5 · c = ±c (3 j n) j − 3 n,i n,j−s 6 6 i=2 n X −1 2s(n − 1)(µ − 3)(µ + n + s − 2) · c = (µ + i + s − 4) n,i µ(n + s)(µ + n − 3)(µ + s − 2) i=2 2
ε · Cofi,1(n − 1) I Abandon the original definition cn,i := lim . ε→0 Cof1,1(n − 1) I Use the conjectured holonomic description for cn,i instead. ε · Cofi,1(n − 1) I The first two identities prove cn,i = lim . ε→0 Cof1,1(n − 1) I The third identity proves the claimed quotient of determinants. 23 / 32 Calculations went close to the limits and required a lot of “human guidance”, for several reasons:
I intermediate results are quite large (several 100 MB) due to the extra two parameters µ and r
I some of the certificates had poles close to the summation boundaries
I additional sums coming from the Kronecker deltas in combination with the row and column operations, some of which having non-natural boundaries
Computational Challenge
24 / 32 I intermediate results are quite large (several 100 MB) due to the extra two parameters µ and r
I some of the certificates had poles close to the summation boundaries
I additional sums coming from the Kronecker deltas in combination with the row and column operations, some of which having non-natural boundaries
Computational Challenge
Calculations went close to the limits and required a lot of “human guidance”, for several reasons:
24 / 32 I some of the certificates had poles close to the summation boundaries
I additional sums coming from the Kronecker deltas in combination with the row and column operations, some of which having non-natural boundaries
Computational Challenge
Calculations went close to the limits and required a lot of “human guidance”, for several reasons:
I intermediate results are quite large (several 100 MB) due to the extra two parameters µ and r
24 / 32 I additional sums coming from the Kronecker deltas in combination with the row and column operations, some of which having non-natural boundaries
Computational Challenge
Calculations went close to the limits and required a lot of “human guidance”, for several reasons:
I intermediate results are quite large (several 100 MB) due to the extra two parameters µ and r
I some of the certificates had poles close to the summation boundaries
24 / 32 Computational Challenge
Calculations went close to the limits and required a lot of “human guidance”, for several reasons:
I intermediate results are quite large (several 100 MB) due to the extra two parameters µ and r
I some of the certificates had poles close to the summation boundaries
I additional sums coming from the Kronecker deltas in combination with the row and column operations, some of which having non-natural boundaries
24 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
n odd n even
s s
t t
n even n odd
s s
25 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
n odd n even
s s
t t
n even n odd
s s
25 / 32 µ µ Es,t(n) Family Ds,t(n) Family t t
n odd n even
s s
t t
n even n odd
s s
25 / 32 µ µ + 2m + 4r − 1 m−r+1 2 + 2m + r + 1 m−r µ , m − r + 1 m−r+1 2 + m + 2r m−r
m−r µ µ 1 (−1) 2 + 2m + r + 1 m−r 2 + 3r − 2 m−r+1 3 , 2 m−r m − r m−r
m−r 1 (−1) 2 m−r+1 µ + 2m + 4r − 1 m−r+1 µ µ 1 . (2m − 2r + 1) 2 + m + 2r m−r 2 + 3r − 2 m−r+1
Triangle Relations
Corollary: Let µ be an indeterminate, and let m, r ∈ Z. If m > r > 1, then µ E2r,1(2m + 1) µ = E2r,1(2m)
µ E2r,1(2m + 1) µ = E2r+1,1(2m)
µ E2r+1,1(2m) µ = E2r,1(2m)
26 / 32 Triangle Relations
Corollary: Let µ be an indeterminate, and let m, r ∈ Z. If m > r > 1, then Eµ (2m + 1) µ + 2m + 4r − 1 µ + 2m + r + 1 2r,1 = m−r+1 2 m−r , Eµ (2m) µ 2r,1 m − r + 1 m−r+1 2 + m + 2r m−r
Eµ (2m + 1) (−1)m−r µ + 2m + r + 1 µ + 3r − 1 2r,1 = 2 m−r 2 2 m−r+1 , Eµ (2m) 3 2r+1,1 2 m−r m − r m−r
Eµ (2m) (−1)m−r 1 µ + 2m + 4r − 1 2r+1,1 = 2 m−r+1 m−r+1 . Eµ (2m) µ µ 1 2r,1 (2m − 2r + 1) 2 + m + 2r m−r 2 + 3r − 2 m−r+1
26 / 32 µ Conjecture for D1,1(n)