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A336110
Irregular triangle of Catalan-based numbers, read by rows.
1
1, 1, 2, -2, 5, -14, 5, 14, -74, 74, -14, 42, -352, 668, -352, 42, 132, -1588, 4808, -4808, 1588, -132, 429, -6946, 30371, -48540, 30371, -6946, 429, 1430, -29786, 176270, -407810, 407810, -176270, 29786, -1430
OFFSET
1,3
COMMENTS
Calculation of the sum over the partitions of r of products of dimensions of two different representations of a symmetric group S_r gives
Sum_{L |- S_r} f(L)*f(l+q^N) = (r+q^N)! * G[N+1] * G[q+1]/(G[N+q+1]) * B_r(1!c_1, ..., r!c_r) where f(L) is the dimension of the symmetric group S_r, G[x] is Barnes function, and B_r() is the complete exponential Bell polynomial.
In the limit N -> infinity the coefficients [are?]
c_1 = 1/(1+x), c_i = 1/(i*N^(2*(i-1)))*P(i-1), for i >= 2.
Coefficient of x^n in the numerator of P(i) is T(s, i).
This triangle of coefficients was discovered by Borisenko et al. In mathematical physics these coefficients appear as an important ingredient of series that define the free energy of the SU(N) standard lattice model in the large N limit.
They are easily obtained from the g.f.
Some special cases are given by A000108 (first column of the triangle), A138156 (second column of the triangle).
The sum of the numbers in row 2*k+1 is (-1)^k * A000260(k) * 2^(2*k).
LINKS
O. Borisenko, V. Chelnokov, and S. Voloshyn, The large N limit of SU(N) integrals in lattice, arXiv:2008.00773 [hep-lat], 2020. See formula (48).
F. Green and S. Samuel, Calculating the large-N phase transition in gauge and matrix models, Nuclear Physics B 194, Issue 1, 11 January 1982, Pages 107-156, See Appendix A.
FORMULA
Coefficients of x^n in the numerator of P(s) = (x * C[s]* 3F2[ s+ 1/2, s+1, s+3/2; 1/2,3/2; x^2] - x^2 * 4^s * 3F2[ s+1,s+3/2, s+2; 3/2, 2; x^2]), where C(s) are Catalan numbers.
or in a more explicit way (only for k >= 1)
T(s, n) = C(s) * U(s, n) - 4^s * V(s, n), where
U(s, n) = Sum_{a=0..(n-1)/2} (u(s, a)/u(0,a)) * M(s, n-1- 2a),
V(s, n) = Sum_{a=0..(n-2)/2} (v(s, a)/v(0,a)) * M(s, n-2- 2a), and
u(s, n) = Product_{L=1..n} binomial(2s+2L+1, 3),
v(s, n) = Product_{L=1..n} binomial(2s+2L+2, 3), and
M(m, n) = Sum_{L=1..n} L!/n! B_{n,l} ( x_1, ..., x_{n-L+1}), and
x_i = (-1)^{1+i} * (3s+i)_i = (-1)^{1+i} * i! * binomial(3s + i, i), where
B_{n,l} (x_1, ..., x_{n-L+1}) is the n-th partial or incomplete exponential Bell polynomial with monomials sorted into graded lexicographic order.
Sum of numbers in the particular row:
Sum_{n=1..2*k+1} T(2*k+1, n) = 2*(4*k+1)!/((k+1)!*(3*k+2)!) *2^(2*k) (odd s);
Sum_{n=1..2*k} T(2*k, n) = 0 (even s).
From Sergii Voloshyn, Oct 22 2020: (Start)
Formulae for particular columns:
T(s, 1) = C(s);
T(s, 2) = C(s)*(3*s+1) - 4^s;
T(s, 3) = C(s)*(binomial(2*s+3,3) + (3*s+1)^2 - binomial(3*s +2,2)) - 4^s*(3*s+1);
T(s, 4) = C(s)*((2*s+1)binomial(2*s+3,3) +(3*s+1)^3 - 2(3*s+1)* binomial(3*s+2,2)+ binomial(3*s+3, 3)) - 4^s*(binomial(3*s+4, 3)/4 + (3*s+1)^2 - binomial(3*s+2,2));
...
T(s, s) = (-1)^(s+1)*C(s). (End)
From Sergii Voloshyn, Mar 17 2021: (Start)
Recursion ( P[0] = x/(1+x) ):
x*(d^3/dx^3)*P[s] = (1/4)*(2*k+2)*(2*k+3)*(2*k+4)*P[s+1]
for P[s_] := x CatalanNumber[s] HypergeometricPFQ[{s + 1/2, s + 1, s + 3/2}, {1/2, 3/2}, x^2] - x^2*4^s HypergeometricPFQ[{s + 1, s + 3/2, s + 2}, {3/2, 2}, x^2].
(End)
Recursion for array ( T(1,1) = 1 ): T(k, p) = (1/(k*(k+1)*(2k+1)))*[p*(p+1)*(p+2)*T(k-1,p+2) - 3*p*(p+1)*(3*k-p-1)*T(k-1,p+1) + 3*p*(3*k-p)*(3*k-p-1)*T(k-1,p) - (3*k-p+1)*(3*k-p)*(3*k-p-1)*T(k-1,p-1)]; p =[1,...,k]. - Sergii Voloshyn, Apr 14 2021
From Sergii Voloshyn, Apr 17 2021: (Start)
G.f.: Sum_{m>=1} x^m Om(k, m) = (1/(1+x)^(3*k+1))*Sum_{n=1..k} x^k * T(k,n);
Om(k, m) = (12^k/((1+k)!*(2k+1)!)) * Product_{L=1..k} binomial(m+2L, 3).
(End)
From Sergii Voloshyn, Apr 25 2021: (Start)
Differential equation for P[k_]:
x*(x^2-1)*(d^3/dx^3)P[s] + (2*k+2)*3*x^2*(d^2/dx^2)P[s] +(2*k+2)*(2*k+1)*3*x (d/dx)P[s] + (2*k+2)(2*k+1)*2*k*P[s] = 0.
Discrete set equation for T(k,n) (n=-1..k-2) at fixed k:
(k-n-1)*(k-n)*(k-n+1)*T(k,n) - (k-n-1)*(k-n)*(8*k+n+5)*T(k,n+1) - (n+1)*(n+2)*(9-n+3)*T(k,n+2) + (n+1)*(n+2)*(n+3)*T(k,n+3) = 0
and
Sum_{m=1..k} (k*(5+7*k) + 12*n*(n-1-k))*T(k,n) = 0. (End)
P[k_]:= (2^k)/((2*k+1)!*(k+1)!)*(-1/(1+x))^{2k+1}[Sum_{r(1)+...+ r(L)=k} (-x/(1+x))^{k-L+1}* 1^{(3*r(1))}*(1+3*r(1)-1)^{(3*r(2))}*... *(1+Sum_{i=1..L-1} 3 r(i) -L+1)^{(3*r(L))}] where Sum_ r_i = k runs over all integer compositions of k, L is a number of parts of this composition and 1^{(3 r(1))} is a rising factorial. - Sergii Voloshyn, Sep 03 2021
EXAMPLE
1;
1;
2, -2;
5, -14, 5;
14, -74, 74, -14;
42, -352, 668, -352, 42;
132, -1588, 4808, -4808, 1588, -132;
429, -6946, 30371, -48540, 30371, -6946, 429;
1430, -29786, 176270, -407810, 407810, -176270, 29786, -1430;
...
MATHEMATICA
T[L_, n_] := CatalanNumber[L] Sum[u[L, a]/u[0, a] M[n - 1 - 2*a, L], {a, 0, (n - 1)/2}]-4^L Sum[v[L, a]/v[0, a] M[n - 2 - 2*a, L], {a, 0, (n-2)/2}];
M[n_, l_] := Sum[k!/n! BellY[n, k, Table[(-1)^(j-1) j!Binomial[3l+j, j], {j, n}]], {k, 0, n}];
u[k_, n_] := Product[Binomial[2 k + 2 l + 1, 3], {l, 1, n}];
v[k_, n_] := Product[ Binomial[2 k + 2 l + 2, 3], {l, 1, n}];
(* alternate program using coefficients in numerator *)
P[s_] := x CatalanNumber[s] HypergeometricPFQ[{s + 1/2, s + 1, s + 3/2}, {1/2, 3/2}, x^2] - x^2*4^s HypergeometricPFQ[{s + 1, s + 3/2, s + 2}, {3/2, 2}, x^2];
Table[CoefficientList[P[s] // Together // Numerator, x] // Rest, {s, 0, 10}] // Flatten (* amended by Jean-François Alcover, Sep 25 2020 *)
(* another program using coefficients in numerator *)
Needs["Combinatorica`"];
OA[p_, x_]:= (2^p(-(1/(x+1)))^(2p+1))/((2p+1)!(p+1)!) Sum[(-x/(1+x))^(p-r+1)Product[Pochhammer[1+Plus@@Table[3*k[[i]]-1, {i, 1, j-1}], 3*k[[j]]], {j, 1, r}], {r, 1, p}, {k, Compositions[p-r, r]+1}]; Table[CoefficientList[OA[s, x] // Together // Numerator, x] //
Rest, {s, 0, 10}] // Flatten (* Sergii Voloshyn, Sep 03 2021 *)
CROSSREFS
KEYWORD
sign,tabf
AUTHOR
Sergii Voloshyn, Jul 08 2020
STATUS
approved