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Irregular triangle of Catalan-based numbers, read by rows.
1

%I #178 Dec 16 2024 14:54:48

%S 1,1,2,-2,5,-14,5,14,-74,74,-14,42,-352,668,-352,42,132,-1588,4808,

%T -4808,1588,-132,429,-6946,30371,-48540,30371,-6946,429,1430,-29786,

%U 176270,-407810,407810,-176270,29786,-1430

%N Irregular triangle of Catalan-based numbers, read by rows.

%C Calculation of the sum over the partitions of r of products of dimensions of two different representations of a symmetric group S_r gives

%C 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.

%C In the limit N -> infinity the coefficients [are?]

%C c_1 = 1/(1+x), c_i = 1/(i*N^(2*(i-1)))*P(i-1), for i >= 2.

%C Coefficient of x^n in the numerator of P(i) is T(s, i).

%C 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.

%C They are easily obtained from the g.f.

%C Some special cases are given by A000108 (first column of the triangle), A138156 (second column of the triangle).

%C The sum of the numbers in row 2*k+1 is (-1)^k * A000260(k) * 2^(2*k).

%H Sergii Voloshyn, <a href="/A336110/b336110.txt">Table of n, a(n) for n = 1..120</a>

%H O. Borisenko, V. Chelnokov, and Sergii Voloshyn, <a href="https://arxiv.org/abs/2008.00773">The large N limit of SU(N) integrals in lattice</a>, arXiv:2008.00773 [hep-lat], 2020. See formula (48).

%H F. Green and S. Samuel, <a href="https://doi.org/10.1016/0550-3213(82)90515-6">Calculating the large-N phase transition in gauge and matrix models</a>, Nuclear Physics B 194, Issue 1, 11 January 1982, Pages 107-156, See Appendix A.

%F 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.

%F or in a more explicit way (only for k >= 1)

%F T(s, n) = C(s) * U(s, n) - 4^s * V(s, n), where

%F U(s, n) = Sum_{a=0..(n-1)/2} (u(s, a)/u(0,a)) * M(s, n-1- 2a),

%F V(s, n) = Sum_{a=0..(n-2)/2} (v(s, a)/v(0,a)) * M(s, n-2- 2a), and

%F u(s, n) = Product_{L=1..n} binomial(2s+2L+1, 3),

%F v(s, n) = Product_{L=1..n} binomial(2s+2L+2, 3), and

%F M(m, n) = Sum_{L=1..n} L!/n! B_{n,l} ( x_1, ..., x_{n-L+1}), and

%F x_i = (-1)^{1+i} * (3s+i)_i = (-1)^{1+i} * i! * binomial(3s + i, i), where

%F 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.

%F Sum of numbers in the particular row:

%F 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);

%F Sum_{n=1..2*k} T(2*k, n) = 0 (even s).

%F From _Sergii Voloshyn_, Oct 22 2020: (Start)

%F Formulae for particular columns:

%F T(s, 1) = C(s);

%F T(s, 2) = C(s)*(3*s+1) - 4^s;

%F T(s, 3) = C(s)*(binomial(2*s+3,3) + (3*s+1)^2 - binomial(3*s +2,2)) - 4^s*(3*s+1);

%F 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));

%F ...

%F T(s, s) = (-1)^(s+1)*C(s). (End)

%F From _Sergii Voloshyn_, Mar 17 2021: (Start)

%F Recursion ( P[0] = x/(1+x) ):

%F x*(d^3/dx^3)*P[s] = (1/4)*(2*k+2)*(2*k+3)*(2*k+4)*P[s+1]

%F 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].

%F (End)

%F 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

%F From _Sergii Voloshyn_, Apr 17 2021: (Start)

%F 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);

%F Om(k, m) = (12^k/((1+k)!*(2k+1)!)) * Product_{L=1..k} binomial(m+2L, 3).

%F (End)

%F From _Sergii Voloshyn_, Apr 25 2021: (Start)

%F Differential equation for P[k_]:

%F 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.

%F Discrete set equation for T(k,n) (n=-1..k-2) at fixed k:

%F (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

%F and

%F Sum_{m=1..k} (k*(5+7*k) + 12*n*(n-1-k))*T(k,n) = 0. (End)

%F 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

%F Sum_{k} abs(T(n,k)) = A000309(n-1). - _Sergii Voloshyn_, Nov 20 2024

%e 1;

%e 1;

%e 2, -2;

%e 5, -14, 5;

%e 14, -74, 74, -14;

%e 42, -352, 668, -352, 42;

%e 132, -1588, 4808, -4808, 1588, -132;

%e 429, -6946, 30371, -48540, 30371, -6946, 429;

%e 1430, -29786, 176270, -407810, 407810, -176270, 29786, -1430;

%e ...

%t 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}];

%t M[n_, l_] := Sum[k!/n! BellY[n,k,Table[(-1)^(j-1) j!Binomial[3l+j,j], {j,n}]], {k, 0, n}];

%t u[k_, n_] := Product[Binomial[2 k + 2 l + 1, 3], {l, 1, n}];

%t v[k_, n_] := Product[ Binomial[2 k + 2 l + 2, 3], {l, 1, n}];

%t (* alternate program using coefficients in numerator *)

%t 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];

%t Table[CoefficientList[P[s] // Together // Numerator, x] // Rest, {s, 0, 10}] // Flatten (* amended by _Jean-François Alcover_, Sep 25 2020 *)

%t (* another program using coefficients in numerator *)

%t Needs["Combinatorica`"];

%t 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] //

%t Rest, {s, 0, 10}] // Flatten (* _Sergii Voloshyn_, Sep 03 2021 *)

%Y Cf. A000108, A000309, A138156, A000260.

%K sign,tabf

%O 1,3

%A _Sergii Voloshyn_, Jul 08 2020