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A181544 Triangle in which the g.f. for row n is [Sum_{k>=0} C(n+k-1,k)^3*x^k]*(1-x)^(3n+1), read by rows of k=0..2n terms.

%I A181544 #52 Jan 28 2025 10:47:30
%S A181544 1,1,4,1,1,20,48,20,1,1,54,405,760,405,54,1,1,112,1828,8464,13840,
%T A181544 8464,1828,112,1,1,200,5925,52800,182700,273504,182700,52800,5925,200,
%U A181544 1,1,324,15606,233300,1424925,4050864,5703096,4050864,1424925,233300,15606,324,1,1,490,35623,818300,7917371,37215794,91789005,123519792,91789005,37215794,7917371,818300,35623,490,1,1,704,73200,2430400,34657700,246781248,955910032,2116980800,2751843600,2116980800,955910032,246781248,34657700,2430400,73200,704,1
%N A181544 Triangle in which the g.f. for row n is [Sum_{k>=0} C(n+k-1,k)^3*x^k]*(1-x)^(3n+1), read by rows of k=0..2n terms.
%H A181544 Paul D. Hanna, <a href="/A181544/b181544.txt">Table of n, a(n) for n = 0..1088, as a flattened triangle of rows 0..32</a>
%H A181544 Ilia Gaiur, Vladimir Rubtsov, and Duco van Straten, <a href="https://arxiv.org/abs/2405.03015">Product formulas for the Higher Bessel functions</a>, arXiv:2405.03015 [math.AG], 2024. See p. 18.
%H A181544 Feihu Liu, Guoce Xin, and Chen Zhang, <a href="https://arxiv.org/abs/2412.18744">Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS</a>, arXiv:2412.18744 [math.CO], 2024. See p. 14.
%F A181544 Row sums equal A006480(n) = (3n)!/(n!)^3, which is de Bruijn's s(3,n).
%F A181544 From _Yahia Kahloune_, Jan 30 2014: (Start)
%F A181544 Using these coefficients we can obtain formulas for the sums
%F A181544 Sum_{i=1..n} C(e-1+i,e)^3. Let us define b(k,e,3) = sum_{i=0..k-e} (-1)^i*C(3*e+1,i)*C(k-i,e)^3, where k=e+i.
%F A181544 For example:
%F A181544 b(e,e,3) = 1;
%F A181544 b(e+1,e,3) = (e+1)^3-(3*e+1) = e^2*(e+3);
%F A181544 b(e+2,e,3) = C(e+2,2)^3 - (3*e+1)*(e+1)^3 + C(3*e+1,2);
%F A181544 b(e+3,e,3) = C(e+3,e)^3 - (3*e+1)*C(e+2,e)^3 + C(3*e+1,2)*C(e+1,e)^3 - C(3*e+1,3);
%F A181544 b(e+4,e,3) = C(e+4,e)^3 - (3*e+1)*C(e+3,e)^3  + C(3*e+1,2)*C(e+2,e) - C(3*e+1,3)*C(e+1,e)^3 + C(3*e+1,4).
%F A181544 Then we have the formula: Sum_{i=1..n} C(e-1+i,e)^3 = Sum_{i=0..2*e} b(e+i,e,3)*C(n+e+i,3*e+1).
%F A181544 Example: Sum_{i=1..7} C(2+i,3)^3 = C(10,10) + 54*C(11,10) + 405*C(12,10) + 760*C(13,10) + 405*C(14,10) + 54*C(15,10) + C(16,10) = 820260. (End)
%F A181544 Let E be the operator D*x*D*x*D, where D denotes the derivative operator d/dx. Then (1/(n)!^3)  * E^n(1/(1 - x)) = (row n generating polynomial)/(1 - x)^(3*n+1) = Sum_{k >= 0} binomial(n+k, k)^3*x^k. For example, when n = 2 we have (1/2!)^3*E^3(1/(1 - x)) = (1 + 20 x + 48 x^2 + 20 x^3 + x^4)/(1 - x)^7. - _Sergii Voloshyn_, Dec 03 2024
%e A181544 Triangle begins:
%e A181544  1;
%e A181544  1, 4, 1;
%e A181544  1, 20, 48, 20, 1;
%e A181544  1, 54, 405, 760, 405, 54, 1;
%e A181544  1, 112, 1828, 8464, 13840, 8464, 1828, 112, 1;
%e A181544  1, 200, 5925, 52800, 182700, 273504, 182700, 52800, 5925, 200, 1;
%e A181544  1, 324, 15606, 233300, 1424925, 4050864, 5703096, 4050864, 1424925, 233300, 15606, 324, 1; ...
%e A181544 Row g.f.s begin:
%e A181544  (1) = (1-x)*(1 + x + x^2 + x^3 + x^4 +...);
%e A181544  (1 + 4*x + x^2) = (1-x)^4*(1 + 2^3*x + 3^3*x^2 + 4^3*x^3 +...);
%e A181544  (1 + 20*x + 48*x^2 + 20*x^3 + x^4) = (1-x)^7*(1 + 3^3*x + 6^3*x^2 +...);
%e A181544  (1 + 54*x + 405*x^2 + 760*x^3 + 405*x^4 + 54*x^5 + x^6) = (1-x)^10*(1 + 4^3*x + 10^3*x^2 + 20^3*x^3 + 35^3*x^4 +...); ...
%t A181544 t[n_, k_] := SeriesCoefficient[Sum[Binomial[n+j, j]^3*x^j, {j, 0, n+k}]*(1-x)^(3*n+1), {x,0, k}]; Table[t[n, k], {n, 0, 9}, {k, 0, 2*n}] // Flatten (* _Jean-François Alcover_, Feb 04 2014, after PARI *)
%o A181544 (PARI) {T(n,k)=polcoeff(sum(j=0,n+k,binomial(n+j,j)^3*x^j)*(1-x)^(3*n+1),k)}
%o A181544 for(n=0,10,for(k=0,2*n,print1(T(n,k),", "));print(""))
%Y A181544 Cf. A181543, A181545, A006480, A002897, A000172.
%Y A181544 Cf. A183204 (central terms), A183205.
%K A181544 nonn,tabf
%O A181544 0,3
%A A181544 _Paul D. Hanna_, Oct 30 2010