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%I #33 Jan 01 2017 19:47:17
%S 1,1,4,1,1,20,48,20,1,1,54,405,760,405,54,1,1,112,1828,8464,13840,
%T 8464,1828,112,1,1,200,5925,52800,182700,273504,182700,52800,5925,200,
%U 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 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 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>
%F Row sums equal A006480(n) = (3n)!/(n!)^3, which is de Bruijn's s(3,n).
%F From _Yahia Kahloune_, Jan 30 2014: (Start)
%F Using these coefficients we can obtain formulas for the sums
%F 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 For example:
%F b(e,e,3) = 1;
%F b(e+1,e,3) = (e+1)^3-(3*e+1) = e^2*(e+3);
%F b(e+2,e,3) = C(e+2,2)^3 - (3*e+1)*(e+1)^3 + C(3*e+1,2);
%F 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 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 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 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.
%F (End)
%e Triangle begins:
%e 1;
%e 1, 4, 1;
%e 1, 20, 48, 20, 1;
%e 1, 54, 405, 760, 405, 54, 1;
%e 1, 112, 1828, 8464, 13840, 8464, 1828, 112, 1;
%e 1, 200, 5925, 52800, 182700, 273504, 182700, 52800, 5925, 200, 1;
%e 1, 324, 15606, 233300, 1424925, 4050864, 5703096, 4050864, 1424925, 233300, 15606, 324, 1; ...
%e Row g.f.s begin:
%e (1) = (1-x)*(1 + x + x^2 + x^3 + x^4 +...);
%e (1 + 4*x + x^2) = (1-x)^4*(1 + 2^3*x + 3^3*x^2 + 4^3*x^3 +...);
%e (1 + 20*x + 48*x^2 + 20*x^3 + x^4) = (1-x)^7*(1 + 3^3*x + 6^3*x^2 +...);
%e (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 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 (PARI) {T(n,k)=polcoeff(sum(j=0,n+k,binomial(n+j,j)^3*x^j)*(1-x)^(3*n+1),k)}
%o for(n=0,10,for(k=0,2*n,print1(T(n,k),", "));print(""))
%Y Cf. A181543, A181545, A006480, A002897, A000172.
%Y Cf. A183204 (central terms), A183205.
%K nonn,tabf
%O 0,3
%A _Paul D. Hanna_, Oct 30 2010
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