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%I #6 Mar 14 2015 12:04:23
%S 1,7,4,1,131,176,96,16,1,3067,6588,5895,2416,477,36,1,79459,235456,
%T 298816,197824,73120,14656,1504,64,1,2181257,8252300,13668975,
%U 12563200,6966400,2373504,490700,58400,3675,100,1,62165039,286326288,587324232,692965040,516541455,252283968,81432456,17138304,2276145,179440,7632,144,1
%N G.f.: A(x,y,z) = Sum_{n>=0} ((2n)!/n!^2)*[Sum_{k=0..2n} T(n,k)*z^k]*x^(2n)*y^n/(1-x-xy)^(4n+1) where A(x,y,x+xy) = Sum_{n>=0, k=0..n} C(n,k)^4*x^n*y^k at z = x+xy; this is the triangle of coefficients T(n,k), read by rows.
%F Row sums equal A000897(n) = (4n)!/((2n)!*n!^2).
%F Column 0 equals A099601(n) = quotient of de Bruijn sums S(4,n)/S(2,n).
%e G.f.: A(x,y,z) = 1/(1-x-x*y)
%e + 2*(7 + 4*z + z^2)*x^2*y/(1-x-x*y)^5
%e + 6*(131 + 176*z + 96*z^2 + 16*z^3 + z^4)*x^4*y^2/(1-x-x*y)^9
%e + 20*(3067 + 6588*z + 5895*z^2 + 2416*z^3 + 477*z^4 + 36*z^5 + z^6)*x^6*y^3/(1-x-x*y)^13 +...
%e G.f. at z = x+xy yields: A(x,y,x+xy) = 1 + (1 + y)*x
%e + (1 + 16*y + y^2)*x^2
%e + (1 + 81*y + 81*y^2 + y^3)*x^3
%e + (1 + 256*y + 1296*y^2 + 256*y^3 + y^4)*x^4
%e + (1 + 625*y + 10000*y^2 + 10000*y^3 + 625*y^4 + y^5)*x^5 +...
%e which is a series involving binomial coefficients to the 4th power.
%e ...
%e This triangle of coefficients T(n,k) of z^k, k=0..2n, begins:
%e [1];
%e [7, 4, 1];
%e [131, 176, 96, 16, 1];
%e [3067, 6588, 5895, 2416, 477, 36, 1];
%e [79459, 235456, 298816, 197824, 73120, 14656, 1504, 64, 1];
%e [2181257, 8252300, 13668975, 12563200, 6966400, 2373504, 490700, 58400, 3675, 100, 1];
%e [62165039, 286326288, 587324232, 692965040, 516541455, 252283968, 81432456, 17138304, 2276145, 179440, 7632, 144, 1];
%e [1818812387, 9876304172, 24205612067, 34939683632, 32837525567, 21029302364, 9356637759, 2899564224, 619135629, 88879924, 8237341, 461776, 14161, 196, 1];
%e [54257991011, 339398092544, 968547444480, 1655445817088, 1881608595776, 1496188189440, 853911382016, 353544477440, 106191762336, 22927328512, 3492995968, 364541184, 24932320, 1044736, 24192, 256, 1];
%e ...
%Y Cf. A000897, A099601.
%K nonn,tabf
%O 0,2
%A _Paul D. Hanna_, Mar 02 2011
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