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%I #2 Mar 30 2012 17:34:27
%S 1,1,1,1,5,1,1,11,11,1,1,26,26,26,1,1,69,62,62,69,1,1,206,159,148,159,
%T 206,1,1,679,453,371,371,453,679,1,1,2408,1436,1016,902,1016,1436,
%U 2408,1,1,8969,4964,3092,2366,2366,3092,4964,8969,1,1,34442,18221,10360,6866
%N A functionally symmetric Polynomial as a triangle of coefficients: p(x,n)=If[n == 0, 1, (x + 1)^n + 2^(n - 4)*Sum[(2^m + 2*m + 2)*x^m*(1 + x^(n - 2*m)), {m, 1, n - 1}]].
%C Row sums are:{1, 2, 7, 24, 80, 264, 880, 3008, 10624, 38784, 145664}.
%F p(x,n)=If[n == 0, 1, (x + 1)^n + 2^(n - 4)*Sum[(2^m + 2*m + 2)*x^m*(1 + x^(n - 2*m)), {m, 1, n - 1}]]; t(n,m)=coefficients(p(x,n)).
%e {1}, {1, 1}, {1, 5, 1}, {1, 11, 11, 1}, {1, 26, 26, 26, 1}, {1, 69, 62, 62, 69, 1}, {1, 206, 159, 148, 159, 206, 1}, {1, 679, 453, 371, 371, 453, 679, 1}, {1, 2408, 1436, 1016, 902, 1016, 1436, 2408, 1}, {1, 8969, 4964, 3092, 2366, 2366, 3092, 4964, 8969, 1}, {1, 34442, 18221, 10360, 6866, 5884, 6866, 10360, 18221, 34442, 1}
%t Clear[p, x, n]; p[x_, n_] = If[ n == 0, 1, (x + 1)^n + 2^(n - 4)*Sum[(2^m + 2*m + 2)*x^m*(1 + x^(n - 2*m)), {m, 1, n - 1}]]; Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n, 0, 10}]; Flatten[%]
%K nonn
%O 0,5
%A _Roger L. Bagula_, Nov 03 2008
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