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%I #2 Oct 12 2012 14:54:54
%S 1,1,1,1,6,1,1,10,10,1,1,15,20,15,1,1,21,35,35,21,1,1,28,56,70,56,28,
%T 1,1,36,84,126,126,84,36,1,1,45,120,210,252,210,120,45,1,1,55,165,330,
%U 462,462,330,165,55,1,1,66,220,495,792,924,792,495,220,66,1
%N A designed polynomial set that gives a {1,6,1} quadratic and gives a symmetrical triangle of coefficients: p(x,n)=If[n == 2, 1, ((x + 1)^n -If[n == 0, 1, x^n + (n - 1)*x^(n - 1) + (n - 1)*x + 1])/x],.
%C Row sums are:{1, 2, 8, 22, 52, 114, 240, 494, 1004, 2026, 4072}.
%F p(x,n)=If[n == 2, 1, ((x + 1)^n -If[n == 0, 1, x^n + (n - 1)*x^(n - 1) + (n - 1)*x + 1])/x]; t(n,m)=coefficients(p(x,n)).
%e {1},
%e {1, 1},
%e {1, 6, 1},
%e {1, 10, 10, 1},
%e {1, 15, 20, 15, 1},
%e {1, 21, 35, 35, 21, 1},
%e {1, 28, 56, 70, 56, 28, 1},
%e {1, 36, 84, 126, 126, 84, 36, 1},
%e {1, 45, 120, 210, 252, 210, 120, 45, 1},
%e {1, 55, 165, 330, 462, 462, 330, 165, 55, 1},
%e {1, 66, 220, 495, 792, 924, 792, 495, 220, 66, 1}
%t Clear[p, x, n] p[x_, n_] = If[ n == 2, 1, ((x + 1)^n - If[n == 0, 1, x^n + (n - 1)*x^(n - 1) + (n - 1)*x + 1])/x]; Table[ExpandAll[p[x, n]], {n, 2, 12}]; Table[CoefficientList[p[x, n], x], {n, 2, 12}]; Flatten[%]
%K nonn,uned
%O 1,5
%A _Roger L. Bagula_ and _Gary W. Adamson_, Oct 02 2008
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