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%I #2 Mar 30 2012 17:34:33
%S 1,1,2,5,-6,5,19,-13,-13,19,337,-1044,1462,-1044,337,2101,-5073,3092,
%T 3092,-5073,2101,62281,-314222,718559,-931796,718559,-314222,62281,
%U 543607,-2329829,3835365,-2044103,-2044103,3835365,-2329829,543607
%N A triangle of polynomial coefficients: q(x,n)=(1 - x)^(n + 1)*Sum[(k + n)^n*x^k, {k, 0, Infinity}]; p(x,n)=q(x,n)+x^n*q(1/x,n).
%C Row sums are:2*n!
%C {2, 2, 4, 12, 48, 240, 1440, 10080, 80640, 725760, 7257600,...}.
%C The result is related to the Eulerian numbers infinite sum form.
%C This was the result of finding the infinite sum identity:
%C Sum[Binomial[k+n,n]*x^k,{k,0,Infinity}]=1/(1-x)^(n+1).
%F q(x,n)=(1 - x)^(n + 1)*Sum[(k + n)^n*x^k, {k, 0, Infinity}];
%F q(x,n)=(1 - x)^(n + 1)*LerchPhi[x, -n, n];
%F p(x,n)=q(x,n)+x^n*q(1/x,n);
%F t(n,m)=coefficients(p(x,n))
%e {1, 1},
%e {2},
%e {5, -6, 5},
%e {19, -13, -13, 19},
%e {337, -1044, 1462, -1044, 337},
%e {2101, -5073, 3092, 3092, -5073, 2101},
%e {62281, -314222, 718559, -931796, 718559, -314222, 62281},
%e {543607, -2329829, 3835365, -2044103, -2044103, 3835365, -2329829, 543607},
%e {22542017, -158151816, 509366204, -972472504, 1197512838, -972472504, 509366204, -158151816, 22542017},
%e {253202761, -1572381217, 4145530310, -5521116358, 2695127384, 2695127384, -5521116358, 4145530310, -1572381217, 253202761},
%e {13486784401, -121343461986, 506850150853, -1285984548968, 2186943445546, -2599897482092, 2186943445546, -1285984548968, 506850150853, -121343461986, 13486784401}
%t Clear[p, x, n, m];
%t p[x_, n_] = (1 - x)^(n + 1)*Sum[(k + n)^n*x^k, {k, 0, Infinity}];
%t Table[FullSimplify[ExpandAll[p[x, n]]], {n, 0, 10}];
%t Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x]
%t + Reverse[ CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x]], {n, 0, 10}];
%t Flatten[%]
%K sign,tabl,uned
%O 0,3
%A _Roger L. Bagula_, Jan 31 2009
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