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%I #3 Mar 30 2012 17:34:27
%S 2,-1,0,3,-1,2,1,4,-1,0,15,5,5,-1,2,21,76,16,6,-1,0,63,287,322,42,7,
%T -1,2,113,1212,2381,1226,99,8,-1,0,255,4265,15675,15549,4349,219,9,-1,
%U 2,493,14644,88150,156316,88108,14692,466,10,-1,0,1023,47795,455312
%N The PolyLog functional part of A008292 (the Eulerian numbers) is treated as a curvature to give a set of polynomial triangle sequence coefficients: p(x,n)=Sum[A008292(n,m)*x^(m-1),{m,0,n}]; q(x,n)=k from Solve[FullSimplify[ExpandAll[p[x, n]/(x - 1)^n]] - (1 + k/x^2) == 0, k].
%C The concept here is that the increase in curvature causes transformation of Pascal's triangle into the Eulerian numbers and the MacMahon numbers, while leaving the numerical Modulo 2 Sierpinski Self -Similarity intact. The resulting polynomials have a finite Blaschke elliptical structure. The row sums are: {0, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800}.
%D Lennart Carleson and Theodore W. Gamelin, Complex Dynamics, Springer, New York, 1993, pp 103 ( Herman's Rings as Finite Blaschke sets)
%D Kenneth Hoffman, Banach Spaces of Analytic Functions, Dover, New York, 1962, page 66, page 132.
%F p(x,n)=Sum[A008292(n,m)*x^(m-1),{m,0,n}]; q(x,n)=k from Solve[FullSimplify[ExpandAll[p[x, n]/(x - 1)^n]] - (1 + k/x^2) == 0, k]; t(n,m)=Coefficients(((x - 1)^n/x^2)*q(n,x)).
%e {0}, {2, -1}, {0, 3, -1}, {2, 1, 4, -1}, {0, 15, 5, 5, -1}, {2, 21, 76,16, 6, -1}, {0, 63, 287, 322, 42, 7, -1}, {2, 113, 1212, 2381, 1226, 99, 8, -1}, {0, 255, 4265, 15675, 15549, 4349, 219, 9, -1}, {2, 493, 14644, 88150, 156316, 88108, 14692, 466, 10, -1}, {0, 1023, 47795, 455312, 1310144,1310606, 454982, 47960, 968, 11, -1}
%t Clear[q, p, x, n, a]; p[x_, n_] = p[x_, n_] = (1 - x)^(n + 1)*PolyLog[ -n, x]/x; q[x_, n_] := ((x - 1)^n/x^2)*k /. Solve[FullSimplify[ExpandAll[p[x, n]/(x - 1)^n]] - (1 + k/x^2) == 0, k]; Table[FullSimplify[Expand[q[x, n]]], {n, 0, 10}]; Table[Flatten[CoefficientList[FullSimplify[Expand[q[x, n]]], x]], {n, 0, 10}]; Flatten[%]
%Y A060187, A008292
%K sign
%O 0,1
%A _Roger L. Bagula_, Oct 31 2008, Nov 01 2008