%I A142071
%S A142071 0,1,0,1,1,0,1,3,2,0,1,7,12,6,0,1,15,50,60,24,0,1,31,180,390,360,120,0,
%T A142071 1,63,602,2100,3360,2520,720,0,1,127,1932,10206,25200,31920,20160,5040,
%U A142071 0,1,255,6050,46620,166824,317520,332640,181440,40320,0,1,511,18660
%N A142071 A triangle sequence of coefficients of an infinite sum polynomial: p(x,
               n)=Sum[k^n*(x/(1 + x))^k, {k, 0, Infinity}]=PolyLog[ -n,x/(1+x)].
%C A142071 Row sums are:
%C A142071 {1, 2, 6, 26, 150, 1082, 9366, 94586, 1091670, 14174522, 204495126}.
%C A142071 If you divide the PolyLog function by x you get A028246.
%C A142071 I got this polynomial sequence looking for a Eulerian number like infinite 
               sum polynomial.
%F A142071 p(x,n)=Sum[k^n*(x/(1 + x))^k, {k, 0, Infinity}]=PolyLog[ -n,x/(1+x)]; 
               t(n,m)=Coefficients(p(x,n)).
%e A142071 {0, 1},
%e A142071 {0, 1, 1},
%e A142071 {0, 1, 3, 2},
%e A142071 {0, 1, 7, 12, 6},
%e A142071 {0, 1, 15, 50, 60, 24},
%e A142071 {0, 1, 31, 180, 390, 360, 120},
%e A142071 {0, 1, 63, 602, 2100, 3360, 2520, 720},
%e A142071 {0, 1, 127, 1932, 10206, 25200, 31920, 20160, 5040},
%e A142071 {0, 1, 255, 6050, 46620, 166824, 317520, 332640, 181440, 40320},
%e A142071 {0, 1, 511, 18660, 204630, 1020600, 2739240, 4233600, 3780000, 1814400, 
               362880}, {0, 1, 1023, 57002, 874500, 5921520, 21538440, 46070640, 
               59875200, 46569600, 19958400, 3628800}
%t A142071 p[x_, n_] = Sum[k^n*(x/(1 + x))^k, {k, 0, Infinity}]; Table[FullSimplify[ExpandAll[p[x, 
               n]]], {n, 0, 10}]; Table[CoefficientList[FullSimplify[ExpandAll[p[x, 
               n]]], x], {n, 0, 10}]; Flatten[%]
%Y A142071 Cf. A028246 .
%Y A142071 Sequence in context: A081576 A054654 A154477 this_sequence A118972 A171224 
               A145878
%Y A142071 Adjacent sequences: A142068 A142069 A142070 this_sequence A142072 A142073 
               A142074
%K A142071 nonn,uned
%O A142071 1,8
%A A142071 Roger L. Bagula and Gary W. Adamson (rlbagulatftn(AT)yahoo.com), Sep 
               15 2008