A133135 - OEIS

%I #12 Dec 06 2022 02:53:00

%S 1,-2,2,-1,1,-4,4,13,-13,-142,142,1931,-1931,-36296,36296,893273,

%T -893273,-27927346,27927346,1081725559,-1081725559,-50861556172,

%U 50861556172,2854289486309,-2854289486309,-188475382997654,188475382997654,14467150771771043,-14467150771771043

%N Third column of the inverse of the triangle of polynomial coefficients P(0,x)=1, 2P(n,x)=(1+x)*[(1+x)^(n-1)+x^(n-1)].

%C The triangle with [x^k] P(n,x) starts

%C 1;

%C 1, 1;

%C 1/2, 3/2, 1;

%C 1/2, 3/2, 2, 1;

%C 1/2, 2, 3, 5/2, 1;

%C 1/2, 5/2, 5, 5, 3, 1;

%C 1/2, 3,15/2, 10,15/2, 7/2, 1;

%C 1/2, 7/2,21/2,35/2,35/2,21/2, 4, 1;

%C 1/2, 4, 14, 28, 35, 28, 14, 9/2, 1;

%C 1/2, 9/2, 18, 42, 63, 63, 42, 18, 5, 1;

%C The sum of the rows of this triangle is A094373 (previously noticed by _Paul Curtz_). - _Jean-François Alcover_, Jul 22 2013

%C Apparently a(2*n) = A102055(n) and a(2*n+1) = -a(2*n) for n >= 0. - _Georg Fischer_, Dec 05 2022

%e The inverse of the triangle of coefficients starts

%e 1;

%e -1, 1;

%e 1, -3/2, 1;

%e -1, 3/2, -2, 1;

%e 1, -5/4, 2, -5/2, 1;

%e -1, 5/4 -1, 5/2, -3, 1;

%e 1, -7/4, 1, 0, 3, -7/2, 1;

%e -1, 7/4,-4, 0, 2, 7/2,-4, 1;

%e 1, 3/8, 4,-21/2, -2, 21/4, 4,-9/2, 1;

%e -1, -3/8,13, 21/2,-26,-21/4,10, 9/2,-5, 1;

%e and defines the sequence in its third column.

%e Apart from the numbers along the diagonal, the absolute values show up in pairs if read along columns. Conjectures: Starting with the third line, columns are alternatingly fractions and integers. The row sums (1, 0, 1/2, -1/2, 1/4, -1/4, 3/4, -3/4, -11/8, 11/8,..) also show up in pairs from the third row on.

%t max = 28; p[0, _] = 1; p[n_, x_] := (1 + x)*((1 + x)^(n - 1) + x^(n - 1))/2; t = Table[ Coefficient[p[n, x], x, k], {n, 0, max + 2}, {k, 0, max + 2}]; a[n_] := Inverse[t][[All, 3]][[n + 3]]; Table[a[n], {n, 0, max}] (* _Jean-François Alcover_, Jul 22 2013 *)

%K sign

%O 0,2

%A _Paul Curtz_, Sep 21 2007

%E Edited and extended by _R. J. Mathar_, Aug 02 2008