%I #13 Jul 21 2015 05:33:54
%S 1,-1,1,0,-1,1,1,-1,-1,1,1,1,-2,-1,1,-2,5,0,-3,-1,1,-9,5,10,-2,-4,-1,
%T 1,-9,-21,25,15,-5,-5,-1,1,50,-105,-11,62,19,-9,-6,-1,1,267,-141,-301,
%U 56,119,21,-14,-7,-1,1,413,777
%N Triangle by rows, inverse of A208891.
%F Inverse of triangle A208891, Pascal's triangle matrix with an appended right border of 1's.
%e Triangle starts:
%e 1;
%e -1, 1
%e 0, -1, 1
%e 1, -1, -1, 1;
%e 1, 1, -2, -1, 1;
%e -2, 5, 0, -3, -1, 1;
%e -9, 5, 10, -2, -4, -1, 1;
%e -9, -21, 25, 15, -5, -5, -1, 1;
%e 50, -105, -11, 62, 19, -9, -6, -1, 1;
%e 267, -141, -301, 56, 119, 21, -14, -7, -1, 1;
%e 413, 777, -1040, -566, 226, 198, 20, -20, -8, -1, 1;
%e ...
%p A208891 := proc(n,k)
%p if n <0 or k<0 or k>n then
%p 0;
%p elif n = k then
%p 1 ;
%p else
%p binomial(n-1,k) ;
%p end if;
%p end proc:
%p A259456 := proc(n)
%p local A, row, col ;
%p A := Matrix(n, n) ;
%p for row from 1 to n do
%p for col from 1 to n do
%p A[row, col] := A208891(row-1,col-1) ;
%p end do:
%p end do:
%p LinearAlgebra[MatrixInverse](A) ;
%p end proc:
%p A259456(20) ; # _R. J. Mathar_, Jul 21 2015
%Y Cf. A208891, A000587 (first column), A014619 (2nd column), A080956 (4th subdiagonal).
%K tabl,sign
%O 0,13
%A _Gary W. Adamson_, Jun 26 2012