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Triangle by rows, inverse of A208891.
1

%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