%I #11 Jan 01 2013 07:08:09
%S 1,0,1,1,0,1,0,5,0,1,1,0,14,0,1,0,21,0,30,0,1,1,0,147,0,55,0,1,
%T 0,85,0,627,0,91,0,1,1,0,1408,0,2002,0,140,0,1,0,341,0,11440,0,
%U 5278,0,204,0,1,1,0,13013,0,61490,0,12138,0,285,0,1,0
%N Triangle T(n,k) = coefficient of x^n in expansion of ((22*cos(x))/x)^k = sum(n>=k, T(n,k) * x^n * (2*k)!/(n+k)!).
%C Triangle T(n,k)*(2*k)!/(n+k)!)=
%C 1. Riordan Array (1,(22*cos(x))/x) without first column.
%C 2. Riordan Array ((22*cos(x))/x^2,(22*cos(x))/x) numbering triangle (0,0).
%F T(n,k):=((1)^k*2^k*((1)^(n+k)+1)*sum(j=1..k, ((sum(i=0..(j1)/2, (j2*i)^(n+k)*binomial(j,i))) *binomial(k,j)*(1)^((n+k)/2+kj))/2^j))/(2*k)!
%e 1
%e 0, 1
%e 1, 0, 1
%e 0, 5, 0, 1
%e 1, 0, 14, 0, 1
%e 0, 21, 0, 30, 0, 1
%e 1, 0, 147, 0, 55, 0, 1
%o (Maxima)
%o T(n,k):=((1)^k*2^k*((1)^(n+k)+1)*sum(((sum((j2*i)^(n+k)*binomial(j,i),i,0,(j1)/2))*binomial(k,j)*(1)^((n+k)/2+kj))/2^j,j,1,k))/(2*k)!
%K sign,tabl
%O 1,8
%A _Vladimir Kruchinin_, Nov 11 2011
