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%I #10 Jan 21 2019 20:04:37
%S 1,1,1,1,8,1,1,33,33,1,1,112,378,112,1,1,353,2938,2938,353,1,1,1080,
%T 18987,44912,18987,1080,1,1,3265,111051,520523,520523,111051,3265,1,1,
%U 9824,612820,5131040,9998182,5131040,612820,9824,1
%N Triangle T(n,m) read by rows: the matrix product A130595 * A156919.
%C Row sums are A165968(n+1).
%C Because A130595 is the inverse of the Pascal triangle A007318, we have A007318 *(this lower triangular matrix) = A156919.
%F G.f.: 1/(1+x-xy-2x/(1-3xy/(1+x-4x/(1-5xy/(1+x-6x/(1-7xy/(1+x-8x/(1-9xy/(1+ ... (continued fraction).
%e Triangle begins in row n=0 with columns 0 <= m <= n:
%e 1;
%e 1, 1;
%e 1, 8, 1;
%e 1, 33, 33, 1;
%e 1, 112, 378, 112, 1;
%e 1, 353, 2938, 2938, 353, 1;
%e 1, 1080, 18987, 44912, 18987, 1080, 1;
%e 1, 3265, 111051, 520523, 520523, 111051, 3265, 1;
%e 1, 9824, 612820, 5131040, 9998182, 5131040, 612820, 9824, 1;
%p A156919 := proc(n,m) if n=m then 1; elif m=0 then 2^n ; elif m<0 or m>n then 0; else 2*(m+1)*procname(n-1,m)+(2*n-2*m+1)*procname(n-1,m-1) ; end if; end proc:
%p A130595 := proc(n,m) (-1)^(n+m)*binomial(n,m) ; end proc:
%p A185412 := proc(n,m) local a,j; a := 0 ; for j from m to n do a := a+A130595(n,j)*A156919(j,m) ; end do: a ; end proc: # _R. J. Mathar_, Feb 03 2011
%K nonn,easy,tabl
%O 0,5
%A _Paul Barry_, Jan 26 2011
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