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%I #9 Mar 04 2018 03:08:41
%S 1,1,1,1,2,1,1,5,3,1,1,14,12,4,1,1,42,57,22,5,1,1,132,303,148,35,6,1,
%T 1,429,1743,1144,305,51,7,1,1,1430,10629,9784,3105,546,70,8,1,1,4862,
%U 67791,90346,35505,6906,889,92,9,1,1,16796,448023,885868,444225,99156
%N Triangle, read by rows, such that column n equals the row sums of A001263^n, which is the n-th matrix power of the Narayana triangle A001263, for n>=0.
%C Column 1 is the Catalan numbers A000108 (offset 1).
%F From _Paul D. Hanna_, Feb 01 2009: (Start)
%F G.f. of column k = B(x)^(k+1) where B(x) = Sum_{n>=0} x^n/[n!*(n+1)!/2^n];
%F T(n,K) = [x^(n-k)] B(x)^(k+1) * (n-k)!*(n-k+1)!/2^(n-k) for n >= k >= 0. (End)
%e Triangle begins:
%e 1;
%e 1, 1;
%e 1, 2, 1;
%e 1, 5, 3, 1;
%e 1, 14, 12, 4, 1;
%e 1, 42, 57, 22, 5, 1;
%e 1, 132, 303, 148, 35, 6, 1;
%e 1, 429, 1743, 1144, 305, 51, 7, 1;
%e 1, 1430, 10629, 9784, 3105, 546, 70, 8, 1;
%e 1, 4862, 67791, 90346, 35505, 6906, 889, 92, 9, 1;
%e ...
%e From _Paul D. Hanna_, Feb 01 2009: (Start)
%e G.f. for rows n=0..3 are:
%e B(x) = 1 + x + x^2/3 + x^3/18 + x^4/180 + x^5/2700 + ... + x^n/[n!*(n+1)!/2^n] + ...
%e B(x)^2 = 1 + 2*x + 5*x^2/3 + 14*x^3/18 + 42*x^4/180 + ... + A000108(n)*x^n/[n!*(n+1)!/2^n] + ...
%e B(x)^3 = 1 + 3*x +12*x^2/3 + 57*x^3/18 +303*x^4/180 + ... + A103370(n)*x^n/[n!*(n+1)!/2^n] + ...
%e B(x)^4 = 1 + 4*x +22*x^2/3 +148*x^3/18+1144*x^4/180 + 9784*x^5/2700 + 90346*x^5/56700 + ... (End)
%o (PARI) {T(n,k)=local(N=matrix(n+1,n+1,m,j,if(m>=j, binomial(m-1,j-1)*binomial(m,j-1)/j))); sum(j=0,n-k,(N^k)[n-k+1,j+1])}
%o (PARI) {T(n,k)=local(B=sum(j=0,n-k,x^j/(j!*(j+1)!/2^j))+x*O(x^(n-k))); polcoeff(B^(k+1),n-k)*(n-k)!*(n-k+1)!/2^(n-k)} \\ _Paul D. Hanna_, Feb 01 2009
%Y Cf. A001263, A105557 (row sums), A103370 (column 2).
%Y Cf. A155926. - _Paul D. Hanna_, Feb 01 2009
%K nonn,tabl
%O 0,5
%A _Paul D. Hanna_, Apr 14 2005
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