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%I #10 Oct 12 2014 23:23:36
%S 1,1,1,1,3,2,1,6,12,6,1,10,40,60,25,1,15,100,300,375,136,1,21,210,
%T 1050,2625,2856,927,1,18,392,2940,12250,26656,25956,7690,1,36,672,
%U 7056,44100,15993,311472,276840,75913
%N Eigentriangle of A001263, the Narayana triangle.
%C The Narayana triangle begins:
%C 1;
%C 1, 1;
%C 1, 3, 1;
%C 1, 6, 6, 1;
%C 1, 10, 20, 10, 1;
%C ...
%C An eigentriangle of T is generated by taking the termwise product of (n-1)-th row terms of triangle T (in this case the Narayana triangle A001263); and the eigensequence of T = A102812 = (1, 1, 2, 6, 25, 136, 927,...).
%C Sum of n-th row terms of triangle A143778 = rightmost term of (n+1)-th row.
%C Right border of the triangle = the eigensequence of T.
%C Row sums of the triangle = the eigensequence of T shifted one place to the left: (1, 2, 6, 25, 136,...)
%C (A102812 * 0^(n-k)) = an infinite lower triangular matrix with A102812 as the main diagonal and the rest zeros.
%F Triangle read by rows, A001263 * (A102812 * 0^(n-k)); 0<=k<=n
%F Apparently for k<n, a(n,k)= binomial(n+1,k+1)*n!/(n+1-k)!. - _Tom Copeland_, Oct 08 2014
%e Triangle begins:
%e 1;
%e 1, 1;
%e 1, 3, 2;
%e 1, 6, 12, 6;
%e 1, 10, 40, 60, 25;
%e 1, 15, 100, 300, 375, 136;
%e 1, 21, 210, 1050, 2625, 2856, 927;
%e ...
%e Row 3 = (1, 6, 12, 6) = (1*1, 6*1, 6*2, 1*6) = termwise product of row 3 of the Narayana triangle: (1, 6, 6, 1) and the first 4 terms of the eigensequence of the Narayana triangle = (1, 1, 2, 6).
%Y Cf. A001263, A102812.
%K nonn,tabl
%O 0,5
%A _Gary W. Adamson_, Aug 31 2008
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