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%I #9 May 24 2014 18:37:11
%S 1,1,1,1,5,1,1,14,14,1,1,30,85,30,1,1,55,337,337,55,1,1,91,1029,2230,
%T 1029,91,1,1,140,2632,10549,10549,2632,140,1,1,204,5922,39533,73157,
%U 39533,5922,204,1,1,285,12090,124805,384948,384948,124805,12090,285,1,1
%N G.f.: A(x,y) = exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^3*y^k] * x^n/n ) = Sum_{n>=0,k=0..n} T(n,k)*x^n*y^k, as a triangle of coefficients T(n,k) read by rows.
%C Compare g.f. to that of the following triangle variants:
%C * Pascal's: exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)*y^k] * x^n/n );
%C * Narayana: exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^2*y^k] * x^n/n );
%C * A181144: exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^4*y^k] * x^n/n );
%C * A218115: exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^5*y^k] * x^n/n );
%C * A218116: exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^6*y^k] * x^n/n ).
%e G.f.: A(x,y) = 1 + (1+y)*x + (1+5*y+y^2)*x^2 + (1+14*y+14*y^2+y^3)*x^3 + (1+30*y+85*y^2+30*y^3+y^4)*x^4 +...
%e The logarithm of the g.f. equals the series:
%e log(A(x,y)) = (1 + y)*x
%e + (1 + 2^3*y + y^2)*x^2/2
%e + (1 + 3^3*y + 3^3*y^2 + y^3)*x^3/3
%e + (1 + 4^3*y + 6^3*y^2 + 4^3*y^3 + y^4)*x^4/4
%e + (1 + 5^3*y + 10^3*y^2 + 10^3*y^3 + 5^3*y^4 + y^5)*x^5/5 +...
%e Triangle begins:
%e 1;
%e 1, 1;
%e 1, 5, 1;
%e 1, 14, 14, 1;
%e 1, 30, 85, 30, 1;
%e 1, 55, 337, 337, 55, 1;
%e 1, 91, 1029, 2230, 1029, 91, 1;
%e 1, 140, 2632, 10549, 10549, 2632, 140, 1;
%e 1, 204, 5922, 39533, 73157, 39533, 5922, 204, 1;
%e 1, 285, 12090, 124805, 384948, 384948, 124805, 12090, 285, 1;
%e 1, 385, 22869, 345389, 1648478, 2748240, 1648478, 345389, 22869, 385, 1;
%e 1, 506, 40678, 861080, 6016297, 15525056, 15525056, 6016297, 861080, 40678, 506, 1; ...
%e Note that column 1 forms the sum of squares (A000330).
%e Inverse binomial transform of columns begins:
%e [1];
%e [1, 4, 5, 2];
%e [1, 13, 58, 123, 136, 76, 17];
%e [1, 29, 278, 1308, 3532, 5867, 6118, 3914, 1407, 218];
%e [1, 54, 920, 7626, 36916, 114637, 240271, 348354, 350881, 241531, 108551, 28742, 3404]; ...
%e the g.f. of the rightmost coefficients of which form the g.f. exp( Sum_{n>=1} (3*n)!/(3*n!^3) * x^n/n ), and yield the self-convolution of A229452.
%o (PARI) {T(n,k)=polcoeff(polcoeff(exp(sum(m=1,n,sum(j=0,m,binomial(m,j)^3*y^j)*x^m/m)+O(x^(n+1))),n,x),k,y)}
%o for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print(""))
%Y Cf. A000330 (column 1), A166990 (row sums), A166896 (antidiagonal sums), A218139.
%Y Cf. variants: A001263 (Narayana), A181144, A218115, A218116.
%K nonn,tabl
%O 0,5
%A _Paul D. Hanna_, Oct 13 2010
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