A181143 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.
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, 1029, 91, 1, 1, 140, 2632, 10549, 10549, 2632, 140, 1, 1, 204, 5922, 39533, 73157, 39533, 5922, 204, 1, 1, 285, 12090, 124805, 384948, 384948, 124805, 12090, 285, 1, 1
Offset: 0
Examples
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 +... The logarithm of the g.f. equals the series: log(A(x,y)) = (1 + y)*x + (1 + 2^3*y + y^2)*x^2/2 + (1 + 3^3*y + 3^3*y^2 + y^3)*x^3/3 + (1 + 4^3*y + 6^3*y^2 + 4^3*y^3 + y^4)*x^4/4 + (1 + 5^3*y + 10^3*y^2 + 10^3*y^3 + 5^3*y^4 + y^5)*x^5/5 +... Triangle begins: 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, 1029, 91, 1; 1, 140, 2632, 10549, 10549, 2632, 140, 1; 1, 204, 5922, 39533, 73157, 39533, 5922, 204, 1; 1, 285, 12090, 124805, 384948, 384948, 124805, 12090, 285, 1; 1, 385, 22869, 345389, 1648478, 2748240, 1648478, 345389, 22869, 385, 1; 1, 506, 40678, 861080, 6016297, 15525056, 15525056, 6016297, 861080, 40678, 506, 1; ... Note that column 1 forms the sum of squares (A000330). Inverse binomial transform of columns begins: [1]; [1, 4, 5, 2]; [1, 13, 58, 123, 136, 76, 17]; [1, 29, 278, 1308, 3532, 5867, 6118, 3914, 1407, 218]; [1, 54, 920, 7626, 36916, 114637, 240271, 348354, 350881, 241531, 108551, 28742, 3404]; ... 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.
Crossrefs
Programs
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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)} for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print(""))
Comments