A111816 Column 0 of the matrix logarithm (A111815) of triangle A078122, which shifts columns left and up under matrix cube; these terms are the result of multiplying the element in row n by n!.
0, 1, -1, -3, 150, 1236, -2555748, -64342116, 5885700899760, 442646611978752, -1737387344860364226240, -367706581563500487774720, 60788555325888838346137808787840, 34626906551623392401873575206240000, -237458311254822429335982538087618909465992960
Offset: 0
Keywords
Examples
E.g.f.: A(x) = x - 1/2!*x^2 - 3/3!*x^3 + 150/4!*x^4 + 1236/5!*x^5 +... where e.g.f. A(x) satisfies: x/(1-x) = A(x) + A(x)*A(3*x)/2! + A(x)*A(3*x)*A(3^2*x)/3! + A(x)*A(3*x)*A(3^2*x)*A(3^3*x)/4! + ... Let G(x) be the g.f. of A078124 (column 1 of A078122), then G(x) = 1 + 3*A(x) + 3^2*A(x)*A(3*x)/2! + 3^3*A(x)*A(3*x)*A(3^2*x)/3! + 3^4*A(x)*A(3*x)*A(3^2*x)*A(3^3*x)/4! + ...
Crossrefs
Programs
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PARI
{a(n,q=3)=local(A=x/(1-x+x*O(x^n)));for(i=1,n, A=x/(1-x)/(1+sum(j=1,n,prod(k=1,j,subst(A,x,q^k*x))/(j+1)!))); return(n!*polcoeff(A,n))}
Formula
E.g.f. satisfies: x/(1-x) = Sum_{n>=1} Prod_{j=0..n-1} A(3^j*x)/(j+1).
Comments