A111839 Column 0 of the matrix logarithm (A111838) of triangle A111835, which shifts columns left and up under matrix 8th power; these terms are the result of multiplying the element in row n by n!.
0, 1, -6, 142, 31800, -159468264, -2481298801008, 1414130111428687344, 1827317023092830201950080, -89946874545119714361987192509568, -9262235489215916508714844705185660161280
Offset: 0
Keywords
Examples
A(x) = x - 6/2!*x^2 + 142/3!*x^3 + 31800/4!*x^4 - 159468264/5!*x^5 +... where e.g.f. A(x) satisfies: x/(1-x) = A(x) + A(x)*A(8*x)/2! + A(x)*A(8*x)*A(8^2*x)/3! + A(x)*A(8*x)*A(8^2*x)*A(8^3*x)/4! + ... Let G(x) be the g.f. of A111836 (column 1 of A111835), then G(x) = 1 + 8*A(x) + 8^2*A(x)*A(8*x)/2! + 8^3*A(x)*A(8*x)*A(8^2*x)/3! + 8^4*A(x)*A(8*x)*A(8^2*x)*A(8^3*x)/4! + ...
Crossrefs
Programs
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PARI
{a(n,q=8)=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(8^j*x)/(j+1).
Comments