A111834 Column 0 of the matrix logarithm (A111833) of triangle A111830, which shifts columns left and up under matrix 7th power; these terms are the result of multiplying the element in row n by n!.
0, 1, -5, 83, 16110, -40097784, -388036363380, 82804198261002036, 50475967918183333160880, -711988160501968313699728393632, -26438313284970847487368499812182785280, 22571673265500745067336177578868612107537514880
Offset: 0
Keywords
Examples
A(x) = x - 5/2!*x^2 + 83/3!*x^3 + 16110/4!*x^4 - 40097784/5!*x^5 +... where e.g.f. A(x) satisfies: x/(1-x) = A(x) + A(x)*A(7*x)/2! + A(x)*A(7*x)*A(7^2*x)/3! + A(x)*A(7*x)*A(7^2*x)*A(7^3*x)/4! + ... Let G(x) be the g.f. of A111831 (column 1 of A111830), then G(x) = 1 + 7*A(x) + 7^2*A(x)*A(7*x)/2! + 7^3*A(x)*A(7*x)*A(7^2*x)/3! + 7^4*A(x)*A(7*x)*A(7^2*x)*A(7^3*x)/4! + ...
Crossrefs
Programs
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PARI
{a(n,q=7)=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(7^j*x)/(j+1).
Comments