A111829 Column 0 of the matrix logarithm (A111828) of triangle A111825, which shifts columns left and up under matrix 6th power; these terms are the result of multiplying the element in row n by n!.
0, 1, -4, 42, 7296, -7931976, -45557382240, 3064554175021200, 801993619807364206080, -2618439032548254776387771520, -30580166025709706974876961026475520, 4440597519115996836838709580481861376121600
Offset: 0
Keywords
Examples
A(x) = x - 4/2!*x^2 + 42/3!*x^3 + 7296/4!*x^4 - 7931976/5!*x^5 +... where e.g.f. A(x) satisfies: x/(1-x) = A(x) + A(x)*A(6*x)/2! + A(x)*A(6*x)*A(6^2*x)/3! + A(x)*A(6*x)*A(6^2*x)*A(6^3*x)/4! + ... Let G(x) be the g.f. of A111826 (column 1 of A111825), then G(x) = 1 + 6*A(x) + 6^2*A(x)*A(6*x)/2! + 6^3*A(x)*A(6*x)*A(6^2*x)/3! + 6^4*A(x)*A(6*x)*A(6^2*x)*A(6^3*x)/4! + ...
Crossrefs
Programs
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PARI
{a(n,q=6)=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(6^j*x)/(j+1).
Comments