A111824 Column 0 of the matrix logarithm (A111823) of triangle A111820, which shifts columns left and up under matrix 5th power; these terms are the result of multiplying the element in row n by n!.
0, 1, -3, 16, 2814, -1092180, -3603928080, 58978973128440, 5974833380453777520, -3294186866481455009752320, -10279982482873484428390722523200, 175129088125361734252730927280177244800
Offset: 0
Keywords
Examples
A(x) = x - 3/2!*x^2 + 16/3!*x^3 + 2814/4!*x^4 - 1092180/5!*x^5 +... where e.g.f. A(x) satisfies: x/(1-x) = A(x) + A(x)*A(5*x)/2! + A(x)*A(5*x)*A(5^2*x)/3! + A(x)*A(5*x)*A(5^2*x)*A(5^3*x)/4! + ... Let G(x) be the g.f. of A111821 (column 1 of A111820), then G(x) = 1 + 5*A(x) + 5^2*A(x)*A(5*x)/2! + 5^3*A(x)*A(5*x)*A(5^2*x)/3! + 5^4*A(x)*A(5*x)*A(5^2*x)*A(5^3*x)/4! + ...
Crossrefs
Programs
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PARI
{a(n,q=5)=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(5^j*x)/(j+1).
Comments