A111814 Column 0 of the matrix logarithm (A111813) of triangle A078121, which shifts columns left and up under matrix square; these terms are the result of multiplying the element in row n by n!.
0, 1, 0, -2, 0, 216, 0, -568464, 0, 36058658688, 0, -53694310935340800, 0, 1790669979087018171448320, 0, -1280832788659041410080025283840000, 0, 18961468161294510864200732026858464699187200, 0
Offset: 0
Keywords
Examples
E.g.f.: A(x) = x - 2/3!*x^3 + 216/5!*x^5 - 568464/7!*x^7 + ... where A(x) satisfies: x/(1-x) = A(x) + A(x)*A(2*x)/2! + A(x)*A(2*x)*A(2^2*x)/3! + A(x)*A(2*x)*A(2^2*x)*A(2^3*x)/4! + ... also: x/(1-x^2) = A(x) + A(x)*A(2*x)*A(2^2*x)/3! + A(x)*A(2*x)*A(2^2*x)*A(2^3*x)*A(2^4*x)/5! + ... Let G(x) be the g.f. of A002577 (column 1 of A078121), then G(x) = 1 + 2*A(x) + 2^2*A(x)*A(2*x)/2! + 2^3*A(x)*A(2*x)*A(2^2*x)/3! + 2^4*A(x)*A(2*x)*A(2^2*x)*A(2^3*x)/4! + ...
Links
- Paul D. Hanna, Table of n, a(n), n== 0..50.
Crossrefs
Programs
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PARI
{a(n,q=2)=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(2^j*x)/(j+1). E.g.f. satisfies: x/(1-x^2) = Sum_{n>=1}Prod_{j=0..2*n}A(2^j*x)/(j+1).
Comments