A111811 Column 0 of the matrix logarithm (A111810) of triangle A098539, 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, 2, 10, 88, 1096, 11856, -402480, -1891968, 36024603264, 359905478400, -53686393014816000, -644141701131494400, 1790653231402788752593920, 25068910772059830672353280, -1280832036591718248285105113241600
Offset: 0
Keywords
Examples
A(x) = x + 2/2!*x^2 + 10/3!*x^3 + 88/4!*x^4 + 1096/5!*x^5 +... where e.g.f. A(x) satisfies: 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) = A(x) - 2*A(x)*A(2*x)/2! + 2^2*A(x)*A(2*x)*A(2^2*x)/3! - 2^3*A(x)*A(2*x)*A(2^2*x)*A(2^3*x)/4! +... Let G(x) be the g.f. of A002449 (column 1 of A098539), then (G(x)-1)/x = 1 + 2*x + 6*x^2 + 26*x^3 + 166*x^4 + 1626*x^5 +... = 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! +...
Crossrefs
Programs
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PARI
{a(n,q=2)=local(A=x+x*O(x^n));for(i=1,n, A=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 = -Sum_{n>=1} Prod_{j=0..n-1} -A(2^j*x)/(j+1), also: x/(1+x) = Sum_{n>=1} (-2)^(n-1)*Prod_{j=0..n-1} A(2^j*x)/(j+1).
Comments