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

0,3

Comments

Let q=2; the g.f. of column k of A098539^m (matrix power m) is: 1 + Sum_{n>=1} (m*q^k)^n/n! * Product_{j=0..n-1} A(q^j*x).

Links

Table of n, a(n) for n=0..15.

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).

Example

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! +...

Prog

(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))}

Crossrefs

Cf. A111810 (matrix log), A098539 (triangle), A002449, A111814 (variant), A111941 (q=-1), A111843 (q=3), A111848 (q=4).

Sequence in context: A354240 A380014 A306404 * A186448 A377789 A144002

Adjacent sequences: A111808 A111809 A111810 * A111812 A111813 A111814

Keyword

sign

Author

Paul D. Hanna, Aug 22 2005

Status

approved