A111844 Column 0 of the matrix logarithm (A111843) of triangle A111840, which shifts columns left and up under matrix cube; these terms are the result of multiplying the element in row n by n!.

0, 1, 3, 27, 486, 7776, -2423196, -97338996, 5883879500784, 548540050402080, -1737375315124971951360, -405928706169160555680960, 60788545124934395018363657569920, 36207408592259278909089966337224960, -237458310218887960183820317532070376189904640

Offset: 0

Author

Paul D. Hanna, Aug 23 2005

Keywords

• sign

Comments

Let q=3; the g.f. of column k of A111840^m (matrix power m) is: 1 + Sum_{n>=1} (m*q^k)^n/n! * Product_{j=0..n-1} A(q^j*x).

Examples

			E.g.f. A(x) = x + 3/2!*x^2 + 27/3!*x^3 + 486/4!*x^4 + 7776/5!*x^5
- 2423196/6!*x^6 - 97338996/7!*x^7 +...
where A(x) satisfies:
x = A(x) - A(x)*A(3*x)/2! + A(x)*A(3*x)*A(3^2*x)/3!
- A(x)*A(3*x)*A(3^2*x)*A(3^3*x)/4! + ...
also:
Let G(x) be the g.f. of A111841 (column 0 of A111840), then
G(x) = 1 + x + 3*x^2 + 18*x^3 + 216*x^4 + 5589*x^5 + 336555*x^6 +...
= 1 + A(x) + A(x)*A(3*x)/2! + A(x)*A(3*x)*A(3^2*x)/3!
+ A(x)*A(3*x)*A(3^2*x)*A(3^3*x)/4! +...
		

Crossrefs

Cf. A111843 (matrix log), A111840 (triangle), A111841, A111816 (variant), A111941 (q=-1), A111843 (q=3), A111848 (q=4).

Programs

PARI

{a(n,q=3)=local(A=Mat(1),B);if(n<0,0, for(m=1,n+1,B=matrix(m,m);for(i=1,m, for(j=1,i, if(j==i,B[i,j]=1,if(j==1,B[i,j]=(A^q)[i-1,1], B[i,j]=(A^q)[i-1,j-1]));));A=B); B=sum(i=1,#A,-(A^0-A)^i/i);return(n!*B[n+1,1]))}

Formula

E.g.f. satisfies: x = -Sum_{n>=1} Prod_{j=0..n-1} -A(3^j*x)/(j+1).