|
Search: id:A111839
|
|
|
| A111839 |
|
Column 0 of the matrix logarithm (A111838) of triangle A111835, which shifts columns left and up under matrix 8-th power; these terms are the result of multiplying the element in row n by n!. |
|
+0
9
|
|
| 0, 1, -6, 142, 31800, -159468264, -2481298801008, 1414130111428687344, 1827317023092830201950080, -89946874545119714361987192509568, -9262235489215916508714844705185660161280
(list; graph; listen)
|
|
|
OFFSET
|
0,3
|
|
|
COMMENT
|
Let q=8; the g.f. of column k of A111825^m (matrix power m) is: 1 + Sum_{n>=1} (m*q^k)^n/n! * Product_{j=0..n-1} A(q^j*x).
|
|
FORMULA
|
E.g.f. satisfies: x/(1-x) = Sum_{n>=1} Prod_{j=0..n-1} A(8^j*x)/(j+1).
|
|
EXAMPLE
|
A(x) = x - 6/2!*x^2 + 142/3!*x^3 + 31800/4!*x^4 - 159468264/5!*x^5 +...
where e.g.f. A(x) satisfies:
x/(1-x) = A(x) + A(x)*A(8*x)/2! + A(x)*A(8*x)*A(8^2*x)/3! +
A(x)*A(8*x)*A(8^2*x)*A(8^3*x)/4! + ...
Let G(x) be the g.f. of A111836 (column 1 of A111835), then
G(x) = 1 + 8*A(x) + 8^2*A(x)*A(8*x)/2! +
8^3*A(x)*A(8*x)*A(8^2*x)/3! +
8^4*A(x)*A(8*x)*A(8^2*x)*A(8^3*x)/4! + ...
|
|
PROGRAM
|
(PARI) {a(n, q=8)=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))}
|
|
CROSSREFS
|
Cf. A111835 (triangle), A111836, A111838 (matrix log); A110505 (q=-1), A111814 (q=2), A111816 (q=3), A111819 (q=4), A111824 (q=5), A111829 (q=6), A111834 (q=7).
Sequence in context: A059488 A067196 A048863 this_sequence A128785 A010043 A085905
Adjacent sequences: A111836 A111837 A111838 this_sequence A111840 A111841 A111842
|
|
KEYWORD
|
sign
|
|
AUTHOR
|
Gottfried Helms (helms(AT)uni-kassel.de) and Paul D. Hanna (pauldhanna(AT)juno.com), Aug 22 2005
|
|
|
Search completed in 0.002 seconds
|
