|
Search: id:A111834
|
|
|
| A111834 |
|
Column 0 of the matrix logarithm (A111833) of triangle A111830, which shifts columns left and up under matrix 7-th power; these terms are the result of multiplying the element in row n by n!. |
|
+0
9
|
|
| 0, 1, -5, 83, 16110, -40097784, -388036363380, 82804198261002036, 50475967918183333160880, -711988160501968313699728393632, -26438313284970847487368499812182785280, 22571673265500745067336177578868612107537514880
(list; graph; listen)
|
|
|
OFFSET
|
0,3
|
|
|
COMMENT
|
Let q=7; the g.f. of column k of A111830^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(7^j*x)/(j+1).
|
|
EXAMPLE
|
A(x) = x - 5/2!*x^2 + 83/3!*x^3 + 16110/4!*x^4 - 40097784/5!*x^5 +...
where e.g.f. A(x) satisfies:
x/(1-x) = A(x) + A(x)*A(7*x)/2! + A(x)*A(7*x)*A(7^2*x)/3! +
A(x)*A(7*x)*A(7^2*x)*A(7^3*x)/4! + ...
Let G(x) be the g.f. of A111831 (column 1 of A111830), then
G(x) = 1 + 7*A(x) + 7^2*A(x)*A(7*x)/2! +
7^3*A(x)*A(7*x)*A(7^2*x)/3! +
7^4*A(x)*A(7*x)*A(7^2*x)*A(7^3*x)/4! + ...
|
|
PROGRAM
|
(PARI) {a(n, q=7)=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. A111830 (triangle), A111831, A111833 (matrix log); A110505 (q=-1), A111814 (q=2), A111816 (q=3), A111819 (q=4), A111824 (q=5), A111829 (q=6), A111839 (q=8).
Adjacent sequences: A111831 A111832 A111833 this_sequence A111835 A111836 A111837
Sequence in context: A082546 A035512 A054953 this_sequence A006471 A061628 A111911
|
|
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
|
