|
Search: id:A118933
|
|
|
| A118933 |
|
Triangle, read by rows, where T(n,k) = n!/[k!*(n-4*k)!*4^k)] for n>=4*k>=0. |
|
+0 4
|
|
| 1, 1, 1, 1, 1, 6, 1, 30, 1, 90, 1, 210, 1, 420, 1260, 1, 756, 11340, 1, 1260, 56700, 1, 1980, 207900, 1, 2970, 623700, 1247400, 1, 4290, 1621620, 16216200, 1, 6006, 3783780, 113513400, 1, 8190, 8108100, 567567000, 1, 10920, 16216200, 2270268000
(list; table; graph; listen)
|
|
|
OFFSET
|
0,6
|
|
|
COMMENT
|
Row n contains 1+floor(n/4) terms. Row sums yield A118934. Given column vector V = A118935, then V is invariant under matrix product T*V = V, or, A118935(n) = Sum_{k=0..n} T(n,k)*A118935(k). Given C = Pascal's triangle and T = this triangle, then matrix product M = C^-1*T yields M(4n,n) = (4*n)!/(n!*4^n), 0 otherwise (cf. A100861 formula due to Paul Barry).
|
|
FORMULA
|
E.g.f.: A(x,y) = exp(x + y*x^4/4).
|
|
EXAMPLE
|
Triangle begins:
1;
1;
1;
1;
1,6;
1,30;
1,90;
1,210;
1,420,1260;
1,756,11340;
1,1260,56700;
1,1980,207900;
1,2970,623700,1247400; ...
|
|
PROGRAM
|
(PARI) T(n, k)=if(n<4*k, 0, n!/(k!*(n-4*k)!*4^k))
|
|
CROSSREFS
|
Cf. A118934 (row sums), A118935 (invariant vector); variants: A100861, A118931.
Sequence in context: A050300 A118394 A145629 this_sequence A046212 A120105 A120101
Adjacent sequences: A118930 A118931 A118932 this_sequence A118934 A118935 A118936
|
|
KEYWORD
|
nonn,tabl
|
|
AUTHOR
|
Paul D. Hanna (pauldhanna(AT)juno.com), May 06 2006
|
|
|
Search completed in 0.002 seconds
|