|
| |
|
|
A125273
|
|
Eigensequence of triangle A085478: a(n) = Sum_{k=0..n-1} A085478(n-1,k)*a(k) for n>0 with a(0)=1.
|
|
7
| |
|
|
1, 1, 2, 6, 23, 106, 567, 3434, 23137, 171174, 1376525, 11934581, 110817423, 1095896195, 11487974708, 127137087319, 1480232557526, 18075052037054, 230855220112093, 3076513227516437, 42686898298650967, 615457369662333260
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 0,3
|
|
|
LINKS
| Guo-Niu Han, Enumeration of Standard Puzzles
|
|
|
FORMULA
| a(n) = Sum_{k=0..n-1} C(n+k-1,n-k-1)*a(k) for n>0 with a(0)=1.
G.f. satisfies: A(x) = 1 + x*A( x/(1-x)^2 ) / (1-x). - Paul D. Hanna (pauldhanna(AT)juno.com), Aug 15 2007
|
|
|
EXAMPLE
| a(3) = 1*(1) + 3*(1) + 1*(2) = 6;
a(4) = 1*(1) + 6*(1) + 5*(2) + 1*(6) = 23;
a(5) = 1*(1) + 10*(1) + 15*(2) + 7*(6) + 1*(23) = 106.
Triangle A085478(n,k) = C(n+k,n-k) begins:
1;
1, 1;
1, 3, 1;
1, 6, 5, 1;
1, 10, 15, 7, 1;
1, 15, 35, 28, 9, 1; ...
where g.f. of column k = 1/(1-x)^(2*k+1).
|
|
|
PROG
| (PARI) a(n)=if(n==0, 1, sum(k=0, n-1, a(k)*binomial(n+k-1, n-k-1)))
|
|
|
CROSSREFS
| Cf. A085478; A125274 (variant).
Sequence in context: A165489 A192315 A193321 * A130908 A200404 A000772
Adjacent sequences: A125270 A125271 A125272 * A125274 A125275 A125276
|
|
|
KEYWORD
| nonn
|
|
|
AUTHOR
| Paul D. Hanna (pauldhanna(AT)juno.com), Nov 26 2006
|
| |
|
|
