|
| |
|
|
A100335
|
|
An inverse Catalan transform of J(2n).
|
|
5
|
|
|
|
0, 1, 4, 11, 27, 64, 149, 341, 768, 1707, 3755, 8192, 17749, 38229, 81920, 174763, 371371, 786432, 1660245, 3495253, 7340032, 15379115, 32156331, 67108864, 139810133, 290805077, 603979776, 1252698795, 2594876075, 5368709120
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,3
|
|
|
COMMENTS
|
The g.f. is obtained from that of A002450 through the mapping g(x)->g(x(1-x)). A002450 may be retrieved through the mapping g(x)->g(xc(x)), where c(x) is the g.f. of A000108.
|
|
|
LINKS
|
Table of n, a(n) for n=0..29.
Index entries for linear recurrences with constant coefficients, signature (5,-9,8,-4).
|
|
|
FORMULA
|
G.f.: x(1-x)/(1-5x+9x^2-8x^3+4x^4); a(n)=5a(n-1)-9a(n-2)+8a(n-3)-4a(n-4); a(n)=sum{k=0..floor(n/2), binomial(n-k, k)(-1)^k*(4^(n-k)-1)/3}.
a(n) = (1/3) [(n+1)*2^n - A010892(n) ]. - Ralf Stephan, May 15 2007
Binomial transform of A042965: (1, 3, 4, 5, 7, 8, 9, 11, 12, 13,...), also row sums of triangle A133110. - Gary W. Adamson, Sep 12 2007
a(n)=Sum_{k, 0<=k<=n} A109466(n,k)*A002450(k). [From Philippe Deléham, Oct 30 2008]
|
|
|
CROSSREFS
|
Cf. A001045, A100334.
Cf. A133110, A042965.
Sequence in context: A000253 A276691 A047859 * A080869 A137229 A301874
Adjacent sequences: A100332 A100333 A100334 * A100336 A100337 A100338
|
|
|
KEYWORD
|
easy,nonn
|
|
|
AUTHOR
|
Paul Barry, Nov 17 2004
|
|
|
STATUS
|
approved
|
| |
|
|
