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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) = 5*a(n-1) - 9*a(n-2) + 8*a(n-3) - 4*a(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..n} A109466(n,k)*A002450(k). - Philippe Deléham, Oct 30 2008

CROSSREFS

Cf. A001045, A100334.

Cf. A133110, A042965.

Sequence in context: A000253 A276691 A047859 * A340228 A080869 A137229

Adjacent sequences:  A100332 A100333 A100334 * A100336 A100337 A100338

KEYWORD

easy,nonn

AUTHOR

Paul Barry, Nov 17 2004

STATUS

approved

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Last modified March 1 17:55 EST 2021. Contains 341740 sequences. (Running on oeis4.)