A100335 An inverse Catalan transform of J(2n).

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

Offset: 0

Author

Paul Barry, Nov 17 2004

Keywords

• easy
• nonn

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(x*c(x)), where c(x) is the g.f. of A000108.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (5,-9,8,-4).

Crossrefs

Cf. A000108, A001045, A002450, A010892.

Cf. A042965, A100334, A109466, A133110.

Programs

Magma

I:=[0,1,4,11]; [n le 4 select I[n] else 5*Self(n-1) -9*Self(n-2) +8*Self(n-3) -4*Self(n-4): n in [1..41]]; // G. C. Greubel, Jan 24 2023

Mathematica

LinearRecurrence[{5,-9,8,-4}, {0,1,4,11}, 41] (* G. C. Greubel, Jan 24 2023 *)

SageMath

def A100335(n): return (1/3)*((n+1)*2^n - chebyshev_U(n,1/2))
[A100335(n) for n in range(41)] # G. C. Greubel, Jan 24 2023

Formula

G.f.: x*(1-x)/(1 - 5*x + 9*x^2 - 8*x^3 + 4*x^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