A138003 Binomial transform of 1, 1, 0, -1, -1 (periodically continued).

1, 2, 3, 3, 0, -8, -21, -34, -34, 0, 89, 233, 377, 377, 0, -987, -2584, -4181, -4181, 0, 10946, 28657, 46368, 46368, 0, -121393, -317811, -514229, -514229, 0, 1346269, 3524578, 5702887, 5702887, 0, -14930352, -39088169, -63245986, -63245986

Offset: 0

Author

Paul Curtz, May 01 2008

Keywords

• sign
• easy

Comments

Shares many elements with A103311, as already indicated by the similarity of the two generating functions. First differences are essentially in A105371. - R. J. Mathar, May 02 2008

The longer of the two recurrences ensures that the sequence (like A133476) equals its 5th differences. - R. J. Mathar, May 02 2008

Links

• Paolo Xausa, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (3,-4,2,-1).

Crossrefs

Cf. A129929.

Programs

Mathematica

LinearRecurrence[{3,-4,2,-1},{1,2,3,3},50] (* Paolo Xausa, Dec 05 2023 *)

PARI

a=[1,2,3,3];for(i=1,99,a=concat(a,3*a[#a]-4*a[#a-1]+2*a[#a-2]-a[#a-3]));a \\ Charles R Greathouse IV, Jun 02 2011

Formula

From R. J. Mathar, May 02 2008: (Start)

O.g.f.: (x^2-x+1)/(x^4-2*x^3+4*x^2-3*x+1).

a(n) = 5a(n-1)-10a(n-2)+10a(n-3)-5a(n-4)+2a(n-5).

a(n) = 3a(n-1)-4a(n-2)+2a(n-3)-a(n-4). (End)

Extensions

Edited by R. J. Mathar, May 02 2008