cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A135055 Pentanacci numbers: a(n) = a(n-1) + a(n-2) + a(n-3) + a(n-4) + a(n-5) for n>4 and with a(0)=-2, a(1)=-1, a(2)=0, a(3)=1, a(4)=2.

Original entry on oeis.org

-2, -1, 0, 1, 2, 0, 2, 5, 10, 19, 36, 72, 142, 279, 548, 1077, 2118, 4164, 8186, 16093, 31638, 62199, 122280, 240396, 472606, 929119, 1826600, 3591001, 7059722, 13879048, 27285490, 53641861, 105457122, 207323243, 407586764, 801294480, 1575303470, 3096965079, 6088473036, 11969622829
Offset: 0

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Author

Artur Jasinski, Nov 15 2007

Keywords

Crossrefs

Programs

  • Magma
    I:=[-2,-1,0,1,2]; [n le 5 select I[n] else Self(n-1)+Self(n-2)+Self(n-3)+Self(n-4)+Self(n-5): n in [1..40]]; // Vincenzo Librandi, Sep 22 2016
  • Mathematica
    a[n_] := a[n] = a[n - 1] + a[n - 2] + a[n - 3] + a[n - 4] + a[n - 5]; a[0] = -2; a[1] = -1; a[2] = 0; a[3] = 1; a[4] = 2; Table[a[n], {n, 0, 50}] (* Artur Jasinski, Nov 15 2007 *)
    LinearRecurrence[{1, 1, 1, 1, 1}, {-2, -1, 0, 1, 2}, 50] (* G. C. Greubel, Sep 21 2016 *)

Formula

From R. J. Mathar, Nov 18 2007: (Start)
G.f.: (1-2*x)*(x+1)*(2*x^2+x+2)/(-1+x+x^2+x^3+x^4+x^5).
a(n) = -2*A001591(n+4) + A001591(n+3) + 3*A001591(n+2) + 4*A001591(n+1) + 4*A001591(n). (End)

A251653 5-step Fibonacci sequence starting with 0,0,1,0,0.

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 2, 4, 7, 14, 28, 55, 108, 212, 417, 820, 1612, 3169, 6230, 12248, 24079, 47338, 93064, 182959, 359688, 707128, 1390177, 2733016, 5372968, 10562977, 20766266, 40825404, 80260631, 157788246, 310203524, 609844071, 1198921876, 2357018348, 4633776065, 9109763884, 17909324244, 35208804417
Offset: 0

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Author

Arie Bos, Dec 06 2014

Keywords

Comments

Doubling the entries > 1 as 1, 2, 2, 4, 4, 7, 7, 14, 14, 28, 28, 55, 55,... (offset 0) gives Nyblom's palindromic binary strings having no 5-runs of 1's. - R. J. Mathar, Mar 28 2025

Crossrefs

Programs

  • J
    (see www.jsoftware.com) First construct the generating matrix
    1  1  1  1  1
    1  2  2  2  2
    2  3  4  4  4
    4  6  7  8  8
    8 12 14 15 16
    Given that matrix one can produce the first 5*200 numbers by
    , M(+/ . *)^:(i.250) 0 0 1 0 0x
  • Mathematica
    LinearRecurrence[{1, 1, 1, 1, 1}, {0, 0, 1, 0, 0}, 100] (* G. C. Greubel, May 27 2016 *)

Formula

a(n+5) = a(n) + a(n+1) + a(n+2) + a(n+3) + a(n+4).
G.f.: x^2*(x^2 + x - 1)/(x^5 + x^4 + x^3 + x^2 + x - 1). - Chai Wah Wu, May 27 2016
Showing 1-2 of 2 results.