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.

A190996 Fibonacci sequence beginning 10, 7.

Original entry on oeis.org

10, 7, 17, 24, 41, 65, 106, 171, 277, 448, 725, 1173, 1898, 3071, 4969, 8040, 13009, 21049, 34058, 55107, 89165, 144272, 233437, 377709, 611146, 988855, 1600001, 2588856, 4188857, 6777713, 10966570, 17744283, 28710853, 46455136, 75165989, 121621125
Offset: 0

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Comments

For n >= 5, the number a(n-3) is the dimension of a commutative Hecke algebra of affine type D_n with independent parameters. See Theorem 1.4, Corollary 1.5, and the table on page 524 in the link "Hecke algebras with independent parameters". - Jia Huang, Jan 20 2019
From Greg Dresden and Yiming Wu, Sep 10 2023: (Start)
For n >= 3, a(n) is the number of ways to tile this shape of length n+2 with squares and dominos:
||___________________||
|||_|||_|||_|||_|||
|| ||. (End)
For n >= 3, a(n) is the number of edge covers of the kayak paddle graphs KP(3,3,n-3), where we interpret KP(3,3,0) as two C_3's with one common vertex. - Feryal Alayont, Sep 28 2024

Crossrefs

Programs

  • Magma
    [n le 2 select 13-3*n else Self(n-1)+Self(n-2): n in [1..50]]; // Vincenzo Librandi, Feb 15 2012
    
  • Maple
    seq(coeff(series((10-3*x)/(1-x-x^2),x,n+1), x, n), n = 0 .. 40); # Muniru A Asiru, Jan 22 2019
  • Mathematica
    LinearRecurrence[{1, 1}, {10, 7}, 100]
  • PARI
    a(n)=7*fibonacci(n)+10*fibonacci(n-1) \\ Charles R Greathouse IV, Jun 08 2011
    
  • SageMath
    [7*fibonacci(n+1) +3*fibonacci(n-1) for n in range(51)] # G. C. Greubel, Oct 26 2022

Formula

a(n) = (5 + 2*sqrt(5)/5)*((1 + sqrt(5))/2)^n + (5 - 2*sqrt(5)/5)*((1 - sqrt(5))/2)^n. - Antonio Alberto Olivares, Jun 07 2011
a(n) = 7*Fibonacci(n) + 10*Fibonacci(n-1). - Charles R Greathouse IV, Jun 08 2011
G.f.: (10-3*x)/(1-x-x^2). - Colin Barker, Jan 11 2012
a(n) = 4*Fibonacci(n+1) + 3*LucasL(n). - G. C. Greubel, Oct 26 2022
a(n) = A000285(n)+3*A000285(n-1). - Feryal Alayont, Sep 28 2024
a(2*n) = A000285(n)^2 + A000285(n-1)^2. - Greg Dresden, Feb 28 2025