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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A298271 Expansion of x/((1 - x)*(1 - 322*x + x^2)).

Original entry on oeis.org

0, 1, 323, 104006, 33489610, 10783550415, 3472269744021, 1118060074024348, 360011871566096036, 115922704584208899245, 37326750864243699460855, 12019097855581887017496066, 3870112182746503375934272398, 1246164103746518505163818216091
Offset: 0

Views

Author

Bruno Berselli, Jan 16 2018

Keywords

Crossrefs

Programs

  • Mathematica
    CoefficientList[x/((1 - x) (1 - 322 x + x^2)) + O[x]^20, x]
  • Maxima
    makelist(coeff(taylor(x/((1-x)*(1-322*x+x^2)), x, 0, n), x, n), n, 0, 20);
    
  • PARI
    a(n)=([0,1,0; 0,0,1; 1,-323,323]^n*[0;1;323])[1,1] \\ Charles R Greathouse IV, Jan 18 2018
    
  • PARI
    concat(0, Vec(x / ((1 - x)*(1 - 322*x + x^2)) + O(x^15))) \\ Colin Barker, Jan 19 2018
  • Sage
    gf = x/((1-x)*(1-322*x+x^2))
    print(taylor(gf, x, 0, 20).list())
    

Formula

G.f.: x/((1 - x)*(1 - 322*x + x^2)).
a(n) = a(-n-1) = 323*a(n-1) - 323*a(n-2) + a(n-3).
a(n) = (1/5760)*((2 + sqrt(5))^(4*n+2) + (2 + sqrt(5))^-(4*n+2) - 18).
a(n) = A298101(n) - A298101(n-1) + A298101(n-2) - A298101(n-3) + ..., hence:
a(n) + a(n-1) = A298101(n).
a(n) - a(n-1) = (1/144)*Fibonacci(12*n).
a(n) - a(n-2) = (1/8)*Fibonacci(12*n-6).
a(n)*a(n-2) = a(n-1)*(a(n-1) - 1).
Sum_{j>1} 1/a(j) = 161 - 72*sqrt(5) = A094214^12.
a(n) = A157459(n+1)/72. - Greg Dresden, Dec 02 2021

A292443 a(n) = (5/32)*A000045(6*n)^2.

Original entry on oeis.org

0, 10, 3240, 1043290, 335936160, 108170400250, 34830532944360, 11215323437683690, 3611299316401203840, 1162827164557749952810, 374426735688279083601000, 120564246064461307169569210, 38821312806020852629517684640, 12500342159292650085397524884890
Offset: 0

Views

Author

Felix Fröhlich, Sep 16 2017

Keywords

Comments

Every term is a triangular number. [Problem B-967 in Euler and Sadek, 2003; solution in Euler and Sadek, 2004]

Crossrefs

Subsequence of A000217.

Programs

  • GAP
    List([0..20], n-> 5*Fibonacci(6*n)^2/32); # G. C. Greubel, Feb 03 2019
  • Magma
    [5*Fibonacci(6*n)^2/32: n in [0..20]]; // G. C. Greubel, Feb 03 2019
    
  • Mathematica
    Table[(5/32) Fibonacci[6 n]^2, {n, 0, 13}] (* Michael De Vlieger, Sep 16 2017 *)
    LinearRecurrence[{323,-323,1},{0,10,3240},20] (* Harvey P. Dale, Aug 31 2024 *)
  • PARI
    a(n) = (5/32)*fibonacci(6*n)^2
    
  • Sage
    [5*fibonacci(6*n)^2/32 for n in (0..20)] # G. C. Greubel, Feb 03 2019
    

Formula

From Colin Barker, Sep 16 2017: (Start)
G.f.: 10*x*(1 + x) / ((1 - x)*(1 - 322*x + x^2)).
a(n) = ((161+72*sqrt(5))^(-n)*(-1+(161+72*sqrt(5))^n)^2) / 32.
a(n) = 323*a(n-1) - 323*a(n-2) + a(n-3) for n > 2.
(End)
a(n) = A000217(A132584(n)). - Amiram Eldar, Jan 11 2022
a(n) = 10*A298101(n). - Pontus von Brömssen, Jan 06 2025
Showing 1-2 of 2 results.