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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A136201 a(n) = 2*a(n-1) + 4*a(n-2) - 6*a(n-3) - 3*a(n-4).

Original entry on oeis.org

0, 0, 0, 1, 2, 8, 18, 53, 124, 328, 780, 1969, 4718, 11648, 28014, 68405, 164824, 400240, 965304, 2337409, 5640122, 13637336, 32914794, 79525973, 191966740, 463636600, 1119239940, 2702647921, 6524535782, 15753313808, 38031163398
Offset: 0

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Author

Paul Curtz, Mar 16 2008

Keywords

Comments

Based on a Pell recurrence.

Programs

  • Maple
    a:=proc(n) options operator, arrow: expand((1/8)*(1+sqrt(2))^n+(1/8)*(1-sqrt(2))^n+(1/24)*3^((1/2)*n)*(-3-sqrt(3)-3*(-1)^n+(-1)^n*sqrt(3))) end proc: seq(a(n),n=0..30); # Emeric Deutsch, Mar 31 2008
  • Mathematica
    LinearRecurrence[{2, 4, -6, -3}, {0, 0, 0, 1}, 50] (* G. C. Greubel, Feb 23 2017 *)
    CoefficientList[Series[x^3/(1-2 x-4 x^2+6 x^3+3 x^4),{x,0,50}],x] (* Harvey P. Dale, Apr 21 2022 *)
  • PARI
    x='x+O('x^50); Vec(x^3/(3*x^4 + 6*x^3 - 4*x^2 - 2*x + 1)) \\ G. C. Greubel, Feb 23 2017

Formula

A137255(n+1) - 2*A137255(n), same recurrence.
a(n) = (-A108411(n) + A001333(n))/4. - R. J. Mathar, Apr 01 2008
a(n) = (1/8)*(1+sqrt(2))^n + (1/8)*(1-sqrt(2))^n + (1/24)*3^(n/2)*(-3 - sqrt(3) - 3(-1)^n + (-1)^n*sqrt(3)). - Emeric Deutsch, Mar 31 2008
G.f.: x^3/(3*x^4 + 6*x^3 - 4*x^2 - 2*x + 1). - Alexander R. Povolotsky, Mar 31 2008
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