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.

Showing 1-2 of 2 results.

A077943 Expansion of 1/(1 - 2*x + 2*x^2 - 2*x^3).

Original entry on oeis.org

1, 2, 2, 2, 4, 8, 12, 16, 24, 40, 64, 96, 144, 224, 352, 544, 832, 1280, 1984, 3072, 4736, 7296, 11264, 17408, 26880, 41472, 64000, 98816, 152576, 235520, 363520, 561152, 866304, 1337344, 2064384, 3186688, 4919296, 7593984, 11722752, 18096128, 27934720, 43122688
Offset: 0

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Author

N. J. A. Sloane, Nov 17 2002

Keywords

Comments

a(n) gives the lower independence number of the (n+3)-halved cube graph up to at least n = 7. - Eric W. Weisstein, Dec 14 2023

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[1/(1 - 2 x + 2 x^2 - 2 x^3), {x, 0, 50}], x] (* Harvey P. Dale, Nov 30 2011 *)
    LinearRecurrence[{2, -2, 2}, {1, 2, 2}, 50] (* Harvey P. Dale, Nov 30 2011 *)
    Table[RootSum[-2 + 2 # - 2 #^2 + #^3 &, 4 #^n - 6 #^(n + 1) + 7 #^(n + 2) &]/22, {n, 0, 20}] (* Eric W. Weisstein, Dec 14 2023 *)
  • PARI
    Vec(1/(1-2*x+2*x^2-2*x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

Formula

a(n) = 2*a(n-1) - 2*a(n-2) + 2*a(n-3); a(0)=1, a(1)=2, a(2)=2. - Harvey P. Dale, Nov 30 2011
From R. J. Mathar, Mar 13 2021: (Start)
a(n) + a(n+1) = |A078071(n+1)|.
a(n) = (-1)^n*A077993(n). (End)

A078071 Expansion of (1-x)/(1+2*x+2*x^2+2*x^3).

Original entry on oeis.org

1, -3, 4, -4, 6, -12, 20, -28, 40, -64, 104, -160, 240, -368, 576, -896, 1376, -2112, 3264, -5056, 7808, -12032, 18560, -28672, 44288, -68352, 105472, -162816, 251392, -388096, 599040, -924672, 1427456, -2203648, 3401728, -5251072, 8105984, -12513280, 19316736, -29818880, 46030848
Offset: 0

Views

Author

N. J. A. Sloane, Nov 17 2002

Keywords

Crossrefs

First differences of A077993.

Programs

  • PARI
    Vec((1-x)/(1+2*x+2*x^2+2*x^3) + O(x^50)) \\ Jinyuan Wang, Apr 07 2020

Formula

a(0) = 1, a(1) = -3, a(2) = 4, a(n) = -2*(a(n-1) + a(n-2) + a(n-3)) for n > 2. - Jinyuan Wang, Apr 07 2020
Showing 1-2 of 2 results.