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.

Previous Showing 11-13 of 13 results.

A286910 Number of independent vertex sets and vertex covers in the n-antiprism graph.

Original entry on oeis.org

3, 1, 5, 10, 21, 46, 98, 211, 453, 973, 2090, 4489, 9642, 20710, 44483, 95545, 205221, 440794, 946781, 2033590, 4367946, 9381907, 20151389, 43283149, 92967834, 199685521, 428904338, 921243214, 1978737411, 4250128177, 9128846213, 19607839978, 42115660581
Offset: 0

Views

Author

Andrew Howroyd, May 15 2017

Keywords

Comments

Sequence extrapolated to n=0 using recurrence.

Crossrefs

Programs

  • Magma
    I:=[3,1,5]; [n le 3 select I[n] else Self(n-1)+2*Self(n-2)+Self(n-3): n in [1..33]]; // Vincenzo Librandi, May 16 2017
  • Mathematica
    CoefficientList[Series[(- 2 x^2 - 2 x + 3) / (- x^3 - 2 x^2 - x + 1), {x, 0, 40}], x] (* Vincenzo Librandi, May 16 2017 *)
    LinearRecurrence[{1, 2, 1}, {3, 1, 5}, 40] (* Vincenzo Librandi, May 16 2017 *)
    Table[RootSum[-1 - 2 # - #^2 + #^3 &, #^n &], {n, 20}] (* Eric W. Weisstein, Aug 16 2017 *)
    RootSum[-1 - 2 # - #^2 + #^3 &, #^Range[20] &] (* Eric W. Weisstein, Aug 16 2017 *)
  • PARI
    Vec((-2*x^2 - 2*x + 3)/(-x^3 - 2*x^2 - x + 1)+O(x^30))
    

Formula

a(n) = a(n-1) + 2*a(n-2) + a(n-3) for n>=3.
G.f.: (2*x^2 + 2*x - 3)/(x^3 + 2*x^2 + x - 1).
a(n) = n*Sum_{k=1..n} C(2*k,n-k)/k, a(0)=3. - Vladimir Kruchinin, Jun 13 2020

A136161 a(n) = 2*a(n-3) - a(n-6), starting a(0..5) = 0, 5, 2, 1, 3, 1.

Original entry on oeis.org

0, 5, 2, 1, 3, 1, 2, 1, 0, 3, -1, -1, 4, -3, -2, 5, -5, -3, 6, -7, -4, 7, -9, -5, 8, -11, -6, 9, -13, -7, 10, -15, -8, 11, -17, -9, 12, -19, -10, 13, -21, -11, 14, -23, -12, 15, -25, -13, 16, -27, -14
Offset: 0

Views

Author

Paul Curtz, Mar 16 2008

Keywords

Comments

Consider the general recurrence a(n) = k*a(n-1) + (5-2*k)*a(n-2) + (2-k)*a(n-3). The coefficients, in k, can be used to form the triple (k, 5-2*k, 2-k). Each triple is associated with a sequence, for example (0, 5, 2) leads to A111108, A112685, ..., (1, 3, 1) leads to A051927, A097075, ..., and so on. This sequence is formed from the triples {(0, 5, 2), (1, 3, 1), (2, 1, 0), (3, -1, -1), (4, -3, -2), ...}, for k >= 0. (Comment modified by G. C. Greubel, Dec 31 2023).

Crossrefs

Programs

  • Magma
    I:=[0,5,2,1,3,1]; [n le 6 select I[n] else 2*Self(n-3) - Self(n-6): n in [1..60]]; // G. C. Greubel, Dec 26 2023
    
  • Mathematica
    LinearRecurrence[{0,0,2,0,0,-1},{0,5,2,1,3,1},60] (* Harvey P. Dale, Aug 16 2012 *)
    Table[PadRight[{n, 5-2*n, 2-n}], {n,0,20}]//Flatten (* _G. C. Greubel, Dec 26 2023 *)
  • PARI
    Vec(x*(5+2*x+x^2-7*x^3-3*x^4)/((1-x)^2*(1+x+x^2)^2+O(x^99))) \\ Charles R Greathouse IV, Jul 06 2011
    
  • SageMath
    def a(n): # a = A136161
        if n<6: return (0,5,2,1,3,1)[n]
        else: return 2*a(n-3) - a(n-6)
    [a(n) for n in range(61)] # G. C. Greubel, Dec 26 2023

Formula

G.f.: x*(5+2*x+x^2-7*x^3-3*x^4) / ( (1-x)^2*(1+x+x^2)^2 ). - R. J. Mathar, Jul 06 2011
a(3n) = n.
a(3n+1) = 5 - 2*n.
a(3n+3) = 2 - n.
a(n) = (1/9)*( 27 - 2*(n+1) - 34*ChebyshevU(n, -1/2) + (-1)^n*(9*A099254(n) - 6*A099254(n-1)) ). - G. C. Greubel, Dec 26 2023

A201222 Number of ways to place k non-attacking knights on a 2 X n horizontal cylinder, summed over all k>=0.

Original entry on oeis.org

3, 9, 18, 81, 123, 324, 843, 2401, 5778, 15129, 39603, 104976, 271443, 710649, 1860498, 4879681, 12752043, 33385284, 87403803, 228886641, 599074578, 1568397609, 4106118243, 10750371856, 28143753123, 73681302249, 192900153618, 505022001201, 1322157322203
Offset: 1

Views

Author

Vaclav Kotesovec, Nov 28 2011

Keywords

Crossrefs

Programs

  • Mathematica
    Table[If[Mod[n,4]==0,LucasL[n/2]^4,LucasL[2n]+1+(-1)^n],{n,1,50}]
Previous Showing 11-13 of 13 results.