A054885 Number of walks of length n along the edges of an icosahedron between two opposite vertices.

0, 0, 0, 10, 40, 260, 1240, 6510, 32240, 162760, 812240, 4069010, 20337240, 101725260, 508587240, 2543131510, 12715462240, 63578287760, 317890462240, 1589457194010, 7947281087240, 39736429850260

Offset

0,4

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (4,10,-20,-25).

Formula

G.f.: (1/12)*(1/(1-5*t) + 5/(1+t) - 6/(1-5*t^2)).

a(n) = (5^n + 5*(-1)^n - 3*(1 + (-1)^n)*sqrt(5)^n)/12.

a(n+1) = 5 * A030518(n) for n > 0.

a(n) = 4*a(n-1) + 10*a(n-2) - 20*a(n-3) - 25*a(n-4). - François Marques, Jul 10 2021

E.g.f.: (1/12)*(5*exp(-x) + exp(5*x) - 6*cosh(sqrt(5)*x)). - G. C. Greubel, Feb 07 2023

Mathematica

LinearRecurrence[{4, 10, -20, -25}, {0, 0, 0, 10}, 41] (* G. C. Greubel, Feb 07 2023 *)

Prog

(Magma) [Floor((5^n+(-1)^n*5-3*(1+(-1)^n)*Sqrt(5)^n)/12): n in [0..30]]; // Vincenzo Librandi, Aug 24 2011
(PARI) a(n) = if(n%2, 5^n-5, 5^n+5-6*5^(n/2))/12; \\ François Marques, Jul 11 2021
(SageMath)
def A054885(n): return (5^n +5*(-1)^n -3*(1+(-1)^n)*5^(n/2))/12
[A054885(n) for n in range(41)] # G. C. Greubel, Feb 07 2023

Crossrefs

Cf. A030517, A030518, A054881, A054882, A054883, A054884.

Sequence in context: A061991 A060580 A118266 * A000449 A027274 A253674

Adjacent sequences: A054882 A054883 A054884 * A054886 A054887 A054888

Keyword

nonn,walk,easy

Author

Paolo Dominici (pl.dm(AT)libero.it), May 23 2000

Status

approved