A358934 a(n) = Fibonacci(n+1)^5 - Fibonacci(n-1)^5.

0, 1, 31, 242, 3093, 32525, 368168, 4051333, 45064131, 499200274, 5538624025, 61414079849, 681135796944, 7553728681433, 83772910243607, 929052526388050, 10303364319347757, 114266002348885717, 1267229634537217144, 14053790947047408701, 155858934437282250075

Offset

0,3

Comments

a(n) is the number of edge covers of a spider graph with five branches where each branch has n vertices besides the center vertex. The idea is each branch is treated as a path P_(n+2). Each branch acts independently then and has F_(n+1) covers (P_n has F(n-1) covers), hence F_(n+1)^5 total. Except we remove the cases where each branch is missing the connecting edge to the center, which is when that edge cover comes from P_n , hence the minus F_(n-1)^5.

An edge cover of a graph is a subset of edges for which each vertex is incident to at least one edge in the subset.

Links

Michael De Vlieger, Table of n, a(n) for n = 0..957

Feryal Alayont and Evan Henning, Edge Covers of Caterpillars, Cycles with Pendants, and Spider Graphs, J. Int. Seq. (2023) Vol. 26, Art. 23.9.4.

Index entries for linear recurrences with constant coefficients, signature (8,40,-60,-40,8,1).

Formula

From Stefano Spezia, Dec 07 2022: (Start)

G.f.: x*(1 + 23*x - 46*x^2 - 23*x^3 + x^4)/((1 + 4*x - x^2)*(1 - x - x^2)*(1 - 11*x - x^2)).

a(n) = 8*a(n-1) + 40*a(n-2) - 60*a(n-3) - 40*a(n-4) + 8*a(n-5) + a(n-6) for n > 5. (End)

5*a(n) = 2*A000045(n)+11*A049666(n)+8*(-1)^n*A001076(n). - R. J. Mathar, May 07 2024

Example

Case n=1 is a star graph with five branches and one edge cover (all edges).
   *   *
    \ /
  *__C__*
     |
     *
For n=2, there are 31 edge covers of the graph obtained by gluing five P_3 paths at one single vertex. Each of the pendant edges of the P_3's have to be in the edge cover for the pendants to be incident with an edge. The middle vertices are then automatically incident with at least one edge. There remains the center vertex. We then need at least one of the remaining five edges to be in the subset, giving us 2^5-1 choices.
   *__ *   *__*
       \ /
  *__*__C__*__*
        |
        *__*

Mathematica

LinearRecurrence[{8, 40, -60, -40, 8, 1}, {0, 1, 31, 242, 3093, 32525}, 20] (* Amiram Eldar, Dec 07 2022 *)

Prog

(Python)
from sympy import fibonacci
def a(n):
    return fibonacci(n+1)**5-fibonacci(n-1)**5
(Python)
from gmpy2 import fib2
def A358934(n): return sum(f:=fib2(n))**5-f[1]**5 # Chai Wah Wu, Jan 04 2023

Crossrefs

Cf. A000045, A056572.

Sequence in context: A059378 A024003 A258807 * A221848 A344723 A284926

Adjacent sequences: A358931 A358932 A358933 * A358935 A358936 A358937

Keyword

nonn,easy,changed

Author

Feryal Alayont, Dec 06 2022

Status

approved