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

Author

Feryal Alayont, Dec 06 2022

Keywords

• nonn
• easy

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.

Examples

			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__*__*
        |
        *__*
		

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).

Crossrefs

Cf. A000045, A056572.

Programs

Mathematica

LinearRecurrence[{8, 40, -60, -40, 8, 1}, {0, 1, 31, 242, 3093, 32525}, 20] (* Amiram Eldar, Dec 07 2022 *)
Join[{0},#[[3]]-#[[1]]&/@Partition[Fibonacci[Range[0,30]]^5,3,1]] (* Harvey P. Dale, Aug 05 2024 *)

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

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