A358917 a(n) = Fibonacci(n+1)^4 - Fibonacci(n-1)^4.

0, 1, 15, 80, 609, 4015, 27936, 190385, 1307775, 8956144, 61405905, 420831071, 2884553280, 19770670945, 135511114479, 928804587920, 6366127657281, 43634071586575, 299072419071840, 2049872742473489, 14050037090947935, 96300386075488816, 660052667580788145

Offset: 0

Author

Feryal Alayont, Dec 05 2022

Keywords

• nonn
• easy

Comments

a(n) is the number of edge covers of a spider graph with four branches where each branch has n vertices besides the center vertex.

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.

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)^4 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)^4.

Examples

			Case n=1 is a star graph with four branches and one edge cover (all edges).
   *   *
    \ /
  *__C__*
For n=2, there are 15 edge covers of the graph obtained by gluing four 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 four edges to be in the subset, giving us 2^4-1 choices.
    *       *
     \     /
      *   *
       \ /
  *__*__C__*__*
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..1196
• 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 (4,19,4,-1).

Crossrefs

Cf. A000045, A056571, A358934.

Programs

Mathematica

LinearRecurrence[{4, 19, 4, -1}, {0, 1, 15, 80}, 25] (* Amiram Eldar, Dec 06 2022 *)

Python

from sympy import fibonacci
def a(n): return fibonacci(n+1)**4-fibonacci(n-1)**4

Formula

From Stefano Spezia, Dec 06 2022: (Start)

G.f.: x*(1 + 11*x + x^2)/((1 - 7*x + x^2)*(1 + 3*x + x^2)).

a(n) = 4*a(n-1) + 19*a(n-2) + 4*a(n-3) - a(n-4) for n > 3. (End)

5*a(n) = 9*A004187(n) + 4*(-1)^n*A001906(n) . - R. J. Mathar, May 07 2024