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