A350473 a(n) = Fibonacci(n+1)^3 - Fibonacci(n-1)^3.

0, 1, 7, 26, 117, 485, 2072, 8749, 37107, 157114, 665665, 2819609, 11944368, 50596649, 214331663, 907922170, 3846022173, 16292007901, 69014058568, 292348234421, 1238407008795, 5245976249306, 22222312038857, 94135224351601, 398763209531232, 1689188062337425

Offset: 0

Author

Greg Dresden, Jan 01 2022

Keywords

• nonn
• easy

Comments

See A346513 for Fibonacci(n+1)^3 - Fibonacci(n)^3.

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..1595
• 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 (3,6,-3,-1).

Crossrefs

Cf. A000045, A110224, A346513.

Programs

Mathematica

Differences[Fibonacci[Range[-1, 26]]^3, 1, 2]

Python

from sympy import fibonacci
def A350473(n): return fibonacci(n+1)**3-fibonacci(n-1)**3 # Chai Wah Wu, Jan 05 2022

Formula

a(n) = 3*a(n-1) + 6*a(n-2) - 3*a(n-3) - a(n-4).

G.f.: x*(1 + 4*x - x^2)/(1 - 3*x - 6*x^2 + 3*x^3 + x^4).

a(n) = (4/5)*Fibonacci(3*n) + (-1)^(n)*(3/5)*Fibonacci(n).

a(n) is the numerator of the continued fraction [1,...,1, 2 ,1,...,1, 2 ,1,1,...,1] with the first two runs of 1's of length n-2 and the last run of length n-1. For example, a(4)=117 which is the numerator of the continued fraction [1,1, 2 ,1,1, 2 ,1,1,1].