A350473 - OEIS

%I #20 Nov 17 2023 12:22:29

%S 0,1,7,26,117,485,2072,8749,37107,157114,665665,2819609,11944368,

%T 50596649,214331663,907922170,3846022173,16292007901,69014058568,

%U 292348234421,1238407008795,5245976249306,22222312038857,94135224351601,398763209531232,1689188062337425

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

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

%H Michael De Vlieger, <a href="/A350473/b350473.txt">Table of n, a(n) for n = 0..1595</a>

%H Feryal Alayont and Evan Henning, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL26/Alayont/ala4.html">Edge Covers of Caterpillars, Cycles with Pendants, and Spider Graphs</a>, J. Int. Seq. (2023) Vol. 26, Art. 23.9.4.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3,6,-3,-1).

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

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

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

%F 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].

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

%o (Python)

%o from sympy import fibonacci

%o def A350473(n): return fibonacci(n+1)**3-fibonacci(n-1)**3 # _Chai Wah Wu_, Jan 05 2022

%Y Cf. A000045, A110224, A346513.

%K nonn,easy

%O 0,3

%A _Greg Dresden_, Jan 01 2022