A346513 a(n) = Fibonacci(n+1)^3 - Fibonacci(n)^3.

1, 0, 7, 19, 98, 387, 1685, 7064, 30043, 127071, 538594, 2281015, 9663353, 40933296, 173398367, 734523803, 3111498370, 13180509531, 55833549037, 236514685384, 1001892323411, 4244083925895, 17978228112962, 76156996238639, 322606213292593, 1366581849044832

Offset: 0

Author

Lamine Ngom, Jul 21 2021

Keywords

• nonn
• easy

Comments

The version related to sum of consecutive Fibonacci numbers cubed is given by A110224.

a(n+1) is divisible by Fibonacci(n). The related quotient sequence is provided by A061646, from its 3rd term.

Links

• Index entries for linear recurrences with constant coefficients, signature (3,6,-3,-1).

Crossrefs

Cf. A056570 (partial sums).

Cf. A000045, A110224, A061646.

Programs

Mathematica

Differences[Fibonacci[Range[0, 26]]^3] (* Amiram Eldar, Jul 22 2021 *)

PARI

a(n) = fibonacci(n+1)^3 - fibonacci(n)^3; \\ Michel Marcus, Jul 22 2021

Formula

a(n) = F(n-1)*(2*F(n+1)^2+(-1)^(n+1)), n>0.

a(n) = F(n-1)*A061646(n+1).

G.f.: (x-1)*(x^2+2*x-1)/((x^2+4*x-1)*(x^2-x-1)). - Alois P. Heinz, Jul 21 2021

For n >= 2, a(n) is the numerator of the continued fraction [1,...,1, 3 ,1,...,1, 2 ,1,...,1] with three runs of 1's each of length n-2. For example, a(5)=387 which is the numerator of the continued fraction [1,1,1, 3 ,1,1,1, 2 ,1,1,1]. - Greg Dresden, Jan 01 2022