A128935 -id:A128935 - cp's OEIS frontend

A145232 a(n) = Fibonacci(5^n).

1, 5, 75025, 59425114757512643212875125, 18526362353047317310282957646406309593963452838196423660508102562977229905562196608078556292556795045922591488273554788881298750625

Offset: 0

Artur Jasinski, Oct 05 2008

Seiichi Manyama, Table of n, a(n) for n = 0..5
Robert Frontczak, Problem B-1341, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 62, No. 1 (2024), p. 84.
Thomas Koshy and Zhenguang Gao, Polynomial Extensions of a Diminnie Delight, Fibonacci Quart. 55 (2017), no. 1, 13-20.
Achilleas Sinefakopoulos, Solution to Problem 1909, Crux Mathematicorum, 20 (1994), 295-296.

Cf. A000045.

Cf. (k^n)-th Fibonacci number: A058635 (k=2), A045529 (k=3), A145231 (k=4), this sequence (k=5), A145233 (k=6), A145234 (k=7), A250487 (k=8), A250488 (k=9), A250489 (k=10).

a := proc(n) option remember; if n = 0 then 1 else 25*a(n-1)^5 - 25*a(n-1)^3 + 5*a(n-1) end if; end:
seq(a(n), n = 0..5); # Peter Bala, Nov 24 2022

G = (1 + Sqrt[5])/2; Table[Expand[(G^(5^n) - (1 - G)^(5^n))/Sqrt[5]], {n, 1, 6}]
Table[Round[N[(2/Sqrt[5])*Cosh[5^n*ArcCosh[Sqrt[5]/2]], 1000]], {n, 1, 4}]
Fibonacci[5^Range[0,4]] (* Harvey P. Dale, Nov 29 2018 *)

a(n) = (G^(5^n) - (1 - G)^(5^n))/sqrt(5) where G = (1 + sqrt(5))/2.
a(n) = (2/sqrt(5))*cosh((2*k+1)^n*arccosh(sqrt(5)/2)).
a(n) = (2/sqrt(5))*cosh(5^n*arccosh(sqrt(5)/2)).
a(n) = (5^n)*A128935(n). - R. J. Mathar, Nov 04 2010
a(n) = A000045(A000351(n)). - Michel Marcus, Nov 07 2013
a(n+1) = 25*a(n)^5 - 25*a(n)^3 + 5*a(n) with a(0) = 1. - Peter Bala, Nov 24 2022
a(n) = 5^n * Product_{k=0..n-1} (5*a(k)^4 - 5*a(k)^2 + 1) (Frontczak, 2024). - Amiram Eldar, Feb 29 2024

A113471 Lucas(k)/(3k) for k = 2*3^n, where Lucas(k) is k-th Lucas number (A000032).

Original entry on oeis.org

1, 107, 1190741689, 14769352340699478579719327005523, 253650450218391594062880777243777017638488805917392303113120204411172926964476779033181303378188721

Offset: 1

Alexander Adamchuk, May 13 2007

nonn

a(n) divides a(n+1). a(n+1)/a(n) = {107, 11128427, 12403489755282666163307, 17174107866559209832245996776509546318861182768126017871860347845227, ...}. a(n+1)/a(n) is prime for n = {1, 2, 4}.

Cf. A000032, A016089 = numbers n such that n divides n-th Lucas number. Cf. A128935 = Fibonacci(5^n) / 5^n.

Table[ ( Fibonacci[ 2*3^n - 1 ] + Fibonacci[ 2*3^n + 1 ] ) / ( 2*3^(n+1) ), {n,1,5} ]

a(n) = ( Fibonacci[ 2*3^n - 1 ] + Fibonacci[ 2*3^n + 1 ] ) / ( 2*3^(n+1) ). a(n) = A000032[ 2*3^n ] / ( 2*3^(n+1) ).

cp's OEIS Frontend

A145232 a(n) = Fibonacci(5^n).

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