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

1, 5, 75025, 59425114757512643212875125, 18526362353047317310282957646406309593963452838196423660508102562977229905562196608078556292556795045922591488273554788881298750625

Offset

0,2

Links

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.

Formula

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

Maple

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

Mathematica

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 *)

Crossrefs

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).

Sequence in context: A247845 A050816 A171981 * A263174 A123591 A133381

Adjacent sequences: A145229 A145230 A145231 * A145233 A145234 A145235

Keyword

nonn,easy

Author

Artur Jasinski, Oct 05 2008

Status

approved