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+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
KEYWORD
nonn,easy
AUTHOR
Artur Jasinski, Oct 05 2008
STATUS
approved