OFFSET
0,1
COMMENTS
LINKS
Amiram Eldar, Table of n, a(n) for n = 0..4
FORMULA
a(n) = Fibonacci(5^(n+1))/Fibonacci(5^n).
Recurrence equation: a(n+1) = 5/2*(a(n)^4 - a(n)^2)*sqrt(4*a(n) + 5) + a(n)^5 + 15/2*a(n)^4 - 25/2*a(n)^2 + 5 with initial condition a(0) = 5.
a(n) = Lucas(4*5^n) - Lucas(2*5^n) + 1. - Ehren Metcalfe, Jul 29 2017
MATHEMATICA
a[n_] := LucasL[4*5^n] - LucasL[2*5^n] + 1; Array[a, 3, 0] (* Amiram Eldar, Jul 20 2025 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Nov 09 2012
STATUS
approved
