A145231 a(n) = Fibonacci(4^n).

1, 3, 987, 10610209857723, 141693817714056513234709965875411919657707794958199867

Offset: 0

Author

Artur Jasinski, Oct 05 2008

Keywords

• nonn
• easy

Comments

This sequence has the property that a(n+1) is divisible by a(n). Conjecture: each prime divisor can occur only once (i.e., all terms are squarefree). - Artur Jasinski, Oct 05 2008

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..6

Crossrefs

Cf. A000045, A341601, A341603.

Cf. (k^n)-th Fibonacci number: A058635 (k=2), A045529 (k=3), this sequence (k=4), A145232 (k=5), A145233 (k=6), A145234 (k=7), A250487 (k=8), A250488 (k=9), A250489 (k=10).

Programs

Maple

a := proc(n) option remember; if n = 1 then 3 else a(n-1)*(5*a(n-1)^2 + 2)*sqrt(5*a(n-1)^2 + 4) end if; end: seq(a(n), n = 1..5); # Peter Bala, Nov 14 2022

Mathematica

G = (1 + Sqrt[5])/2; Table[Expand[(G^(4^n) - (1 - G)^(4^n))/Sqrt[5]], {n, 1, 6}]
Table[Round[(4/5)^(1/2)*Cosh[4^n*ArcCosh[((5/4)^(1/2))]]], {n, 1, 7}]
Fibonacci[4^Range[5]] (* Harvey P. Dale, Mar 28 2012 *)

Formula

a(n) = (G^(4^n) - (1-G)^(4^n) )/sqrt(5) where G = (1 + sqrt 5)/2 = A001622.

a(n) = round( sqrt(4/5) *cosh( 4^n*arccosh (sqrt(5/4)) )).

a(n)= A000045(A000302(n)). - Michel Marcus, Nov 07 2013

From Peter Bala, Nov 11 2022: (Start)

a(n+1) = a(n)*(5*a(n)^2 + 2)*sqrt(5*a(n)^2 + 4) for n >= 1.

a(n) == 3 (mod 4) for n >= 1.

a(n+1) == a(n) (mod 2^(2*n+1)).

A341601(n) == a(n) (mod 2^n) for n >= 2.

In the ring of 2-adic integers, the sequence {Fibonacci(4^n)} converges to the 2-adic integer A341603. (End)