OFFSET
1,3
COMMENTS
It appears that the sequence consists of the numbers congruent to 0 or 1 mod 3 (A032766) except for 24, 49, 55, 90, 99, 109, 111, ... What are these exceptions?
Also numbers n such that 2*Fibonacci(n+1) is squarefree because Lucas(n) = Fibonacci(n-1)+Fibonacci(n+1). - Michel Lagneau, Jun 04 2015
Numbers n such that Fibonacci(n+1) is odd and squarefree. - Chai Wah Wu, Jun 04 2015
Is it a theorem that this is a subsequence of A032766? - N. J. A. Sloane, Jun 04 2015
This sequence is a subsequence of A032766. Proof: since Fibonacci(0) = 0 and Fibonacci(1) = 1, Fibonacci(n) mod 2 has the pattern: 0, 1, 1, 0, 1, 1, 0, ..., i.e. if n mod 3 = 0, then Fibonacci(n) is even, and n-1 is not a member of this sequence. In other words, members of this sequence must be congruent to 0 or 1 mod 3. - Chai Wah Wu, Jun 04 2015
LINKS
Chai Wah Wu, Table of n, a(n) for n = 1..611 (based on A037918)
MATHEMATICA
Select[Range[0, 200], SquareFreeQ[Fibonacci[#] + LucasL[#]] &]
PROG
(Magma) [n: n in [0..200] | IsSquarefree(Fibonacci(n)+Lucas(n))];
(Python)
from sympy import factorint
A258574_list = []
a, b = 0, 2
for n in range(10**2):
....if max(factorint(b).values()) <= 1:
........A258574_list.append(n)
....a, b = b, a + b # Chai Wah Wu, Jun 04 2015
(PARI) is(n)=n%3<2 && issquarefree(fibonacci(n+1)) \\ Charles R Greathouse IV, Jun 04 2015
(Sage) [n for n in (0..110) if is_squarefree(2*fibonacci(n+1))] # Bruno Berselli,
CROSSREFS
KEYWORD
nonn
AUTHOR
Vincenzo Librandi, Jun 01 2015
EXTENSIONS
Edited by N. J. A. Sloane, Jun 04 2015
STATUS
approved