|
|
A258574
|
|
Numbers n such that Fibonacci(n)+Lucas(n) is squarefree.
|
|
2
|
|
|
0, 1, 3, 4, 6, 7, 9, 10, 12, 13, 15, 16, 18, 19, 21, 22, 25, 27, 28, 30, 31, 33, 34, 36, 37, 39, 40, 42, 43, 45, 46, 48, 51, 52, 54, 57, 58, 60, 61, 63, 64, 66, 67, 69, 70, 72, 73, 75, 76, 78, 79, 81, 82, 84, 85, 87, 88, 91, 93, 94, 96, 97, 100
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
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
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
|
|
|
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
a, b = 0, 2
for n in range(10**2):
....if max(factorint(b).values()) <= 1:
(Sage) [n for n in (0..110) if is_squarefree(2*fibonacci(n+1))] # Bruno Berselli,
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|