OFFSET
1,1
COMMENTS
If p is a prime, then A000045(p)^2==1 (mod p).
This sequence contains the odd composite integers for which the congruence holds.
The generalized Lucas sequence of integer parameters (a,b) defined by U(n+2)=a*U(n+1)-b*U(n) and U(0)=0, U(1)=1, satisfies the identity U^2(p)==1 (mod p) whenever p is prime and b=-1.
For a=1, b=-1, U(n) recovers A000045(n) (Fibonacci numbers).
REFERENCES
D. Andrica, O. Bagdasar, Recurrent Sequences: Key Results, Applications and Problems. Springer (to appear, 2020).
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..1000
Dorin Andrica and Ovidiu Bagdasar, On Generalized Lucas Pseudoprimality of Level k, Mathematics (2021) Vol. 9, 838.
D. Andrica and O. Bagdasar, On some new arithmetic properties of the generalized Lucas sequences, preprint for Mediterr. J. Math. 18, 47 (2021).
MATHEMATICA
Select[Range[3, 30000, 2], CompositeQ[#] && Divisible[Fibonacci[#, 1]*Fibonacci[#, 1] - 1, #] &]
PROG
(PARI) lista(nn) = my(list=List()); forcomposite(c=1, nn, if ((c%2) && (Mod(fibonacci(c), c)^2 == 1), listput(list, c))); Vec(list); \\ Michel Marcus, Sep 29 2023
CROSSREFS
KEYWORD
nonn
AUTHOR
Ovidiu Bagdasar, Aug 20 2020
STATUS
approved