OFFSET
1,1
COMMENTS
If p is a prime, then A052918(p-1)^2 == 1 (mod p).
This sequence contains the even composite integers for which the congruence holds.
The generalized Lucas sequence of integer parameters (a,b) defined by U(m+2) = a*U(m+1)-b*U(m) and U(0)=0, U(1)=1, satisfies the identity U^2(p) == 1 (mod p) whenever p is prime and b=-1,1.
For a=5, b=-1, U(m) recovers A052918(m-1), for m=1,2,....
REFERENCES
D. Andrica, O. Bagdasar, Recurrent Sequences: Key Results, Applications and Problems. Springer (2020)
D. Andrica, O. Bagdasar, On some new arithmetic properties of the generalized Lucas sequences, Mediterr. J. Math. (to appear, 2021)
MATHEMATICA
Select[Range[2, 20000, 2], CompositeQ[#] && Divisible[Fibonacci[#, 5]*Fibonacci[#, 5] - 1, #] &]
CROSSREFS
KEYWORD
nonn
AUTHOR
Ovidiu Bagdasar, Oct 22 2020
EXTENSIONS
More terms from Amiram Eldar, Oct 22 2020
a(31)-a(38) from Daniel Suteu, Oct 22 2020
STATUS
approved