OFFSET
1,1
COMMENTS
If p is a prime, then A001076(p)^2==1 (mod p).
This sequence contains the 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=-1,1.
For a=4, b=-1, U(n) recovers A001076(n).
REFERENCES
D. Andrica, O. Bagdasar, Recurrent Sequences: Key Results, Applications and Problems. Springer (to appear, 2020).
LINKS
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, 20000, 2], CompositeQ[#] && Divisible[Fibonacci[#, 4]*Fibonacci[#, 4] - 1, #] &]
CROSSREFS
KEYWORD
nonn
AUTHOR
Ovidiu Bagdasar, Aug 20 2020
STATUS
approved