A337235 Even composite integers m such that A006190(m)^2 == 1 (mod m).

4, 8, 16, 68, 1208, 1424, 3056, 3824, 3928, 20912, 52174, 63716, 88708, 123148, 161872, 582224, 887566, 17083292, 18900412, 34648888, 39991684, 44884912, 51390736, 103170448, 107825236, 132238514, 279900272, 686071244, 769252508, 3251623346, 3358311986, 3535011826

Offset: 1

Author

Ovidiu Bagdasar, Aug 20 2020

Keywords

• nonn

Comments

If p is a prime, then A006190(p)^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(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=3, b=-1, U(n) recovers A006190(n) ("Bronze" Fibonacci numbers).

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).

Crossrefs

Cf. A337231 (a=1, odd terms), A337232 (a=1, even terms), A337233 (a=2), A337234 (a=3, odd terms).

Programs

Mathematica

Select[Range[3, 25000, 2], CompositeQ[#] && Divisible[Fibonacci[#, 5]*Fibonacci[#, 5] - 1, #] &]

Extensions

More terms from Amiram Eldar, Aug 21 2020

a(18)-a(32) from Daniel Suteu, Aug 29 2020