A337232 Even composite integers m such that F(m)^2 == 1 (mod m), where F(m) is the m-th Fibonacci number.

4, 8, 14, 22, 26, 34, 38, 46, 58, 62, 74, 82, 86, 88, 94, 106, 118, 122, 134, 142, 146, 158, 166, 178, 194, 202, 206, 214, 218, 226, 254, 262, 274, 278, 298, 302, 314, 326, 334, 346, 358, 362, 382, 386, 394, 398, 422, 446, 454, 458, 466, 478, 482, 502, 514, 526

Offset: 1

Author

Ovidiu Bagdasar, Aug 20 2020

Keywords

• nonn

Comments

If p is a prime, then A000045(p)^2 == 1 (mod p).

This sequence contains the even 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=1, b=-1, U(n) recovers A000045(n) (Fibonacci numbers).

No terms divisible by 3. - Robert Israel, Sep 15 2020

References

• D. Andrica, O. Bagdasar, Recurrent Sequences: Key Results, Applications and Problems. Springer (to appear, 2020).

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• 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. A000045, A337231 (odd terms).

Programs

Maple

select(t -> combinat:-fibonacci(t) &^ 2 - 1 mod t = 0, 2*[$2..1000]); # Robert Israel, Sep 15 2020

Mathematica

Select[Range[2, 2000, 2], CompositeQ[#] && Divisible[Fibonacci[#, 1]*Fibonacci[#, 1] - 1, #] &]