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

1,1

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).

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, #] &]

Crossrefs

Cf. A000045, A337231 (odd terms).

Sequence in context: A312705 A312706 A312707 * A004797 A053459 A024398

Adjacent sequences: A337229 A337230 A337231 * A337233 A337234 A337235

Keyword

nonn

Author

Ovidiu Bagdasar, Aug 20 2020

Status

approved