A337234 Odd composite integers m such that A006190(m)^2 == 1 (mod m).

9, 33, 55, 63, 99, 119, 153, 231, 385, 399, 561, 649, 935, 981, 1023, 1071, 1179, 1189, 1199, 1441, 1595, 1763, 1881, 1953, 2001, 2065, 2255, 2289, 2465, 2703, 2751, 2849, 2871, 3519, 3599, 3655, 3927, 4059, 4081, 4187, 5015, 5151, 5559, 6061, 6119, 6215, 6273, 6431

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 odd 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.

For a=3, b=-1, U(n) recovers A006190(n) ("Bronze" Fibonacci numbers).

References

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

Links

• Dorin Andrica and Ovidiu Bagdasar, On Generalized Lucas Pseudoprimality of Level k, Mathematics (2021) Vol. 9, 838.
• 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).

Programs

Mathematica

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