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Even composite integers m such that A006190(m)^2 == 1 (mod m).
5

%I #24 Nov 23 2023 13:18:55

%S 4,8,16,68,1208,1424,3056,3824,3928,20912,52174,63716,88708,123148,

%T 161872,582224,887566,17083292,18900412,34648888,39991684,44884912,

%U 51390736,103170448,107825236,132238514,279900272,686071244,769252508,3251623346,3358311986,3535011826

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

%C If p is a prime, then A006190(p)^2 == 1 (mod p).

%C This sequence contains the even composite integers for which the congruence holds.

%C 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.

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

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

%H D. Andrica and O. Bagdasar, <a href="https://repository.derby.ac.uk/item/92yqq/on-some-new-arithmetic-properties-of-the-generalized-lucas-sequences">On some new arithmetic properties of the generalized Lucas sequences</a>, preprint for Mediterr. J. Math. 18, 47 (2021).

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

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

%K nonn

%O 1,1

%A _Ovidiu Bagdasar_, Aug 20 2020

%E More terms from _Amiram Eldar_, Aug 21 2020

%E a(18)-a(32) from _Daniel Suteu_, Aug 29 2020