A366982 a(n) is the smallest odd k > 1 such that n^((k+1)/2) == n (mod k).

3, 3, 7, 3, 3, 5, 3, 3, 7, 3, 3, 5, 3, 3, 5, 3, 3, 9, 3, 3, 5, 3, 3, 7, 3, 3, 5, 3, 3, 5, 3, 3, 7, 3, 3, 5, 3, 3, 11, 3, 3, 5, 3, 3, 5, 3, 3, 11, 3, 3, 5, 3, 3, 7, 3, 3, 5, 3, 3, 5, 3, 3, 9, 3, 3, 5, 3, 3, 13, 3, 3, 5, 3, 3, 5, 3, 3, 7, 3, 3, 5, 3, 3, 17, 3, 3

Offset: 0

Author

Thomas Ordowski, Oct 30 2023

Keywords

• nonn

Comments

If this sequence is bounded, then it is periodic with period P = LCM(A), where A is the set of all (pairwise distinct) terms.

Note that n^((1729+1)/2) == n (mod 1729) for every n >= 0, where 1729 is the smallest absolute Euler pseudoprime (A033181).

Thus a(n) <= 1729. So, as said, this sequence is periodic.

What is its period?

If the largest term of this sequence is indeed 1729, it should be expected that its period P may be longer than the period of Euler primary pretenders (A309316), namely P > 41#*571#/4 (248 digits).

Links

• Antti Karttunen, Table of n, a(n) for n = 0..65537

Crossrefs

Cf. A033181, A309316, A366930, A366973.

Programs

Mathematica

a[n_] := Module[{k = 3}, While[PowerMod[n, (k + 1)/2, k] != Mod[n, k], k += 2]; k]; Array[a, 100, 0] (* Amiram Eldar, Oct 30 2023 *)

PARI

a(n) = my(k=3); while (Mod(n, k)^((k+1)/2) != n, k+=2); k; \\ Michel Marcus, Oct 31 2023

Extensions

More terms from Amiram Eldar, Oct 30 2023