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

0,1

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

Table of n, a(n) for n=0..85.

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 *)

Prog

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

Crossrefs

Cf. A033181, A309316, A366930, A366973.

Sequence in context: A333339 A089488 A367034 * A366973 A076560 A359048

Adjacent sequences: A366979 A366980 A366981 * A366983 A366984 A366985

Keyword

nonn

Author

Thomas Ordowski, Oct 30 2023

Extensions

More terms from Amiram Eldar, Oct 30 2023

Status

approved