

A366982


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


2



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
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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



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



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



