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%I #27 Nov 05 2023 09:02:48
%S 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,
%T 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,
%U 3,13,3,3,5,3,3,5,3,3,7,3,3,5,3,3,17,3,3
%N a(n) is the smallest odd k > 1 such that n^((k+1)/2) == n (mod k).
%C If this sequence is bounded, then it is periodic with period P = LCM(A), where A is the set of all (pairwise distinct) terms.
%C Note that n^((1729+1)/2) == n (mod 1729) for every n >= 0, where 1729 is the smallest absolute Euler pseudoprime (A033181).
%C Thus a(n) <= 1729. So, as said, this sequence is periodic.
%C What is its period?
%C 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).
%t 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 *)
%o (PARI) a(n) = my(k=3); while (Mod(n, k)^((k+1)/2) != n, k+=2); k; \\ _Michel Marcus_, Oct 31 2023
%Y Cf. A033181, A309316, A366930, A366973.
%K nonn
%O 0,1
%A _Thomas Ordowski_, Oct 30 2023
%E More terms from _Amiram Eldar_, Oct 30 2023
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