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A366930 a(n) is the smallest odd composite k such that n^((k+1)/2) == n (mod k). 2
9, 9, 341, 121, 341, 65, 15, 21, 9, 9, 9, 33, 33, 21, 21, 15, 15, 9, 9, 9, 21, 15, 21, 33, 25, 15, 9, 9, 9, 21, 15, 15, 25, 33, 21, 9, 9, 9, 57, 39, 15, 21, 21, 21, 9, 9, 9, 65, 21, 21, 21, 15, 39, 9, 9, 9, 21, 21, 57, 145, 15, 15, 9, 9, 9, 33, 15, 33, 25, 21 (list; graph; refs; listen; history; text; internal format)
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?
The period P of this sequence may be longer than the period of Euler primary pretenders (A309316), namely P > 41#*571#/4 (248 digits).
LINKS
FORMULA
a(n) >= A309316(n).
MATHEMATICA
a[n_] := Module[{k = 9}, While[PrimeQ[k] || PowerMod[n, (k + 1)/2, k] != Mod[n, k], k += 2]; k]; Array[a, 100, 0] (* Amiram Eldar, Nov 01 2023 *)
PROG
(PARI) a(n) = my(k=3); while (isprime(k) || Mod(n, k)^((k+1)/2) != n, k+=2); k; \\ Michel Marcus, Nov 01 2023
CROSSREFS
Sequence in context: A353182 A171738 A309316 * A124116 A213154 A226050
KEYWORD
nonn
AUTHOR
Thomas Ordowski, Nov 01 2023
STATUS
approved

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Last modified April 28 09:58 EDT 2024. Contains 372037 sequences. (Running on oeis4.)