

A366973


Smallest odd prime p such that n^((p+1)/2) == n (mod p).


4



3, 3, 7, 3, 3, 5, 3, 3, 7, 3, 3, 5, 3, 3, 5, 3, 3, 13, 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, 13, 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

a(n) is the smallest odd prime p for which the Legendre symbol (n / p) >= 0.
For any set S of odd primes, by Chinese Remainder Theorem, there is n such that n is a primitive root mod each prime p in S, and then n^((p1)/2) =/= 1 (mod p). Since n is invertible mod p, n^((p1)/2) =/= 1 (mod p) implies n^((p+1)/2) =/= n (mod p). So this sequence is unbounded.  Robert Israel, Oct 31 2023


LINKS



MAPLE

f:= proc(n) local p;
p:= 2;
do
p:= nextprime(p);
if n &^ ((p+1)/2)  n mod p = 0 then return p fi
od
end proc:


MATHEMATICA

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


PROG

(PARI) a(n) = my(p=3); while(Mod(n, p)^((p+1)/2) != n, p=nextprime(p+1)); p; \\ Michel Marcus, Oct 30 2023


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



