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^((p-1)/2) != 1 (mod p). Since n is invertible mod p, n^((p-1)/2) != 1 (mod p) implies n^((p+1)/2) != n (mod p). So this sequence is unbounded. - Robert Israel, Oct 31 2023
From Charles L. Hohn, Sep 27 2024: (Start)
Smallest odd prime p for which n is a square mod p.
Smallest odd prime p for which n mod p is a member of row A096008(p). (End)
LINKS
Robin Visser, Table of n, a(n) for n = 0..10000
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:
map(f, [$0..100]); # Robert Israel, Oct 30 2023
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
(PARI) a(n) = for(i=2, oo, my(p=prime(i)); for(j=0, (p-1)/2, if(n%p==j^2%p, return(p)))) \\ Charles L. Hohn, Sep 27 2024
CROSSREFS
KEYWORD
nonn,changed
AUTHOR
Thomas Ordowski, Oct 30 2023
EXTENSIONS
More terms from Amiram Eldar, Oct 30 2023
STATUS
approved