A377212 a(n) is the least number k that is not a quadratic residue modulo prime(n) but is a quadratic residue modulo all previous primes.

2, 3, 6, 21, 15, 91, 246, 429, 1005, 399, 3094, 3045, 21099, 41155, 43059, 404754, 214230, 569130, 182919, 2190279, 860574, 9361374, 8042479, 33440551, 36915670, 11993466, 287638530, 182528031, 697126530, 78278655, 3263415285, 6941299170, 25856763139, 32968406926, 13803374706

Offset: 2

Author

Robert Israel, Oct 19 2024

Keywords

• nonn

Comments

a(n) = A000037(j) for the least j such that A144294(j) = prime(n).

Such numbers k exist for all n >= 2: for example, if x is a quadratic nonresidue modulo prime(n), by the Chinese Remainder Theorem there exists k such that k == x (mod prime(n)) and k == 1 (mod prime(j)) for 1 <= j < n.

Examples

			a(4) = 6 because 6 is not a quadratic residue modulo 7, but is a quadratic residue modulo 2, 3, and 5, and no smaller number works.
		

Crossrefs

Cf. A000037, A096636, A144294, A144295.

Programs

Maple

f:= proc(n) local k,p;
  if issqr(n) then return -1 fi;
  p:= 1;
  for k from 1 do
      p:= nextprime(p);
      if numtheory:-quadres(n,p) = -1 then return k fi
  od
end proc:
V:= Array(2..32): count:= 0:
for k from 2 while count < 31 do
  v:= f(k);
if v > 0 and v <= 32 and V[v] = 0 then
  V[v]:= k; count:= count+1
fi
od:
convert(V,list);

Python

from itertools import count
from math import isqrt
from sympy.ntheory import prime, nextprime, legendre_symbol
def A377212(n):
    p = prime(n)
    for r in count(1):
        k, q = r+(m:=isqrt(r))+(r>=m*(m+1)+1), 2
        while (q:=nextprime(q)):
            if q>p or legendre_symbol(k,q)==-1:
                break
        if p==q:
            return k # Chai Wah Wu, Oct 20 2024

Extensions

a(33)-a(36) from Chai Wah Wu, Oct 21 2024