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

2,1

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.

Links

Table of n, a(n) for n=2..36.

Example

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.

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);

Prog

(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

Crossrefs

Cf. A000037, A096636, A144294, A144295.

Sequence in context: A002078 A000372 A333445 * A123930 A238895 A125601

Adjacent sequences: A377209 A377210 A377211 * A377213 A377214 A377215

Keyword

nonn

Author

Robert Israel, Oct 19 2024

Extensions

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

Status

approved