A237436 Least prime p > prime(n+1) such that p is a square mod the first n odd primes 3, 5, 7, 11, ..., prime(n+1).

7, 19, 79, 331, 751, 1171, 7459, 10651, 18379, 78439, 78439, 399499, 644869, 1427911, 1427911, 4355311, 5715319, 43030381, 43030381, 163384621

Offset: 1

Author

Jonathan Sondow, Feb 15 2014

Keywords

• nonn

Comments

Least prime p > prime(n+1) such that p is a quadratic residue mod the first n odd primes 3, 5, 7, 11, ..., prime(n+1).

Least odd prime p such that the Legendre symbol (p|q) = 1 for q = 3, 5, 7, 11, ..., prime(n+1).

Examples

			Let f(p) = list of Legendre (p|q) for q = 3, 5, 7, 11, 13, 17, 19, 23, ...
Then f(p) is
p=3: 0, -1, -1, 1, 1, -1, -1, 1, ...
p=5: -1, 0, -1, 1, -1, -1, 1, -1, ...
p=7: 1, -1, 0, -1, -1, -1, 1, -1, ...
p=11: -1, 1, 1, 0, -1, -1, 1, -1, ...
p=13: 1, -1, -1, -1, 0, 1, -1, 1, ...
p=17: -1, -1, -1, -1, 1, 0, 1, -1, ...
p=19: 1, 1, -1, -1, -1, 1, 0, -1, ...
f(7) is the first list that begins with 1, so a(1) = 7.
f(19) is the first list that begins with 1, 1, so a(2) = 19.
		

Links

• Wikipedia, Legendre symbol
• Wikipedia, Quadratic residue

Crossrefs

Cf. A222756 (p and q switched), A237437.

Programs

Mathematica

Table[p = Prime[n+2]; While[Length[Select[Prime[Range[2, n + 1]], JacobiSymbol[p, #] == 1 &]] < n, p = NextPrime[p]]; p, {n, 1, 18}]

Formula

a(n) = a(n+1) if and only if Legendre (a(n)|prime(n+2)) = 1.

a(n) <= A096636(n).

a(n) < A096636(n) if and only if a(n) = a(n+1).

a(n) = A096636(n) if and only if Legendre (a(n)|prime(n+2)) = -1.