A241472 a(n) = |{0 < k < sqrt(prime(n)): k^2 + 1 is a quadratic residue modulo prime(n)}|.

0, 0, 0, 1, 1, 1, 1, 2, 1, 1, 3, 2, 4, 2, 3, 3, 3, 1, 5, 5, 4, 6, 5, 5, 3, 5, 5, 3, 3, 4, 7, 3, 7, 4, 6, 7, 7, 4, 4, 3, 6, 8, 8, 6, 5, 8, 7, 8, 6, 6, 8, 11, 8, 6, 7, 7, 5, 9, 2, 8, 3, 11, 10, 8, 6, 8, 7, 10, 5, 8, 8, 9, 13, 10, 10, 6, 6, 10, 11, 10

Offset: 1

Author

Zhi-Wei Sun, Apr 23 2014

Keywords

• nonn

Comments

a(n) > 0 for all n > 3. In fact, for any prime p > 5 one of 1^2 + 1 = 2, 2^2 + 1 = 5 and 3^2 + 1 = 10 is a quadratic residue modulo p. Similarly, if p > 7 is a prime not equal to 19, then k^2 - 1 is a quadratic residue modulo p for some positive integer k < sqrt(p). In fact, for any prime p > 7, one of 2^2 - 1 = 3, 3^2 - 1 = 8 and 5^2 - 1 = 24 is a quadratic residue modulo p.

See also A239957 for a similar conjecture involving primitive roots modulo primes. Note that a primitive root modulo an odd prime p must be a quadratic nonresidue modulo p.

Examples

			a(6) = 1 since 3^2 + 1 = 10 is a quadratic residue modulo prime(6) = 13.
a(7) = 1 since 1^2 + 1 = 2 is a quadratic residue modulo prime(7) = 17.
a(9) = 1 since 1^2 + 1 = 2 is a quadratic residue modulo prime(9) = 23.
a(10) = 1 since 2^2 + 1 = 5 is a quadratic residue modulo prime(10) = 29.
a(18) = 1 since 2^2 + 1 = 5 is a quadratic residue modulo prime(18) = 61.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000040, A002522, A239957, A241492.

Programs

Mathematica

a[n_]:=Sum[If[JacobiSymbol[k^2+1,Prime[n]]==1,1,0],{k,1,Sqrt[Prime[n]-1]}]
Table[a[n],{n,1,80}]