A185150 Number of odd primes p between n^2 and (n+1)^2 with (n/p) = 1, where (-) is the Legendre symbol.

1, 1, 2, 3, 2, 2, 2, 3, 3, 1, 4, 2, 4, 3, 5, 7, 2, 3, 4, 6, 5, 3, 3, 4, 8, 5, 4, 5, 4, 4, 6, 6, 6, 4, 9, 9, 7, 7, 5, 6, 7, 5, 9, 5, 7, 3, 9, 6, 10, 6, 10, 6, 8, 8, 7, 7, 10, 3, 12, 8, 7, 10, 8, 14, 11, 7, 10, 10, 5, 9, 11, 8, 7, 9, 9, 18, 11, 11, 12, 9, 20, 6, 13, 6, 10, 9, 13, 9, 8, 10, 10, 12, 12, 6, 13, 9, 12, 12, 8, 23

Offset: 1

Author

Zhi-Wei Sun, Dec 29 2012

Keywords

• nonn

Comments

Conjecture: a(n)>0 for all n>0.

This is a refinement of Legendre's conjecture that for each n=1,2,3,... the interval (n^2,(n+1)^2) contains a prime.

We have verified the conjecture for n up to 10^9.

Zhi-Wei Sun also made some similar conjectures involving primes and Legendre symbols, below are few examples:

(1) If n>10 then there is a prime p between n^2 and (n+1)^2 with (n/p) = ((1-n)/p) = 1. If n>2 is different from 7 and 17, then there is a prime p between n^2 and (n+1)^2 with (n/p) = ((n+1)/p) = 1. If n>1 is not equal to 27, then there is a prime p between n^2 and (n+1)^2 with (n/p) = ((n+2)/p) = 1.

(2) If n>2 is different from 6, 12, 58, then there is a prime p between n^2 and n^2+n such that (n/p) = 1. If n>20 is not a square, and different from 37 and 77, then there is a prime p between n^2 and n^2+n such that (n/p) = -1.

(3) For each n=15,16,... there is a prime p between n and 2n such that (n/p) = 1. If n>0 is not a square, then

there is a prime p between n and 2n such that (n/p) = -1.

Examples

			a(10)=1 since 107 is the only prime p between 10^2 and 11^2 with (10/p) = 1.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Zhi-Wei Sun, Conjectures involving primes and quadratic forms, arXiv:1211.1588.

Crossrefs

Cf. A014085, A185636.

Programs

Mathematica

a[n_]:=a[n]=Sum[If[n^2+k>2&&PrimeQ[n^2+k]==True&&JacobiSymbol[n,n^2+k]==1,1,0],{k,1,2n}]
Do[Print[n," ",a[n]],{n,1,100}]