A185150 - OEIS

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

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OFFSET

1,3

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.

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.

EXAMPLE

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

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}]

CROSSREFS

Cf. A014085, A185636.