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A185150
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Number of odd primes p between n^2 and (n+1)^2 with (n/p) = 1, where (-) is the Legendre symbol
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4
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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
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1,3
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COMMENTS
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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.
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LINKS
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EXAMPLE
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a(10)=1 since 107 is the only prime p between 10^2 and 11^2 with (10/p) = 1.
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MATHEMATICA
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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}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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