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A321858
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a(n) = Pi(12,5)(n) + Pi(12,7)(n) - Pi(12,1)(n) - Pi(12,11)(n) where Pi(a,b)(x) denotes the number of primes in the arithmetic progression a*k + b less than or equal to x.
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14
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0, 0, 0, 0, 1, 1, 2, 2, 2, 2, 1, 1, 0, 0, 0, 0, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 3, 3, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 3, 3, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 1
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OFFSET
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1,7
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COMMENTS
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a(n) is the number of odd primes <= n that have 3 as a quadratic nonresidue minus the number of primes <= n that have 3 as a quadratic residue.
The first 10000 terms are nonnegative. a(p) = 0 for primes p = 2, 3, 13, 433, 443, 457, 479, 491, 503, 3541, ... The earliest negative term is a(61463) = -1. Conjecturally infinitely many terms should be negative.
In general, assuming the strong form of the Riemann Hypothesis, if 0 < a, b < k are integers, gcd(a, k) = gcd(b, k) = 1, a is a quadratic residue and b is a quadratic nonresidue mod k, then Pi(k,b)(n) > Pi(k,a)(n) occurs more often than not. This phenomenon is called "Chebyshev's bias". (See Wikipedia link and especially the links in A007350.) [Edited by Peter Munn, Nov 19 2023]
Here, although 11 is not a quadratic residue modulo 12, for most n we have Pi(12,7)(n) + Pi(12,11)(n) > Pi(12,1)(n) - Pi(12,5)(n), Pi(12,5)(n) + Pi(12,11)(n) > Pi(12,1)(n) + Pi(12,7)(n) and Pi(12,5)(n) + Pi(12,7)(n) > Pi(12,1)(n) + Pi(12,11)(n).
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LINKS
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Andrew Granville and Greg Martin, Prime number races, Amer. Math. Monthly, 113 (No. 1, 2006), 1-33.
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FORMULA
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a(n) = -Sum_{primes p<=n} Kronecker(12,p) = -Sum_{primes p<=n} A110161(p).
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EXAMPLE
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Pi(12,1)(100) = 5, Pi(12,5)(100) = Pi(12,7)(100) = Pi(12,11)(100) = 6, so a(100) = 6 + 6 - 5 - 6 = 1.
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PROG
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(PARI) a(n) = -sum(i=1, n, isprime(i)*kronecker(12, i))
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CROSSREFS
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Let d be a fundamental discriminant.
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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