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A069004
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Number of times n^2 + s^2 is prime for positive integers s < n.
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5
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0, 1, 1, 1, 2, 2, 1, 3, 1, 4, 2, 1, 4, 3, 3, 3, 4, 3, 4, 6, 2, 4, 5, 3, 7, 6, 4, 4, 4, 4, 7, 6, 5, 6, 8, 5, 6, 7, 3, 9, 5, 5, 8, 8, 7, 9, 6, 7, 10, 8, 6, 9, 10, 5, 8, 8, 6, 10, 11, 8, 11, 10, 6, 9, 15, 5, 10, 11, 4, 11, 13, 6, 12, 10, 12, 11, 9, 8, 11, 19, 10, 15, 9, 8, 19, 11, 8, 11, 14, 15, 13
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OFFSET
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1,5
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COMMENTS
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The graph of this sequence inspires the following conjecture: A > a(n)/pi(n) > B, where A and B are constants and pi(n) is the prime counting function (A000720). - T. D. Noe, Feb 26 2007
Stronger conjecture: Let pi(n) be the prime counting function (A000720). Then pi(n) >= a(n) >= pi(n)/5 for n>1, with the following equalities: pi(2)=a(2), pi(10)=a(10) and a(12)=pi(12)/5. - T. D. Noe, Feb 26 2007
Records in a(n) are for n = 1, 2, 5, 8, 10, 20, 25, 35, 40, 49, 59, 65, 80, 115, 125, 130, 158, 200, 250, 265, 310, ... - Thomas Ordowski, Mar 05 2017
Number of primes p = (x^2 + y^2)/2 with 0 < x < y such that x + y = 2n. - Thomas Ordowski, Mar 06 2017
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LINKS
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FORMULA
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a(n) = O(n/log(n)). a(n) <= phi(n), a(n) = phi(n) for n = 2, 6, and 10. a(n) <= phi(2n)/2, a(n) = phi(2n)/2 for n = 2, 3, 5, 6, and 10. - Thomas Ordowski, Mar 01 2017
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EXAMPLE
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a(5)=2 because there are 2 values of s (2 and 4) such that 5^2 + s^2 is a prime number.
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MATHEMATICA
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maxN=100; lst={}; For[n=1, n<=maxN, n++, cnt=0; For[d=1, d<n, d=d+2, p=n^2+(n-d)^2; If[PrimeQ[p], cnt++ ] ]; AppendTo[lst, cnt]; ]; lst
Table[Count[n^2+Range[n-1]^2, _?PrimeQ], {n, 100}] (* Harvey P. Dale, Mar 01 2023 *)
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PROG
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(PARI) a(n) = sum(s=1, n-1, isprime(n^2+s^2)); \\ Michel Marcus, Jan 15 2017
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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