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A238703
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a(n) = |{0 < k < n: floor(k*n/3) is prime}|.
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1
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0, 0, 1, 1, 2, 1, 3, 3, 1, 3, 4, 0, 4, 2, 1, 3, 5, 0, 4, 4, 1, 4, 5, 0, 3, 4, 0, 3, 6, 0, 5, 4, 1, 6, 6, 0, 7, 4, 1, 5, 4, 0, 7, 6, 0, 8, 5, 0, 8, 7, 1, 6, 7, 0, 9, 9, 1, 9, 8, 0, 6, 7, 0, 7, 12, 0, 9, 7, 1, 11, 10, 0, 6, 8, 0, 7, 9, 0, 7, 12
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OFFSET
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1,5
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COMMENTS
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Conjecture: If n > m > 0 with n not divisible by m, then floor(k*n/m) is prime for some 0 < k < n.
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LINKS
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EXAMPLE
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a(4) = 1 since floor(2*4/3) = 2 is prime.
If p is a prime, then a(3*p) = 1 since floor(k*3p/3) = k*p is prime only for k = 1. If m > 1 is composite, then a(3*m) = 0 since floor(k*3m/3) = k*m is composite for all k > 0.
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MATHEMATICA
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p[n_, k_]:=PrimeQ[Floor[k*n/3]]
a[n_]:=Sum[If[p[n, k], 1, 0], {k, 1, n-1}]
Table[a[n], {n, 1, 80}]
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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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