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A237819
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Number of primes p < n such that floor(sqrt(n-p)) is a Sophie Germain prime.
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1
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0, 0, 0, 0, 0, 1, 2, 2, 3, 3, 4, 4, 4, 4, 5, 5, 6, 5, 4, 4, 4, 4, 4, 4, 4, 4, 5, 6, 5, 6, 6, 7, 7, 7, 7, 8, 8, 8, 6, 6, 6, 7, 6, 7, 7, 7, 6, 7, 6, 6, 7, 7, 5, 6, 5, 6, 6, 6, 4, 4, 4, 5, 5, 5, 5, 6, 5, 6, 5, 5, 6, 7, 6, 6, 6, 6, 5, 6, 5, 5
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OFFSET
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1,7
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COMMENTS
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Conjecture: (i) a(n) > 0 for all n > 5.
(ii) For any integer n > 10, there is a prime p < n such that q = floor(sqrt(n-p)) and q + 2 are both prime.
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LINKS
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EXAMPLE
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a(6) = 1 since 2, floor(sqrt(6-2)) = 2 and 2*2 + 1 = 5 are all prime.
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MATHEMATICA
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f[n_]:=Floor[Sqrt[n]]
q[n_]:=PrimeQ[f[n]]&&PrimeQ[2*f[n]+1]
a[n_]:=Sum[If[q[n-Prime[k]], 1, 0], {k, 1, PrimePi[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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