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A237837
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Number of primes p < n such that the number of Sophie Germain primes among 1, ..., n-p is a cube.
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1
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0, 0, 1, 2, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 2, 2, 2, 3, 3, 5, 5, 5, 5, 5, 5, 6, 6, 7, 7, 7, 7, 9, 9, 9, 9, 9, 9, 9, 9, 10
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OFFSET
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1,4
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COMMENTS
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Conjecture: a(n) > 0 for all n > 53.
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LINKS
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EXAMPLE
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a(55) = 2 since 53 is prime and there is exactly 1^3 = 1 Sophie Germain prime not exceeding 55 - 53 = 2, and 2 is prime and there are exactly 2^3 = 8 Sophie Germain primes not exceeding 55 - 2 = 53 (namely, they are 2, 3, 5, 11, 23, 29, 41, 53).
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MATHEMATICA
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sg[n_]:=Sum[If[PrimeQ[2*Prime[k]+1], 1, 0], {k, 1, PrimePi[n]}]
CQ[n_]:=IntegerQ[n^(1/3)]
a[n_]:=Sum[If[CQ[sg[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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