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A224989
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Let p = prime(n). a(n) = number of primes q less than p, such that both p-q+1 and p-q-1 are primes.
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0
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0, 0, 0, 1, 2, 1, 3, 2, 4, 3, 2, 3, 4, 3, 5, 4, 5, 3, 3, 6, 5, 6, 6, 6, 3, 7, 5, 5, 6, 9, 4, 8, 3, 7, 7, 5, 6, 5, 7, 6, 8, 8, 9, 5, 10, 9, 12, 8, 6, 8, 9, 13, 10, 12, 9, 8, 12, 9, 7, 14, 9, 10, 8, 13, 9, 9, 11, 10, 6, 13, 12, 11, 8, 9, 17, 9, 12, 6, 11, 14
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OFFSET
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1,5
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LINKS
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EXAMPLE
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For n=3, p=5, there are no primes q(<5) such that both 5-q+1 and 5-q-1 are primes and hence a(3)=0. Also for n=5, p=11, there are a(5)=2 solutions 5,7 since 11-5+1=7, 11-5-1=5 and 11-7+1=5, 11-7-1=3.
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MATHEMATICA
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Table[p = Prime[n]; c = 0; i = 1; While[i < n, p1 = p - Prime[i]; If[PrimeQ[p1 + 1] && PrimeQ[p1 - 1], c = c + 1]; i++]; c, {n, 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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