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A071414
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Number of times 4n+r and 2n+r are both prime, with 0 < r < 2n.
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0
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1, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 4, 2, 2, 6, 2, 2, 7, 3, 4, 5, 3, 2, 6, 3, 3, 5, 4, 3, 8, 2, 5, 9, 5, 5, 6, 4, 3, 7, 5, 4, 10, 3, 5, 11, 6, 5, 9, 5, 7, 10, 5, 5, 5, 5, 6, 11, 3, 6, 15, 3, 6, 12, 4, 6, 11, 8, 4, 11, 9, 4, 10, 6, 6, 16, 6, 7, 16, 7, 8, 11, 6, 6, 15, 9, 8, 12, 8, 8, 13, 9, 5, 14, 7, 12
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OFFSET
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1,3
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COMMENTS
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Related to the even terms of A070897. If a(n)>0 then there is a simple pairing whose sums involve just 1 (if r=1) or 2 primes. For the case n=2, which corresponds to A070897(4), the numbers 1,2,3,4 can be paired with 5,6,7,8 so that the sums are 11 and 7 (the primes obtained when r=3).
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LINKS
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EXAMPLE
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a(3)=2 because 12+r and 6+r are both prime for only 2 values of r: 1 and 5.
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MATHEMATICA
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For[lst={}; n=1, n<=100, n++, cnt=0; For[k=1, k<2n, k=k+2, If[PrimeQ[4n+k]&&PrimeQ[2n+k], cnt++ ]]; AppendTo[lst, cnt]]; lst
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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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STATUS
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approved
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