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A127499
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The number of times that binomial(2n,n) has two prime factors that add to 2n.
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2
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0, 0, 0, 0, 1, 0, 1, 2, 1, 0, 1, 1, 1, 1, 0, 1, 3, 2, 1, 2, 3, 2, 2, 1, 3, 0, 2, 0, 3, 3, 1, 3, 4, 1, 2, 2, 2, 3, 3, 1, 3, 3, 2, 3, 4, 2, 1, 4, 2, 4, 4, 2, 2, 5, 3, 2, 1, 2, 1, 6, 1, 4, 4, 0, 4, 3, 3, 2, 4, 3, 4, 6, 3, 3, 6, 3, 5, 6, 2, 5, 5, 1, 4, 5, 4, 2, 4, 3, 3, 5, 2, 5, 7, 3, 4, 4, 3, 4, 4, 3, 4, 7, 3, 4, 8
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OFFSET
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1,8
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COMMENTS
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In general, a(3n) is much greater than a(3n-1) and a(3n+1), which is apparent in the graph of this sequence.
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LINKS
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EXAMPLE
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Consider n=8. Then binomial(16,8)=12870, which has prime factors 2,3,5,11,13. There are two pairs of prime factors that sum to 16: (3,13) and (5,11). Hence a(8)=2.
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MATHEMATICA
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Table[p=Rest[Transpose[FactorInteger[Binomial[2n, n]]][[1]]]; cnt=0; i=1; While[i<=Length[p] && p[[i]]<=n, If[MemberQ[p, 2n-p[[i]]], cnt++ ]; i++ ]; cnt, {n, 100}]
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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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