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A127499
The number of times that binomial(2n,n) has two prime factors that add to 2n.
2
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
OFFSET
1,8
COMMENTS
In general, a(3n) is much greater than a(3n-1) and a(3n+1), which is apparent in the graph of this sequence.
EXAMPLE
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.
MATHEMATICA
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}]
CROSSREFS
Cf. A070542 (n such that a(n)=0).
Sequence in context: A339885 A106799 A212210 * A198068 A358194 A121361
KEYWORD
nonn
AUTHOR
T. D. Noe, Jan 19 2007
STATUS
approved