A127499 The number of times that binomial(2n,n) has two prime factors that add to 2n.

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.

Links

T. D. Noe, Table of n, a(n) for n=1..10000

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

Adjacent sequences: A127496 A127497 A127498 * A127500 A127501 A127502

Keyword

nonn

Author

T. D. Noe, Jan 19 2007

Status

approved