A112801 Number of ways of representing 2n-1 as sum of three integers, each with two distinct prime factors.

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 2, 2, 2, 4, 4, 4, 8, 7, 8, 11, 11, 13, 15, 16, 18, 23, 23, 26, 30, 31, 33, 40, 40, 45, 51, 53, 56, 62, 66, 66, 76, 79, 82, 88, 94, 96, 105, 111, 111, 124, 127, 132, 141, 145, 148, 164, 166, 170, 180, 187, 187, 206, 204, 208

Offset: 1

Author

Jonathan Vos Post and Ray Chandler, Sep 19 2005

Keywords

• nonn
• look

Comments

Meng proves a remarkable generalization of the Goldbach-Vinogradov classical result that every sufficiently large odd integer N can be partitioned as the sum of three primes N = p1 + p2 + p3. The new proof is that every sufficiently large odd integer N can be partitioned as the sum of three integers N = a + b + c where each of a, b, c has k distinct prime factors for the same k.

See A243751 for the range of this sequence, and A243750 for the indices of record values. - M. F. Hasler, Jun 09 2014

Examples

			a(14) = 1 because the only partition into three integers each with 2 distinct prime factors of (2*14)-1 = 27 is 27 = 6 + 6 + 15 = (2*3) + (2*3) + (3*5).
a(16) = 1 because the only partition into three integers each with 2 distinct prime factors of (2*16)-1 = 31 is 31 = 6 + 10 + 15 = (2*3) + (2*5) + (3*5).
a(17) = 2 because the two partitions into three integers each with 2 distinct prime factors of (2*17)-1 = 33 are 33 = 6 + 6 + 21 = 6 + 12 + 15.
		

Links

• R. J. Mathar, Table of n, a(n) for n = 1..1020
• Xianmeng Meng, On sums of three integers with a fixed number of prime factors, Journal of Number Theory, Vol. 114 (2005), pp. 37-65.

Crossrefs

Cf. A000961, A112799, A112800, A112802.

Programs

PARI

A112801(n)={n=n*2-1;sum(a=6,n\3,if(omega(a)==2,sum(b=a,(n-a)\2, omega(b)==2 && omega(n-a-b)==2)))} \\ M. F. Hasler, Jun 09 2014

Formula

Number of ways of representing 2n-1 as a + b + c where a<=b<=c are elements of A007774.