%I
%S 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,
%T 23,23,26,30,31,33,40,40,45,51,53,56,62,66,66,76,79,82,88,94,96,105,
%U 111,111,124,127,132,141,145,148,164,166,170,180,187,187,206,204,208
%N Number of ways of representing 2n1 as sum of three integers, each with two distinct prime factors.
%C Meng proves a remarkable generalization of the GoldbachVinogradov 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.
%C See A243751 for the range of this sequence, and A243750 for the indices of record values.  _M. F. Hasler_, Jun 09 2014
%H R. J. Mathar, <a href="/A112801/b112801.txt">Table of n, a(n) for n = 1..1020</a>
%H Xianmeng Meng, <a href="http://dx.doi.org/10.1016/j.jnt.2005.04.013">On sums of three integers with a fixed number of prime factors</a>, Journal of Number Theory, Vol. 114 (2005), pp. 3765.
%F Number of ways of representing 2n1 as a + b + c where a<=b<=c are elements of A007774.
%e 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).
%e 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).
%e 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.
%o (PARI) A112801(n)={n=n*21;sum(a=6,n\3,if(omega(a)==2,sum(b=a,(na)\2, omega(b)==2 && omega(nab)==2)))} \\ _M. F. Hasler_, Jun 09 2014
%Y Cf. A000961, A112799, A112800, A112802.
%K nonn,look
%O 1,17
%A _Jonathan Vos Post_ and _Ray Chandler_, Sep 19 2005
