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A112801
Number of ways of representing 2n-1 as sum of three integers, each with two distinct prime factors.
7
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
1,17
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
LINKS
Xianmeng Meng, On sums of three integers with a fixed number of prime factors, Journal of Number Theory, Vol. 114 (2005), pp. 37-65.
FORMULA
Number of ways of representing 2n-1 as a + b + c where a<=b<=c are elements of A007774.
EXAMPLE
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.
PROG
(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
CROSSREFS
KEYWORD
nonn,look
AUTHOR
STATUS
approved