A112802 Number of ways of representing 2n-1 as sum of three integers with 3 distinct prime factors.

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 2

Offset: 1

Author

Jonathan Vos Post and Ray Chandler, Sep 19 2005

Keywords

• nonn

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.

Examples

			a(83) = 1 because the only partition into three integers each with 3 distinct prime factors of (2*83)-1 = 165 is 165 = 30 + 30 + 105 = (2*3*5) + (2*3*5) + (3*5*7). Coincidentally, 165 itself has three distinct prime factors 165 = 3 * 5 * 11.
a(89) = 1 because the only partition into three integers each with 3 distinct prime factors of (2*89)-1 = 177 = 30 + 42 + 105 = (2*3*5) + (2*3*7) + (3*5*7).
a(107) = 2 because the two partitions into three integers each with 3 distinct prime factors of (2*107)-1 = 213 are 213 = 30 + 78 + 105 = 42 + 66 + 105.
		

Links

• R. J. Mathar, Table of n, a(n) for n = 1..1290
• 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, A007304, A112799, A112800, A112801.

Programs

Maple

isA033992 := proc(n)
    numtheory[factorset](n) ;
    if nops(%) = 3 then
        true;
    else
        false;
    end if;
end proc:
A033992 := proc(n)
    option remember;
    local a;
    if n = 1 then
        30;
    else
        for a from procname(n-1)+1 do
            if isA033992(a) then
                return a;
            end if;
        end do:
    end if;
end proc:
A112802 := proc(n)
    local a,i,j,p,q,r,n2;
    n2 := 2*n-1 ;
    a := 0 ;
    for i from 1 do
        p := A033992(i) ;
        if 3*p > n2 then
            return a;
        else
            for j from i do
                q := A033992(j) ;
                r := n2-p-q ;
                if r < q then
                    break;
                end if;
                if isA033992(r) then
                    a := a+1 ;
                end if;
            end do:
        end if ;
    end do:
end proc:
for n from 1 do
    printf("%d %d\n",n,A112802(n));
end do: # R. J. Mathar, Jun 09 2014

Formula

Number of ways of representing 2n-1 as sum of three members of A033992. Number of ways of representing 2n-1 as a + b + c where omega(a) = omega(b) = omega(c) = 3, where omega=A001221.