A112800 Number of ways of representing 2n-1 as sum of three integers with 1 distinct prime factor.

0, 0, 0, 1, 3, 4, 6, 8, 9, 10, 12, 14, 14, 16, 18, 18, 20, 23, 25, 26, 28, 30, 30, 32, 32, 34, 37, 36, 40, 43, 42, 44, 46, 46, 46, 50, 51, 53, 59, 57, 57, 61, 62, 62, 66, 68, 69, 71, 72, 71, 73, 76, 74, 81, 81, 78, 87, 90, 87, 91, 93, 90, 94, 97, 94, 100, 107, 103, 114, 115

Offset

1,5

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.

Links

R. J. Mathar, Table of n, a(n) for n = 1..1655

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 sum of three primes (A000040) or powers of primes (A000961 except 1). Number of ways of representing 2n-1 as a + b + c where omega(a) = omega(b) = omega(c) = 1.

Example

a(4) = 1 because the only partition into nontrivial prime powers of (2*4)-1 = 7 is 7 = 2 + 2 + 3.
a(5) = 3 because the 3 partitions into nontrivial prime powers of (2*5)-1 = 9 are 9 = 2 + 2 + 5 = 2 + 3 + 4 = 3 + 3 + 3. The middle one of those partitions has "4" which is not a prime, but is a power of a prime.
a(6) = 4 because the 4 partitions into nontrivial prime powers of (2*6)-1 = 11 are 11 = 2 + 2 + 7 = 2 + 4 + 5 = 3 + 3 + 5.
a(7) = 6 because the 6 partitions into nontrivial prime powers of (2*7)-1 = 13 are 13 = 2 + 2 + 9 = 2 + 3 + 8 = 2 + 4 + 7 = 3 + 3 + 7 = 3 + 5 + 5 = 4 + 4 + 5.

Maple

isA000961 := proc(n)
    if n = 1 then
        return true;
    end if;
    numtheory[factorset](n) ;
    if nops(%) = 1 then
        true;
    else
        false;
    end if;
end proc:
A000961 := proc(n)
    option remember;
    local a;
    if n = 1 then
        1;
    else
        for a from procname(n-1)+1 do
            if isA000961(a) then
                return a;
            end if;
        end do:
    end if;
end proc:
A112800 := proc(n)
    local a, i, j, p, q, r, n2;
    n2 := 2*n-1 ;
    a := 0 ;
    for i from 2 do
        p := A000961(i) ;
        if 3*p > n2 then
            return a;
        else
            for j from i do
                q := A000961(j) ;
                r := n2-p-q ;
                if r < q then
                    break;
                end if;
                if isA000961(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, A112800(n));
end do: # R. J. Mathar, Jun 09 2014

Crossrefs

Cf. A000040, A112799, A112801, A112802.

Sequence in context: A134745 A183867 A182829 * A294456 A062969 A175035

Adjacent sequences: A112797 A112798 A112799 * A112801 A112802 A112803

Keyword

nonn

Author

Jonathan Vos Post and Ray Chandler, Sep 19 2005

Status

approved