A112800 -id:A112800 - cp's OEIS frontend

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

Jonathan Vos Post and Ray Chandler, Sep 19 2005

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

			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.

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.

Cf. A000961, A112799, A112800, A112802.

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

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

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

Original entry on oeis.org

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

Jonathan Vos Post and Ray Chandler, Sep 19 2005

nonn

			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.

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.

Cf. A000961, A007304, A112799, A112800, A112801.

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

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.

A112799 Least odd number such that all greater odd numbers can be represented as sum of three integers with n distinct prime factors (conjectured).

Original entry on oeis.org

5, 29, 283, 4409, 95539, 2579897, 88149143

Offset: 1

Jonathan Vos Post and Ray Chandler, Sep 19 2005

Strangely, the first 5 values of this sequence are all primes. 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.
a(5) = 95539; all odd numbers up to 200000 checked, no larger term found that could not be represented as sum of three integers each with 5 distinct prime factors.
a(1)-a(3): checked odd numbers < 10^5. a(4): checked odd numbers < 10^6. a(5): checked odd numbers < 3*10^6. a(6): checked odd numbers < 3*10^7. a(7): checked odd numbers between 8*10^7 and 2*10^8. [From Donovan Johnson, Feb 04 2009]

Xianmeng Meng, On sums of three integers with a fixed number of prime factors, Journal of Number Theory, Vol. 114 (2005), pp. 37-65.

Cf. A112800, A112801, A112802.

a(6)-a(7) from Donovan Johnson, Feb 04 2009

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

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

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A112799 Least odd number such that all greater odd numbers can be represented as sum of three integers with n distinct prime factors (conjectured).

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