A112802 -id:A112802 - cp's OEIS frontend

A033992 Numbers that are divisible by exactly three different primes.

30, 42, 60, 66, 70, 78, 84, 90, 102, 105, 110, 114, 120, 126, 130, 132, 138, 140, 150, 154, 156, 165, 168, 170, 174, 180, 182, 186, 190, 195, 198, 204, 220, 222, 228, 230, 231, 234, 238, 240, 246, 252, 255, 258, 260, 264, 266, 270, 273, 276, 280, 282, 285, 286

Offset: 1

Labos Elemer

nonn

This sequence and A000977 are identical through their first 32 terms, but A000977(33) = 210. [Comment edited by Jon E. Schoenfield, Dec 30 2014]

			220 = 2*2*5*11 is here but 210 = 2*3*5*7 is not; compare A000977.

T. D. Noe, Table of n, a(n) for n = 1..1000
Hans Montanus and Ron Westdijk, Cellular Automation and Binomials, Green Blue Mathematics (2022), p. 90.

A225228 is a subsequence.
Row 3 of A125666.
Cf. A000977, A000961, A001221, A112802.

a033992 n = a033992_list !! (n-1)
a033992_list = filter ((== 3) . a001221) [1..]
-- Reinhard Zumkeller, May 03 2013

A033992 := proc(n)
if (nops(numtheory[factorset](n)) = 3) then
   RETURN(n)
fi: end:  seq(A033992(n), n=1..500); # Jani Melik, Feb 24 2011

Select[Range[300],PrimeNu[#]==3&] (* Harvey P. Dale, May 01 2013 *)

is(n)=omega(n)==3 \\ Charles R Greathouse IV, Apr 28 2015

A246655(lim)=my(v=List(primes([2,lim\=1]))); for(e=2,logint(lim,2), forprime(p=2,sqrtnint(lim,e), listput(v,p^e))); Set(v)
list(lim,pr=3)=if(pr==1, return(A246655(lim))); my(v=List(),pr1=pr-1,mx=prod(i=1,pr1,prime(i))); forprime(p=prime(pr),lim\mx, my(u=list(lim\p,pr1)); for(i=1,#u,listput(v,p*u[i]))); Set(v) \\ Charles R Greathouse IV, Feb 03 2023

omega(a(n)) = A001221(a(n)) = 3. - Jonathan Vos Post, Sep 20 2005

a(n) ~ 2n log n / (log log n)^2. - Charles R Greathouse IV, Jul 28 2016

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

Original entry on oeis.org

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.

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

Original entry on oeis.org

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

Jonathan Vos Post and Ray Chandler, Sep 19 2005

nonn

			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.

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.

Cf. A000040, A112799, A112801, A112802.

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

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.

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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A033992 Numbers that are divisible by exactly three different primes.

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

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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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