This site is supported by donations to The OEIS Foundation.

 Annual appeal: Please make a donation to keep the OEIS running! Over 6000 articles have referenced us, often saying "we discovered this result with the help of the OEIS". Other ways to donate

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A112802 Number of ways of representing 2n-1 as sum of three integers with 3 distinct prime factors. 5
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,107 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..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. 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. EXAMPLE 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. 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 CROSSREFS Cf. A000961, A007304, A112799, A112800, A112801. Sequence in context: A229878 A235145 A266342 A285936 A037281 A143241 A258825 Adjacent sequences:  A112799 A112800 A112801 * A112803 A112804 A112805 KEYWORD nonn AUTHOR Jonathan Vos Post and Ray Chandler, Sep 19 2005 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent | More pages
The OEIS Community | Maintained by The OEIS Foundation Inc.