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

%I A112800 #12 Mar 13 2019 23:40:11
%S A112800 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,
%T A112800 37,36,40,43,42,44,46,46,46,50,51,53,59,57,57,61,62,62,66,68,69,71,72,
%U A112800 71,73,76,74,81,81,78,87,90,87,91,93,90,94,97,94,100,107,103,114,115
%N A112800 Number of ways of representing 2n-1 as sum of three integers with 1 distinct prime factor.
%C A112800 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.
%H A112800 R. J. Mathar, <a href="/A112800/b112800.txt">Table of n, a(n) for n = 1..1655</a>
%H A112800 Xianmeng Meng, <a href="https://doi.org/10.1016/j.jnt.2005.04.013">On sums of three integers with a fixed number of prime factors</a>, Journal of Number Theory, Vol. 114 (2005), pp. 37-65.
%F A112800 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.
%e A112800 a(4) = 1 because the only partition into nontrivial prime powers of (2*4)-1 = 7 is 7 = 2 + 2 + 3.
%e A112800 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.
%e A112800 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.
%e A112800 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.
%p A112800 isA000961 := proc(n)
%p A112800     if n = 1 then
%p A112800         return true;
%p A112800     end if;
%p A112800     numtheory[factorset](n) ;
%p A112800     if nops(%) = 1 then
%p A112800         true;
%p A112800     else
%p A112800         false;
%p A112800     end if;
%p A112800 end proc:
%p A112800 A000961 := proc(n)
%p A112800     option remember;
%p A112800     local a;
%p A112800     if n = 1 then
%p A112800         1;
%p A112800     else
%p A112800         for a from procname(n-1)+1 do
%p A112800             if isA000961(a) then
%p A112800                 return a;
%p A112800             end if;
%p A112800         end do:
%p A112800     end if;
%p A112800 end proc:
%p A112800 A112800 := proc(n)
%p A112800     local a,i,j,p,q,r,n2;
%p A112800     n2 := 2*n-1 ;
%p A112800     a := 0 ;
%p A112800     for i from 2 do
%p A112800         p := A000961(i) ;
%p A112800         if 3*p > n2 then
%p A112800             return a;
%p A112800         else
%p A112800             for j from i do
%p A112800                 q := A000961(j) ;
%p A112800                 r := n2-p-q ;
%p A112800                 if r < q then
%p A112800                     break;
%p A112800                 end if;
%p A112800                 if isA000961(r) then
%p A112800                     a := a+1 ;
%p A112800                 end if;
%p A112800             end do:
%p A112800         end if ;
%p A112800     end do:
%p A112800 end proc:
%p A112800 for n from 1 do
%p A112800     printf("%d %d\n",n,A112800(n));
%p A112800 end do: # _R. J. Mathar_, Jun 09 2014
%Y A112800 Cf. A000040, A112799, A112801, A112802.
%K A112800 nonn
%O A112800 1,5
%A A112800 _Jonathan Vos Post_ and _Ray Chandler_, Sep 19 2005