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
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
KEYWORD
nonn
AUTHOR
Jonathan Vos Post and Ray Chandler, Sep 19 2005
STATUS
approved