

A112802


Number of ways of representing 2n1 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
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OFFSET

1,107


COMMENTS

Meng proves a remarkable generalization of the GoldbachVinogradov 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. 3765.


FORMULA

Number of ways of representing 2n1 as sum of three members of A033992. Number of ways of representing 2n1 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(n1)+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*n1 ;
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 := n2pq ;
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: A216579 A229878 A235145 A266342 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



