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May 9, 2014 at 7:51 comment added Thomas Bloom And trivially $G(n)\geq n$, so the hard part is in improving the log term if possible. The Goldbach problem for large $k$ (i.e. writing $n$ as the sum of $k(n)$ primes) would probably be a good place to look.
May 9, 2014 at 7:48 comment added Thomas Bloom Using the fact that for large odd $m$ there are $\gg m^{2}(\log m)^{-3}$ ways to write $m$ as the sum of $3$ primes, and $\ll (n/\log n)(m/\log m)$ ways to write it as the sum of $3$ primes where one is at most $n$, it follows that $G(n)\ll n\log n$ (even restricting $G$ to just consider primes, not prime powers).
May 9, 2014 at 7:14 history edited G. Flowers CC BY-SA 3.0
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May 9, 2014 at 5:54 history asked G. Flowers CC BY-SA 3.0