1. The Proposition in the post is almost equivalent to the Conjecture, namely it implies that $$\frac{g_n}{\log g_n}\leq (2+o(1))\log n.$$ In particular, the Proposition (hence also the Conjecture) implies that $$g_n\ll\log n\,\log\log n.\tag{$\ast$}$$ It is expected that $g_n\gg (\log n)^2$ holds for infinitely many $n$'s, contradicting $(\ast)$.

2. On the other hand, we don't know that $(\ast)$ is false. The best result in this direction is due to Ford-Green-Konyagin-Maynard-Tao (2014), and it states that $$g_n\gg\frac{\log n \,\log \log n\,\log\log\log\log n}{\log \log \log n}$$ holds for infinitely many $n$'s.

1. The Proposition in the post is almost equivalent to the Conjecture, namely it implies that $$\frac{g_n}{\log g_n}\leq (2+o(1))\log n.$$ In particular, the Proposition (hence also the Conjecture) implies that $$g_n\ll\log n\,\log\log n.\tag{$\ast$}$$ This would contradict the common expectation that $g_n\gg (\log n)^2$ holds for infinitely many $n$'s.

2. On the other hand, we don't know that $(\ast)$ is false. The best result in this direction is due to Ford-Green-Konyagin-Maynard-Tao (2014), and it states that $$g_n\gg\frac{\log n \,\log \log n\,\log\log\log\log n}{\log \log \log n}$$ holds for infinitely many $n$'s.

The Proposition in the post is almost equivalent to the Conjecture, namely it implies that $$\frac{g_n}{\log g_n}\leq (2+o(1))\log n.$$ In particular, the Proposition (hence also the Conjecture) implies that $$g_n\ll\log n\,\log\log n.\tag{$\ast$}$$ On the other hand, it is expected that $g_n\gg (\log n)^2$ holds for infinitely many $n$'s, in which case $(\ast)$ is false. On the other hand, we don't know that $(\ast)$ is false. The best result in this direction is due to Ford-Green-Konyagin-Maynard-Tao (2014), and it states that $$g_n\gg\frac{\log n \,\log \log n\,\log\log\log\log n}{\log \log \log n}$$ holds for infinitely many $n$'s.

1. The Proposition in the post is almost equivalent to the Conjecture, namely it implies that $$\frac{g_n}{\log g_n}\leq (2+o(1))\log n.$$ In particular, the Proposition (hence also the Conjecture) implies that $$g_n\ll\log n\,\log\log n.\tag{$\ast$}$$ It is expected that $g_n\gg (\log n)^2$ holds for infinitely many $n$'s, contradicting $(\ast)$.

2. On the other hand, we don't know that $(\ast)$ is false. The best result in this direction is due to Ford-Green-Konyagin-Maynard-Tao (2014), and it states that $$g_n\gg\frac{\log n \,\log \log n\,\log\log\log\log n}{\log \log \log n}$$ holds for infinitely many $n$'s.