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Feb 4, 2015 at 11:36 comment added Omer That depends on your definition of $A\ll B$. If it mean $A<cB$, then as Noam said, this is true. If it means $A/B\to 0$ then the answer is less clear. For example, if we change the problem to count solutions of $s_1+2s_2=n$ then it is easy to get $S(n) \ge n^{1/2}$, even with $g=1$.
Jan 31, 2015 at 0:44 comment added Noam D. Elkies Looks like this is just counting in two ways solutions of $k=s_1+s_2$ with $s_1 \leq s_2 \leq n$ and $k \leq 2n$ to get $S(n)^2 \leq 4gn$.
Jan 30, 2015 at 23:23 history asked SJY CC BY-SA 3.0