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I've been playing around with numerical semigroups lately. I'm pretty new to this stuff, so I apologize in advance if my notation is non-standard. Fix positive integers $x_1,\dots,x_r$ with $\gcd(x_1,\dots,x_r) = 1$, and let $S = \lbrace a_1x_1 + \cdots a_rx_r\mid a_1,\dots,a_r\in\mathbb{N}_0\rbrace$ be the numerical semigroup generated by the $x_i$. The representation number of $n\in\mathbb{N}_0$ over $S$ is given by $R_S(n) = \vert\lbrace (a_1,\dots,a_r)\in\mathbb{N}_0^r\mid n = \sum a_ix_i\rbrace\vert$. Now define $m(k) = \min\lbrace n\in\mathbb{N}\mid R_S(n)\geq k\rbrace$.

I am trying to show that there exists some $B \geq r-1$ so that $k^{1/B}\leq m(k)$ for all $k\in\mathbb{N}$.

I suspect that what I am trying to prove is elementary, and that someone else has already produced a proof, but I have not been able to find it or something similar yet. Any suggestions as to where I should look or ideas that will help me along with this proof would be greatly appreciated.

Thanks in advance.

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We have $$ \sum_{n\geq 0} R_S(n)t^n = \frac{1}{(1-t^{x_1})\cdots(1-t^{x_r})}. $$ It follows that $R_S(n) \sim \frac{1}{(r-1)!x_1\cdots x_r} n^{r-1}$. This shows that for any $B>r-1$ we have $k^{1/B}\leq m(k)$ for $k$ sufficiently large. By taking $B$ large enough this will hold for all $k\in\mathbb{N}$.

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