关于数值半群商的弗罗贝尼乌斯数
On the Frobenius Number of Quotients of Numerical Semigroups
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中文总结 AI 辅助
研究数值半群商的弗罗贝尼乌斯数问题，通过受柯蒂斯定理启发，利用狄利克雷定理，得到该数不能用有限多项式公式统一表示、特定函数性质及不存在相关非零多项式等结果。
中文摘要 AI 辅助
给定一个数值半群\(S\)和正整数\(p\)，商\(\frac{S}{p}=\{n\in \mathbb{N} \mid pn\in S\}\)也是一个数值半群。当\(S=\langle a,b\rangle\)且\(\gcd(a,b)=1\)时，求\(g\!\left(\frac{\langle a,b\rangle}{p}\right)\)的封闭形式公式是一个著名的开放问题，即使在\(b=a + 1\)的特殊情况下也未解决。受柯蒂斯关于\(g(\langle s_1,s_2,s_3\rangle)\)不存在多项式公式定理的启发，我们在某种意义上对这个开放问题给出了否定答案。具体而言，我们得到三个主要结果：(i) \(g\!\left(\frac{\langle a,b\rangle}{p}\right)\)不能由任何有限的多项式（或有理）公式统一表示；(ii) 对于每个固定的\(p\)，函数\(a\mapsto g\!\left(\frac{\langle a,a + 1\rangle}{p}\right)\)是周期整除\(p\)的二次拟多项式；(iii) 不存在非零多项式\(F\in \mathbb{C}[X_1,X_2,X_3]\)使得对于所有满足\(2<p<a\)的素数\(a,p\)有\(F\left(a,p,g\!\left(\frac{\langle a,a + 1\rangle}{p}\right)\right)=0\)。狄利克雷关于算术级数中素数的定理在论证中起关键作用。
英文摘要
Given a numerical semigroup $S$ and a positive integer $p$, the quotient $\frac{S}{p}=\{n\in \mathbb{N} \mid pn\in S\}$ also forms a numerical semigroup. When $S=\langle a,b\rangle$ with $\gcd(a,b)=1$, a well-known open problem is to find a closed-form formula for the Frobenius number $g\!\left(\frac{\langle a,b\rangle}{p}\right)$, which remains open even in the special case $b=a+1$. Inspired by Curtis's theorem on the non-existence of polynomial formulas for the Frobenius number $g(\langle s_1,s_2,s_3\rangle)$, we provide a negative answer to this open problem in a certain sense. Concretely, we obtain the following three main results. (i): The Frobenius number $g\!\left(\frac{\langle a,b\rangle}{p}\right)$ cannot be represented, uniformly in $a,b,p$, by any finite collection of polynomial (or rational) formulas. (ii): For each fixed $p$, the function $a\mapsto g\!\left(\frac{\langle a,a+1\rangle}{p}\right)$ is a quadratic quasi-polynomial with period dividing $p$. (iii): There is no nonzero polynomial $F\in \mathbb{C}[X_1,X_2,X_3]$ satisfying $F\left(a,p,g\!\left(\frac{\langle a,a+1\rangle}{p}\right)\right)=0$ for all primes $a,p$ with $2<p<a$; the same conclusion already holds if only $p$ is required to be prime and $a$ ranges over all integers greater than $p$. While (iii) is stronger than (i), the proofs of the two results reveal different insights. Dirichlet's theorem on primes in arithmetic progressions plays a crucial role in our arguments.
