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Preprint Jul 2026

On the Frobenius Number of Quotients of Numerical Semigroups

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.

Feihu Liu · 0 citations