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Let $a,b,c$ be three pairwise coprime positive integers, and $\Gamma=\langle a,b,c\rangle$ be the corresponding numerical semigroup. Consider the linear equations:

$n_1a=m_{12}b+m_{13}c$

$n_2b=m_{21}a+m_{23}c$

$n_3c=m_{31}a+m_{32}b$

Where $n_i,m_{ij}$ are positive integers. Also, let $n_1=\min\{x\in\mathbb{Z}_{>0}|xa\in\langle b,c\rangle\}$, i.e, we choose the smallest $n_1$ that satisfies the equation, and $n_2$ and $n_3$ are chosen similarly.

Then it's not hard to prove that $n_1=m_{21}+m_{31}$, $n_2=m_{12}+m_{32}$, and $n_3=m_{13}+m_{23}$. For the prove see http://hera.ugr.es/doi/15773139.pdf.

My question is, does there exist $a,b,c$ such that $n_1a<F(\langle b,c\rangle)=bc-b-c$, $n_2b<F(\langle a,c\rangle)=ac-a-c$, $n_3c<F(\langle a,b\rangle)=ab-a-b$where $F$ is the Frobenius number?

Edit: thanks for the example, the answer is positive. Furthermore, does there exist $a,b,c$ such that $\max\{n_1a,n_2b,n_3c\}<F(\Gamma)?$

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  • $\begingroup$ The formula for the Frobenius number should be $F(\langle a,b \rangle)=ab-a-b$. $\endgroup$ – Francesco Nov 8 at 11:03
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Let $a=30$, $b=31$ and $c=37$. Then

\begin{array}{ll} 7a=2b+4c=210 \\ 7b=6a+c=217 \\ 5c=a+5b=185 \\ \end{array}

and the Frobenius number of $\langle a,b,c \rangle$ is 267.

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  • $\begingroup$ Thanks for the example! See the edits, and the edit is the utilmate question I want to ask. $\endgroup$ – J.Doe Nov 8 at 15:14
  • $\begingroup$ I changed example. Clearly this one works also for the previous question. $\endgroup$ – Francesco Nov 8 at 20:24

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