The old formula for the Frobenius number of a numerical semigroup generated by two elements can be stated as follows: assume $\gcd\{a+1,b+1\}=1$, then the Frobenius number of $S= \left<a+1,b+1\right>$ is $ab-1$.
Now let $n\geq 2$ be some fixed integer. Let $A= \{a_1, a_2,\dots, a_n\}$ be a list of positive integers and $x_i = 1+ \sum_{j=0}^{n-2}{a_ia_{i+1}...a_{i+j}}$ and let $S_A =\left< x_1,\dots, x_n\right>$ be the semigroup generated by the $x_i$s.
For example, when $n=3$, for a triplet $A=\{a,b,c\}$ we have the generators $ab+a+1, bc+b+1,ca+c+1$ for $S_A$. My problem is:
Problem: suppose $\gcd\{x_1,\dots,x_n\}=1$ and they minimally generate $S_A$. Then show that the Frobenius number of $S_A$ is $(n-1)(a_1a_2...a_n-1)$.
If correct, this is an obvious generalization of the old formula for two generators. I imagine in three variables this can be shown by some brute force, but any comments/proof would be appreciated.
Motivation: from trying to solve this question. A counter example can be built by starting with a triplet $a,b,c>1$, build $x_1,x_2,x_3$ as above, and throw in $x_4= (abc-1)$. With $\{a,b,c\}= \{2,3,4\}$ we get $R=k[t^9,t^{13},t^{16},t^{23}]$ as in my example. When $\{a,b,c\}= \{2,2,3\}$, we get $R=k[t^7,t^{9},t^{10},t^{11}]$ as mentioned in the comment.
