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= <a+1,b+1>$ 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 = <x_1,\dots, x_n>$ 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.