I'm given $(n_1,n_2,n_3)$, with $\operatorname{gcd}(n_1,n_2,n_3)=1$. Then, I need to find $c_1$, the least positive integer such that $c_1n_1=n_2\mathbb{N}+n_3\mathbb{N}$. I additionally need the specific coefficients $r_{12}$ and $r_{13}$ such that $c_1n_1=r_{12}n_2+r_{13}n_3$.
Note that every number mentioned so far is non-negative.
So far, I'm calculating the Apery set of $n_2\mathbb{N}+n_3\mathbb{N}$ (with respect to it's multiplicity), and checking for membership through this. While this works to find my $c_1$, it leaves me nowhere with finding $r_{12}$ or $r_{13}$. While I could immediatally solve the diophantine equation $c_1n_1=r_{12}n_2+r_{13}n_3$, I can't help but think this isn't a very efficient way to do it.
Should I just solve the linear diophantine equation, or is there some better way to find the particular way that $c_1n_1$ is written in terms of $n_2$ and $n_3$?
A numerical semigroup is a subset of $\mathbb{I}=\mathbb{N}\cup\{0\}$ which is a subsemigroup of $(\mathbb{I},+)$ containing $0$ and its complement is finite. The only cyclic numerical semigroup is the trivial one generated by $1$ which is the whole $\mathbb{I}$. For any other element $x\neq 1$, the semigroup generated by $\{x\}$ plus zero has an infinite complement since it is precisely $x\mathbb{I}$ or you can write $\{0\}\cup x\mathbb{N}$. If $x=0$ then the complement is $\mathbb{N}$, infinite. If $x$ is nonzero, since it is not $1$, it is greater than or equal to $2$, and there are infinite natural numbers not multiple of $x$.
Now knowing this, the only cases that $\langle n_2,n_3\rangle\cup\{0\}$ can be cyclic is when one of $n_2$ and $n_3$ is $1$ or $(n_2,n_3)=(2,3)$. Otherwise it won't be the whole $\mathbb{I}$ and hence it is not possible to generate it by one element. No matter what is $n_1$ and what you choose for $c_1$. Very simple example $(n_1,n_2,n_3)=(2,3,4)$. The set $(2c_1)\mathbb{I}$ never contains an odd number, no matter of what $c_1$ is! And $3\mathbb{I}+4\mathbb{I}=\{0,3,4,6,7,8,9,\cdots\}$. Of course you have $g.c.d(2,3,4)=1$ but it won't make the statement in your question to be true.