Stationary distribution of a Markov chain on the nonnegative integers Let $\lambda_k,\mu_k\in\mathbb R_{\ge0}$ $(k\ge1)$ be nonnegative real numbers such that $\sum_{k=1}^\infty k\lambda_k<\infty,$ let $S=\mathbb Z_{\ge0}$ be the nonnegative integers, let $T=\mathbb R_{\ge0}$ be the nonnegative real numbers and consider the continuous-time Markov chain $X=(X_t)_{t\in T}$ on $S$ with rates
$$Q(n,n+k)=(n+1)\lambda_k\quad(k\ge1),\qquad Q(n,n-k)=(n+1-k)\mu_k\quad(1\le k\le n).$$
(This Markov chain appears in biology as a model of the length of an evolving DNA sequence (Miklós et. al. 2004). A MathOverflow post contains more information about this process.) For example, if $0=\lambda_k=u_k$ for all integers $k\ge2,$ then we recover the linear birth-death process with immigration with birth rate $\lambda_1,$ death rate $\mu_1$ and immigration rate $\lambda_1,$ whose nonzero rates are
$$Q(n,n+1)=(n+1)\lambda_1\quad(n\ge0,k\ge1),\qquad Q(n,n-1)=n\mu_1\quad(n\ge1).$$
Now, assume this chain is reversible and irreducible, and let $\nu$ be the stationary measure. Then detailed balance implies $\forall n\in\mathbb Z_{\ge0}~\forall k\in\mathbb Z_{\ge1}\quad \nu_n\cdot(n+1)\lambda_k=\nu_{n+k}\cdot(n+1)\mu_k,$ so that $\nu_n\lambda_k=\nu_{n+k}\mu_k.$ From reversibility and irreducibility it follows that $\nu_i>0$ for all $i\in S,$ so that $\forall k\in\mathbb Z_{\ge1}\quad\lambda_k=0\leftrightarrow\mu_k=0.$
Now, let $A=\{k\in\mathbb Z_{\ge1}:\mu_k\ne0\}~(=\{k\in\mathbb Z_{\ge1}:\lambda_k\ne0\}),$ and let $d=\gcd A.$ If $d\ne1,$ then $d>1,$ and so considering things mod $d$ we see that $X$ is not irreducible, contradiction; thus $\gcd A=d=1.$ And if $1\in A,$ then for all $n\in S$ we have $\nu_n\lambda_1=\nu_{n+1}\mu_1,$ and since $\mu_1\ne0$ we can say $\nu_{n+1}=\nu_n\lambda_1/\mu_1,$ so that $\forall k\in S\quad\nu_k=\nu_0(\lambda_1/\mu_1)^k.$ Normalizing, we have $\forall k\in S\quad\nu_k=(1-\lambda_1/\mu_1)(\lambda_1/\mu_1)^k.$ Note that detailed balance also yields $\forall k\in A\quad\lambda_k/\mu_k=(\lambda_1/\mu_1)^k.$
Now, my question is: What is the stationary distribution, if $1\not\in A?$ I tried the following example: Take $A=\{2,3\}.$ In addition, let ($r$ for ratio) $r_2=\lambda_2/\mu_2,$ $r_3=\lambda_3/\mu_3.$ From detailed balance, we have for all $n\in S$ that $\nu_nr_3=\nu_3=\nu_{n+1}r_2,$ so that $\nu_{n+1}=\nu_nr_3/r_2.$ It follows that $\forall n\in S\quad\nu_n=(1-r_3/r_2)(r_3/r_2)^n.$ (Note that $r_3/r_2=\lambda_1/\mu_1$ if $\{1,2,3\}\subseteq A.$) However, I cannot think of how to generalize this example to arbitrary sets $A.$

Miklós, I., Lunter, G. A., & Holmes, I. (2004). A “long indel” model for evolutionary sequence alignment. Molecular Biology and Evolution, 21(3), 529-540.
 A: The reversibility assumption (and detailed balance) are doing a lot of heavy lifting here. If it holds, then we have $\nu_{n+k} = r_k \nu_n$ for all $n \in \mathbb Z_{\ge 0}$ and all $k \in A$, which essentially tells us $\nu$, we just have to massage it into the right form.
Since $\gcd A = 1$, there is some integer linear combination $\sum_{k \in A} c_k \cdot k = 1$, and it follows from $\nu_{n+k} = r_k \nu_n$ that 
$$
   \nu_{n+1} = \nu_n \prod_{k \in A} r_k^{c_k}.
$$
So we have $\nu_n = (1-\rho) \rho^n$, where $\rho = \prod_{k \in A} r_k^{c_k}$. In your example, we can take $c_2 = -1$ and $c_3 = 1$, so $\rho = \frac{r_3}{r_2}$, and we get back your formula. Of course, we can also take $c_2 = 2$ and $c_3 = -1$, and get a formula with $\rho = \frac{r_2^2}{r_3}$. Detailed balance only holds if all of these expressions for $\rho$ are equal; for any $j,k \in A$ we must have $r_j^k = r_k^j$. In that case, we actually just have $r_k = \rho^k$ for all $k \in A$, and the process is not meaningfully different in the end from  the $A = \{1\}$ case.
(In other words, assuming detailed balance holds, we can just pick any $k \in A$ and say that $\nu_n = (1-r_k^{1/k}) r_k^{n/k}$, ignoring all other elements of $A$.)
