Consider a random walk on $\{1,2,\dots,N\}$, that in state $i$ has probability $p_i$ of going left/right and probability $1-2p_i$ of staying in place, so the transition matrix is \begin{align*} \Lambda=\left( \begin{array}{cccccc} 1-p_1 & p_1 & 0 &\dots\\ p_2 & 1-2p_2 & p_2 & 0 & \dots\\ 0 & p_3 & 1-2p_3 &p_3 & 0 & \dots\\ \vdots& & \\ && \dots & 0 & p_{N-2} & 1-2p_{N-2} & p_{N-2} & 0\\ && & \dots & 0 & p_{N-1} & 1-2p_{N-1} & p_{N-1}\\ && & &\dots & 0 & p_N & 1-p_N \end{array} \right), \end{align*} In this case one can use the detailed balance equations \begin{align*} \pi_ip_{i,i+1}=\pi_{i+1}p_{i+1,i},\quad i=1,2,\dots,N, \end{align*} to find the stationary distribution $\pi$, and it turns out that \begin{align*} \pi_i \propto \frac{p_1}{p_i},\quad i=1,\dots,N, \end{align*} where $\propto$ denotes "proportional to".
Now assume that at the boundary states, the random walk can make jumps of size two. Then the detailed balance equations can no longer be used, as the random walk is not reversible, but is there a clever procedure way to find the stationary distribution?
The transition matrix of the new random walk is \begin{align*} \hat\Lambda=\left( \begin{array}{cccccc} 1-2p_1 & p_1 & p_1 & 0 &\dots\\ p_2 & 1-2p_2 & p_2 & 0 & \dots\\ 0 & p_2 & 1-2p_3 &p_3 & 0 & \dots\\ \vdots& & \\ && \dots & 0 & p_{N-2} & 1-2p_{N-2} & p_{N-2} & 0\\ && & \dots & 0 & p_{N-1} & 1-2p_{N-1} & p_{N-1}\\ && & \dots& 0& p_N & p_N & 1-2p_N \end{array} \right), \end{align*} and the relation between the two transition matrices is \begin{align*} \hat\Lambda = \Lambda + \left( \begin{array}{cccccc} -p_1 & 0 & p_1 & 0 &\dots\\ 0 & 0 & 0 & 0 & \dots\\ \vdots& & \\ && & \dots & 0 & 0 & 0 & 0\\ && & \dots& 0& p_N & 0 & -p_N \end{array} \right). \end{align*}
Did you try to compute the stationary distribution (with computer) for some instances? Very often there is no nice, closed formula. But I have the feeling that the solution to the first problem gives a very good approximation to the solution of the second. This happens sometimes when the transition matrices are close, see for example a paper of mine (OK, this is about absorbing chains, but you get the idea):
http://math.unideb.hu/media/pongracz-andras/papers/cyclic_graph_disc.pdf
I would not expect a nice result, but instead, I would try to show that your $\pi_i$ is close to the solution.