$m$ marked R-visible up steps and $m+1$ elevation lines such that there is exactly one marked step between two adjacent elevation lines.

Let $P$ be a partial $2$-Motzkin path of length $i-1$ ending at level $j-1$. Suppose that there are $m$ marked R-visible up steps with initial points at levels $j_1, j_2, \ldots, j_m$. Let $t_1=j_1$ and $t_{i}=j_{i}-j_{i-1}-1$ with $j_{m+1}=j-1$ for $i\geq 2$. Then one see that the number of ways to choose the $m+1$ elevation lines such that there is exactly $m+1$ marked R-visible up step is equals to $$(t_1+1) (t_2+1) \cdots (t_m+1).$$ Note that the $t_i$’s range over $t_1+t_2+\cdots +t_{m+1}=j-m-1$. Thus, the number of partial $2$-Motzkin paths of length $i-1$ ending at level $j-1$ with the required marked steps and elevation lines equal $$\label{c.1}
\sum_{t_1+t_2+\ldots t_{m+1}=j-1-m}(t_1+1)(t_2+1)\cdots
(t_{m+1}+1).$$ Let $g(x)=\sum_{n\geq 0}(n+1)x^n$. It is clear that $g(x)={1\over
(1-x)^2}$. Hence the summation (\[c.1\]) equals the coefficient of $x^{j-1-m}$ in the expansion of ${1\over (1-x)^{2m+2}}$, that is, the binomial coefficient ${j+m\choose 2m+1}$.

In fact, we may give a combinatorial interpretation of the binomial coefficient ${j+m \choose 2m+1}$ in the above proof. Let $P$ be a partial $2$-Motzkin path of length $i-1$ ending at level $j-1$ with $m$ marked R-visible up steps and 