A $1$-step convolution identity involving the Motzkin triangle The Motzkin triangle $T(n,k)$ is built according to the rules:
(1) $T(n,0)=1$;
(2) $T(n,k)=0$ if $k<0$ or $k>n$;
(3) $T(n,k)=T(n-1,k-2)+T(n-1,k-1)+T(n-1,k)$.
After some numerical evidence I ask:

QUESTION. Can you provide a combinatorial proof for the below identity?
  $$\sum_{k=0}^nT(n,k)T(n,k+1)=\sum_{k=0}^n\binom{2n}{2k+1}\binom{2k+1}k\frac1{k+2}.$$ 

Note. It's okay to give other alternative justifications to add variety to the discussion here but I wish for a combinatorial argument.
 A: Using different bound variables on the two sides for clarity in the subsequent discussion, the goal is:
$$\sum_{k=0}^{n-1}T(n,k) \, T(n,k+1)=\sum_{j=0}^{n-1}\binom{2n}{2j+1}\binom{2j+1}{j}\frac1{j+2}
%\sum_{k=0}^{n-1}T(n,k) \, T(n,k+1)=\sum_{j=0}^{n-1}\binom{2n}{j,j+1}\frac1{j+2}$$
$T(n,k)$ counts paths from $(0,0)$ to $(n,n-k)$ with steps from $(1,-1)$, $(1,0)$, $(1,1)$ which don't go below $y=0$. Therefore the natural interpretation of the LHS is that it counts paths with $n$ steps from $(1,-1)$, $(1,0)$, $(1,1)$ followed by a step $(0,-1)$ followed by $n$ steps from $(-1,-1)$, $(-1,0)$, $(-1,1)$ which return to the origin without going below $y=0$.
Then the "obvious" interpretation of the RHS would be that $j$ counts the number of steps (not including the $(n+1)$th) which decrease $y$, $j+1$ counts the number of steps which increase $y$, and only $\frac{1}{j+2}$ orderings thereof avoid going below $y=0$. Note for comparison that without the forced position of the $(0,-1)$ step in the sequence we would be looking at $\binom{2n+1}{2j+2} \frac{1}{j+2}\binom{2j+2}{j+1}$: select $2(j+1)$ steps for the vertical movement and then we have the Catalan number $C_{j+1}$ giving their assignments to $j+1$ increments and $j+1$ decrements.
To prove that this interpretation works, first fix $j$.
Choose the $(n+1)$th step and $2j+1$ of the other $2n$ steps, preserving their order. Then consider Dyck words of length $2j+2$. It suffices to show that exactly half of the (step subsequence, Dyck word) pairs place a closing parenthesis with the $(n+1)$th step, since $$\frac12 \binom{2j+2}{j+1}\frac{1}{j+2} = \binom{2j+1}{j}\frac{1}{j+2}$$
Consider the statistics $s(n,k)$ and $\overline{s}(n,k)$ which count the number of Dyck words on $2n$ symbols which have respectively an opening or a closing parenthesis at index $k$. Since those are the only possibilities, $s(n,k) + \overline{s}(n,k) = C_n$. Looking at a Dyck word in a physical mirror is equivalent to reversing the symbols and applying a substitution which exchanges opening and closing parentheses, so $\overline{s}(n, 2n+1-k) = s(n,k)$.
The involution $x \to 2n+2-x$ with fixpoint $n+1$ induces a fixpointless involutive bijection $B$ between the selected subsequences of $2j+2$ steps. Moreover, if subsequence $S$ has $n+1$ at index $i$ then $B(S)$ has $n+1$ at index $2j+3-i$. Then summing over $s \in \{S, B(S)\}$ the Dyck words which have a closing parenthesis at the index of $n+1$ we get $$\overline{s}(j+1, i) + \overline{s}(j+1, 2j+3-i) = \overline{s}(j+1, i) + s(j+1, i) = C_{j+1}$$
which is exactly half of the (step subsequence, Dyck word) pairs for those step subsequences. Sum over all pairs of step subsequences to obtain the desired result.
