\ldots, P_j)$ of length $2(i-1)$ with $j$ segments. Case 4: $P_1=Q_1UQ_2D$, where $Q_1$ and $Q_2$ are not empty. Then we get a composition $(Q_1, Q_2,
P_2, \ldots, P_j)$ of length $2(i-1)$ with $j+1$ segments. Adding up the terms in the above cases, we obtain the desired recursion (\[eq.1\]).

From Lemma \[dotheom\] one sees that the entry $a_{i,j}$ in the triangular matrix of the identity (\[eq2.2\]) can be explained as the number compositions of Dyck paths of length $2i$ that contain $j$ segments. We remark that this combinatorial interpretation can also be derived from the generating function of the entries in the $j$-th column in of the matrix in (\[eq2.2\]). The following formula for $a_{i,j}$ has been derived by Cameron and Nkwanta [@cn]: $$a_{i,j}= {j \over i} \, {2i \choose i-j} .$$

Let us rewrite the matrix identity (\[eq2.2\]) as follows $$\label{eq.2}
\sum_{j= 1}^i ja_{i,j}=4^{i-1}.$$ A combinatorial formulation of the above identity is given by Callan [@callan].

We are now ready to give a combinatorial proof of the above identity. Clearly, $4^n$ is the number of sequences of length $n$ on four letters, say, $\{1, 2, 3, 4\}$. The term $ja_{i,j}$ suggests that we should specify a segment in a composition 