I am looking for a combinatorial proof (or a reference to such) of the following fact related to Catalan and Motzkin numbers.
Consider the extended Motzkin number sequence, where the $n$th term ($n\ge 0$) counts the number of nonempty Motzkin paths (see A001006) either consisting of a single point or starting with a level step. Its generating function is $$ \tilde{m}(z)=1+zm(z)=\frac{1+z-\sqrt{1-2z-3z^2}}{2z}. $$ It appears that the coefficient $$[x^n]\bigl(\tilde{m}(z)\tilde{m}(z\tilde{m}(z))\bigr)\ge 2C_n \qquad \text{for}\ n\ge 1,$$where $C_n$ is the $n$th Catalan number (A000108) and $$C(x)=\frac{1-\sqrt{1-4z}}{2z}$$ is the Catalan generating function. In fact, $$[x^n]\bigl(\tilde{m}(z)\tilde{m}(z\tilde{m}(z))\bigr)=[x^n]\bigl(2C(x)-1\bigr) \qquad \text{for}\ 0\le n\le 7,$$ and it appears that $$[x^n]\bigl(\tilde{m}(z)\tilde{m}(z\tilde{m}(z))\bigr)>[x^n]\bigl(2C(x)-1\bigr) \qquad \text{for}\ n\ge 8.$$ (Cf. OEIS sequences A348197 and A068875.) I understand that the left-hand side grows asymptotically faster than the right-hand side (with rates of growth of $\frac{3}{2}(1+\sqrt{3})\approx 4.098$ vs. $4$) and so should be eventually greater. It is less clear to me why the weak inequality holds from the very beginning.