If $P'\in \mathcal{D}_n(s)$ is another Dyck path with peaks at length $s_1,\dots,s_p$, we have $s_i\geq 2i-1,\ i=1,\dots,p$ and $p\leq k$. Thus $$\label{3}\text{comaj}(P)=\sum_{i=1}^k(s-2i+1)\geq \sum_{i=1}^p(s-s_i)+\sum_{i=p+1}^k(s-2i+1)\geq\text{comaj}(P')$$ with equality if and only if $P=P'$.

\[conint\]We have that $\mathcal{V}^{1\rightarrow m}_{s,n}(q)=0$ if $s-n\notin 2{\mathbb Z}_+$ and otherwise it is a monic polynomial of degree $(s-n)(s+n)/4$. Moreover, the powers of $q$ which appear in the polynomial $\mathcal{V}^{1\rightarrow m}_{s,n}(q)$ form an interval of consecutive integers iff one of the following conditions hold (i) $n\neq 0$,    (ii) $m=2$,   (iii) $s=n$.

The first part of the corollary follows from the discussion preceding the corollary, see and . We start by proving the backward direction of the second part. Let $n\neq 0$ and $\tilde{P}\neq P$ an arbitrary path in $\mathcal{D}_{m,n}(s)$, where $P$ is as in . We shall show that $q^{\text{comaj}(\tilde{P})+1}$ appears with a nonzero coefficient in the polynomial $\mathcal{V}^{1\rightarrow m}_{s,n}(q)$. Since $\tilde{P}\neq P$, we must have two consecutive up-steps in $\tilde{P}$ followed by at least one down-step. Thus $\tilde{P}$ is of the form $$\tilde{P}=\cdots \underbrace{11\cdots 11}_{\ell}\underbrace{00\cdots 00}_{r}1\cdots,\ \ell\geq 2,\ \ r\geq 1,$$ and we can assume without loss of generality that the path preceding the $\ell$–up steps is contained in $\mathcal{D}_0$, i.e. it ends at the $x$–axis. *Case 1:* Suppose 