$R(x)=1+xR(x)+xR(x{)}^2$, and $2xR(x)=1-x-(1-6x+x^2{)}^{1\mathrm{/}2}$

First-return decomposition. Let $v\in {\mathrm{S}\mathrm{c}\mathrm{h}}_{n+1}$, of length $\mathrm{\ell }$, with step word $w$ and heights $h$. Its first step is not $D$, so it is $L_2$ or $U$. If it is $L_2$ then translating the remaining path by $(-2,0)$ gives a member of ${\mathrm{S}\mathrm{c}\mathrm{h}}_n$, since its endpoints become $(0,0)$ and $(2n,0)$ and its heights are unchanged. If it is $U$ then $h(1)=1$; the set of positive indices $j\le \mathrm{\ell }$ with $h(j)=0$ contains $\mathrm{\ell }$, so by [L11] it has a least element $\tau \ge 2$, and $h(\tau -1)\ge 1$ by minimality while $h(\tau )=0$, so the step at $\tau$ lowers the height and is therefore $D$, with $h(\tau -1)=1$. Translating the portion of $v$ from the index $1$ to the index $\tau -1$ by $(-1,-1)$ gives a path from $(0,0)$ whose heights are $h(1+j)-1\ge 0$ and which returns to height $0$; its numbers of up and down steps are therefore equal, so its horizontal extent is even, say $2i$, and it lies in ${\mathrm{S}\mathrm{c}\mathrm{h}}_i$. Since the $D$ step at $\tau$ has width $1$, the first coordinate at $\tau$ is $2i+2$, so translating the portion from $\tau$ to $\mathrm{\ell }$ by $(-2i-2,0)$ gives a member of ${\mathrm{S}\mathrm{c}\mathrm{h}}_{n-i}$, and $0\le i\le n$. Conversely, prepending $L_2$ to a member of ${\mathrm{S}\mathrm{c}\mathrm{h}}_n$, and sending $(i,P,Q)$ to the path with step word $U$, that of $P$, $D$, that of $Q$, produce members of ${\mathrm{S}\mathrm{c}\mathrm{h}}_{n+1}$ whose first-return data are the ones started from, because the heights strictly inside the first block are at least $1$; the two constructions are two-sided inverses, so by [L1] and [L2] the set ${\mathrm{S}\mathrm{c}\mathrm{h}}_{n+1}$ is in bijection with the disjoint union of ${\mathrm{S}\mathrm{c}\mathrm{h}}_n$ and the sets $\{i\}\times {\mathrm{S}\mathrm{c}\mathrm{h}}_i\times {\mathrm{S}\mathrm{c}\mathrm{h}}_{n-i}$ for $0\le i\le n$.

The index range is where this differs from the Motzkin case. A $U$ and its matching $D$ have width $1$ each, so together they consume two units of horizontal extent and therefore exactly one unit of half-length; the inner and outer blocks then carry half-lengths $i$ and $n-i$ with $i+(n-i)=n$, so the convolution index runs over all of $\{0,\dots ,n\}$ and not only over $\{0,\dots ,n-1\}$. Counting the two sides of step 1.1 with [F1], [L3], [L4], [L5] and [L10] gives, in $\mathbb{N}$, $R_{n+1}=R_n+{\sum }_{i=0}^n{R}_i{R}_{n-i},$ the sum being over the finite index set $\{0,\dots ,n\}$. At $n=0$ the sum has the single term $R_0{R}_0=1$, so $R_1=R_0+1=2$, matching the two paths with step words $L_2$ and $UD$. At $n=1$ it gives $R_2=R_1+R_0{R}_1+R_1{R}_0=6$, at $n=2$ it gives $R_3=R_2+R_0{R}_2+R_1{R}_1+R_2{R}_0=22$, and at $n=3$ it gives $R_4=R_3+R_0{R}_3+R_1{R}_2+R_2{R}_1+R_3{R}_0=90$.

Comparing coefficients gives the functional equation. At the index $0$: $[x^0](1+xR+x{R}^2)=1$ by [L6], and $[x^0]R=R_0=1$. At an index $n+1$: $[x^{n+1}](xR)=[x^n]R=R_n$ and $[x^{n+1}](x{R}^2)=[x^n](R^2)={\sum }_{i=0}^n{R}_i{R}_{n-i}$ by the shift and Cauchy-product clauses of [L6], and the sum of the two is $R_{n+1}$ by step 2.1. Extensionality in [L6] gives $R=1+xR+x{R}^2$.

Rearranging step 3.1 in the commutative ring $\mathbb{Q}\mathrm{⟦}x\mathrm{⟧}$ gives $x{R}^2+(x-1)R+1=0$, and hence $(1-x-2xR{)}^2=(x-1{)}^2+4x(x{R}^2+(x-1)R)=(x-1{)}^2-4x=1-6x+x^2\mathrm{.}$ The series $1-x-2xR$ has coefficient $1$ at the index $0$, so it lies in $1+x\mathbb{Q}\mathrm{⟦}x\mathrm{⟧}$, and $1-6x+x^2=1+u$ with $u=-6x+x^2\in x\mathbb{Q}\mathrm{⟦}x\mathrm{⟧}$; by the uniqueness clause of [L7] with $k=2$ it is therefore $(1-6x+x^2{)}^{1\mathrm{/}2}$ as defined in [L8], which gives $2xR=1-x-(1-6x+x^2{)}^{1\mathrm{/}2}$. No division by $2x$ occurs, and none is available.