A comment to show the connection between $B(x)$ and ordinary generating functions, inspired by Odlyzko-Wilf's [1].

Namely, I claim that $$ B(x)=\frac{1}{2-F(x,1)},\quad\text{where $F(x,y)$ satisfies } F(x,y)=\frac{1}{1-\frac{xy}{1-xyF(x,xy)}}. $$ Indeed, substititing in the latter $y$ with $x^ky$ gives $$ F(x,x^ky)=\frac{1}{1-\frac{x^{k+1}y}{1-x^{k+1}yF(x,x^{k+1}y)}}, \quad k\geq 0, $$ allowing to expand $$ F(x,y)=\frac{1}{1-\frac{xy}{1-xyF(x,xy)}}= \frac{1}{1-\frac{xy}{1-\frac{xy}{1-\frac{x^{2}y}{1-x^{2}yF(x,x^2y)}}}}=\dots, $$

With, as in [1], denoting $G=G(x,y):=xyF(x,xy)$, one obtains $$ F(x,y)=F=1-G+xyF+FG. $$ (in [1] one has a simpler relation between $F$ and $G$, namely $F=1+FG$.)

At the moment I don't know a good interpretation of $F$ and $G$.

Here is a sketch of proof. First, show the connection between $B(x)$ and ordinary generating functions, inspired by Odlyzko-Wilf's [1].

Namely, I claim that $$ B(x)=\frac{1}{2-F(x,1)},\quad\text{where $F(x,y)$ satisfies } F(x,y)=\frac{1}{1-\frac{xy}{1-xyF(x,xy)}}. $$ Indeed, substititing in the latter $y$ with $x^ky$ gives $$ F(x,x^ky)=\frac{1}{1-\frac{x^{k+1}y}{1-x^{k+1}yF(x,x^{k+1}y)}}, \quad k\geq 0, $$ allowing to expand $$ F(x,y)=\frac{1}{1-\frac{xy}{1-xyF(x,xy)}}= \frac{1}{1-\frac{xy}{1-\frac{xy}{1-\frac{x^{2}y}{1-x^{2}yF(x,x^2y)}}}}=\dots, $$

With, as in [1], denoting $G=G(x,y):=xyF(x,xy)$, one obtains $$ F(x,y)=F=1-G+xyF+FG. $$ (in [1] one has a simpler relation between $F$ and $G$, namely $F=1+FG$.)

EDIT. $F(x,1)$ (with an offset of 1) is the generating function for A227309, which is a standard transformation from A161492. There is a reference to DOI 10.1016/0012-365X(93)90224-H. In the latter, however, one doesn't see continued fractions, and although it's probably matter of enough calculus to see.

Finally, using the language of Flajolet and Sedgewick book "Analytic combinatorics", $B(x)$ and $F(x,1)$ are related by a sequence construction (again a kind of standard transformation, not sure how one calls these on OEIS), as $$ B(x)=\frac{1}{1-(F(x,1)-1)}=F(x,1)+(F(x,1)-1)^2+((F(x,1)-1)^3+\dots. $$ This shows that $B(x)$ indeed counts the entries of A225114.