Number of subgraphs in the ladder graph Assume you have the usual (in both directions infinite) ladder graph. I can try to provide a picture if needed.
Further assume the vertices are labelled and I have one distinct vertex (call it the "origin" or whatever you like) Now my question is, how many connected subgraphs (not necessarily trees) containing the origin are there containing exactly $n$ edges?
There is a somewhat similar question here Number of spanning trees in a ladder graph
(that also has a picture in it) but I don't quite see how to apply this to my case.
 A: First I assume to have a ladder graph semi-infinite (only infinite in one direction). And I denote $A(n)$ the number of connected sub-graph containing the vertex $(0,0)$. I compute it by summing different cases. The idea is to split the ladder in tranche of 3 edges $(0,0)-(0,1)$; $(0,0)-(1,0)$ and $(0,1)-(1,1)$ that I will call transversal, lower and upper edge:


*

*for all the graphs that contains only the lower edge in the first tranche:
$$A(n-1)$$ 

*for all the graphs that contains only the lower edge and the transversal edge in the first tranche:
$$A(n-2)$$

*for all the graph that contains the transversal and the upper edge
$$A(n-2)$$ 

*For the graphs that contains the three edges, we know that they are connected by the two nodes $(1,0)$ and $(1,1)$. The idea is to count the number of tranches the contains both the upper and lower edges (we denote it $k$) and the number of transversal on this $k$ tranches (we denote it $t$). And in the last tranche the is only the upper or lower part (hence 2 possibility).
$$ \sum_{k=0}^{(n-3)/2}\sum_{t=0}^{k+1}2{k+1 \choose t} A(n-3-t-2k-1) $$

*Same case but where from now all the tranches contain both the upper and lower part (where $\delta_{x,y}=1$ iff $x=y$ and $0$ otherwise):
$$ \sum_{k=0}^{(n-3)/2}\sum_{t=0}^{k+1}{k+1 \choose t} \delta_{n,3+2k+t} $$

*The only reaming case is the case where both the upper and lower edge are present but not the transversal edge. We proceed as the former case but we have to be sure that there is at least on transversal in the tranches containing both the upper and lower edge. 
$$ \sum_{k=0}^{(n-3)/2}\sum_{t=1}^{k+1}2{k+1 \choose t} A(n-2-t-2k-1) $$

*Same case but where from now all the tranches contain both the upper and lower part:
$$ \sum_{k=0}^{(n-3)/2}\sum_{t=1}^{k+1}{k+1 \choose t} \delta_{n,2+2k+t} $$


We obtain thus that $$A(n)=A(n-1)+2A(n-2)+ \sum_{k=0}^{(n-3)/2}\sum_{t=0}^{k+1}(2{k+1 \choose t} A(n-3-t-2k-1)+ \delta_{n,3+2k+t})+\sum_{k=0}^{(n-3)/2}\sum_{t=1}^{k+1}(2{k+1 \choose t} A(n-2-t-2k-1)+ \delta_{n,2+2k+t})$$
Note that the sum are not well define because $n-2-t-2k-1$ could be lower than $0$ but you can add the condition $A(n)=0$ for $n<0$.
Also you need the base cases: $A(0)=1$, $A(1)=2$ and $A(2)=4$.
Unfortunately I don't have access to  a software that could solve this. but if you do post the result in the comment :) May be we can use that to do an approximation ... (I'm a bit to lazy right now ...)
As for the case of bi-infinite ladder a good approximation is $\sum_{k=0}^n A(k).A(n-k)$ but it's just a bit more complex since for example in the case where in the first tranche there is only the upper and lower edge you have to consider the case where the connectedness comes from the right or the case when in comes form the left (on not count twice the graph where it's both connected on the right and left).
I hope it help a bit and that someone will be able to give a closed form :)
