Number non self avoiding closed walks surrounding some point While studying some Peierls-like arguments in statistical physics I thought about the following problem: We have some 2d-integer lattice like this, for simplicity infinite in all directions. 
Now fix some point $x$ between four lattice sites (i.e. in the middle of some white square). I'm interested in the number of self avoiding closed curves on the lattice surrounding the point $x$ of a given length $n$. Self avoiding and closed here means that the curve is closed but does not intersect or is tangent to itsself in any point (I hope it's clear what I mean).
Obviously the length has to be even and $\geq 4$. 


*

*For $n=4$ there is obviously only $1$ such curve. 

*For $n=6$ there are $4$ such curves (two horizontal and two vertical rectangles where $x$ is in the lower/upper/left/right part of the rectangle). 

*For $n=8$ things start to be more interesting: There are $6$ rectangles of size $3*1$, $4$ squares of size $2*2$ and $12$ L-shaped figures, in total $22$. 

*For $n=10$ there are $8$ rectangles of size $4*1$, $12$ rectangles of size $2*3$, $32$ L-shaped figures and $16$ tetris-like .:.-figures, in total $68$


Continuing like this it seems to be a good idea to sort the figures by their volume but I don't know how to use this in order to find a closed form. Obviously for given $n$ the possible volumes of curves are exactly $\frac{n-2}{2},\frac{n-2}{2}+1,\dots,\lfloor \left(\frac{n}{4}\right)^2\rfloor$. I tried to find an integer sequence starting like this but didn't find one.
 A: These are called self avoiding polygons.There are no known explicit formulas for the number of them. Your restriction to surround a point seems minimal as you can just count the number of self avoiding polygons of length $n$ and translate each to surround your point. 
What is known or conjectured is that if you look at the number of such polygons of size $n$, call it $N(n)$, then $N(n)\approx $const$\times n^ac^n$ where $a$ is a constant that is independent of the lattice type and $c$ is the "connectivity constant" which depends on the lattice. It is highly nontrivial that this $a$ is actually the same for usual self avoiding walks. One can show the existence of $c$, in other words $\lim_{n\rightarrow\infty}N^{1/n}(n)=c$ using subadditivity (Fekete's Lemma). This is because roughly, $N(n+m)\leq N(n)N(m)$ since taking two self avoiding polygons of length $n,m$ can potentially be combined (by cutting them at one point, unfolding them and gluing them together. This is a very loose argument which is used for usual self avoiding walls just fine, it probably needs some extra care for polygons. Hammersley was the first to prove these results).. 
