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.
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)..