TITLE: Number of spanning trees in a grid
QUESTION [7 upvotes]: Given a $\sqrt{n}\times\sqrt{n}$ piece of the integer $\mathbb{Z}^2$ grid, define a graph by joining any two of these points at unit distance apart. How many spanning trees does this graph have (asymptotically as $n\to\infty$)?
Can you also say something about the triangular grid generated by $(1,0)$ and $(1/2,\sqrt{3}/2)$?

REPLY [4 votes]: Expanding on Steve Huntsman's answer, call the product which appears in A007341 f(n).  That is, 
$$f(n) = \prod_{k=0}^{n-1} {\prod_{l=0}^{n-1}}^\prime \left(2 - \cos {\pi k \over n} - \cos {\pi l \over n } \right)$$
where the $\prime$ on the second product indicates that we start at $l=1$ in the case $k = 0$.  The number of interest here is $a(n) = 2^{n^2-1} f(n)/n^2$ .
The product is the exponential of a sum, so
$$\log f(n) = \sum_{k=0}^{n-1} {\sum_{l=0}^{n-1}}^\prime \log \left(2 - \cos {\pi k \over n} - \cos {\pi l \over n } \right).$$
This sum is, in turn, $n^2$ times a Riemann sum for the integral
$$  C = \int_0^1 \int_0^1 \log(2-\cos x\pi - \cos y\pi) \: dx \: dy $$
which I believe converges, although actually evaluating it numerically is tricky.  If you believe that, then $\log f(n) \sim Cn^2$ as $n \to \infty$, and $\log a(n) \sim (C+\log 2) n^2$ as $n \to \infty$.  From evaluating $f(n)$ for various $n$, it appears that $C$ is near $0.473$, $e^C$ is near $1.605$ and so we have
$$ a(n) \approx 3.21^{n^2} $$
where I write $p(n) \approx q(n)$ for $\log p(n)/\log  q(n) \to 1 $ as $n \to \infty$, i. e. $\log p(n) \sim \log q(n)$.