A follow up on [an earlier MO question](https://mathoverflow.net/questions/258386/kasteleyns-formula-for-domino-tilings-generalized).


Kasteleyn's formula for the number of domino tilings of a $2n\times 2n$ square 
$\prod\_{j=1}^n\prod\_{k=1}^n \left( 4\cos^2(\pi j/(2n+1))+4\cos^2(\pi k/(2n+1))\right)$ is known to be $\sim e^{\frac{4G}{\pi}n^2}$; where
$G$ is [Catalan's constant](https://en.wikipedia.org/wiki/Catalan%27s_constant).


By way of generalization, define the $r$-fold product
$$K\_r(n):=\prod\_{\ell\_1=1}^n\cdots\prod\_{\ell\_r=1}^n\left(
4\cos^2\left(\frac{\pi\ell\_1}{2n+1}\right)+\cdots+4\cos^2\left(\frac{\pi\ell\_r}{2n+1}\right)\right).$$



> 
> **Question.** Can one give an asymptotic estimate for $K\_3(n)$? More generally, for $K\_r(n)$?
> 
> 
> 



Asymptotics for 'generalized" Kasteleyn's formula
Take the logarithm and note that for large $n$ we may replace the sum by an integral:


$$\lim\_{n\rightarrow\infty}\frac{1}{n^r}\log K\_r(n)=2\log 2+C\_r$$
with an $r$-dependent numerical coefficient $C\_r$ given by an integral over the $r$-dimensional unit cube:
$$C\_r=\int\_0^1 dx\_1 \cdots\int\_0^1 dx\_r \log\left[\sum\_{p=1}^r \cos^2(\pi x\_p/2)\right]$$
$$\qquad=(2/\pi)^r\int\_0^1 \frac{dx\_1}{\sqrt{1-x\_1^2}} \cdots\int\_0^1 \frac{dx\_r}{\sqrt{1-x\_r^2}}\;\log\left[\sum\_{p=1}^r x\_p^2\right]$$


*Mathematica* tells me $C\_2=(4/\pi)G-2\log 2$, which gives the Catalan constant asymptotics of Kasteleyn's formula (who, incidentally, was my [colleague](http://ilorentz.org/research/kasteleyn/kasteleyn.html) here in Leiden). 


A numerical integration gives $C\_3=0.287095$, $C\_4=0.613413$, $C\_5=0.856191$, all of which may or may not be expressable in terms of some named constants unknown to *Mathematica*.




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For $r\gg 1$ we can argue as follows: consider $r$ i.i.d. random variables $u\_1,u\_2,\ldots u\_r\in(0,1)$, each with distribution
$$P(u)=\frac{1}{\pi\sqrt{u(1-u)}},$$
normalized to unity. The coefficient $C\_r$ equals the expectation value
$$C\_r=E\left(\log\left[\sum\_{p=1}^r u\_p\right]\right).$$
Note that $E(u)=1/2$. By concentration of measure we have
$$C\_r\rightarrow\log(r/2),\;\;\text{for}\;\;r\rightarrow\infty.$$
I note that $\log(5/2)=0.92$, already not too far from the exact value of $0.86$.


Collecting results, the large-$n$, large-$r$ asymptotics of $K\_r(n)$ is
$K\_r(n)\mapsto(2r)^{\displaystyle{n^r}}.$ This final answer could have been obtained directly from the definition of $K\_r(n)$ as an $n^r$-term product, simply by substituting $1/2$ for each $\cos^2$.


