It seems a marvel when a bunch of irrational numbers "conspire" to become rational, even better an integer. An elementary example is $\prod_{j=1}^n4\cos^2\left(\pi j/(2n+1)\right)=1$.

Kasteleyn's formula reveals $\prod_{j=1}^n\prod_{k=1}^n \left( 4\cos^2(\pi j/(2n+1))+4\cos^2(\pi k/(2n+1))\right)$ is an integer because it enumerates the domino tilings of a $2n$-by-$2n$ square.

These results prompt me to ask for more. First, let's introduce the $r$-product $$K_r(n):=\prod_{\ell_1=1}^n\cdots\prod_{\ell_r=1}^n\left( 4\cos^2\left(\frac{\pi\ell_1}{2n+1}\right)+\cdots+4\cos^2\left(\frac{\pi\ell_r}{2n+1}\right)\right).$$

Questions. This is based on experimental assesment.

(a) Is $K_r(n)$ always an integer?

(b) Is there perhaps a higher-dimensional combinatorial interpretation of Kasteleyn for $K_r(n)$?

(c) Why do $K_r(n)$ feature "small primes" with high-power factorizations? For example, $K_3(2)=3^2(19)^3,\,\, K_3(3)=3^35^6(83)^3(97)^3\,\, K_3(4)=2^63^{34}(17)^6(19)^6(37)^6\,\,$ and $$K_3(5)=3^5(43)^6(1409)^3(2267)^3(2707)^3(3719)^6(3389)^6.$$

I'm not aware of such a generalization, but any reference would be appreciated. Thanks.

UDPATE The comments have answered (a). Is there a more direct (elementary) proof? Any suggestions for parts (b) and (c)?

  • $\begingroup$ For $r=3$ the series starts with 1, 3, 61731, 220157391087140625. For $r=4" it starts with 1, 4, 689404732416 $\endgroup$ – Andreas Rüdinger Dec 30 '16 at 14:30
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    $\begingroup$ I don't know about (b), but (a) seems clear by just expanding out the product. Eventually one needs that $\sum_{j=1}^{n} (2+ e(j/(2n+1))+e(-j/(2n+1))^r$ is an integer for natural numbers $r$, which also follows by just expanding out. $\endgroup$ – Lucia Dec 30 '16 at 17:02
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    $\begingroup$ The formula for $K_r(n)$ is invariant under the Galois group of the extension of $\mathbb{Q}$ by the $\cos(\pi j/(2n+1))$'s, $1\leq j\leq n$. Thus $K_r(n)\in\mathbb{Q}$. Since $2\cos(j\pi/(2n+1)$ is an algebraic integer, so is $K_r(n)$, so $K_r(n)\in\mathbb{Z}$. $\endgroup$ – Richard Stanley Dec 30 '16 at 17:08
  • $\begingroup$ I guess in (c) the 2nd term is $K_3(3)$. Have you noticed that essentially, $K_3(n)$ seems to be a cube, more precisely $K_3(n)=3^n\cdot x_{3,n}^3$? Should we have similarly that $K_r(n)$ is essentially a $r$th power? $\endgroup$ – Wolfgang Dec 30 '16 at 21:33
  • $\begingroup$ @Wolfgang: Thanks, typo fixed. I agree with you and believe the trend continues: $r^n,\vert K_r(n)$. This property has been noticed already when $r=2$. $\endgroup$ – T. Amdeberhan Dec 30 '16 at 21:53

Have you considered using the matrix-tree theorem [2, 3] which counts spanning trees instead of domino tilings?

Let $\kappa(G)$ count the number of spanning trees of $G$.

Theorem $\kappa(G) = \frac{1}{n}\lambda_1\lambda_2\dots \lambda_{n-1}$ where $\lambda_k$ are the eigenvalues of the graph Laplacian $L$ of $G$.

For a rectangular graph (or any product of intervals) we can engineer a graph whose eigenvalues multiply to $K_r$ (not that I have worked it out).

There is a bijection by Temperley which connects domino tilings and spanning trees (since both involve planar graphs). I don't know of higher dimensional Temperley bijection.

  • As Richard Stanley explains, the Galois group for $[\mathbb{Q}(e^{2\pi i / (2n+1)}): \mathbb{Q}] = 2n+1$ should act on the product and preserve it. There may even be an action on the spanning trees themselves. These numbers must have number-theoretic properties.
  • The log of the number tilings, divided by area (aka entropy) tends to Catalan's constant, $L(2, \chi_4)$. And L-functions appear as entropies for other lattice as well.

Tzeng + Wu [Spanning Trees on Hypercubic Lattices and Non-Orientable Surfaces] (also works out Möbius strip and Klein bottle).

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For a modern discussion of Kasteleyn's theorem look at

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Here is an explanation why all prime factors are small.

Following the well-known motto "A mathematician who doesn't succeed proving a conjecture tries to generalize it", I have asked this question, where the $K_r(n)$ can in fact be interpreted as constant terms of certain polynomials. (See the edit at the end, initially I didn't realize that!)

Now, these polynomials are 'highly reducible', meaning that all their irreducible factors have degrees at most $n$ (and moreover very small coefficients), their CTs also consist of small prime factors.

This is because the irreducible factors are nothing else than the minimal polynomials of $$4\cos^2\left(\frac{\pi\ell_1}{2n+1}\right)+\cdots+4\cos^2\left(\frac{\pi\ell_r}{2n+1}\right),$$ and those quantities only involve the $(2n+1)^{th}$ roots of unity. So their algebraic degrees divide $\phi(2n+1)$, and by some symmetry consideration it should follow that moreover they are strictly smaller than $\phi(2n+1)$, i.e. $\le n$.

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  • $\begingroup$ More precisely we're dealing with $2(2n+1)^{th}$ roots of unity. But that doesn't change anything, as $\phi(2n+1)=\phi(2(2n+1))$. :) $\endgroup$ – Wolfgang Jan 19 '17 at 11:28
  • $\begingroup$ Yes, I saw that and it is cool. $\endgroup$ – T. Amdeberhan Jan 19 '17 at 17:06

The fact that the values $K_2(n)$ in Kastelyn's formula are rational integers is not so amazing (once we see the Galois Theory proof). It is amazing that that they count what they do. However that same integrality proof applies to simpler formulas.

It is worth mentioning that from $$\prod_{j=1}^n4\cos^2\left(\pi j/(2n+1)\right)=1$$ we also have $$\prod_{j=1}^n2\cos\left(\pi j/(2n+1)\right)=1$$ We can't simply use the fact that $$K_r(n):=\prod_{\ell_1=1}^n\cdots\prod_{\ell_r=1}^n\left( 4\cos^2\left(\frac{\pi\ell_1}{2n+1}\right)+\cdots+4\cos^2\left(\frac{\pi\ell_r}{2n+1}\right)\right) \in {\large \mathbb{Z}}$$ to see that also $$J_r(n):=\prod_{\ell_1=1}^n\cdots\prod_{\ell_r=1}^n\left( 2\cos\left(\frac{\pi\ell_1}{2n+1}\right)+\cdots+2\cos\left(\frac{\pi\ell_r}{2n+1}\right)\right) \in {\large \mathbb{Z}}.$$ However the same reasoning which gives $K_r(n) \in \mathbb{Z}$ also gives $J_r(n) \in \mathbb{Z}.$ And we have yet to see that there is any reason to favor $K_r(n)$ over $J_r(n)$ othet than that we know an amazing fact about $K_2(n).$

The factors $J_r(n)$ would seem to have as much reason to be small as those of $K_r(n).$. Of course one would be more confident asserting that after computing some of the values, which I have not done.

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