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Suppose I fill a rectangle with regular hexagons assuming periodic boundary conditions (each vertex/edge located at the boundary of the rectangle is equivalent to the opposed boundary if there is one).

Suppose moreover that we can only cut the hexagons along a line that passes trough edges or vertices. I am not sure if there is a unique way to build such a rectangle over an hexagonal tiling of the plane.

As far as I know it is not possible to have an integer number of hexagons inside a rectangle (nor a square).

What I'm looking for is how many vertices there are in such a rectangle -without multiple counting due to the periodic boundary conditions- as a function of its size !

I know that Euler's formula holds : $$v - e + f = 2$$ for $v$ vertices, $e$ edges and $f$ hexagons (faces). But I can't manage to derive a formula that takes the boundary conditions into account.

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    Can you explain *"each vertex/edge located at the boundary of the rectangle is equivalent to the opposed boundary if there is one"* and *"we can only cut the hexagons along a line that passes trough edges or vertices"* more? I'm not sure that I understand. – 2017-02-25
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    A picture is worth a thousand words : http://www.physics-in-a-nutshell.com/article/4 Section 1.3 there is figure of the honeycomb lattice. What I basically meant is that I want a honeycomb lattice with periodic boundary conditions in a rectangle, and count the number of atoms inside. The boundaries should pass trough the vertices or the edges of the hexagons because it seems more natural and it is easier to account for the periodic boundary conditions. – 2017-02-26

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