A configuration is made of congruent regular hexagons,where each hexagon shares a side with another hexagon. What is the largest integer $k$, such that the figure cannot have $k$ vertices ? For example this figure has $13$ vertices.
That's what I've been able to do so far.
A hexagon built on one hexagon adds up $4$ vertices to the configuration.
A hexagon between two hexagons adds up $3$ vertices.
A hexagon between $3$ hexagons adds up $2$ vertices
A hexagon between $4$ hexagons adds $1$ vertice
A hexagon between $5$ hexagons adds $0$ vertices
A hexagon between $6$ hexagons adds $0$ vertices
Now it's clear that what ever number of vertices I have, I get it through the above combinations.
However I don't know what I have to do now.
It seems to me that I can add an infinite number of hexagons constructed one on the other and have an infinite number of vertices...
Can you guys help ?
Making a chain of $\geq2$ hexagons you can realize all $k$ of the form $k=10+4m$, $m\geq0$. Starting with your figure and attaching a chain of hexagons you can realize all $k=13+4m$ vertices, and starting with a blob of $4$, resp. $5$, hexagons one shows that all $k$ of the form $k=16+4m$ or $k=19+4m$, $m\geq0$, can be realized in this way. Note that $10$, $13$, $16$, $19$ represent all four remainders modulo $4$. Given that, we can realize all of $$10,13,14,16,17,18,19, 20,21,22,23,24,\ldots\ .$$
Therefore it seems that the largest non-realizable $k$ is $19-4=15$. In order to make sure that $k=15$ cannot be realized an impossibility proof is required. Such proofs are notoriously difficult.
Any configuration can be built up one for one, so that it remains connected all the time. Start with two hexagons glued together, making $10$ vertices. Then show that you can never reach $15$ vertices using your list of simple principles.