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What formula would find the number of vertices within a 'normal' hexagonal graph, based on its radius (number of hexagons from center to edge)?

I've figured with pseudo code:

for (int i = 0; i < r; i++) { vertices += ((r + i) * 2) + 1; } vertices = vertices * 2;

Given the graph below, with a radius of 2, the above results in:

([((2 + 0) *2) +1] + [((2 + 1) *2) +1]) *2 = 24

enter image description here

So... Is there a formula that can do this for a radius of n?

Or, given V - F + E = 2, and knowing F = (3* r^2) - (3* r) + 1; a formula to derive the number of edges would work just as well.

Thanks in advance!!

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  • $\begingroup$ Although I can probably guess what you mean by a hexagonal graph, I shouldn't have to guess. Please define exactly what you mean. $\endgroup$
    – Casteels
    May 6, 2015 at 8:59
  • $\begingroup$ @Casteels - Well, I'm not sure what the correct term is. But I mean that it would consist of normal hexagons, and its overall shape would resemble a hexagon as well. Like the image. Sorry if I'm not explaining it very well. $\endgroup$
    – SomeoneLost
    May 6, 2015 at 11:10
  • $\begingroup$ Ok so for example, the next "larger" one would be obtained by surrounding your existing image by $12$ hexagons, and then the next larger one would look like the image here? $\endgroup$
    – Casteels
    May 6, 2015 at 11:20
  • $\begingroup$ oeis.org/A033581 $\endgroup$
    – Perry Elliott-Iverson
    May 6, 2015 at 16:34
  • $\begingroup$ @Casteels - Yes, exactly. $\endgroup$
    – SomeoneLost
    May 6, 2015 at 17:51

1 Answer 1

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There are $6n^2$ vertices in such a configuration with $n$ rings of hexagons. See for example, here.

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