Roman C. answered 05/11/17
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You can only give bounds for this number.
Let's assume, the n hexagons must form a polyhex (like a polyomino but with regular hexagons).
Maximum:
The first hexagon contributes 6 vertices and each new hexagon after that contributes at most 4 vertices.
MAX = 6 + 4(n-1) = 4n+2
Minimum:
You need to maximize the number of hexagons which are completely surrounded. See sequence A121149 in the Online
Encyclopedia of Integer sequences for the minimum. There is unlikely to be a nice formula for the minimum.
As we add hexagons, we add more and more hexagons, we find that the vast majority of the new hexagons will contribute 2 new vertices each. Specifically, the new hexagons that contribute more than two become exceedingly rare and there are never new ones contributing only 1 new vertex.
Thus:
limn→∞ MAX(n) / MIN(n) = 2