# Calculating the hexagons in a hexagon of hexagons

I want to make a hex based board game but need to figure out how many 'tiles' I need to print to complete the board. I don't have a set size yet, so I wanted to see if anyone had a formula for finding the hexagons in a hexagon of hexagons.

If that makes no sense, let me explain. Imagine Catan, it's hex based, and assuming your playing Settler of Catan, it looks something like this. It's a hexagon of hexagons. Now, I want to be able to calculate the number of individual hexagons in that board given the radius from the center hexagon to the outer hexagons. In this case, it's 2, because there are 2 hexagons between the center desert tile and the outer tiles.(Although if you counted the center hexagon, you could argue it's 3.)

Does anybody know a formula to do this?

As clearly seen here: the n'th ring contains 6n hexagons for n > 0; and 1 hexagon for n=0.

Thus H(n), the number of hexagons contained within ring n, is obtained as

``````H(n) = 1 + 6 * SUM (i) with i ranging from 1 to n.
``````

Using the well known formula for the sum of the integers from 1 to n then yields:

``````H(n) = 1 + 6 * [ n * (n+1) / 2 ]
= 1 + 3 * n * (n+1)
``````

Test:

• H(0) = 1 - Check
• H(1) = 7 - Check
• H(2) = 19 - Check
• H(3) = 37 - Check
• ....

Also:

``````H(n+1) - H(n) = [ 3 * (n+1) * (n+2) ] - [ 3 * ()* (n+2) ]
= 3 * (n+1) * [ (n+2) - n ]
= 6 * (n+1)
``````

as required. Combined with the calculated values above as base class, this constitutes the guts of a proof by induction.

• @Andrew: So you claim that H(3) = 1 + 6 + 36 + 216 = 259? Clearly not. Might I recommend actually testing your hypotheses prior to publication. Jun 2, 2021 at 0:56