n | sq total | sq ext. | sq int. |
---|---|---|---|
1 | 1 | 0 | 1 |
2 | 5 | 4 | 1 |
3 | 13 | 8 | 5 |
4 | 25 | 12 | 13 |
5 | 41 | 16 | 25 |
6 | 61 | 20 | 41 |
n^2+(n-1)^2
- If the current n-interesting polygon is
n
, then its area isn^2
. - If the previous n-interesting polygon is
(n-1)
, then its area is(n-1)^2
- Thus the current area plus the previous area equals the next...
n^2+(n-1)^2
4(n-1)
- Consider each n-interesting polygon as a solid object
- Draw the current polygon by scooting the previous polygon from its anchored center to each of the 4 different cardinal directions by 1 square (up, right, down, left) always returning to center before taking the next direction
- The exterior squares are those added to the previous polygon which create the current
- Visually counting just the exterior squares in each polygon (as compared to the original previous polygon) reveals a pattern also shown in the table above:
2n+2(n-2)
2n+2n-4
4n-4
4(n-1)
- Thus the exterior area is 4 times that of the previous polygon
2n^2-6n+5
- The interior number of squares is simply the total minus the exterior
[n^2+(n-1)^2]-[4(n-1)]
n^2+n^2-2n+1-4n+4
2n^2-6n+5
I apologize that it's not clear at all but it's a pattern I noticed visually and wish I could express it here in pictures. For now I'll just try with my words.
Look at each polygon like it's a square of varying resolution balancing precariously on a point. Like any reasonable square, you have two pairs of parallel sides. The first pair of sides draw at a diagonal from left to right like so,
/ /
. These sides aren
long and there are two of them. Thus this first pair can be represented as2n
. The second pair of sides draw at a diagonal from right to left like so,\ \
. However where they intersect with the previous sides, we don't count (almost like an 'inner' side without which we would be left with a gap as the square increases in size). These current 'sub-sides', if you will, have a length equal to the previous polygon two spots back, son-2
. When I look two polygons back at the\ \
sides, I consider the entire length which is the length of that polygon (again,n-2
). There are two of them thus we represent this pair as2(n-2)
.Combine the two pairs of sides
/ / + \ \
and you have2n + 2(n-2)
.n = '/'
(n-2) = '\'
2n = '/ /'
2(n-2) = '\ \'
2n + 2(n-2) = '/ / + \ \'
Does that make any sense?