Ring Counts for Equilateral Triangles

The bottom line:

For an equilateral triangle of Japanese 12-in-2 with n large-ring pairs on a side, you will need

n(n+1)   large rings and
3(n-1)n  small rings

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By the numbers:

1. Consider an equilateral triangle of Japanese 12-in-2, with n large-ring pairs on each side. Such a structure can be seen as a series of rows, the first of which contains one large-ring pair, the second two, and so on. There will be n large-ring pairs in the last row (which is the bottom edge of the triangle).

   The total number of large-ring pairs, therefore, is the sum of the first n whole numbers. This is the nth so-called triangular number, which equals n(n+1)/2.

   So the total number of large rings in such a triangle is double that sum, or n(n+1).

2. In Japanese 12-in-2, the number of small-ring pairs attached to a given large-ring pair depends on the angle it forms with its neighbors:

   60 degrees: 2 small-ring pairs  (an acute corner)
   120         3                   (an obtuse corner)
   180         4                   (an edge ring)
   240         5                   (an indentation)

   A ring fully in pattern (surrounded on all sides) attaches to 6 small-ring pairs. (Hence the name of the weave.)

3. Every small ring goes through exactly two large-ring pairs. So if you look at every large pair in a design in turn, and count the small ring-pairs through it, each small pair will be counted exactly twice. Adding up those ring-counts for the whole design, you get twice the total number of small-ring pairs—which is the same as the total number of small rings.

4. Consider once more a triangle with n large-ring pairs on each side. Adding a row of n+1 rings to the bottom edge of such a triangle creates a new triangle with n+1 large-ring pairs on each side.

   Let's look at each added or changed ring, and count the number of small-ring pairs added for each:

   
   
   • Two 60-degree corners become edge-rings, adding 2x(4 - 2) = 4 pairs.
   • Two new corners are formed, adding 4 pairs.
   • n-2 edge-rings become locked into pattern, adding (6 - 4)(n-2) = 2n - 4 pairs.
   • n-1 edge-rings are added, adding 4(n-1) pairs.
   
   

   So the total is 4 + 4 + (2n - 4) + (4n - 4) = 6n pairs. But remember that we have double-counted. So we have actually added 6n small rings.

5. Looking at the total number of small rings in an equilateral triangle as a function sm of n (where n is the number of large-ring pairs on each side), we have established the following recurrence relation:

   sm(1) = 0
   sm(n+1) = sm(n) + 6n

   Solving the recurrence yields sm(n) = 3(n-1)n.