Looking at the problem differently, Jon writes Pattern 1, Pattern 2, Pattern 3, and so on.
Looking at these separate patterns individually will yield different rules for each Pattern.
To begin with you have one triangle with the base at the bottom and a vertex at the top(3 segments and
1 triangle). Notice I'm looking at ordered pairs involving segments and triangles; that is (s,t) where
s=# of segments and t=# of triangles.
/\(this is the first triangle)
Now make a second triangle under the first; 5 segments, 2 triangles.
Make a third triangle to the right of the first; 7 segments, 3 triangles
Make a fourth triangle under the third; 9 segments, 4 triangles
Make a fifth triangle at the top to the right of the third one you made; 11 segments, 5 triangles
This is the first Pattern(not including the first triangle because these patterns occur in groups of four.)
(5,2), (7,3), (9,4), (11,5)
The sixth triangle in the pattern only needs one more segment giving us 12 segments, 6 triangles.
The seventh triangle in the pattern yields 14 segments and 7 triangles.
This is the second Pattern of four ordered pairs:
(12,6), (14,7), (16,8), (18,9)
The third Pattern yields the ordered pairs (19,10), (21,11), (23,12), (25,13)
The fourth Pattern yields (26,14), (28,15), (30,16), (32,17)
The fifth Pattern yields (33,18), (35,19), (37,20), (39,21)
You can continue this pattern of triangles for as long as you like.
The sequence for the ordered pairs (s,t) in Pattern 1 would be t=(s-1)/2
The sequence for the ordered pairs (s,t) in Pattern 2 would be t=s/2
Pattern 3 you would have t=(s+1)/2
Pattern 4 you would have t=(s+2)/2
Pattern 5 you would have t=(s+3)/2
These rules themselves have a pattern to them(s is being increased by 1 every four newly created triangles.)
Now if we could take all these rules and put them together into one rule we would have another solution
to the problem that would not be changed or interrupted every four triangles.
I discovered these patterns looking at the problem in terms of segments and triangles.