All possible formulas for the amount of triangles are as follows.

Let A(n) be the total number white of triangles in the n’th run.

Let B(n) be the total number of triangles in the n’th run in the structure that does not include the two corner triangles.

Let C(n) be the total numbers of triangles in the n’th run including the corner triangles.

Then we the following starting numbers and recursive formulas hold:

A(1)=1 and A(n+1)=3A(n) for all integers n>0

Reason: previous structure is embedded exactly 3 times

B(1)=1 and B(n+1)=3B(n) + 2 for all integers n>0

Reason: previous structure is embedded 3 times and this creates 1 new center triangle and 1 new greater triangle

C(1)=3 and C(n+1)=3(C(n) - 2) + 4 for all integers n>0

Reason: previous structure with the corner triangles removed is embedded 3 times, similar to previous case, and we add the 2 corner triangles at the end.

For this run (n=4) that means A(4)=27 B(4)=53 C(3)=55