Total area of infinite circles nested in an equilateral triangle. 
Given that the radius of the bigger circle is 1, what is the total area of the infinite circles in the picture above?
I know how to solve part of the problem, following the steps of this site.
But the problem is the remaining circles. I tried to make some algebra using a special case (where one of the circles is a line) of Descarte´s theorem, but I found no pattern to write a series and then find the sum.
How can I find the area of the remaining circles, indicated in red in the picture below?

By the theory of Ford circles, touching circles satisfy $$\frac{1}{\sqrt{r_M}}=\frac{1}{\sqrt{r_L}}+\frac{1}{\sqrt{r_R}}$$
In the case of the given problem, each circle touches two unique larger circles. If we concentrate on just one branch of the set (a third of the circles), the central circle has radius $1$ and the next largest circle has radius $1/3$ by similarity. Their touching circle has radius $1/(1+\sqrt3)^2$ by the above formula.
Each circle can be represented by a pair of integers $(m,n)$ which is the sum of its parents's indices, and has radius $r_{n,m}$ given by $\frac{1}{(m+n\sqrt{3})^2}$, using the above formula. The following diagram represents just one family of circles generated by the largest one $(1,0)$ and the next largest $(0,1)$. Each vertex in the tree represents a space between the circles and each edge represents the tangent touching two circles.
$\hspace{2cm}$
The next family to the left is generated by $(0,1)$ and $(3,0)$ because each circle, with center on the line going from the center of the triangle to the left vertex, has radius $1/3^n$ (represented by $(3^{n/2},0)$ or $(0,3^{(n-1)/2})$).
Tabulating $1/\sqrt{r_{n,m}}$ for the first family of circles gives:
Family 1: $$\begin{matrix}
1\\
1+\sqrt3\\
1+2\sqrt3&2+\sqrt3\\
1+3\sqrt3&2+3\sqrt3&3+2\sqrt3&3+\sqrt3\\
\cdots\end{matrix}
$$
Following is a Mathematica script for generating these pairs:
level[n_] := level[n] = Riffle[level[n - 1], Most@level[n - 1] + Rest@level[n - 1]]
level[1]={{1,0},{0,1}}
sum[n_] := Plus @@ ((1/(#[[1]] + #[[2]] Sqrt[3.])^4) & /@ level[n])
area1 = Pi(sum[25] - 1)

(The central circle is subtracted off.)
A numerical value for area of the first family is $A_1\approx0.4550$.
The rest of the families are similar to the first family because they are scaled versions of them. For example, the second family is generated by $(3,0)$ and $(0,1)$, hence is a third of family one in size (and ninth in area).
Thus the total area of one branch is $B=A_1(1+\frac{1}{9}+\frac{1}{9^2}+\cdots)=\frac{9}{8}A_1\approx0.5119$.
The required answer for the total area is $3B+\pi$, adding the central circle. A numerical approximation of this area is $4.68$, which is just over $90\%$ of the whole triangle.
