Tree of Primitive Pythagorean Triples graph scale infinite series Consider the tree of primitive Pythagorean triples as seen here:
https://en.wikipedia.org/wiki/Tree_of_primitive_Pythagorean_triples
Consider the values for c in the triples (a, b, c) and their density in each generation layer of the tree. Taking the number of c's in any generation and dividing by the difference between the maximum and minimum values for c in that generation gives the density of that generation.
The 'scale' of the graph is given by the ratio of densities as the generations go on to infinity. I have calculated the maximum, minimum and number of c's in increasing generations of the tree and the 'scale' asymptotes to 0.5147186257… by calculation.
Note: This is a graph scaling question and is not the same question as the simple density of primitive triples as solved by Lehmer (1900).
I postulate that this graph scaling value is in fact 3(3-2sqrt(2)). Can this be proved?
 A: If $(a,b,c)$ is a primitive Pythagorean triple, then the $c$ values of its descendants (by the matrix expression) are $$2a - 2b + 3c, \quad 2a + 2b + 3c, \quad {\rm and} \quad -2a + 2b + 3c.$$
Clearly the middle one is the largest. Hence the middle one is the largest in any generation.
With the starting value $(3,4,5)$, the first is the smallest. Hence the first is the smallest in any generation.
In the notation of Wikipedia and by diagonalization we have
$A^n = \begin{pmatrix}1 & -2 n & 2 n \\
2 n & 1-2 n^2 & 2 n^2 \\
2 n & -2 n^2 & 2 n^2+1
\end{pmatrix}.$
Hence the minimum $c$ value in the $n$th generation is $c^{\rm min}_n = 2n^2 + 6n + 5$.
Similarly but more horrifically (diagonalizing $B$), the maximum $c$ value in the $n$th generation is 
$$c^{\rm max}_n = \frac{1}{2}\left(5 -\frac{7}{\sqrt{2}}\right)(3 - 2 \sqrt{2})^n + \frac{1}{2}\left(5 + \frac{7}{\sqrt{2}}\right)(3 + 2 \sqrt{2})^n$$
The density of the $n$th generation is $d_n = \frac{3^n}{c^{\rm max}_n - c^{\rm min}_n}$.
The quotient of two consecutive densities is $q_n = \frac{d_{n+1}}{d_n} = 3\frac{c^{\rm max}_n - c^{\rm min}_n}{c^{\rm max}_{n+1} - c^{\rm min}_{n+1}}$.
Finally, we have $$q = \lim_{n\to\infty} q_n = 3\lim_{n\to\infty} \frac{c^{\rm max}_n - c^{\rm min}_n}{c^{\rm max}_{n+1} - c^{\rm min}_{n+1}} = 3(3-2\sqrt{2}).$$
The last limit was computed using a CAS:
http://www.wolframalpha.com/input/?i=limit%2B((5-7%252fsqrt(2))*(3-2*sqrt(2))%255en%252b(5%252b7%252fsqrt(2))*(3%252b2*sqrt(2))%255en-2n%255e2-6n-5)%2B%252f%2B((5-7%252fsqrt(2))*(3-2*sqrt(2))%255e(n%252b1)%252b(5%252b7%252fsqrt(2))*(3%252b2*sqrt(2))%255e(n%252b1)-2(n%252b1)%255e2-6(n%252b1)-5)%2Bto%2Binfinity%26incParTime%3Dtrue

If someone sees how to compute this by hand, feel free to edit this answer.
A: Ricardo's analysis for Berggren's Tree can be applied to Price's Tree also, but in the later case the c values of the descendants are:
$ - 2a + b + 3c = {c_1},\quad  + 2a - b + 3c = {c_2},\quad 2a + b + 3c = {c_3}$.
