# 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?

• So density(G) = given a generation G, take difference of the maximum value of C for that generation, minimum value of C, then divide by the number of nodes in that generation, which is of course a power of 3, yielding, $\frac{max(C)-min(C)}{3^g})$? – frogeyedpeas Dec 29 '14 at 7:18
• Actually, I defined it the other way up ... as the count (a power of 3) divided by the range(max(c)-min(c)). That is, #occurances per width of range. – Peter Russell Dec 29 '14 at 12:18
• Given the calculation method of the triples I suspect this may involve an infinite matrix series. – Peter Russell Dec 29 '14 at 12:38
• So now what is scale exactly? Is it the ratio of successive generations? that is density(G)/density(G+1)? Or is it just the density of the graph as g approaches infinity? It wasn't made formally clear – frogeyedpeas Dec 29 '14 at 12:45
• Thanks for requesting the clarification, it is the ratio density(G)/density(G+1). As G->infinity this ratio settles quickly as the numbers get large quite early, but the ratio stays solid between generations and asymptotes. I could not run my calculations long enough without hitting precision issues, but there is a theoretical basis to a belief that the calculated value tends to 3(3-2sqrt(2)) as this is the scale factor of a classical 4-vertex octahedral graph based on octahedral expansion, so by extension also the scale factor of a 'rolled' out one like the pythagorean triple graph. – Peter Russell Dec 29 '14 at 13:01

## 2 Answers

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.

• Thank-you for a great answer. It seems then that my speculated relationship between the simple pythagorean triples, 4-vertex graph with scale 3(3-2sqrt(2)) and octahedral geometry is taking shape. Much appreciated. – Peter Russell Dec 29 '14 at 13:54
• To compute this limit by hand I would do the following: 1. $3-2*sqrt(2)$ is < 1 therefore the entire LHS of $c^{max}_n$ and its $n+1$ version drop out to zero 2. $c^{min}_n$ and its $n+1$ version also drop out as they are at most polynomial and are swamped in the quotient of the power n's 3. simplify by cancelling leaves $1/(3+2*sqrt(2))$ which is the inverse of $3-2*sqrt(2)$ – Peter Russell Jan 6 '15 at 23:24

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:

$- 2a + b + 3c = {c_1},\quad + 2a - b + 3c = {c_2},\quad 2a + b + 3c = {c_3}$.