The article implies that, asymptotically, the spiral takes $F_n$ (the $n$-th Fibonacci number) branches to go $F_{n+2}$ times around the trunk of the tree. The fraction $2/5$ is only a first approximation to this phenomenon (the other cited approximations being $1/2,1/3,3/8,5/13$). The limiting ratio is $1/\phi^2$, where $\phi$ is the golden ratio $(1+\sqrt{5})/2$. Hence the rotation angle between branches should be approximately $360^\circ / \phi^2\approx137.507764^\circ$.
To see why the ratio is what it is, first let's define the ratio of successive Fibonacci numbers, in the limit, to be $$\phi = \lim_{n\to\infty}\frac{F_{n+1}}{F_n}.$$
(We'll leave aside the question of why this limit exists, i.e. why the Fibonacci numbers exhibit exponential growth.) Then, by the recurrence definition of Fibonacci numbers, we have
$$\frac{F_{n+1}}{F_n}=\frac{F_n+F_{n-1}}{F_n}=1+\left(\frac{F_n}{F_{n-1}}\right)^{-1}\to1+\phi^{-1}.$$
Hence $\phi$ solves the equation $x=1+1/x$. This can be multiplied through for the quadratic equation $$x^2-x-1=0.$$
The golden ratio $\phi$ must be the positive solution, $(1+\sqrt{5})/2$. Now note that $$\frac{F_n}{F_{n+2}}=\left(\frac{F_{n+2}}{F_{n+1}}\right)^{-1}\cdot\left(\frac{F_{n+1}}{F_n}\right)^{-1}\to\phi^{-1}\cdot\phi^{-1}=1/\phi^2.$$
If we set $\varphi = \frac{1+\sqrt{5}}{2}$ and $\bar{\varphi}=\frac{1-\sqrt{5}}{2}$, then the Binet formula for Fibonacci numbers says:
$F_n=\frac{\varphi^n-(\bar\varphi)^{ n}}{\sqrt{5}}$.
Because $\mid \bar\varphi \mid < 1$, the term $(\bar\varphi)^n$ decays fairly quickly toward $0$. This points to why successive ratios of Fibonacci numbers get close to $\varphi$.
Note: Here is a reference for the Binet formula: http://mathworld.wolfram.com/BinetsFibonacciNumberFormula.html
Best Answer
The Padovan sequence has a 3-D analogue that fills the space with prisms that grow in linear dimension as the sequence. That's all I remember of the top of my head.