Lorenz system is
\begin{equation*} \begin{cases} x' = -\sigma x + \sigma y \\ y' = x \left (r - z \right ) - y \\ z' = xy - bz \end{cases} \end{equation*} where $\sigma = 10, b = \frac{8}{3}$
I am trying to find out what occurs with the Lorenz system when $r$ changes from $r < 13.926$ to $r >13.926.$ In some literature, it is said that the attractor of the curve should change: if the curve was attracted to point $\left (\sqrt{b\left(r -1\right )}, \sqrt{b\left(r -1\right )}, r - 1 \right )$ then now it should attract to $\left (-\sqrt{b\left(r -1\right )}, -\sqrt{b\left(r -1\right )}, r - 1 \right).$ Nevertheless my experiments shows that it is not true. For example, for if start point is $\left(-10, -10, 10\right )$ then the moment of attractor change is between $r=7$ and $r=8.$ So, what actually occurs when $r$ changes from $r < 13.926$ to $r >13.926$
It's not true that every orbit changes attractor when $r>13.926$. The "change" is mostly noticed around the origin.
These are the (approximate) basins of attraction of the two non trivial equilibrium points for $r=12$, in the section of the plane $\pi:\{(x, y, z): z=0, |x|<30,|y|<30\}$. The origin is at the center of the picture. Points in red are attracted by one equilibrium and points in blue are attracted by the other. Since the whole Lorenz system is symmetric around the origin, the basins are too.
And these are the basins in the same section for $r=14$. Note that most points are still attracted to the same equilibrium, but around the origin all points have switched attractor.