Zeros of higher derivatives of $\zeta(s)$ Zeros of successive higher order derivatives of the Riemann zeta function seem to cluster along roughly horizontal lines.
Is there a heuristic explanation of why this happens (especially inside the critical strip)?
R. Spira, in Zero free regions of $\zeta^{(k)}(s)$ wrote "The zeros of $\zeta^{\prime\prime}$ have imaginary part almost exactly equal to those of $\zeta^\prime$, and lie to the right of them."  Figure 1 from his paper shows zeros of $\zeta^\prime$ marked with triangles, and zeros of $\zeta^{\prime\prime}$ marked with squares.

Below is a Mathematica plot of the argument of $\zeta^{\prime\prime}/\zeta^\prime(s)$ for $1/2\le \sigma\le 1$ and $10^6\le t\le 10^6+20$ (Five strips of height $4$.). The poles at the zeros of $\zeta^\prime(s)$ have to opposite orientation of colors that the zeros of $\zeta^{\prime\prime}(s)$ have.

S.L Skorokhodov, in "Pade Approximants and Numerical Analysis of the Riemann zeta function", Computational Mathematics and Mathematical Physics vol 43 (2003) pp. 1277-1299, formula (7.7) differentiates the Dirichlet series expansion for $\zeta(s)$ term by term (for $\sigma>1$), and deduces "Therefore, it is again natural to expect a zero of the derivative $\zeta^\prime(s)$ to be located on the right [sic] of each finite zero of $\zeta^{\prime\prime}(s)$."
I don't understand this, particularly in regards to zeros in the critical strip.
Binder, Pauli, and Saidak, in "Zeros of High Derivatives of the Riemann Zeta Function", Rocky Mountain J. Math, vol 45 (2015), pp. 903-926 have similar results (Theorem 2.3), again using the Dirichlet series.  They see chains of zeros up to order 90:

From Theorem 3 in the seminal paper of Levinson and Montgomery, we can see this happens on average over an interval of size, say, $U=\log T$:
$$
\frac{2\pi}{\log T}\left(\sum_{T<\gamma^{(k+1)}<T+\log T}(\beta^{(k+1)}-1/2)-\sum_{T<\gamma^{(k)}<T+\log T}(\beta^{(k)}-1/2)\right)
=\log\log T/2\pi+O(1).
$$
The right side is independent of $k$.

Edit/Correction:  The corollary of the Levinson-Montgomery result speaks to the horizontal spacing of consecutive higher derivatives, on average independent of $k$.  But it does not address Spira's observation about the vertical spacing: consecutive higher derivatives at very nearly the same height.
 A: Consider the absolute value of the $n$th derivative $\zeta^{(n)}$.
It is large (in magnitude) to the left and it slopes
down to be asymototic to 0 as you move to the right.  A zero of that
function is a sharp valley interrupting the overall trend.  A zero
of its derivative is a saddle point of that surface.  So one should
expect the saddle point to be directly to the right of the valley (because from the left the land slopes down, and as you emerge moving to the right from the valley it slopes up before resuming its overall downward trend).
That might not happen if there were other nearby valleys, causing
the topography to be weird.  But that happens rarely.
The above picture might not be completely accurate in the critical strip,
but all those pictures of zeros of successive derivatives are in the region
of absolute convergence.
A: Here's another approach.  Assume the Riemann Hypothesis for simplicity.  As $\sigma\to +\infty$, $\zeta^{\prime\prime}/\zeta^\prime(s)\to -\log(2)$.  Meanwhile, on the critical line, $\zeta^{\prime\prime}/\zeta^\prime(s)$ will be dominated by terms in the sum $\sum_{\rho^\prime} 1/(s-\rho^\prime)$ with $\rho^\prime$ near $s$, and this implies the real part of $\zeta^{\prime\prime}/\zeta^\prime(s)$ will again be negative.
Now fix a zero $\rho^\prime$ of $\zeta^\prime(s)$, and consider the level curves $\arg(\zeta^{\prime\prime}/\zeta^\prime(s))=C$ exiting the pole at $\rho^\prime$, in particular the curve $C=0$ (i.e. Im($\zeta^{\prime\prime}/\zeta^\prime(s)$)=0 and Re($\zeta^{\prime\prime}/\zeta^\prime(s)$)$>0$.).  By the above observations, this contour can't cross the critical line, nor extend too far into the right half plane.  The only possibility is that it terminates in a zero of $\zeta^{\prime\prime}(s)$.  Since
$$
\frac{\zeta^{\prime\prime}}{\zeta^\prime}(s)=\frac{1}{s-\rho^\prime}+O(1),
$$
the contour has to exit the pole to the right, and so the zeros of $\zeta^{\prime\prime}$ will typically be to the right of the zeros of $\zeta^\prime$.
The graphics below shows, for $1/2\le\sigma\le 5/2$ and $10^6\le t\le 10^6+10$, the argument of $\zeta^{\prime\prime}/\zeta^\prime(s)$ on the left, and only those curves corresponding to the four coordinate axes on the right.  We are focusing on the red curves.  (NB the graphic on the left is the same as in the original question, but I've fixed a bug in the Mathematica function ComplexPlot[] which shifts the phase by $\pi$)

This same argument works for higher derivatives as well.
