= Solution
For a large aperture in the standard turbulence model, <astronomical seeing> is approximately $s=0.98\lambda/r_0$ radians, with <Fried parameter> $r_0$. Maintaining comparable correction requires a <deformable mirror> actuator pitch of order $r_0$, so the number of actuators across a diameter is proportional to $D/r_0$ and the total illuminated actuator count is proportional to $(D/r_0)^2$. Therefore
$$
\boxed{N_{\rm across}\propto\lambda^{-6/5},\qquad N_{\rm total}\propto\lambda^{-12/5},\qquad s\propto\lambda^{-1/5}.}
$$
Longer wavelengths require fewer actuators and have slightly smaller uncorrected atmospheric angular blur, even though the telescope's <diffraction-limited resolution> scale $\lambda/D$ increases with wavelength.
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