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Just for fun here's the diffraction pattern from Hubble's circular aperture and 4-vane secondary support. Most of the time the concentric rings are not visible because images are broadband and they get washed out, but they are here because the exposure is through a narrow-band filter. From What is the cause of all of these sharp, concentric rings around bright stars in this HST image?

HST diffraction pattern through narrow-band filter

This is a partial answer.

Hexagonal aperture. I take the following function: $$ \begin{split} h(x,y) = \left[\theta(2y+\sqrt{3})-\theta(2y-\sqrt{3})\right]\cdot \left[\theta(y-\sqrt{3}(x-1))-\theta(y-\sqrt{3}(x+1))\right]\cdot\\ \left[\theta(y-\sqrt{3}(-x-1))-\theta(y-\sqrt{3}(1-x))\right] \end{split}$$ With $\theta(x)$ -- is a [Heaviside step function][2]Heaviside step function.
While it is in principle possible to calculate the Fourier transform of this by hand, I've just fed it to CAS and I've got something like this: $$\frac{2\sqrt{3}\omega_x\left(\cos\frac{\omega_x}{2}\cos\frac{\sqrt{3}\omega_y}{2} -\cos\omega_x\right)-6\omega_y\sin\frac{\omega_x}{2}\sin\frac{\sqrt{3}\omega_y}{2}} {\pi \omega_x(\omega_x^2-3\omega_y^2)}$$ ...plus some singular terms that I've dropped.
Plotting square of the result (from -100 to 100)

This is a partial answer.

Hexagonal aperture. I take the following function: $$ \begin{split} h(x,y) = \left[\theta(2y+\sqrt{3})-\theta(2y-\sqrt{3})\right]\cdot \left[\theta(y-\sqrt{3}(x-1))-\theta(y-\sqrt{3}(x+1))\right]\cdot\\ \left[\theta(y-\sqrt{3}(-x-1))-\theta(y-\sqrt{3}(1-x))\right] \end{split}$$ With $\theta(x)$ -- is a [Heaviside step function][2].
While it is in principle possible to calculate the Fourier transform of this by hand, I've just fed it to CAS and I've got something like this: $$\frac{2\sqrt{3}\omega_x\left(\cos\frac{\omega_x}{2}\cos\frac{\sqrt{3}\omega_y}{2} -\cos\omega_x\right)-6\omega_y\sin\frac{\omega_x}{2}\sin\frac{\sqrt{3}\omega_y}{2}} {\pi \omega_x(\omega_x^2-3\omega_y^2)}$$ ...plus some singular terms that I've dropped.
Plotting square of the result (from -100 to 100)

Just for fun here's the diffraction pattern from Hubble's circular aperture and 4-vane secondary support. Most of the time the concentric rings are not visible because images are broadband and they get washed out, but they are here because the exposure is through a narrow-band filter. From What is the cause of all of these sharp, concentric rings around bright stars in this HST image?

HST diffraction pattern through narrow-band filter

This is a partial answer.

Hexagonal aperture. I take the following function: $$ \begin{split} h(x,y) = \left[\theta(2y+\sqrt{3})-\theta(2y-\sqrt{3})\right]\cdot \left[\theta(y-\sqrt{3}(x-1))-\theta(y-\sqrt{3}(x+1))\right]\cdot\\ \left[\theta(y-\sqrt{3}(-x-1))-\theta(y-\sqrt{3}(1-x))\right] \end{split}$$ With $\theta(x)$ -- is a Heaviside step function.
While it is in principle possible to calculate the Fourier transform of this by hand, I've just fed it to CAS and I've got something like this: $$\frac{2\sqrt{3}\omega_x\left(\cos\frac{\omega_x}{2}\cos\frac{\sqrt{3}\omega_y}{2} -\cos\omega_x\right)-6\omega_y\sin\frac{\omega_x}{2}\sin\frac{\sqrt{3}\omega_y}{2}} {\pi \omega_x(\omega_x^2-3\omega_y^2)}$$ ...plus some singular terms that I've dropped.
Plotting square of the result (from -100 to 100)

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uhoh
uhoh
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  • 98
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This is a partial answer.

Each small mirror will impose its own strong six-spiked diffraction pattern in the telescopes point-spread function. For telescope apertures build from a large number of hexagonal mirrors this will dominate the point-spread function.

The choice of pattern for the tiling of small mirrors to fill the full aperture will have a much smaller effect.

JWST simulated point spread function from James Webb Space Telescope User Documentation's NIRCam Point Spread Functions

WST simulated point spread function

I found an analytical expression for the diffraction pattern from (or at least the Fourier transform of) a single hexagonal aperture (or flat mirror) in this answer to How does Fraunhofer diffraction depend on the orientation of the sides of a lens? in Physics SE.

Hexagonal aperture. I take the following function: $$ \begin{split} h(x,y) = \left[\theta(2y+\sqrt{3})-\theta(2y-\sqrt{3})\right]\cdot \left[\theta(y-\sqrt{3}(x-1))-\theta(y-\sqrt{3}(x+1))\right]\cdot\\ \left[\theta(y-\sqrt{3}(-x-1))-\theta(y-\sqrt{3}(1-x))\right] \end{split}$$ With $\theta(x)$ -- is a [Heaviside step function][2].
While it is in principle possible to calculate the Fourier transform of this by hand, I've just fed it to CAS and I've got something like this: $$\frac{2\sqrt{3}\omega_x\left(\cos\frac{\omega_x}{2}\cos\frac{\sqrt{3}\omega_y}{2} -\cos\omega_x\right)-6\omega_y\sin\frac{\omega_x}{2}\sin\frac{\sqrt{3}\omega_y}{2}} {\pi \omega_x(\omega_x^2-3\omega_y^2)}$$ ...plus some singular terms that I've dropped.
Plotting square of the result (from -100 to 100)

Here's my plot and a python script based on that expression:

diffraction from a hexagonal aperture

import numpy as np
import matplotlib.pyplot as plt

def hexamp(omx, omy):
    """diffraction amplitude hexagonal aperture from
    https://physics.stackexchange.com/a/9910/83380"""
    C, S = np.cos, np.sin
    rt3 = np.sqrt(3)
    
    term_1 = 2 * rt3 * omx * (C(omx/2.)*C(rt3*omy/2.) - C(omx))
    
    term_2 = -6 * omy * S(omx/2.) * S(rt3*omy/2.)

    bottom = np.pi * omx * (omx**2 - 3.*omy**2)

    return (term_1 + term_2)/bottom

omega = np.arange(-100, 100, 0.1)
omx, omy = np.meshgrid(omega, omega)

amplitude = hexamp(omx, omy)

I = np.abs(amplitude)**2

if True:
    plt.figure()
    plt.imshow(np.log10(I), cmap='gray', vmin=-6)
    plt.gca().set_aspect('equal')
    plt.gca().axes.xaxis.set_ticklabels([])
    plt.gca().axes.yaxis.set_ticklabels([])
    plt.colorbar()
    plt.title('log10(I)', fontsize=14)
    plt.show()