Olbers' paradox: why is the night sky dark?

But I've shown in my mathematics that every pixel (or rod or cone) would end up recording the same value, so each bit of space would look the same. If I'm wrong, point out the flaw in the mathematics.

As i've repeatedly said, the shells have equal luminance and flux. The distribution of light in individual shells varies, but that is somewhat irrelevant. Olbers' paradox doesn't say that each shell will have a uniform distribution of light, but that the light of the night sky i.e. when you view all the shells together, will be uniformly distributed. This distinction seems to be what you're forgetting.

I did the mathematics differently, using the same initial condition as Olbers' paradox and ended up with the result that the light is uniformly distributed. Where is the mistake in my mathematics?

Then you're either lying or an amnesiac, because you repeatedly ask if two shells have equal brightness.

Yes, which I have answered for you by giving a fairly extensive mathematical treatment of the subject. You refuse to address the mathematics, presumably because they debunk your absurd notions. If you were right then you would be gleefully tearing my mathematics apart and exposing the flaws.

The distribution, which is why I mathematically modelled the distribution and lo and behold, each pixel/rod/cone ends up emasuring the same value of light – a uniform distribution.

You said that you aren't asking anything, so what is this? Did you mean to type a question mark? It's just that every time I answer this question you insist that you weren't asking a question. So is this a question that you're asking, or a typo?

As I've repeatedly said, the shells all have equal luminance and flux, but a different distribution. But the eye sees all the shells together, not individually. Modelling just two shells is inadequate. If you look at the mathematics I've repeatedly given you, you will see that I model what the eye sees by using several shells and find that although the distribution of light varies from shell to shell, the end result when you combine the shells is a uniform distribution.

1. You suggested that a 100x100 pixel detector would be enough to judge the brightness of the sky. Let's assume that this detector has a lens that allows it to see an area of sky subtended by a solid angle of 1.024x10^-10 steradians. Each pixel therefore covers 1.024x10^-14 steradians. Happy with that?
2. Let's also imagine a shell of stars at a distance 1r. These stars are close enough to occupy 1.024x10^-14 steradians, so they are exactly imaged by 1 pixel. There are so many stars at this distance that our detector can pick up 5000 of them. This means that there is a star every other pixel. And those stars are measured like point sources. Happy with that?
3. Photons arrive from one of these stars at a rate of 1024 photons per second. (That gives a flux of 1024 photons per second per pixel.) If our detector takes a 10 second exposure, that means that a pixel illuminated by a star from the 1r shell will have a brightness of 10,240. Happy with that?
4. hat happens with the rest of the pixels? Well let's look at the shell of stars at 1.41r. These stars are 1.41x further away, so there will be twice as many i.e. 10,000. This means that each pixel will have one of these stars and they are measured like point sources. Happy with that?
5. These stars will only occupy 5.12x10^15 radians (half a pixel) because they're 1.41x further away and photons will come in at a rate of 512 photons per second. (That gives a flux of 1024 photons per second per pixel). If our detector takes a 10 second exposure then one of these stars will contribute 5120 photons to the brightness of the pixel. Happy with that?
6. We're not done yet though, because Olbers' paradox states that every sight line ends in a star. At the moment we've still got 5.12x10^15 steradians to deal with for this pixel. Happy with that?
7. So let's go to the shell at a distance of 2r. This will have 20,000 stars, which is 2 per pixel. Happy with that?
8. Each of these stars occupies 2.56x10^-15 steradians (quarter of a pixel) and photons will come in at a rate of 256 per second. (That gives a flux of 1024 photons per second per pixel). If our detector takes a 10 second exposure then each of these stars will contribute 2560 photons to the brightness of the pixel. Happy with that?
9. At this point, the whole pixel is covered, meaning that every line of sight ends in a star. We can therefore turn to working out the brightness of the pixel. Happy with that?
10. There is a total of 5120 photons from the one star in the 1.41r shell and 2*2560=5120 photons from the two stars in the 2r shell, giving a total value for the pixel of 10,240 photons. Happy with that?
11. This is the same value as the pixels which are illuminated by the stars from the 1r shell, so every pixel has the same brightness value, meaning that the sky is uniformly bright. Happy with that?
12. This result isn't surprising because the flux from every star is the same. So if each pixel sees the same flux, has the same area, and is exposed for the same length of time, then of course each pixel will have the same brightness. Happy with that?
13. You can change the numbers, but as long as you ensure at every line of sight ends in a star, you'll find that every pixel ends up with the same brightness. What that brightness is will depend on the values you put in, but the brightness will be uniform. Happy with that?