As a mathematical biologist, I find that very often people choose the most difficult line between two points when they are trying to develop a mathematical description. I'd say the same thing applies here. With the exception of Michael Briggs, who has given us a new result that is different than all the others, all of these equations are functionally identical, though not all identical in application (enlarger height versus print dimension).

In terms of Michael's equation, there is no reason that you should have to add one to the two terms. As far as the physics of the situation, there is no basis for it. If it helps to obtain better prints, then sure, go ahead, and since I've never tried it that way perhaps it works reasonably well. As Michael said, for larger prints, the difference between his equation and the others in terms of the results provided becomes smaller and smaller. For small prints, though, it could provide pretty different prints at the different sizes.

The basic physics of the problem is, as others have said, that light falls off at an exponential rate with distance. This does not mean that the problem is particularly difficult, though. A simple way to look at it is to think of it solely in terms of the light put out by the enlarger. There is a certain quantity of light that is coming through the enlarger lens. That same amount is going to come out of the lens whether the image on the print is covering 4x5 or 8x10 or 16x20. So, by simply taking the ratio of the areas in the prints (or the ratio of the squares of the enlarger heights) you can accurately measure the ratio of exposure times. So...a 4x5 is 20 inches square, an 8x10 is 80 inches square, an 8x10 takes 4 times the exposure of a 4x5. The mathematical version of this is

Tn = To*(An/Ao)

where Tn = new exposure time, To = old exposure time, An = new print area, and Ao = old print area. To alter this equation for enlarger distance, it simply becomes

Tn = To*(Dn^2/Do^2)

where Dn = new enlarger distance and Do = old enlarger distance. Neither of these are new, others above have already provided these results in different forms, but this presentation makes more sense to me, so I hope it is helpful to somebody else.

The missing link in the above description is the light scattered during transmission through air. The greater the distance to the print, the greater the potential light loss, affecting our assumption above that there is a uniform amount of light falling on the prints of various sizes. In practice, the only way this would be important is if you were printing over exceptional distances or the air in the darkroom was particularly dirty...

The overall line of reasoning laid out above is related to the reason that teleconverters reduce the effective aperture of your lenses, and why the physical aperture of a lens is larger to achieve the same F-stop (which is a ratio) for a longer focal length...but if people are interested in that I'll post it separately...and otherwise I'll happily keep it to myself.