Set up a series of reference frames all at rest with each other, at a linear distance ##a## from one another. The origin is our observer, and each of the other reference frames can be labeled by their distance from the origin: ##ka, k\in \mathbb{Z}##. Let's say a light beam sweeps over the surface at a local speed of ##bc, b>1##. This is (in the limit where ##a\to 0##) roughly equivalent to a system where a light pulse is emitted from ##-ka## at ##t=0## in the observer's frame, a second pulse is emitted from ##-(k-1)a## at ##t=\frac{a}{bc}## in the observer's frame...an ##n##th pulse is emitted from ##(n-k-1)a## at ##t=\frac{na}{bc}## in the observer's frame.
The observer will see the pulse at the origin (the ##(k+1)##th pulse) first, at a local time of ##t=\frac{(k+1)a}{bc}##. At this time, the ##k##th pulse will have traveled a distance of ##\frac{a}{b} < a##, and the prior pulses (that is, the ##n<k+1## pulses) will have traveled a distance of ##\frac{(k-n)a}{b} < (k-n)a##. The observer will thus see these pulses in reverse order, with the ##n##th pulse arriving at ##t_{n<k+1} = \frac{na -(n-k-1)ab}{bc}##, or ##\frac{a(n-k-1)(1-b)}{bc}## after the first pulse (this quantity is positive since ##n-k-1## and ##1-b## are both negative for ##n<k+1##).
For the ##n>k+1## pulses, all of the above formulas remain valid, but with the sign switched: ##t_{n>k+1} = \frac{na +(n-k-1)ab}{bc}##, or ##\frac{a(n-k-1)(1+b)}{bc}## after the first pulse. Thus, the observer sees the ##(k+1)##th pulse first, then we can compare, e.g., the ##k##th and ##(k+2)##th pulses to determine the behavior of the recession of the pulses. The ##k##th pulse arrives at ##\frac{a(b-1)}{bc}## after the ##(k+1)##th pulse, and the ##(k+2)##th pulse arrives at ##\frac{a(b+1)}{bc}## after the ##(k+1)##th pulse. Extrapolating linearly, the ##(k+1+p)##th pulses arrive at ##p\frac{a(b\pm 1)}{bc}## after the initially observed pulse, with the plus holding for ##p>0## and the minus for ##p<0##. So the "backward receding" pulses (##n<k+1##) will appear to recede faster than the corresponding "forward receding" pulses by a factor of ##\frac{b-1}{b+1}## (that is, the backward receding pulses appear to the observer to take ##\frac{b-1}{b+1}## as long to complete as an equal number of corresponding forward receding pulses).