They don't really offer any argument for this though. Apparently one infected cell can infect 10k new cells after a day or two[1], the mutation rate is 4 per kbp per cell[2], and the HIV genome is 9 kbp long[3]. That would mean 36 new mutations for each newly infected cell.

I'm not positive this is right... but I would think a rough estimate on the upper bound could be arrived at assuming each mutation is equally likely and independent. Then the probability of a single mutation at any given site would follow a Poisson distribution with k=1 and r=.004:

  r^k*exp(-r)/k! = r*exp(-r)= 0.00398
Then the probability of a mutation at n=2 sites in the same cell would be:

  (r*exp(-r))^n= 1.58 x 10^-5
If each infected cell infects N=10,000 new cells each generation g, after one generation (g=1) the expected number of cells containing a set of two specific mutations would be:

  N^g*(r*exp(-r))^n= 0.158
However after two generations there would be 10^8 infected cells and 1587 would be mutants at any two given sites. Then for any n=3 sites there would be about 6 cells containing mutations at each.

As I said, that would definitely be an upper bound. Some sites will be less likely to mutate than others, eventually you run out of new cells, etc.

Also, this ignores that cutting the DNA may be killing the cells.

  [1] http://journals.plos.org/ploscompbiol/article?id=10.1371/journal.pcbi.1000906
  [2] http://journals.plos.org/plosbiology/article?id=10.1371/journal.pbio.1002251
  [3] http://www.ncbi.nlm.nih.gov/nuccore/AF033819