Let's go to Wolfram Alpha and extrapolate:

FindFit[{{32603,28815},{326034,247675},{3260339,1976676},{32603387,14344391}},a x^b,{a,b},x] FindFit[{{32603,28815},{326034,247675},{3260339,1976676},{32603387,14344391}},a x^b log x,{a,b},x] FindFit[{{32603,0.883814},{326034,0.75966},{3260339,0.606279},{32603387,0.439966}},{(1+log(1+x/a))^b,b<0,(1+log(1+32603387/a))^b=0.439966},{a,b},x] FindFit[{{32603,0.883814},{326034,0.75966},{3260339,0.606279},{32603387,0.439966}},{(1+log(1+x/a)/10)^b,b<0,(1+log(1+3260339/a)/10)^b=0.606279,(1+log(1+32603387/a)/10)^b=0.439966},{a,b},x] Table[{x((1+log(1+x/242428))^-0.462189),x((1+log(1+x/197913)/10)^-1.98901),4.76888 x^0.862246,0.812062 x^0.799794 log x},{x,{1,1000,32603,10^5,326034,3260339,32603387,10^9,10^10,10^11}}] Plot[{x,x((1+log(1+x/242428))^-0.462189),x((1+log(1+x/197913)/10)^-1.98901),4.76888 x^0.862246,0.812062 x^0.799794 log x},{x,1,10^5}] # ... and so on for other plot scales, up to x=10^11

Update: the longest FindFit above may be simplified to:

FindFit[{{3260339,0.606279},{32603387,0.439966}},{(1+log(1+x/a)/10)^b,b<0},{a,b},x]

which gives almost the same result (a=197787 b=-1.9888). With either, we're essentially solving two equations with two unknowns.

Further, when fitting the curve to only 3 points, Wolfram Alpha is able to find three parameters at once, which lets us do:

FindFit[{{326034,0.75966},{3260339,0.606279},{32603387,0.439966}},{(1+a log(1+x/b))^c,c<0},{a,b,c},x] FindFit[{{326034,0.75966},{3260339,0.606279},{32603387,0.439966}},{(1+log(1+x/b)/a)^c,c<0},{a,b,c},x] Table[{x((1+0.00799391 log(1+x/35208.8))^-14.8322),x((1+log(1+x/1891.9)/209.283)^-15.0619)},{x,{1,1000,32603,10^5,326034,3260339,32603387,10^9,10^10,10^11}}]

This looks like it'd match our data really well (including in the 1 to 326k range not explicitly included in the FindFit queries), but unfortunately at these parameter values (a power of -15) things get somewhat less stable (precision loss?) We might need to use a better tool for this. The table and graphs below do not use these 3-parameter formulas.

In plain English: we got four pretty good approximation formulas, which produce values matching RockYou's nearly perfectly in the ~300k to 32.6M range. (The simpler two formulas produce impossible values in the 1 to 100k range, though. The more complicated two are chosen such that they can't produce theoretically impossible values, but they nevertheless do produce higher than RockYou's actual numbers of unique passwords in the ~100 to ~200k range.) Then we use these formulas to extrapolate to up to 100 billion of total passwords.