sample uniformly x in (0,1] then compute floor(1/x). this approximately samples the Zipf distribution with exponent 2.
let s > 1. sample uniformly x in (0,1/s] then compute floor(1/x). this samples between s and infinity. subtract s to sample between 0 to infinity, because infinity - finite = infinity. this distribution has a fatter tail -- it less heavily favors smaller values -- than s=1 above.
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