[jacobolus]

This is relying on the https://en.wikipedia.org/wiki/Chinese_remainder_theorem

The idea is that you will find a place in the tree for your given number at a relatively shallow depth, because every number in the ring Z/p1×p2×..pn×Z has a unique list of remainders in the separate rings Z/pnZ. As long as the product of the primes is larger than the size of your key space, then you won’t get any collisions.

I’m not sure using consecutive primes starting from 2 is actually the best choice though. You could use any primes you like (or for that matter any collection of factors which are all coprime), so long as their product is more than the largest possible key. The choice of factors should probably be tweaked based on benchmarks in real-world use. For instance, the factors could be 25=5², 29, 31, 32=2⁵, 33=3×11, 37.

[Excuse the lack of subscripts in the notation here; no universally installed fonts contain a subscript n, as far as I know. Likewise I’d prefer to use ℤ rather than Z, but I’m afraid it might show up as little boxes for some readers.]