Is the set of logarithms of $N$-almost primes equidistributed?

Question

Given the set of all primes. From this one builds subsets of $N$-almost primes according to a certain partition, e.g. $\lambda=(3,1)$, so the set is $$ M=\{2^33,2^35,2^37,...,3^32,3^35,...\}. $$ Is the set of natural logarithms of elements of $M$ equidistributed$\mod 1$?

Answer 1

Equidistribution is a property of sequences, not of sets. von Neumann proved that any set that is dense in $[0,1)$ can be ordered in such a way as to be a uniformly distributed sequence. The sets you are talking about are surely dense in $[0,1)$ (although I wouldn't be able to give you a proof of that off the top of my head), so it comes down to a question of how you order them.

The order implicit in the way you've written the question - all the ones with $2^2$ first, then, after that infinity of numbers, all the ones with $3^2$, etc. - well, that's not a sequence, so it gives us nothing to go on.

I guess the simplest order is just increasing order of the numbers, so, $$\log24,\log40,\log54,\log56,\log88,\log104,\log135,\dots$$

Now it's an exercise to show that the sequence $\log1,\log2,\log3,\log4,\dots$ is not uniformly distributed mod $1$, and that's basically because it grows too slowly. Once it gets into a subinterval $[a,b)$ it just stays there too long. Now I'm confident that the same is true for the sequence of logs of primes, and the Prime Number Theorem should be strong enough to show that (although again that's not a proof I'm prepared to write down on the spur of the moment). If that's true, then it's also true for the sequence of squares of primes, of cubes of primes, of any fixed power of primes, because multiplying all the terms in a sequence that isn't u.d. by a fixed number can't make the sequence u.d.

So the next thing is to show that the numbers $p^3q$ grow more slowly than, say, the numbers $p^4$. That ought to be enough to conclude (if everything else I've written here holds up to careful scrutiny) that the sequence in question is not u.d., and once you've done $\lambda=(3,1)$ you should be able to generalize to any $\lambda$ you like.

Sorry for being so sketchy and leaving all the details as exercises.