1
$\begingroup$

This is a variation on the theme of a rather flawed question that I asked months ago.

Imagine a doubly infinite sequence, i.e. each member has a successor and a predecessor.

Grab one term of the sequence and toss a coin to decide which congruence class mod $2$ it belongs to; then all the terms are alternately even or odd. Then randomly assign it to a congruence class mod $2^2$, each having probability $1/2^2$ of being chosen, then similarly with $2^3$, then $2^4$, and so on.

Then randomly assign it to a congruence class mod $3$, each having probability $1/3$ of being chosen, then mod $3^2$, each having probability $1/3^2$ of being chosen, then $3^3$, etc.

Then do the same with powers of $5$.

And so on: do this with each prime number.

Notice that with probability $1$, there is no member $n$ of this sequence for which there is some prime number $p$ such that $n$ is divisible by all powers of $p$. I.e. the multiplicity of every prime factor as a divisor of every member of the sequence is finite.

Because the harmonic series diverges to $\infty$, the expected number of prime factors of any one of these objects is $\infty$, so they are like "very very big" (infinite) positive integers, each having a prime factorization.

What interesting results are known about this random process?

Later edit: Everything should be construed the way I obviously meant it, when possible. So, e.g.

  • If one member of the sequence is congruent to 7 mod 9, then its successor is congruent to 8 mod 9, and so on.
  • The choice of a congruence class mod 9 is not independent of the choice of a congruence class mod 3, and so on. So the probability that the "number" we're looking at is congruent to 7 mod 9 is of course $1/9$, but the conditional probability that it's congruent to 7 mod 9, given that it's congruent to 2 mod 3, is $0$. And so on . . . . . .
  • 0
    @Gerry : I'm not sure why$a$question about a specfic random process needs to be accompanied by an example of another random process that resembles it. This is of course supposed to be a sort of limiting case: what happens with very big but finite numbers approximates this, and the approximation can be made as close as you want by making the numbers big enough, provided "close" means what "close to infinity" usually means when you talk about something approaching $\infty$.2012-07-17

0 Answers 0