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 . . . . . .