Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 . . . . . .
share|improve this question
    
We are looking at one term of the sequence, and doing some assignments. What is happening to the other terms? And if we have decided that the term $a$ is $\equiv 0\pmod{3}$, can we decide that $a\equiv 4\pmod{9}$? If so, how often can we change our minds? I am having trouble grasping the process. –  André Nicolas Jul 17 '12 at 2:26
    
The other terms follow. If the one we're looking at is congruent to 6 mod 9, then its successor is congruent to 7 mod 9, etc..... –  Michael Hardy Jul 17 '12 at 3:04
    
OK, so we are woking with $a_0$. What about the consistency question. If we decide that $a_0$ is divisible by $5$, can we later decide it is congruent to $16$ modulo $25$? –  André Nicolas Jul 17 '12 at 3:13
    
See my later edit above. –  Michael Hardy Jul 17 '12 at 3:22
2  
I suspect you already know this, since it's probably part of the motivation for this construction: Since you've basically modeled the standard heuristic treatment of primes, you can transfer all sorts of results from that. For instance, the probability of a given term still being "prime" after you've tossed coins up to $p$ goes as $1/\log p$, the density of twin "primes" follows the first Hardy–Littlewood conjecture, and so on... –  joriki Jul 17 '12 at 6:29
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.