# Does Chaitin's constant have infinitely many prime prefixes?

Define $f(n) = \lfloor 2^n \cdot \Omega \rfloor$, that is, $f(n)$ is the first $n$ bits of Chaitin's constant interpreted as a number written in binary.

I am trying to figure out if $f(n)$ can have infinitely many prime values. To show that it cannot, it would suffice to find a way of compressing these prime values to descriptions arbitrarily shorter than $\log_2{p}$ bits. This seems to be a tricky task as it runs up against the edge cases of the definitions.

Let $L(x) = \lfloor \log_2{x} \rfloor + 1$. Any number $n$ can be encoded in a prefix-free fashion by first emitting the length of the length of its length in unary, then the length of its length in binary, then its length in binary, then the number itself in binary. This uses $2 \cdot L(L(L(n))) + L(L(n)) + L(n)$ bits.

If we have a prime number $p$, we can improve on this by encoding its index $\pi(p)$, and decoding it with a program that finds the $\pi(p)^{th}$ prime. By the prime number theorem, we save $L(L(p)) + O(1)$ bits this way. But the pesky $2 \cdot L(L(L(p)))$ term remains.

There is more information we can take advantage of: the $k^{th}$ prime prefix of $\Omega$ is a special type of prime, because it itself has $k-1$ prime prefixes. So we can count only those primes when decoding it, and heuristically this gives an encoding of size $2 \cdot L(L(L(p))) + L(p) - O(k)$ bits. Or roughly, prime prefixes of $\Omega$ must be triply-exponentially sparse, as long as for some $c \gt 1$ there are always $O(\frac{n}{c^k \cdot \log{n}})$ primes less than $n$ with exactly $k$ prime prefixes. (I still believe something like this is true but my reasoning here is a bit off.)

This approach succeeds on sets with less density than the primes: only a finite number of prefixes of $\Omega$ are primes with prime index (we save $2 \cdot L(L(p))$ bits, bringing the total description length under $L(p)$ by an increasing margin), and similarly only a finite number of prefixes of $\Omega$ are part of a twin prime pair (using Brun's theorem). Generally, any computable set with density $O((\log{n})^{-1-\epsilon})$ for $\epsilon \gt 0$ has a finite intersection with the prefixes of $\Omega$.

• (Also, of course your prefix-free encoding principle can be extended deeper still; you can use $L(n)+L(L(n))+L(L(L(n)))+2L(L(L(L(n))))$ bits, etc. ) – Steven Stadnicki May 17 '17 at 19:19
• $\Omega$ is not a number - it depends on the model of computation and how programs are encoded. I suppose you are trying to prove something for all values of $\Omega$. – Reinstate Monica May 17 '17 at 19:41
• @Solomonoff'sSecret: I think $\Omega$ is mainly used for advertisement purposes; the question really seems to be if a ML random number can have initial pieces equal to a prime infinitely often. – user138530 Jun 1 '17 at 5:57
• In fact, this can be made rigorous quite easily: if $f(n)=p_k$, then $K(f(n))=K(k)+O(1)$, and we will have the required $K(f(n))\ge n-O(1)$ if $K(k)\ge\log k + \log\log k-O(1)$. Since there are infinitely many such $k$, we can certainly have a sequence of primes such $K(p)\ge \log p -O(1)$ for all primes in the sequence. – user138530 Jun 1 '17 at 6:51