On super-logarithmic Collatz numbers

308 Views Asked by At

The Collatz conjecture (or 3n + 1 conjecture) is a conjecture in mathematics that concerns a sequence defined as follows: start with any positive integer n. Then each term is obtained from the previous term as follows: if the previous term is even, the next term is one half the previous term. Otherwise, the next term is three times the previous term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

We denote by $collatz(n)$ the number of steps it takes to reach $1$ from $n$ (its delay).

Clearly, $collatz(n)\ge \log_2 n$ -- but by how much?

While the conjecture is still open, do we know any integer sequences with a super-logarithmic delay? Namely:

Is there a sequence $\{a_n\}$ such that $\mathit{Collatz}(a_n)=\omega(\log a_n)$?

What about the weakest bound, which is probably true:

Is there a sequence $\{a_n\}$ such that $\mathit{Collatz}(a_n)=\log_2 a_n + \omega(1)$?


From simulating many ($> 10^{12}$) random numbers (not in a uniform manner) between $1$ and $2^{9999}$, it seems that $\mathit{Collatz}(n)$ may be bounded by $50\log_2 n$.

Is there an $n$ for which $\mathit{Collatz}(n) \ge 50\log_2 n$?