Let $h(x)$ be the number of steps^ needed for $x$ to reach $1$ in the Collatz/3n+1 problem. I found that

$$h(238!+n)=h(238!+1), \;\; \forall 1 < n \leq 690,000,000$$ Here "!" is the standard factorial.

This is a lot of consecutive terms with the same height and beats the current record by far. Now I am wondering:

What is the smallest $m > 1$, such that $h(238!+m) \neq h(238!+1)$?

I don't know of an efficient way of finding it. We know that $h(2\cdot(238!+1))=h(238!+1)+1$, so $m \leq 238!+1$, but that's a rather large upper bound.

UPDATE 16/08/2021: Martin Ehrenstein found that $10^9 < m < 10^{94}$. See A346775. Later user mjqxxxx improved the upper bound to $m \leq 2^{64}$.

UPDATE 21/08/2021: Martin Ehrenstein improved the upper bound to $m < 11442739136455298475$.

^ $3x+1$ is considered one step and $x/2$ is one step.

  • 1
    $\begingroup$ Thank you for the edits! $\endgroup$ Aug 12, 2021 at 8:58
  • 1
    $\begingroup$ I think you should call a mathematician. There may actually be more interest in HOW you found this. $\endgroup$ Aug 12, 2021 at 10:32
  • 1
    $\begingroup$ @DanielWainfleet my method is not very exciting. I used a simple brute force program that iterates through all the factorials. I got stuck at $238!$ and realized I need help. I am hoping that this information could be useful for someone working on the Collatz problem. $\endgroup$ Aug 12, 2021 at 12:04
  • 2
    $\begingroup$ I’m going to conjecture that $m=2^{64}$ is the smallest. (Certainly it’s a much improved upper bound.) $\endgroup$
    – mjqxxxx
    Aug 16, 2021 at 3:18
  • 1
    $\begingroup$ @SeekingAnswers great question. So far I have only investigated powers of two (oeis.org/A277109) and factorials (oeis.org/A346775). Powers of two seem to have more patterns and are perhaps more predictable, while factorials look more chaotic. Both give many consecutive terms with the same height. $\endgroup$ Aug 17, 2021 at 4:48


You must log in to answer this question.

Browse other questions tagged .