# Smallest $m>1$ such that the number of Collatz steps needed for $238!+m$ to reach $1$ differs from that for $238!+1$.

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.

• Thank you for the edits! Aug 12, 2021 at 8:58
• I think you should call a mathematician. There may actually be more interest in HOW you found this. Aug 12, 2021 at 10:32
• @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. Aug 12, 2021 at 12:04
• I’m going to conjecture that $m=2^{64}$ is the smallest. (Certainly it’s a much improved upper bound.) Aug 16, 2021 at 3:18
• @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. Aug 17, 2021 at 4:48