@amWhy Please do not delete / move on-topic messages to trash. This is an abuse of room owner privileges (esp. since the message critiques one of your actions).
@user21820 Well, it is of the form “If P(n) holds for n=k then it holds for n=k+1” which can be called sloppy. Why don't you leave a comment at the answer and ask for clarification/improvement?
The problem that I see is that the answer essentially repeats the proof from the question, and that the qestion is a request for solution verifications without pointing out a specific concern, therefore I voted to close as “needs more focus.”
@MartinR I agree , at most a sloppy formulation. And also , I cannot detect a sign for a bot having written this. But apparently there are software tools that cen that without a chance of a fallacy :)
@MartinR No that's not the problem! Let Q be the induction property. The line I specified only holds if you know Q(k+1). But that is what you are supposed to prove from Q(k)! It didn't say that. Instead it just asserted Q(k+1). In the second error, it is asserting "Q(k+1) because A^(k+1)=A^k·A^1", but that's nonsense too.
Let me just emphasize that a wrong proof is a wrong proof. It's not enough to write down a list of true statements; that isn't a proof.
And nobody said anything about software detecting bots. Some humans can do it very well. See the message linked by Martin Sleziak.
This answer has been flagged as low-quality (and possibly bot-written). It does not match enough of the network heuristics to merit deletion (and the area of mathematics is not one where I can really judge without spending some real time trying to figure it out). More qualified eyes would be helpful.
I think that the most famous and beautiful trajectory of the $3x+1$ problem is without doubt that starting from $n=27$ and having a maximum at $9232$.
The thing that I find very beautiful is that:
$$19\cdot 3^3=513\equiv 1\pmod {2^k}$$
And
$$\frac{9232}{2}=19\cdot 3^5-1$$
Is it chance or there is...
