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Is there a formula to construct a Collatz (3x + 1) sequence of arbitrary length that is strictly increasing? Obviously one can do this with a strictly decreasing sequence by just taking $2^n$ but I haven't seen a way to do it with a strictly increasing sequence.

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    $\begingroup$ Presumably, you mean a $\frac{3x + 1}{2}$-sequence, since if $x$ is odd then $3x + 1$ is even. In that case: start with $x = 2^n - 1$, or more generally $x = m*2^n - 1$. $\endgroup$
    – user44191
    Commented Aug 3, 2023 at 23:51

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Start with a number equal to $-1$ modulo $2^n$. Then, after one step, the number is $\frac{3(-1)+1}{2}=-1$ modulo $\frac{2^n}{2}=2^{n-1}$, so inductively it will increase $n$ steps.

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