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\begin{center}
\vskip 1cm{\Large\bf 
Corrigendum to Article 23.3.5\\ 
There are no Collatz-$m$-Cycles with $m\leq 91$
}
\vskip 1cm
\large
Christian Hercher\\
Abteilung f\"{u}r Mathematik und ihre Didaktik\\
Europa-Universit\"{a}t Flensburg\\
Auf dem Campius 1b\\
24943 Flensburg\\
Germany \\
\href{mailto:christian.hercher@uni-flensburg.de}{\tt christian.hercher@uni-flensburg.de} \\
\end{center}

The derivation of the inequalities $T(n_{m-m_2+1+\ell})<\frac{3}{(2^v-1)^\delta}$
in the proof of Theorem~21 at the bottom of page~14 contains an error. To show the statement
\[
T(n_{m-m_2+1})+\sum_{i=m-m_2+2}^{m} T(n_i)
< \frac{3}{2^v-1} + \frac{3\cdot (m_2-1)}{(2^v-1)^\delta},
\]
which is used later in the proof, we can proceed as follows.

For a local minimum \(n_i\) in the hypothetically existing Collatz cycle, define
\[
x_i:=\frac{\log (n_i+1)}{\log 2}.
\]
Then, by Lemma~8, we know that \(1\leq k_i\leq x_i\), and by Lemma~20 that \(n_{i+1}<n_i^\delta\), which implies \(x_{i+1}<\delta x_i\). Using the same argument as on page~13 in the original proof of Theorem~21, we know that there are \(m_2\) consecutive local minima
$n_{m-m_2+1}, \dots, n_m$ such that $x_{m-m_2+1} + \cdots + x_m \geq k_{m-m_2+1} + \cdots + k_m \geq \frac{m_2}{m}\cdot K$.
Without loss of generality, let the cycle be entered in such a way that this sum is maximal. Then \(x_{m-m_2}\leq x_m\); otherwise, we would have chosen the shifted interval of local minima. Since $x_{m-m_2+1}<\delta x_{m-m_2}$, this implies $x_{m-m_2+1}<\delta x_m$.

By Remark~7, we have $T(n_i)<\frac{3}{n_i}$. Thus, 
\[
\sum_{i=m-m_2+1}^{m} T(n_i) < \sum_{i=m-m_2+1}^{m} \frac{3}{2^{x_i}-1} =:S(x_{m-m_2+1},\dots,x_m).
\]
Since all \(x_i\) are positive, the function \(S\) is convex and attains its maximum value at a vertex of the polytope determined by the constraints
$x_i\geq 1$, $x_{i+1}\leq \delta x_i$, $x_{m-m_2+1}<\delta x_m$, and $x_{m-m_2+1} + \cdots + x_m \geq \frac{m_2}{m}\cdot K$. Thus, at least \(m_2\) of these constraints must be active, i.e., must hold with equality.

Suppose there is a value \(x_i\) such that \(x_i=1\). Then $x_{i+1}\leq \delta$, $\dots$, $x_{i-1}\leq \delta^{m_2-1}$.
Hence,
\[
x_{m-m_2+1} + \cdots + x_m \leq \frac{\delta^{m_2}-1}{\delta-1} < \frac{m_2}{m}\cdot K,
\]
contradicting the sum constraint. The last inequality follows from the definition of \(m_2\) in the statement of Theorem~21.

Thus, \(m_2\) of the remaining \(m_2+1\) constraints must be active. If all constraints $x_{i+1}\leq \delta x_i,$ $x_{m-m_2+1}<\delta x_m$ were active simultaneously, this would lead to the contradiction $x_m=\delta^{m_2}x_m$, and hence \(\delta=1\), contradicting $\delta=\frac{\log 3}{\log 2}$. Therefore, the sum condition must be active, and we have
\[
x_{m-m_2+1} + \cdots + x_m =\frac{m_2}{m}\cdot K,
\]
together with all but one of the equations $x_{i+1}= \delta x_i,$ and $x_{m-m_2+1}=\delta x_m$. Consequently, \(S\) attains its maximum at a point whose coordinates are a cyclic permutation of
$x_{m-m_2+1}, \delta x_{m-m_2+1}, \dots, \delta^{m_2-1}x_{m-m_2+1}$, and
\[
\frac{m_2}{m}\cdot K = x_{m-m_2+1} +\delta x_{m-m_2+1} +\dots+ \delta^{m_2-1}x_{m-m_2+1} = x_{m-m_2+1}\cdot \frac{\delta^{m_2}-1}{\delta-1}.
\]
Hence,
\[
x_{m-m_2+1} = \frac{m_2}{m}\cdot K\cdot \frac{\delta-1}{\delta^{m_2}-1}=v,
\]
and we obtain
\begin{align*}
T(n_{m-m_2+1})+\sum_{i=m-m_2+2}^{m} T(n_i)
&< S(x_{m-m_2+1},\dots,x_m)\\
&\leq S(v,\delta v,\dots,\delta^{m_2-1}v)\leq S(v,\delta v,\dots,\delta v)\\
&=\frac{3}{2^v-1}+\frac{3\cdot (m_2-1)}{2^{\delta v}-1}<\frac{3}{2^v-1}+\frac{3\cdot (m_2-1)}{(2^v-1)^\delta},
\end{align*}
which is exactly what we wanted to prove.

\medskip

The author is highly grateful to Xinjun Wang, ORCID 0009-0007-5895-5984,  for pointing out this mistake and for helpful discussions regarding its correction.
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