Does it follow from Lemma 8 in the paper by Simons and de Weger (2005) that if a non-trivial cycle exists: $$2^{K+L} - 3^K < (10^{-5})3^K$$
Let:
- $x\ge 1$ be an integer
- $T(x) = \begin{cases} \frac{1}{2}(3x + 1), && \text{if }x\text{ is odd}\\ \frac{1}{2}x, && \text{if }x\text{ is even}\\ \end{cases}$
- $T_1(x) = T(x)$
- $T_{c+1}(x) = T(T_c(x))$
- $\delta = \dfrac{\log 3}{\log 2}$
- A non-trivial cycle is any case where $x > 2$ and there exists integer $c > 0$ where $T_c(x) = x$
- $[x_1, x_2, \dots, x_{c-1}]$ be the distinct numbers which make up a non-trivial cycle with:
- $K = $ the number of odd integers
- $L = $ the number of even integers
- $K+L = c-1$
Here's my thinking:
(1) From Lemma 8 on page 9:
$$K + L < 1.000001\delta K$$
(2) $2^{K+L} - 3^K < 2^{1.000001\delta K} - 2^{\delta K} = 2^{\delta K}\left(2^{.000001\delta K} - 1\right) < 3^K(3^{.000001K} - 1) < 3^K(1.00001 - 1) = (10^{-5})3^K$
Am I understanding the implications of the lemma correctly? Did I make a mistake?
Update: The last part is wrong. For larger $K$, it is incorrect.
You could use this instead: If a non-trivial m-cycle exists,
$$K+L-K\log_23<\frac{m}{X_0\log2}$$ The current $X_0=2,95...10^{20}$, so $$(K+L)\log2-K\log3<\frac{m}{2,95\cdot 10^{20}}$$ $$\log \frac{2^{K+L}}{3^K}<\frac{m}{2,95\cdot 10^{20}}$$ $$\frac{2^{K+L}}{3^K}-1<e^\frac{m}{2,95\cdot 10^{20}}-1$$ or $$2^{K+L}-3^K<3^K(e^\frac{m}{2,95\cdot 10^{20}}-1)$$