Collatz Conjecture

Comment:     \begin {asettheorem} 
        
    Let    
    \[ M = 3^n = 2^{[\alpha] + \{\alpha\}} = \sum_{i=1}^{n^*} \gamma_i 2^i, \]
    \begin{equation}
        1 - \{\alpha\} > \frac{1}{2}, \quad n^* = n \times \left[\frac{\ln(3)}{\ln(2)}\right],
    \end{equation}
    then 
    \[ \sum_{\gamma_i = 0} 1 \ge \frac{n^*}{2} - 5. \]
\end{asettheorem}

\begin{proof}
    Suppose
    \[
    3^{n} = 2^{\alpha} \Rightarrow \alpha = \frac{n}{\ln(3) / \ln(2)} \Rightarrow 3^{n} = 2^{[\alpha] + \{\alpha\}}
    \]
    Using Theorem 1, we create a sequence
    \[
    \epsilon_i, \, m_i, \, \epsilon_1 = \{\alpha\}
    \]
    \[
    2^{\epsilon_1} = \sum_{k=0}^{i-1} 2^{[\alpha_k] - \alpha_1} + 2^{\alpha_i - \alpha_1}
    \]
    Assume $\delta_j = 1$, and in such cases, we have
    \[
    \Rightarrow \sigma_{j+1}\ln 2 = \frac{2\sigma_{j}\ln 2}{1 - \sigma_{j+1}\ln 2/2} + o\left(\ln 2 \frac{\sigma^2_{j+1}}{4}\right)
    \]
    \[
    2^{-1}\sigma_{j+1}\ln 2 = \frac{\sigma_{j}\ln 2}{1 - \sigma_{j+1}\ln 2/2} + 2^{-1} \cdot o\left(\ln 2\frac{\sigma^2_{j+1}}{4}\right)
    \]
    \[
    2^{-1}\sigma_{j}\ln 2 = \frac{\sigma_{j-1}\ln 2}{1 - \sigma_{j}\ln 2/2} + 2^{-1} \cdot o\left(\ln 2\frac{\sigma^2_{j}}{4}\right)
    \]
    Repeating j times, we get
    \[
    2^{-j}\sigma_{j+1}\ln 2 = \frac{\sigma_{1}\ln 2}{\prod_{k=1}^{j} (1 - \sigma_{k+1}\ln 2/2)} + \sum_{k=1}^{j} 2^{-k} \cdot o\left(\ln 2 \frac{\sigma^2_{k+1}}{4}\right)
    \]

    From Theorem 1 and the conditions of this theorem, we obtain
    \[
    (1-\alpha)\ln 2 = \sigma_{1}\ln 2 < o\left(\ln 2\frac{\sigma^2_{k+1}}{4}\right)
    \]
    From the smallness of $\alpha$ and the last estimate,
    \[
    \Rightarrow \exists \delta_j > 1
    \]
    Now let's consider the question of the number of $\delta_j > 1$ elements. For this, we consider the general case where there are $\delta_j > 1$ and $\delta_k = 1$. Let's introduce the following notations:
    \[
    \text{for } \delta_j = 1: \alpha_j = 0, \, \beta_j = \frac{1}{1 - \sigma_{j+1}\ln 2/2}
    \]
    \[
    \text{for } \delta_j > 1: \alpha_j = -2^{\delta_j - 1} \frac{\ln 2 - 2^{-\delta_j + 1}}{1 - \sigma_{j+1}\ln 2/2}, \, \beta_j = \frac{2^{\delta_j - 1}}{1 - \sigma_{j+1}\ln 2 / 2}
    \]
    \[
    \ln 2 \sigma_{j+1} = \alpha_j + \ln 2 \beta_j \sigma_{j}
    \]
    Solving the last equation for $\sigma_j$, we get
    \[
    \sigma_{j+1}\ln 2 = \alpha_j + \sum_{m=1}^{j-1}\alpha_{j-m}\prod_{l=1}^{m} \beta_{j-l+1} + \ln 2\prod_{l=0}^{j} \beta_{j-l} \sigma_{1}
    \]

    \[
    \frac{\sigma_{j+1}\ln 2}{\prod_{l=0}^{j-1} \beta_{j-l}} = 
    \frac{\alpha_j}{\prod_{l=0}^{j-1} \beta_{j-l}} +
    \sum_{m=0}^{j-1} \frac{\alpha_{j-m}}{\prod_{l=m+1}^{j-1} \beta_{j-l}} + \ln 2\sigma_{1}
    \]

    According to the condition $\alpha_j \leq 0$, we have
    \[
    \frac{\sigma_{j+1}\ln 2}{\prod_{l=0}^{j-1} \beta_{j-l}} - \sum_{m=0}^{j-1} \frac{\alpha_{j-m}}{\prod_{l=m}^{j-1} \beta_{j-l}} \geq \sigma_{1}\ln 2
    \]

    Denoting by $$ T = \inf_{k<j}\{\beta_k\} $$, we obtain
    \[
    \frac{\sigma_n\ln 2}{T^n} - \sum_{m=0}^{j-1} \frac{\alpha_{j-m}}{\prod_{l=m}^{j-1} \beta_{j-l}} \geq \sigma_{1}\ln 2
    \]
    We come to evaluating the most important term:
    \[
    I_j = -\sum_{m=1}^{j-1} \frac{\alpha_{j-m}}{\prod_{l=m}^{j-1}\beta_{j-l}}
    \]
    The sum is defined by non-zero terms $\alpha_{j-m}$:
    \[
    \xi_* \leq \xi_{j-m} = \frac{\ln 2 - 2^{-\delta_{j-m} + 1}}{1 - \sigma_{j+1-m}\ln 2/2} \leq \xi^*,
    \quad
    \alpha_{j-m} = -2^{\delta_{j-m} - 1} \xi_{j-m}
    \]
    \[
    \mu_* \leq \mu_j = \frac{1}{1 - \sigma_{j+1}\ln 2/2} \geq \mu^*,
    \quad
    \beta_j = 2^{\delta_j - 1} \mu_j
    \]

    \[
    I_j = \sum_{m=1}^{j-1} \frac{2^{\delta_{j-m} - 1} \xi_{j-m}}{\prod_{l=m}^{j-1}2^{\delta_{j-l} - 1} \mu_j}          
    ,\quad
    I_j = \sum_{m=1}^{j-1} \frac{\xi_{j-m}}{\prod_{l=m+1}^{j-1}2^{\delta_{j-l} - 1} \mu_j}             
    \]

    \[
    I_*(n) = \frac{\xi_*}{\mu^*} \sum_{m=1}^{n-1}2^{A_m},
    \quad
    A_m = \sum_{l=m+1}^{n-1}[-\delta_{j-l} + 1],
    \quad
    S^*(n) = \frac{\xi^*}{\mu_*} \sum_{m=1}^{n-1}2^{A_m},
    \]
    \[
    S(n) = 2^{A_n} \sum_{m=1}^{n-1}2^{A_m-A_n},
    \quad
    B_n = A_m - A_n
    \]
    Consider the cases \( B_m = i \leq \frac{n}{2} \) according to the definition \( B_n \)
By definition, each Collatz sequence corresponds to a set \( A_m \); therefore, by iterating over all the sets, we iterate over all corresponding variants of the Collatz sequence.
Consider now all possible variants of reaching the maximum level \( B_{m_*} = i \), due to the monotonicity of the sequence \( B_{m_*+k} = i, k > 0 \) and for each variant, calculate 
\[
 S(n) = 2^{A_n} \sum_{m=1}^{m_*} 2^{B_m} + 2^{A_n} \sum_{m=m_*}^{n} 2^{B_m},
\]

\[
 S(n) = 2^{-i} \sum_{m=1}^{m_*} 2^{m\frac{i}{m_*} } + 2^{i} \sum_{m=m_*}^{n} 2^{i},
\]

\[
 J(n) = 2^{-i} \sum_{m=1}^{n} 2^{m\frac{i}{m_*} } \leq S(n)
\]

\[
 J(n) = n\frac{ 2^{-i}(2^{i}-1 + O(1) )}{i\ln 2}
\]

\[
 \frac{n}{i\ln 2} \leq \sigma_1,\,\,\, \Rightarrow \text{the theorem statement}
\]

    \end{proof}

    
\begin{asettheorem}
    Let
    $$ a_n = \sum_{i=0}^{n} \gamma_{i} 2^{i}, \quad n > 1000, \quad \gamma_{i} \in \{0,1\}. $$
    Then
    $$ \exists j^* < 0.1n $$ and 
    $$ a_{4n - j^*} < a_n. $$
\end{asettheorem}
\begin{proof}
    Let's introduce operators defined by the formulas
    $$ P f = \frac{f}{2}, \quad T f = 3f + 1, \quad Z f = 3f. $$
    $$ T_{i} \in \{P, T\}, R_{i} \in \{Z, P\}. $$
    Consider all possible scenarios of the behavior of the Collatz sequence, which can be written in the following form:
    $$ \begin{gathered}
    a_{n+n} = T_1 T_2 \ldots T_n a_n \\
    \end{gathered} $$
    We need to estimate each 2n-th term of the Collatz sequence based on the number of P, T, Z operators applied during n steps.
    $$ \begin{gathered}
    a_{n+n} = T_n T_{n-1} \ldots T_1 a_n \\
    \end{gathered} $$
    Let $a_n$ have m ones in its binary representation, then we count the number of Z operator applications by the formula:
    $$ m = \sum_{R_{i} = Z, \, i \le n} 1 $$
    and count the number of P operator applications by the formula:
    $$ \sum_{R_{i} = P, \, i \le n} 1 = m + n - m. $$
    Since each application of Z is accompanied by operator P, and the number of P applications corresponds to the number of zeros in $a_n$, which equals n - m. According to the rules of Collatz, after n steps we have:
    $$ a_{n+n} = \frac{3^m}{2^n} a_n + T_n T_{n-1} \ldots T_1 1 = \frac{3^m}{2^n} a_n + B_n. $$
    $$ B_n \leq 2^{-n+m} \sum_{j=1}^{m} \frac{3^j}{2^j} a_n < 2^{2-n+m} \frac{3^m}{2^m} a_n \leq 2^{-2n+1} 3^m a_n. $$
    According to the last formula, we see that the growth of each term of the sequence depends on the number of ones in the binary representation. Next, we will show that a large number of ones at the 2n-th step leads to an increase in the number of zeros at the 3n-th step in the binary representation according to previous theorems, from which it follows that subsequent terms of the sequence decrease:
    $$ \begin{gathered}
    a_{2n} = 3^m a_n \times 2^{-n} + B_n = 3^m + 3^m (a_n - 2^n) + B_n \\
    \end{gathered} $$
    Repeating the reasoning of Theorem 2, we consider the equation
    $$ 2^x = a_{2n} = 3^m a_n \times 2^{-n} + B_n = 3^m + 3^m (a_n - 2^n) 2^{-n} + B_n. $$
    $$ x \ln 2 = m \ln(3) + \ln\left(1 + (a_n - 2^n) 2^{-n} + B_n \times 3^{-m}\right). $$
    From the last equation, to apply the results of Theorem 2, we need $\sigma_1 = 1 - \{x\} > 0.5$.
    To meet the last inequality, consider $m_j = m - j,$
    $$ \{x\} = \inf_{j < 0.1 \times n} \left\{ \frac{(m - j) \ln(3)}{\ln(2)} + \frac{\ln(1 + \theta)}{\ln 2} + O\left(\frac{1}{2^n \ln 2}\right) \right\}. $$
    Depending on the behavior of $\gamma_{n-1}, \gamma_{n-2}, \ldots$ we can always choose a variant where the fractional part of x meets the conditions of Theorem 2.
    Denoting
    $$ \begin{aligned}
    & m^* \text{ is the number of non-zero } \gamma_{i}, \\
    & l^* \text{ is the number of zero } \gamma_{i},
    \end{aligned} $$
    according to Theorem 2 we get
    $$ \begin{gathered}
    m^* \leq (2n - j^*) \frac{\ln 3}{\ln 2} / 2 + 5, \\
    l^* \geq (2n - j^*) \frac{\ln 3}{\ln 2} / 2 - 5.
    \end{gathered} $$
    After $3n - j^*$ steps of applying the Collatz rules, we have
    $$ a_{4n - j^*} = \frac{3^{m^*}}{2^{2n - j^*}} a_{2n} + T_{3n - j^*} T_{3n - 1 - j^*} \ldots T_1 1 = \frac{3^{m^*}}{2^{2n}} a_{2n} + B_{3n}. $$
    $$ a_{4n - j^*} = \frac{3^{m^*}}{2^{2n}} a_{2n} + T_{3n - j^*} T_{3n - j^* - 1} \ldots T_1 1 = \frac{3^{m^*}}{2^{2n}} \left( \frac{3^m}{2^{n - j^*}} a_n + B_n \right) + B_{3n - j^*}. $$
    $$ a_{4n - j^*} = 3^{m^* + m} 2^{-3n - j^*} a_n + 3^{m^*} 2^{-2n - j^*} B_n + B_{3n - j^*}. $$
    $$ B_{3n - j^*} = 3^{m^*} 2^{-2n} - j^* B_n = 3^{m^* + m} 2^{-2n - n - j^*} a_{n}. $$
    $$ a_{4n - j^*} \leq q_1 \times a_n. $$
    By definition of $m^*, l^*, B_n$, we get
    $$ q_1 < 1. $$
    Using $n > 1000 \Rightarrow q_1 < 1 \Rightarrow a_{4n} < a_n$.
\end{proof}

\begin{asettheorem}
    
  Let
$$
a_{n}=\sum_{i=0}^{n} \gamma_{i} 2^{i}, \quad n>1000, \quad \gamma_{i} \in\{0,1\}
$$
then for $a_{n}$ Collatz conjecture is true
\end{asettheorem}