Is $\lim\limits_{k \to \infty}\left[ \lim\limits_{p \to \infty} \frac{M}{1+3+5+\cdots+ [2^{p(k-1)}-2^{p(k-2)}-2^{p(k-3)}-\cdots-1]}\right]=1$?

Question

Firstly, my $\LaTeX$, Mathematics and English knowledge is very limited. It is extremely difficult for me to ask this question. Now I am improving myself.I hope you understand me...

Look at this function: $$f(n) = \begin{cases} n/2 &\text{if } n \equiv 0 \pmod 2 \\ n+1 & \text{if } n\equiv 1 \pmod 2.\end{cases}$$

We know that for any positive number, there is a number $\text{“}k\text{''}$, which that $f^k(n)=1$

For function $f(n)$ go backward from number $1$.

Let step number is $k$ $$[2^{\sum_{z=1}^{k-1} m_z}-2^{\sum_{z=2}^{k-1} m_z}-2^{\sum_{z=3}^{k-1} m_z}-\cdots-1]\stackrel{k\to \infty}{\longleftarrow}\mathbf{\cdots} \stackrel{k=5}{\longleftarrow} \mathbf{[2^{m_4+m_3+m_2+m_1}-2^{m_4+m_3+m_2}-2^{m_4+m_3}-2^{m_4}-1]}\stackrel{k=4}{\longleftarrow} \mathbf{[{2^{m_1+m_3+m_2}-2^{m_2+m_3}-2^{m_3}-1}]}\stackrel{k=3}{\longleftarrow} \mathbf{[{2^{m_1+m_2}-2^{m_2}-1}]}\stackrel{k=2}{\longleftarrow} \mathbf{[{2^{m_1}-1}]}\stackrel{k=1}{\longleftarrow} \mathbf1$$

Then, $2^{\sum_{z=1}^{k-1} m_z}-2^{\sum_{z=2}^{k-1} m_z}-2^{\sum_{z=3}^{k-1} m_z}-\cdots-1=F_{10}(m_1,m_2,\ldots,m_{k-1})$ and $2^{\sum_{z=1}^{k-i} m_z}-2^{\sum_{z=2}^{k-i} m_z}-2^{\sum_{z=3}^{k-i} m_z}-\cdots-1 =F_{ij}(m_1,m_2,\ldots,m_{k-i})$

Let, for $\max [F_{10}]$ , we can write

$m_1=m_2=m_3=\cdots=m_{k-1}=p \Rightarrow \max[F_{10}]=2^{p(k-1)}-2^{p(k-2)}-2^{p(k-3)}-\cdots-1$ and for each $F_{ij}$ must be $\max [F_{ij}]<\max [F_{10}]$.

Then, Let's write all possible sums:

$$\sum_{m_{k-1}=1}^p \sum_{m_{k-2}=1}^p\cdots\sum_{m_{1}=1}^p F_{10}(m_1,m_2,\ldots,m_{k-1})+\sum \sum\cdots\sum F_{11}(m_1,m_2,\ldots,m_{k-1})+\sum\sum\cdots\sum F_{12}(m_1,m_2,\ldots,m_{k-1})+\sum\sum\cdots\sum F_{13}(m_1,m_2,\ldots,m_{k-1})+\cdots+\sum\sum\cdots\sum F_{20}(m_1,m_2,\ldots,m_{k-2})+\sum\sum\cdots\sum F_{21}(m_1,m_2,\ldots,m_{k-2})+\sum\sum\cdots\sum F_{22}(m_1,m_2,\ldots,m_{k-2})+\sum\sum\cdots\sum F_{23}(m_1,m_2,\ldots,m_{k-2})+\cdots+\sum\sum\cdots\sum F_{30}(m_1,m_2,\ldots,m_{k-3})+\sum\sum\cdots\sum F_{31}(m_1,m_2,\ldots,m_{k-3})+\sum\sum\cdots\sum F_{32}(m_1,m_2,\ldots,m_{k-3}) +\sum\sum\cdots\sum F_{33}(m_1,m_2,\ldots,m_{k-3}) + \dots + \cdots + \sum_{m_1=1}^{[log_2{(2^{p(k-1)}-2^{p(k-2)}-2^{p(k-3)}-\cdots-1+1)}]} (2^{m_1}-1) = M$$

What is $\sum\sum\cdots\sum F_{ij}(m_1,m_2,\ldots,m_{k-i})$ ?

Example: Let,$\sum\sum\cdots\sum F_{11}(m_1,m_2,\ldots,m_{k-1}) = \sum_{m_1=1}^1 \sum_{m_2=1}^1 \cdots \sum_{m_{k-2}=1}^1 \sum_{m_{k-1}=p+1}^{p+1}F_{11}(m_1,m_2,\ldots,m_{k-1})$

which that, $\max [F_{10}]>\max [F_{11}]$

İf $k\to \infty$ and $p\to \infty$ I think $\text{“}M\text{''}$ must be sum of all odd numbers, which that the last number equal to $2^{p(k-1)}-2^{p(k-2)}-2^{p(k-3)}-\cdots-1.$

The Question: Is this limit equal to $1$ ?

$$\lim_{k \to \infty}\left[ \lim_{p \to \infty} \frac{M}{1+3+5+7+\cdots+ [2^{p(k-1)} - 2^{p(k-2)}-2^{p(k-3)}-\cdots-1]}\right]=1$$

Answer 1

(too long for a comment, too)

Although I deeply understand the pain of writing in a second language, you can still try to make every idiosyncratic notation or terminology clear to everyone. It is still very difficult to follow, due to lots of notations which are left unexplained.

Let me give an example by refining part of your question which I think that I am following.

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Consider the function $f:\mathbb{N} \to \mathbb{N}$ on the set of positive integers $\mathbb{N} = \{1,2,\cdots\}$ defined by

$$ f(n) = \begin{cases} n/2 & \text{if $n$ is even} \\ n+1 & \text{if $n$ is odd} \end{cases} $$

We know that for each $n\in\mathbb{N}$, there exists $k\geq1$ such that $f^{k}(n)=1$ where $f^k$ is the $k$-fold composition of $f$.

Now we would like to track this dynamics backward. This may be formally described by considering sequences of the form $\mathbf{a}=(1=a_0,a_1,a_2,\cdots)$ that satisfy $a_i=f(a_{i+1})$ for all $i$. We focus on the indices $i$ that satisfy $a_{i-1}=a_{i}+1$. Enumerating all such indices in increasing order, we obtain a sequence $(i_k)$. (Here, we suppress the dependence on $\mathbf{a}$ from notation for brevity.) We call $i_k$ the $k$-th step of $\mathbf{a}$. Then the locations of steps are determined by specifying the gaps $m_k = i_k - i_{k-1} - 1$ for $k \geq 1$, where we adopt the convention that $i_0 = 0$. Then

\begin{align*} &a_{i_0} = 1\\ \xrightarrow{\text{next step}} \quad& a_{i_1} = 2^{m_1} - 1 \\ \xrightarrow{\text{next step}} \quad& a_{i_2} = 2^{m_1+m_2} - 2^{m_2} - 1 \\ \xrightarrow{\text{next step}} \quad& a_{i_3} = 2^{m_1+m_2+m_3} - 2^{m_2+m_3} - 2^{m_3} - 1 \\ \xrightarrow{\text{next step}} \quad& \cdots \end{align*}

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And for the rest of your question, especially regarding quantities $F_{ij}$, I have no idea what you are trying to say. At least I can understand is that (borrowing some of my notations)

\begin{align*} F_{1,0} = F_{1,0}(m_1, \cdots, m_k) &= [\text{$a_{i_k}$ with the first $k$ gaps given by $(m_1, \cdots, m_k)$}] \\ &= 2^{m_1+\cdots+m_k} - \left( \sum_{l=2}^{k} 2^{m_l+\cdots+m_k} \right) - 1 \end{align*}

and that

$$ \max\{ F_{1,0} : m_1, \cdots, m_k \in \{0,\cdots, p\} \} = F_{1,0}(\underbrace{p, \cdots, p}_{\text{$k$-tuple}}). $$

Answer 2

(Too long for a comment.) The question is difficult to follow, even with the additional details in the posted image. To me, it looks like it might have something to do with the "total stopping time" of the Collatz-like function $g(n)$, only defined as $n+1$ in the odd case, instead of $3n+1$. If my guess is wrong, please stop reading and post a comment so that I remove this post. Even if you keep reading, note that it is not an answer, just some related thoughts which may or may be not useful.

The modified $g(n)$ is known to always go to $1$ if repeatedly iterated, see this answer for example.

About how "fast" it goes to $1\,$, let $\gamma(n)=l$ be the lowest positive integer such that $g^{\,l}(n)=1\,$. Some bounds on $\gamma(n)$ can be derived by looking at some particular cases.

The "fastest" descent to $1$ happens when the iterations consist of divisions, only. This means all intermediate values are even, which happens iff $n=2^k\,$, and in that case $g^{\,k}(n)=1$ so $\gamma(2^k) = k$.

The "slowest" descent to $1$ happens when iterations alternate between additions and divisions. This means $n=2^k+1\,$, and in that case $g^{\,2k}(n)=1$ so $\gamma(2^k+1)=2k\,$.

Of course, $\gamma(n)$ is not monotonical in $n$, so the above do not give "hard" bounds for arbitrary $n\,$, but it still follows that $\;\displaystyle \liminf_{n \to \infty} \frac{\gamma(n)}{\log_2(n)} \le 1\;$ and $\;\displaystyle \limsup_{n \to \infty} \frac{\gamma(n)}{\log_2(n)} \ge 2\,$.