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\title{\textbf{Methodological Generalization of the Collatz Sequences to \((1 + 2^k)n + S_k(n)\) with Computational Verification for \(k = 1\) up to \(k = 51\)}}

\author{Ammar HAMDOUS}

\begin{document}

\maketitle



\begin{center}
\section*{Abstract} 

\end{center}


\begin{justify}
Several attempts have been made to generalize the Collatz sequence \(3n + 1\), but many of them produced sequences that lack the essential structural properties of the original Collatz dynamics. Among these, the most promising known generalization is the one proposed in 2022 by Naouel Boulkaboul~\cite{Boulkaboul2022}, which takes the form \(3n + 3^k\) and leads sequences to converge toward \(3^k\). 
In this work, we propose a new methodological generalization of the Collatz conjecture based on the transformation \((1 + 2^k)n + S_k(n)\), where \(S_k(n)\) is a function specifically designed to preserve the singularity and the core behavior of the original sequence. We demonstrate that this generalization retains key features such as singularity patterns and trivial cycles.
Furthermore, we conduct a computational verification for \(k = 1\) to \(k = 51\), supporting the conjecture that all such generalized sequences converge to \textbf{1}. This generalization opens the door to new strategies for understanding and potentially proving the Collatz conjecture in its broader form.
\end{justify}



\section*{1 Introduction}

\begin{justify}
The work that we present in this preprint is the methodical generalization of the Collatz sequences \(3n+1\), such that all Collatz sequences resulting from this generalization exhibit a generalized form of the \(3n+1\) singularity revealed in the previous preprint published on Zenodo ~\cite{Hamdous2025}. Following the theoretical development, and to justify the adoption of this generalization for our further studies, we carry out a computational verification over a moderate range to evaluate the behavior and validity of these generalized sequences within bounded domains.

\end{justify}


\section*{2 Reminder:}
In our previous work ~\cite{Hamdous2025} we revealed a singularity for the \(3n+1\) Collatz sequence as follows:\\
The singularity consists in having or reaching a term in the sequence whose binary form satisfies the following condition:

\begin{itemize}
  \item all bits at even indices are equal to \textbf{1},.
  \item and all bits at odd indices are equal to \textbf{0}.
\end{itemize}

In other words, the binary form is composed of a bit equal to 1 at both the beginning and the end, and every bit equal to 1 is followed by a bit equal to 0.

\begin{center}
    {\fontsize{30pt}{36pt}\selectfont
    \texttt{\textbf{10\ -\ -\ 0101}}
    }
\end{center}

\section*{3 Generalization of the \(3n+1\) Singularity}

We define a \emph{generalized singularity} as the occurrence of a term in a generalized Collatz sequence whose binary representation satisfies the following conditions:
\begin{itemize}
    \item It begins with \( k \) consecutive bits equal to \textbf{1},
    \item It ends with \( k \) consecutive bits equal to \textbf{1}, and
    \item Each group of \( k \) consecutive \textbf{1}s is immediately followed by \( k \) consecutive \textbf{0}s.
\end{itemize}

\[
\Large
\texttt{1\ldots110\ldots00\text{-\,-\,-\,}1\ldots11}
\underbrace{\texttt{0\ldots0}}_{k}
\underbrace{\texttt{1\ldots1}}_{k}
\]

\section*{4 Deduction of \(f_k\) function from the singularity}
In the previous preprint ~\cite{Hamdous2025}, we defined the auxiliary function \(f_3\) as follows:

\[f_3 : \mathbb{N} \rightarrow \mathbb{N}
\] 
\[\quad f_3(n) = 3n + 1 
\]

Now, how can a term \(\textbf{a}\) that qualifies as a generalized singularity be transformed into a power of $\textbf{2}$ by a single iteration of \(f_k\) function, which is the generalization of the function \(f_3\)?

To achieve this, we simply add to the term \( \textbf{a} \) the quantity \(2^k \cdot a\), such that the resulting binary representation has all its bits equal to $\textbf{1}$. That is, the structure of \( \textbf{a} \) and \( 2^k \cdot a \) must be aligned in a way that all \textbf{1-bits} in the binary representation of \( \textbf{a} \) are complemented by \textbf{0-bits} in the binary representation of \( 2^k \cdot a \), and vice versa.

\begin{flushleft}
\hspace*{6.1cm}$a = 1\ldots110\ldots00\ldots\ldots001\ldots11$\\
\hspace*{4.2cm}$2^k \cdot a = 1\ldots110\ldots001\ldots11\ldots\ldots110\ldots00$\\
\hspace*{1.56cm}$a + 2^k a = (1 + 2^k)n = 111111111111111111\ldots\ldots11111111$\\[1ex]
\hspace*{4.5cm}\textit{(the binary number consisting entirely of 1s)}
\end{flushleft}

The addition of \( \textbf{a} \) and \( 2^k \cdot a \) gives a natural number with a binary form composed only by bits equal to $1$
Now to transform \(a + 2^k\ a = (1+2^k)\ a\) into a power of $2$, all we need is adding $1$, but we will expose later the real meaning of adding $1$ to \((1+2^k)\ a\) that we have used here.\\

So we almost finished the definition of the auxiliary function \(f_k\) for the generalized Collatz sequences, as multiplying the odd term by \((1 + 2^k)\), then adding $1$ to the result, but this definition will be adjusted later.\\

We have transformed a generalized singularity term \( \textbf{n} \) by just one iteration of \(f_k\) into a power of $2$ term, so now to make the generalized Collatz sequences converge to $1$, all we need is a successive operation of dividing by $2$, so the second auxiliary function of the generalized Collatz sequences will be the same \(f_2\) as in the \(3n + 1\) original Collatz sequences ~\cite{Hamdous2025}, and we can argue the choice of dividing by \(2^1\) function rather than dividing by \(2^k\) function by these two properties.

\begin{itemize}
    \item Dividing by \(2^1\) function gives us the opportunity to prevent the generalized sequences from diverging to infinity by the ability to divide by $2$ as soon as we have an even term. 
    \item With the dividing by \(2^1\) function \(f_2\), the reduction of the generalized sequences is maximum since the dividing by 2 continues until we get an odd term.
\end{itemize}

These two properties are not available in dividing by \(2^k\) auxiliary function.

\subsection*{4.1 Adjusting the \(f_k\) function}
\subsection*{4.1.1 Reason why we should adjust the \(f_k\) function}

If we consider that \(f_k\) is defined by \(f_k(n) = (1 + 2^k)\ n + 1\),  
then let us take \(k = 3\) as an example.  
We will now examine the behavior of \(f_k(n)\) starting from \(n = 1\).  
The first 114 terms of the sequence are:

\begin{flushright}
\begin{tabular}{lll}
1st: 1                  & 2nd: 10                    & 3rd: 5 \\
4th: 46                & 5th: 23                    & 6th: 208 \\
7th: 104               & 8th: 52                    & 9th: 26 \\
10th: 13               & 11th: 118                  & 12th: 59 \\
13th: 532              & 14th: 266                  & 15th: 133 \\
16th: 1198             & 17th: 599                  & 18th: 5392 \\
19th: 2696             & 20th: 1348                 & 21st: 674 \\
22nd: 337              & 23rd: 3034                 & 24th: 1517 \\
25th: 13654            & 26th: 6827                 & 27th: 61444 \\
28th: 30722            & 29th: 15361                & 30th: 138250 \\
31st: 69125            & 32nd: 622126               & 33rd: 311063 \\
34th: 2799568          & 35th: 1399784              & 36th: 699892 \\
37th: 349946           & 38th: 174973               & 39th: 1574758 \\
40th: 787379           & 41st: 7086412              & 42nd: 3543206 \\
43rd: 1771603          & 44th: 15944428             & 45th: 7972214 \\
46th: 3986107          & 47th: 35874964             & 48th: 17937482 \\
49th: 8968741          & 50th: 80718670             & 51st: 40359335 \\
52nd: 363234016        & 53rd: 181617008            & 54th: 90808504 \\
55th: 45404252         & 56th: 22702126             & 57th: 11351063 \\
58th: 102159568        & 59th: 51079784             & 60th: 25539892 \\
61st: 12769946         & 62nd: 6384973              & 63rd: 57464758 \\
64th: 28732379         & 65th: 258591412            & 66th: 129295706 \\
67th: 64647853         & 68th: 581830678            & 69th: 290915339 \\
70th: 2618238052       & 71st: 1309119026           & 72nd: 654559513 \\
73rd: 5891035618       & 74th: 2945517809           & 75th: 26509660282 \\
76th: 13254830141      & 77th: 119293471270         & 78th: 59646735635 \\
79th: 536820620716     & 80th: 268410310358         & 81st: 134205155179 \\
82nd: 1207846396612    & 83rd: 603923198306         & 84th: 301961599153 \\
85th: 2717654392378    & 86th: 1358827196189        & 87th: 12229444765702 \\
88th: 6114722382851    & 89th: 55032501445660       & 90th: 27516250722830 \\
91st: 13758125361415   & 92nd: 123823128252736      & 93rd: 61911564126368 \\
94th: 30955782063184   & 95th: 15477891031592       & 96th: 7738945515796 \\
97th: 3869472757898    & 98th: 1934736378949        & 99th: 17412627410542 \\
100th: 8706313705271   & 101st: 78356823347440      & 102nd: 39178411673720 \\
103rd: 19589205836860  & 104th: 9794602918430       & 105th: 4897301459215 \\
106th: 44075713132936  & 107th: 22037856566468      & 108th: 11018928283234 \\
109th: 5509464141617   & 110th: 49585177274554      & 111th: 24792588637277 \\
112th: 223133297735494 & 113th: 111566648867747     & 114th: 1004099839809724 \\
\end{tabular}
\end{flushright}

As you can see, the sequence seems to grow to infinity, and there is no trivial cycle as in the original \(3n + 1\) Collatz sequence.\\

\subsection*{4.1.2 Finding the Best Alternative to \(+1\) part in \(f_k\)}

The adjustment of the function \(f_k\), which we have so far defined as \((1 + 2^k) n + 1\), concerns only the \( + 1\) part, and It will be made by replacing this part with the best alternative that satisfies the following four conditions:\\

\noindent \textbf{1\textsuperscript{st} Condition:} Offering the ability to integrate the Collatz sequence \(3n + 1\) for a certain value of \(k\).\\
\noindent \textbf{2\textsuperscript{nd} Condition:} Maintaining the occurrence of the \(+1\) part in \(f_k\), which we previously used to transform \((1 + 2^k)\ a\) into a power of \(2\), when \(a\) is a generalized singularity term.\\
\noindent \textbf{3\textsuperscript{rd} Condition:} There must be a trivial cycle starting from \(1\) for any \(k\in\mathbb{N^*}\).\\
\noindent \textbf{4\textsuperscript{th} Condition:} Preventing the generalized Collatz sequences from diverging to infinity, in such a way that convergence to $1$ remains possible\\

First let \(\textbf{a}\) be a generalized singularity term, and let apply the \(f_k\) iteration, the result is:

\begin{flushleft}
\hspace*{6.1cm}$a = 1\ldots110\ldots00\ldots\ldots001\ldots11$\\
\hspace*{4.15cm}$2^k \cdot a = 1\ldots110\ldots001\ldots11\ldots\ldots110\ldots00$\\
\hspace*{1.6cm}$a + 2^k a = (1 + 2^k)a = 111111111111111111\ldots\ldots11111111$\\[1ex]
\hspace*{4.5cm}\textit{(a binary number consisting entirely of 1s)}
\end{flushleft}

We have that the \((1 + 2^k)\cdot a\) par of \(f_k\) gives a natural number whose binary form is composed only of bits equal to $1$, and the number of these bits is a multiple of \(k\), so the \(+1\) part of \(k\) will transform \((1 + 2^k)\cdot a\) into a power of $2$, with a binary form composed of the same multiple of \(k\) bits equal to $0$, followed by a single bit equal to $1$. 
So, in the case of a generalized singularity term \(a\), we have:
\(f_k(a) \equiv 0 \ (\mathrm{mod}\ 2^k)\)\
That is to say, the \(+1\) part of \(f_k\) allows \(f_k(a)\)\ to be divisible by \(2^k\).\\

But in the general case, where \(\textbf{n}\) is not necessarily a generalized singularity term, we have:
\(n \equiv 1 \ (\mathrm{mod}\ 2)\)\, and \((1+2^k) \equiv 1 \ (\mathrm{mod}\ 2)\)\
So \((1+2^k)n \equiv 1 \ (\mathrm{mod}\ 2)\)\, and by the adding of \(+1\) part of \(f_k\) we get:
\(f_k(n)=(1+2^k)n+1\) such that \(f_k(n) \equiv 0 \ (\mathrm{mod}\ 2)\)\ 
Thus, the \(+1\) part of \(f_k\) allows only to get an even term \(f_k(n)\)\

We consider that the real meaning of \(+1\) part of \(f_k\) is to make \(f_k(n)\equiv 0 \ (\mathrm{mod}\ 2^k)\)\, and we can justify this by the fact that when we make \(f_k(n)\equiv 0 \ (\mathrm{mod}\ 2^k)\)\ we are aiming to satisfy the \( 4 ^{th}\) Condition, since any odd term \(\textbf{n}\) multiplied by \((1+2^k)\) will be divisible by \(2^{k+p}\text{ with}\ p \in \mathbb{N} \) by adding what we call in Modular Arithmetic the smallest non-negative additive inverse of \((1+2^k)n\) \textbf{mod} \(2^k\), and we represent it by the function:\\
\(S_k\)

\subsection*{4.1.2.1 Definition of the \(S_k\) function}
The smallest non-negative additive inverse of \(((1+2^k)n\) \textbf{mod} \(2^k\)) can be defined as the smallest natural number \(s_k(n)\in\mathbb{N^*}\) that we need to add to \((1+2^k)n\) such that \((1+2^k)n+s_k(n)\) will be divisible by \(2^k\).   


In general, we have:
\[
S_k(n) = \left[ 2^k - \left( (1 + 2^k)n \bmod 2^k \right) \right] \bmod 2^k
\]

But since \((1 + 2^k)n \bmod 2^k\) is always strictly positive, because \((1 + 2^k)n\) is always an odd strictly positive natural number, then:
\[
S_k(n) = 2^k - \left( (1 + 2^k)n \bmod 2^k \right)
\]
\(S_k\) function statement:\\

Let \(n \in 2\mathbb{N} + 1\) be the set of all odd natural numbers.\\

We can now define the function \(S_k\), which gives the smallest non-negative additive inverse of \((1 + 2^k)n\) modulo \(2^k\), as follows:

\[
S_k : 2\mathbb{N} + 1 \rightarrow \{1, 2, \dots, 2^k - 1\}
\]
\[
S_k(n) = 2^k - \left((1 + 2^k)n \bmod 2^k\right)
\]


\subsection*{4.1.2.2 Does \(S_k\) satisfy the 1\textsuperscript{st} Condition?}
If \(k=1\), then:\\

\(f_k(n) = (1 + 2^1)n + S_1(n) = 3n + S_1(n)\)\\
\(S_1(n) = 2^1 - \left((1 + 2^1)n \bmod 2^1\right) = 2 - (3n \bmod 2) = 1\)\\

So if \(k=1\), then \(f_k(n) =3n + 1\), and since the second auxiliary function \(f_2\) in the \(3n + 1\) Collatz sequences is the same as in the generalized Collatz sequences, then yes the \(3n + 1\) Collatz sequences is integrated on generalized Collatz sequences for \(k=1\).\\

So yes \(S_k\) satisfies the 1\textsuperscript{st} Condition

\subsection*{4.1.2.3 Does \(S_k\) satisfy the 2\textsuperscript{nd} Condition?}

Let be \(\textbf{a}\) a generalized singularity term, when applying the \(f_k\) iteration \(f_k(a)=(1+2^k)a + S_k(a)\)
We know that the \((1 + 2^k)\ a\) par of \(f_k\) is a natural number whose binary form is composed only of bits equal to $1$, and the number of these bits is a multiple of \(k\)\\

So \(S_k(a)\) is always equal to $1$ when \(\textbf{a}\) is a generalized singularity term, which means that the occurrence of \(+1\) part in \(f_k\) is always maintained when the odd term is a generalized singularity, that is to say \(S_k\) satisfies the 2\textsuperscript{nd} Condition.\\

\subsection*{4.1.2.4 Does \(S_k\) satisfies the 3\textsuperscript{rd} Condition?}

We have for all \(k\in\mathbb{N^*}\)\\

\(f_k(1)=(1+2^k)\cdot 1 + S_k(1)\)\\

In this case we have:\\

\(S_k(1)=2^k - \left((1 + 2^k) \bmod 2^k\right)=2^k-1\)\\

So:\\

\(f_k(1)=(1+2^k)\cdot 1 + S_k(1)=(1+2^k)+(2^k-1)=2^{k+1}\)\\

So after \(k+1\) iterations via the second auxiliary function \(f_2\) of the generalized Collatz sequences applied to \(f_k(1)=2^{k+1}\) we will get the first term $1$.\\ 
\[
\frac{f_k(1)}{2^{k+1}}=\frac{2^{k+1}}{2^{k+1}}=1
\]

We conclude that \(S_k\) function allows the existence of a \textbf{trivial cycle}:
\[
\{1, 2^{k+1}, 2^k, 2^{k-1}, \dots, 2^2, 2^1\} \quad \text{for all } k \in \mathbb{N}^*
\]


So yes \(S_k\) satisfies the 3\textsuperscript{rd} Condition

\subsection*{4.1.2.5 Does \(S_k\) satisfies the 4\textsuperscript{th} Condition?}

By choosing the \(S_k\) function, we are aiming to satisfy the \( 4 ^{th}\) Condition, by preventing the generalized Collatz sequences from diverging to infinity, in such a way that convergence to $1$ remains possible, and to finally proof that the \(S_k\) is the best alternative to \(+1\) part in \(f_k\)
auxiliary function we need to proceed to a Computational Verification over a moderate range of the Generalized Collatz Conjecture \((1+2^k)n +S_k(n)\) to verify if for all \(k\in\mathbb{N^*}\) all the generalized sequences tested \(f_k(n)\) converge to $1$
 
\section*{5 Computational Verification over a moderate range of the Generalized Collatz Conjecture \((1+2^k)n +S_k(n)\)}
\subsection*{5.1 Efficiency Improvements of the algorithm}
\subsection*{5.1.1 Testing only odd natural numbers strategy}

Let \(b\in\mathbb{N^*}\) such that \(b > 1\) and \(b\) is odd.\\

And let  
\[
\{1, 2, \dots, b\} \text{  be the set of the first \(b\) natural numbers in \(\mathbb{N}^*\).}
\]  
If all odd natural numbers in 
\(\{1, 2, \dots, b\}\) converge to \(1\), then all even natural numbers in 
\(\{1, 2, \dots, b+1\}\) also converge to \(1\), because each even number in that set will be transformed via \(f_2\) iterations to an odd number in 
\(\{1, 2, \dots, b\}\), which by assumption converges to \(1\).

\subsection*{5.1.2 Early Convergence Criteria}
Let \(b\in\mathbb{N^*}\) such that \(b > 1\).\\

And let  
\[
\{1, 2, \dots, b\} \text{  be the set of the first \(b\) natural numbers in \(\mathbb{N}^*\).}
\] 
If all odd natural numbers in 
\(\{1, 2, \dots, b\}\) converge to \(1\), then any natural number that converges to an element of that set, will converge to \(\textbf{1}\).\\

This reflects the idea that a sequence doesn't need to reach \(\textbf{1}\) to be considered valid, it's sufficient that it reaches a previously tested smaller value, which ensures it will eventually reach \textbf{1}.\\

To test the convergence to \(\mathbf{1}\) of the generalized Collatz sequences for \(k\in\{{1,2,3,..,51}\}\)\\
\begin{itemize}
  \item \([1,\, 2^{30}]\) for \(0 < k < 17\)
  \item \([1,\, 2^{19}]\) for \(16 < k < 44\)
  \item \([1,\, 2^{18}]\) for \(k= 44\)
  \item \([1,\, 2^{17}]\) for \(k= 45\)
  \item \([1,\, 2^{16}]\) for \(k= 46\)
  \item \([1,\, 2^{15}]\) for \(k= 47\)
  \item \([1,\, 2^{14}]\) for \(k= 48\)
  \item \([1,\, 2^{13}]\) for \(k= 49\)
  \item \([1,\, 2^{12}]\) for \(k= 50\)
  \item \([1,\, 2^{11}]\) for \(k= 51\)
\end{itemize}
we use the improved algorithms \textbf{GC Tester}, but for k=2, the \textbf{GC Tester} detected that there is a sequence with 140 millions iterations without reaching a lower value so the program gives message telling that we must use \textbf{GC Tester2} which dont use the efficiency improvements, and to use this program we need first search the non-trivial cycles for k=2 with a third program \textbf{Non Trivial Cycles}. for generating the histogram, we use the simpler \textbf{Histogram} with no improvement and it tests each natural number, odd and even.

\subsection*{5.2 Hardware Specifications Used for Computational Verification}

The computational verification was performed on a personal computer with the following hardware specifications:

\begin{itemize}
  \item \textbf{Processor:} Pentium(R) Dual-Core E5300 CPU  2.60GHz (2 cores)
  \item \textbf{RAM:} 2.5 GB DDR2
  \item \textbf{Storage:} 512 GB HDD
  \item \textbf{Operating System:} Windows 7 Integral 64-bit
  \item \textbf{Compiler:} Borland Delphi 7
\end{itemize}

\noindent
To ensure that the program can handle very large numbers generated during the iterations of the generalized Collatz sequences, I chose the \texttt{Int64} data type. It allows for values up to \(2^{63} - 1\), which provides a wide enough range to test billions of natural numbers and accommodate exponential growth without immediate risk of overflow.\\

\noindent 
The tests for  
\( k = 1, 3, 4, \ldots, 11 \) took a total of \textbf{8 hours}.\\
The tests for \( k = 12, 13, 14 \) took \textbf{3 hours}.\\
The test for \( k = 15 \) took \textbf{2 hours}.\\
The test for \( k = 16 \) took \textbf{10 hours}.\\
The test for \( k = 17 \) took \textbf{2.5 hours}.\\
The test for \( k = 18 \) took \textbf{9 hours}.\\
The tests for \( k = 19 \) to \( k = 51 \) took only a \textbf{few minutes} in total.

\subsection*{5.3 Data extracted by each program}

The data extracted by \textbf{GC Tester} program:

\begin{itemize}
  \item \textbf{count}: the number of natural numbers tested.
  \item \textbf{count1}: the number of natural numbers validated because they converged to 1.
  \item \textbf{jt}: the total number of iterations via function \( f_2 \) recorded by all the tested numbers in the whole large interval.
  \item \textbf{mt}: the total number of iterations via function \( f_k \) recorded by all the tested numbers in the whole large interval.
  \item \textbf{jmt}: the total number of iterations via functions \( f_2 \) and \( f_k \) recorded by all the tested numbers in the whole large interval.
  \item \textbf{jmax}: the highest number of \( f_2 \) iterations recorded by a number in the large interval.
  \item \textbf{mmax}: the highest number of \( f_k \) iterations recorded by a number in the large interval.
  \item \textbf{mjmax}: the highest number of \( f_2 \) and \( f_k \) iterations recorded by a number in the large interval.
  \item \textbf{njmax}: the number that reached the highest number of \( f_2 \) iterations in the large interval.
  \item \textbf{nmmax}: the number that reached the highest number of \( f_k \) iterations in the large interval.
  \item \textbf{nmjmax}: the number that reached the highest number of \( f_2 \) and \( f_k \) iterations in the large interval.
  \item \textbf{Jav}: the average number of \( f_2 \) iterations over the large interval.
  \item \textbf{mav}: the average number of \( f_k \) iterations over the large interval.
  \item \textbf{jmav}: the average number of \( f_2 \) and \( f_k \) iterations over the large interval.
  \item \textbf{jti}: the total number of iterations via function \( f_2 \) recorded by all the tested numbers in one of the 32 small intervals.
  \item \textbf{mti}: the total number of iterations via function \( f_k \) recorded by all the tested numbers in one of the 32 small intervals.
  \item \textbf{jmti}: the total number of iterations via functions \( f_2 \) and \( f_k \) recorded by all the tested numbers in one of the 32 small intervals.
  \item \textbf{jmaxi}: the highest number of \( f_2 \) iterations recorded by a number in one of the 32 small intervals.
  \item \textbf{mmaxi}: the highest number of \( f_k \) iterations recorded by a number in one of the 32 small intervals.
  \item \textbf{mjmaxi}: the highest number of \( f_2 \) and \( f_k \) iterations recorded by a number in one of the 32 small intervals.
  \item \textbf{njmaxi}: the number that reached the highest number of \( f_2 \) iterations in one of the 32 small intervals.
  \item \textbf{nmmaxi}: the number that reached the highest number of \( f_k \) iterations in one of the 32 small intervals.
  \item \textbf{nmjmaxi}: the number that reached the highest number of \( f_2 \) and \( f_k \) iterations in one of the 32 small intervals.
  \item \textbf{beg}: the starting value of one of the 32 small intervals.
  \item \textbf{Javi}: the average number of \( f_2 \) iterations over one of the 32 small intervals.
  \item \textbf{mavi}: the average number of \( f_k \) iterations over one of the 32 small intervals.
  \item \textbf{jmavi}: the average number of \( f_2 \) and \( f_k \) iterations over one of the 32 small intervals.
\end{itemize}

The data extracted by \textbf{GC Tester2} program:

\begin{itemize}
  \item \textbf{Beg}: the start of each of the 32 small intervals.
  \item \textbf{Count}: the total number of even and odd integers tested in each of the 32 small intervals and in the large interval.
  \item \textbf{count1}: the total number of even and odd integers tested that have converged to 1 in each of the 32 small intervals and in the large interval.
\end{itemize}

\noindent The data extracted by \textbf{Non Trivial Cycles} program:\\

Each time it has tested 100 odd natural numbers, it will display the k, the number of cycles found, and each following line contains the number of elements in the cycle followed by all the elements of the cycle.
The program runs continuously, but if you see that no new cycles are being found, you can choose to copy the result and paste it into the graphical interface of the \textbf{GC Tester2} program.

\subsection*{5.4 Computational Results Report}
\begin{itemize}
  \item All the generalized Collatz sequences for \(k\in\{{1,3,4,5,..,16}\}\) over the moderate ranges \(1\) to \(2^{30}\) converges to $1$, and the program \textbf{GC Tester} have not detected any non-trivial cycle since there were no sequences with 140 millions iterations.

  \item All the generalized Collatz sequences for \(k\in\{{17,18,..,51}\}\) converges to $1$, and the program \textbf{GC Tester} have not detected any non-trivial cycle.

  \item For \(k=2\) the only Non-Trivial cycles that have an odd element less than $3000$ are:\\
  \{{31,156,78,39,196,98,49,248,124,62}\}\\
  \{{37,188,94,47,236,118,59,296,148,74}\}\\
  \{{61,308,154,77,388,194,97,488,244,122}\}\
  
  \item For \(k=2 \)\\                                      Out of the first \( 1,\!073,\!741,\!823 \) natural numbers tested, \( 673,\!597,\!093 \) produced a sequence that converges to \(1\), which represents \(62.73\%\) of the total.
  \end{itemize}

\noindent All the data from the Computational Verification are available in the file (Output Data.xlsx) joined with the preprint.\\

\noindent You can reproduce all these data using the Delphi programs joined to this preprint.

\subsection*{5.5 Conclusion of the Computational Verification}

Based on the computational verification, we can conclude that the generalization of Collatz sequences via \((1 + 2^k)n + S_k(n)\) is justified.

\section*{6 Generalized Collatz conjecture \((1 + 2^k)n + S_k(n)\)}

\subsection*{6.1 Generalized Collatz sequences \((1 + 2^k)n + S_k(n)\) Statment}

We first define the two auxiliary functions:

\[
f_2 : 2\mathbb{N} \rightarrow \mathbb{N}
\]
\[
\quad f_2(n) = \frac{n}{2} \quad \text{where } 2\mathbb{N} \text{ is the set of even natural numbers}
\]

\[f_k : \mathbb{N} \rightarrow \mathbb{N}
\] 
\[\quad f_k(n) = (1 + 2^k)n + S_k(n)\quad \text{where } S_k(n) = 2^k - \left((1 + 2^k)n \bmod 2^k\right)
\]
\(\textbf{k}\) is called the \textbf{Exponent parameter}.\\

The generalized Collatz sequence is then defined as follows:
\[
C_k : \mathbb{N}^* \rightarrow \mathbb{N}
\]
\[
C_0 \in \mathbb{N}^*
\]
\[
C_{n+1} =
\begin{cases}
f_k(C_n) & \text{if } C_n \equiv 1 \ (\mathrm{mod}\ 2) \\
f_2(C_n) & \text{if } C_n \equiv 0 \ (\mathrm{mod}\ 2)
\end{cases}
\]

\subsection*{6.2 Generalized Collatz conjecture Statment}

The generalized Collatz conjecture states that, whatever the exponent parameter of the sequence \( k \in \mathbb{N}^* \setminus \{2\} \), and regardless of the initial natural number \( C_0 \in \mathbb{N}^* \), the Collatz sequence always reaches the value $\textbf{1}$ after a finite number of iterations.\\

It can be formally expressed as:
\[
\forall k \in \mathbb{N}^* \setminus \{2\}\text{ and } \forall C_0 \in \mathbb{N}^*,\ \exists j \in \mathbb{N}^* : C_j = 1
\]

This is a new conjecture ( generalized Collatz conjecture ) and to this day, it remains unproven.

 

\section*{7 Conclusion}

\begin{justify}

In this work, we introduced a methodological generalization of the Collatz sequence using the transformation $(1 + 2^k)n + S_k(n)$, where $S_k(n)$ is the smallest non-negative additive inverse of \((1 + 2^k)n\) modulo $2^k$ ensuring that the result is divisible by $2^k$. This generalization was designed to preserve key dynamical features of the original $3n + 1$ sequence, including the appearance of generalized singularities and the existence of a trivial cycle.
Importantly, this work emphasizes the role of the singularity revealed in our previous preprint~\cite{Hamdous2025}, by extending its structural logic to the generalized sequences. The transformation was carefully constructed so that the generalized singularities characterized by specific binary patterns remain central to the convergence dynamics across all tested values of $k$.
A comprehensive computational verification was conducted for $k$ ranging from $1$ to $51$. The results confirm that, for all $k \ne 2$, every tested sequence eventually converges to 1, without encountering any non-trivial cycles within the tested ranges. In the special case of $k = 2$, a small number of non-trivial cycles were detected, suggesting a fundamental difference in behavior for this specific value.
These findings support the validity of the proposed generalized Collatz conjecture:
\[
\forall k \in \mathbb{N}^* \setminus \{2\},\ \forall C_0 \in \mathbb{N}^*,\ \exists j \in \mathbb{N}^*:\ C_j = 1
\]
While not a formal proof, this study provides strong empirical evidence in favor of the conjecture and lays the groundwork for future theoretical exploration of the underlying structure of generalized Collatz sequences.

\end{justify}

\begin{thebibliography}{9}
\bibitem{Hamdous2025}
Ammar Hamdous,
\textit{Revealing a Singularity in Collatz Sequences, Offering New Perspectives for Proving the Conjecture},
Zenodo, 2025.

\bibitem{Boulkaboul2022}
Naouel Boulkaboul,
\textit{3n + 3\textasciicircum k : New Perspective on Collatz Conjecture},
arXiv, 2022.
\end{thebibliography}

\begin{figure}[h]
    \centering
    \includegraphics[width=1\textwidth]{figure1a.jpg}
    \caption{Number of \(f_k\) iterations in red and number of \(f_2\) iterations in blue 
For $k = \text{1}$, for $n \in [1, 100]$ (top).}
    \label{figure1a.jpg}
    \end{figure}

\begin{figure}[h]
    \centering
    \includegraphics[width=1\textwidth]{figure1b.jpg}
    \caption{Number of \(f_k\) iterations in red and number of \(f_2\) iterations in blue 
For $k = \text{1}$, $n \in [501, 600]$ (middle).}
    \label{figure1b.jpg}
    \end{figure}

\begin{figure}[h]
    \centering
    \includegraphics[width=1\textwidth]{figure1c.jpg}
    \caption{Number of \(f_k\) iterations in red and number of \(f_2\) iterations in blue 
For $k = \text{1}$, and $n \in [901, 1000]$ (bottom).}
    \label{figure1c.jpg}
    \end{figure}

\begin{figure}[h]
    \centering
    \includegraphics[width=1\textwidth]{figure11a.jpg}
    \caption{Number of \(f_k\) iterations in red and number of \(f_2\) iterations in blue 
For $k = \text{11}$, for $n \in [1, 100]$ (top).}
    \label{figure11a.jpg}
    \end{figure}

\begin{figure}[h]
    \centering
    \includegraphics[width=1\textwidth]{figure11b.jpg}
    \caption{Number of \(f_k\) iterations in red and number of \(f_2\) iterations in blue 
For $k = \text{11}$, $n \in [501, 600]$ (middle).}
    \label{figure11b.jpg}
    \end{figure}

\begin{figure}[h]
    \centering
    \includegraphics[width=1\textwidth]{figure11c.jpg}
    \caption{Number of \(f_k\) iterations in red and number of \(f_2\) iterations in blue 
For $k = \text{11}$, and $n \in [901, 1000]$ (bottom).}
    \label{figure11c.jpg}
    \end{figure}

\begin{figure}[h]
    \centering
    \includegraphics[width=1\textwidth]{figure21a.jpg}
    \caption{Number of \(f_k\) iterations in red and number of \(f_2\) iterations in blue 
For $k = \text{21}$, for $n \in [1, 100]$ (top).}
    \label{figure21a.jpg}
    \end{figure}

\begin{figure}[h]
    \centering
    \includegraphics[width=1\textwidth]{figure21b.jpg}
    \caption{Number of \(f_k\) iterations in red and number of \(f_2\) iterations in blue 
For $k = \text{21}$, $n \in [501, 600]$ (middle).}
    \label{figure21b.jpg}
    \end{figure}

\begin{figure}[h]
    \centering
    \includegraphics[width=1\textwidth]{figure21c.jpg}
    \caption{Number of \(f_k\) iterations in red and number of \(f_2\) iterations in blue 
For $k = \text{21}$, and $n \in [901, 1000]$ (bottom).}
    \label{figure21c.jpg}
    \end{figure}

\begin{figure}[h]
    \centering
    \includegraphics[width=1\textwidth]{figure31a.jpg}
    \caption{Number of \(f_k\) iterations in red and number of \(f_2\) iterations in blue 
For $k = \text{31}$, for $n \in [1, 100]$ (top).}
    \label{figure31a.jpg}
    \end{figure}

\begin{figure}[h]
    \centering
    \includegraphics[width=1\textwidth]{figure31b.jpg}
    \caption{Number of \(f_k\) iterations in red and number of \(f_2\) iterations in blue 
For $k = \text{31}$, $n \in [501, 600]$ (middle).}
    \label{figure31b.jpg}
    \end{figure}

\begin{figure}[h]
    \centering
    \includegraphics[width=1\textwidth]{figure31c.jpg}
    \caption{Number of \(f_k\) iterations in red and number of \(f_2\) iterations in blue 
For $k = \text{31}$, and $n \in [901, 1000]$ (bottom).}
    \label{figure31c.jpg}
    \end{figure}

\begin{figure}[h]
    \centering
    \includegraphics[width=1\textwidth]{figure41a.jpg}
    \caption{Number of \(f_k\) iterations in red and number of \(f_2\) iterations in blue 
For $k = \text{41}$, for $n \in [1, 100]$ (top).}
    \label{figure41a.jpg}
    \end{figure}

\begin{figure}[h]
    \centering
    \includegraphics[width=1\textwidth]{figure41b.jpg}
    \caption{Number of \(f_k\) iterations in red and number of \(f_2\) iterations in blue 
For $k = \text{41}$, $n \in [501, 600]$ (middle).}
    \label{figure41b.jpg}
    \end{figure}

\begin{figure}[h]
    \centering
    \includegraphics[width=1\textwidth]{figure41c.jpg}
    \caption{Number of \(f_k\) iterations in red and number of \(f_2\) iterations in blue 
For $k = \text{41}$, and $n \in [901, 1000]$ (bottom).}
    \label{figure41c.jpg}
    \end{figure}

\begin{figure}[h]
    \centering
    \includegraphics[width=1\textwidth]{figure51a.jpg}
    \caption{Number of \(f_k\) iterations in red and number of \(f_2\) iterations in blue 
For $k = \text{51}$, for $n \in [1, 100]$ (top).}
    \label{figure51a.jpg}
    \end{figure}

\begin{figure}[h]
    \centering
    \includegraphics[width=1\textwidth]{figure51b.jpg}
    \caption{Number of \(f_k\) iterations in red and number of \(f_2\) iterations in blue 
For $k = \text{51}$, $n \in [501, 600]$ (middle).}
    \label{figure51b.jpg}
    \end{figure}

\begin{figure}[h]
    \centering
    \includegraphics[width=1\textwidth]{figure51c.jpg}
    \caption{Number of \(f_k\) iterations in red and number of \(f_2\) iterations in blue 
For $k = \text{51}$, and $n \in [901, 1000]$ (bottom).}
    \label{figure51c.jpg}
    \end{figure}




\end{document}