Partitions of the odd integers Understanding the nature of the odd integers is a necessity to prepare oneself to work on the unsolved problems in number theory, such as the Collatz $3n+1$ problem.     I hope to demonstrate how the odd integers can be represented as a sequence of sets which fit together like a glove with infinite fingers.  First, some definitions are required.
Define a sequence $(A_k)$ of ordered sets by:
$$ A_0 = \{3, 7, 11\} $$ $$ A_1 = \{1, 9, 17\} $$ $$ A_2 = \{13,29,45\} $$
$$ A_3 = \{5, 37, 69\} $$   and
$$ A_{k+2} = A_k + 4 ( A_k – A_{k-2}) \forall k>1, $$
Note that the operations implied are matrix addition, subtraction and scalar multiplication on the $1 \times 3$ matrices formed from sets $A_k.$  
For example, when  $k=2$ we have:
$$ A_2 = \{13,29,45\}, \mbox{ in matrix form is }\begin{pmatrix}13 \\29 \\45\end{pmatrix} $$
$$ A_0 = \{3, 7, 11\}, \mbox{ in matrix form is }\begin{pmatrix}3\\ 7\\ 11\end{pmatrix} $$
By definition: $$A_4 = A_2 + 4 ( A_2 - A_0) $$ 
$$ A_4 = \begin{pmatrix}13\\ 29\\ 45\end{pmatrix} + 4\left(\begin{pmatrix}13\\ 29\\ 45\end{pmatrix} - \begin{pmatrix}3\\ 7\\ 11\end{pmatrix}\right)$$
$$ A_4 = \begin{pmatrix}13\\ 29\\ 45\end{pmatrix} + 4\begin{pmatrix}10\\ 22\\ 34\end{pmatrix} $$
$$ A_4 = \begin{pmatrix}13\\ 29\\ 45\end{pmatrix} + \begin{pmatrix}40\\ 88\\ 136\end{pmatrix} $$
$$ A_4 = \begin{pmatrix}53\\ 117\\ 181\end{pmatrix} $$ 
Converting back to set notation gives: 
$$ A_4 = \{53, 117, 181\} $$
The next step is to extend the finite sets $\,A_k\,$ to infinite sets $\,M_k\,$ as follows:
Define: $$ M_k = \left\{p \in \mathcal{N}, p \equiv 1 \pmod 2 : \exists a \in A_k, p\equiv a\pmod {3\left(2^{k+2}\right)} \right\} $$
By definition of $M_0$: if $(a \in A_0)$, then
$$ a + 12i \in M_0, \forall i \in \mathcal{N},$$ 
By definition of $M_k$: if $ (a \in A_k)$ then 
$$ a+3i(2^{k+2}) \in M_k, \forall i \in \mathcal{N}. $$
Assertion: 
$$ \forall n \in \mathcal{N}, n\equiv 1\pmod 2,\,\exists k \in \mathcal{N}, k\geq0 \text { such that } n \in M_k $$
Since this analysis was developed informally, is there a better way to express the assertion in order to search for prior solutions?
 A: This is not an answer and shall be deleted later, it is just too long for a comment and I'm asking, whether I got the understanding of the S-sets correct. If I made errors, please feel free to correct them inline if possible 
That's what I think to have understood, a bit simplified in notation
Let $k=2j$ and $j \in N$ then we define sets $S_{k,m=1..4}$ by
$$ \begin{array} {lll}
 S_{2j,0} &=&  f (2j) &                  & \pmod {2^{2j+3}}  \\
 S_{2j,1} &=& f (2j) &+ 1 \cdot 2^{2j+1} & \pmod {2^{2j+3}}  \\
 S_{2j,2} &=& f (2j) &+ 2 \cdot 2^{2j+1} & \pmod {2^{2j+3}} \\
 S_{2j,3} &=& f (2j) &+ 3 \cdot 2^{2j+1} & \pmod {2^{2j+3}}
 \end{array}$$
$$ \text { where } f \left(k\right) = \sum_{i=0}^k \,4^i  = {4^{k+1}-1 \over 4-1} $$ 
With this I got numerically $$ \begin{array} {lll}
S_{j,0} &\underset{j \ge 0}{=}& \{1,5, 21,85,...\} \\
S_{j,1} &\underset{j \ge 0}{=}& \{3,13, 53,213,...\} \\ 
S_{j,2} &\underset{j \ge 0}{=}& \{5,21, 85,341,...\} \\ 
S_{j,3} &\underset{j \ge 0}{=}& \{7,29,117,469,...\} \\ 
\end{array} $$                
Of course, if this is correct so far, then it can be further simplified and straightened in that the $\pmod {2^{2j+3}}$ is not needed because the coefficients to be evaluated are always smaller than that modulus.
Answer: The modulus is needed to fully extend the partition for $S_{k,0}$, $S_{k,1}$ and $S_{k,3}$.   It is only for $\,S_{k,2}\,$ that I subdivide further by setting $\,S_{k+2}\,$ = $\,S_{k,2}$.     
And while I might have understood the sets $S_{j,0},S_{j,1}$ correctly (they would then be the $L$-sets in my scheme) I do not understand the function of the $S_{j,2},S_{j,3}$ and also I do not understand why we use only even $k$ (resp. $2j$ in my proposed notation) ?
ANSWER:   I use only even k for S because each level corresponds to two levels of my original sets $A_k$ and $M_k$.  Indeed:  
$$S_{k,0} \equiv M_{k+1}$$
$$\left(S_{k,1} \cup S_{k,3}\right) \equiv M_k \, and $$
$$ S_{k,2} = S_{k+2}.$$  
