The Octets of Odd Numbers

In this paper we present a coherent network of concepts, theorems, propositions, and corollaries that arise from a theorem. A linear combination of powers of two uniquely represents every odd number except 1. The powers of 2 are consecutive with the coefficients −1 or +1. From this representation, a set of mathematical concepts leads to a mathematical object, the octets of odd numbers. We studied the properties of octets and their relationship to the structure of odd numbers. Thus, a new framework for the study of natural numbers emerges.

Introduction

Every odd number $N\ne 1$ is written in a specific form, as a sum of successive powers of $2$ (Theorem 1). The coefficients of $2$ in this combination take the values of $-1$ or $+1$, a possibility that does not exist in other number systems arithmetic systems (binary, decimal etc.). This difference leads to the mathematical concepts [1] “the conjugate”, “the complementary”, “the $l/r$ symmetry” and “the transpose” of an odd number. These concepts are related to the factors of composite odd numbers (Proposition 4 and Corollaries 2, 8, 9, 10).

Symmetry $l/r$ (Definition 3) divides odd numbers into four categories, $A$, $W$, $Y$ and $B$ (Definition 5). The transpose (Definition 4) categorizes the odd numbers into “symmetric” $A$, $B$ and “asymmetric” $W$, $Y$. The properties of the transpose are given by Theorems to 2–9. One consequence of these theorems is the relationship between symmetric and asymmetric numbers. For every symmetric number of the binary interval ${\mathrm{\Omega }}_n=[2^{n+1},2^{n+2}]$, $n\in \mathbb{N}$, there are infinite asymmetric numbers, that belong to the intervals ${\mathrm{\Omega }}_m$, $m=n+1,n+2,n+3,\dots$, whose transpose is equal to the symmetric number (Corollary 13).

From the combination of the conjugate (Definition 1) and transpose we obtain [1] the “octets” of odd numbers [1] (Definition 6), the mathematical object that is at the core of the theory we present in this article. We present and codify the structure of an octet and the possible relationships between different octets. Symmetric numbers belong to “symmetric” octets. Asymmetric numbers produce symmetric octets, while they themselves belong to “asymmetric” octets.

With $\left[a\right]$ we denote the integer part and with $\left|a\right|$ the absolute value of a real number $a$. With ${\mathbb{N}}^{\ast }$ we denote the set of natural numbers that do not contain zero as: ${\mathbb{N}}^{\ast }=\{1,2,3,\dots \}$. By ${\mathrm{\Omega }}_n$ we denote the binary interval ${\mathrm{\Omega }}_n=[2^{n+1},2^{n+2}]$ and; $n=-1,0,1,\dots$.

A Representation of Natural Numbers with Powers of 2

The following Theorem gives a representation of odd numbers with powers of 2.

Theorem 1. [1], [2] Every odd number $N\ne 1$ is written uniquely in the form

(1) $$\begin{array}{r}N=2^{n+1}+2^n+\sum _{k=0}^{n-1}{b}_k{2}^k,\end{array}$$

where $n\in \mathbb{N}$, $n+1=\left[\frac{\ln N}{\ln 2}\right]$, $b_k=\pm 1$,$k=0,1,2,\dots ,n-1$.

Proof. If $N=1$ we have $n+1=\left[\frac{\ln 1}{\ln 2}\right]=0⟶n=-1$ and $N=1\in {\mathrm{\Omega }}_{-1}=[2^0,2^1]=[1,2]$.

If $N=3$ we have $n+1=\left[\frac{\ln 3}{\ln 2}\right]⟶n=0$ and from (1) we obtain

$N=3=2^1+1\in {\mathrm{\Omega }}_0=[2^1,2^2]$.

Here, we consider the case $n\in {\mathbb{N}}^{\ast }$. From (1) we obtain the minimum $N_{min}$ and the maximum value $N_{max}$ of $N$,

(2) $$\begin{array}{r}{N}_{min}=N_{min}\left(n\right)=2^{n+1}+2^n-2^{n-1}-2^{n-2}-\dots -2^1-1=2^{n+1}+1,\end{array}$$

(3) $$\begin{array}{r}{N}_{max}=N_{max}\left(n\right)=2^{n+1}+2^n+2^{n-1}+2^{n-2}+\dots +2^1+1=2^{n+2}-1.\end{array}$$

For the odd $N$ of (1), from (2) and (3) we obtain;

(4) $$\begin{array}{r}{N}_{min}=2^{n+1}+1\le N\le 2^{n+2}-1=N_{max}.\end{array}$$

The number of odd numbers in the closed interval $[2^{n+1}+1,2^{n+2}-1]$ is

(5) $$\begin{array}{r}\frac{N_{max}-N_{min}}{2}+1=\frac{(2^{n+2}-1)-(2^{n+1}+1)}{2}+1=2^n.\end{array}$$

Equation (1) represents $2^n$ odd numbers for $b_k=\pm 1$, $k=0,1,2,\dots ,n-1$. Therefore (1) represents the $2^n$ odd numbers in (5). Hence, (1) represents all odd numbers in the interval ${\mathrm{\Omega }}_n$, for each $n\in {\mathbb{N}}^{\ast }$.

Now, from Inequality (4) we obtain

$2^{n+1}+1\le N\le 2^{n+2}-1$

so we have $2^{n+1}<N<2^{n+2}$ or equivalent $(n+1)\ln 2<\ln N<(n+2)\ln 2$ from which we get

$\frac{\ln N}{\ln 2}-1<n+1<\frac{\ln N}{\ln 2}$

and finally we obtain,

(6) $$\begin{array}{r}n+1=\left[\frac{\ln N}{\ln 2}\right].\end{array}$$

We now that every odd number $N\ne 1$ can be uniquely written in the form of (1). We write the odd $N$ as

(7) $$\begin{array}{r}N=2^{n+1}+2^n+\sum _{k=0}^{n-1}{b}_k{2}^k,\end{array}$$

where $n\in \mathbb{N}$, $n+1=\left[\frac{\ln N}{\ln 2}\right]$, $b_k=\pm 1$, $k=0,1,2,\dots ,n-1$, and

(8) $$\begin{array}{r}N=2^{n+1}+2^n+\sum _{k=0}^{n-1}{c}_k{2}^k,\end{array}$$

where $n+1=\left[\frac{\ln N}{\ln 2}\right]$ and $c_k=\pm 1$, $k=0,1,2,\dots ,n-1$.

From (7) and (8) we get

$(b_0-c_0)\cdot 2^0+(b_1-c_1)\cdot 2^1+(b_2-c_2)\cdot 2^2+\dots +(b_{n-1}-c_{n-1})\cdot 2^{\nu -1}=0$.

From this Equation, considering that

$b_k=\pm 1,k=0,1,2,\dots ,n-1$, $k\in \{0,1,2,\dots ,n-1\}$ and $c_k=\pm 1,k=0,1,2,\dots ,n-1$, $k\in \{0,1,2,\dots ,n-1\}$

we get $b_k=c_k$ for every $k=0,1,2,\dots ,n-1$. ■

[1] Algorithm for representing an odd number of Theorem 1. Let $N\ne 1$ be an odd number. First, we determin the binary interval ${\mathrm{\Omega }}_n$ to which it belongs via (6), and the sum

$2^{n+1}+2^n$.

If $2^{n+1}+2^n<N$ we added $2^{n-1}$, whereas if $2^{n+1}+2^n>N$ we subtract $2^{n-1}$. Repeating the process, after $n$ steps we obtain $N$ as given by Theorem 1.

Example 1. For 25 we have

$n+1=\left[\frac{\ln 25}{\ln 2}\right]$ so $n=3$.

We run the algorithm and get

$2^{n+1}+2^n=2^4+2^3=24<25$ ($2^2$)

$2^4+2^3+2^2=28>25$ (thus $2^1$ was subtracted)

$2^4+2^3+2^2-2^1=26>25$ (thus $2^0=1$ was subtracted)

$2^4+2^3+2^2-2^1-1=25$.

For Fermat numbers [3]–[6] t $F_s$ we have

(9) $$\begin{array}{r}{F}_s=2^{2^s}+1=N_{min}(2^s-1)=2^{2^s}+2^{2^s-1}-2^{2^s-2}-2^{2^s-3}-\dots -2^1-1,S=0,1,2,\dots .\end{array}$$

For Mersenne numbers [7], [8] $M_p$ we have

(10) $$\begin{array}{r}{M}_p=2^p-1=N_{max}(p-2)=2^{p-1}+2^{p-2}+2^{p-3}+\dots +2^1+1,\text{p is a prime number}\end{array}$$

Now we define the conjugate of an odd number.

Definition 1. [1], [2] 1. The conjugate $N^{\ast }$ of $N\ne 1$ is defined as

(11) $$\begin{array}{r}{N}^{\ast }=N^{\ast }(n,c_j)=2^{n+1}+2^n+\sum _{j=0}^{n-1}{c}_j{2}^j,\end{array}$$

where $c_j=-b_j$, $j=0,1,2,\dots ,n-1$.

(12) $$\begin{array}{r}\text{2. We set}{1}^{\ast }=1.\end{array}$$

Proposition 1. [1], [2] For conjugate numbers $N$ and $N^{\ast }$ we have the following.

(13) $$\begin{array}{r}\text{1.}{\left(N^{\ast }\right)}^{\ast }=N.\end{array}$$

2. If $N\in {\mathrm{\Omega }}_n$ then

(14) $$\begin{array}{r}{N}^{\ast }=3\cdot 2^{n+1}-N.\end{array}$$

3. The only common factor that two conjugates can have is 3.

(15) $$\begin{array}{r}4.IfN\in {\mathrm{\Omega }}_n,then\left|N-3\cdot 2^n\right|=\left|3\cdot 2^n-N^{\ast }\right|\end{array}$$

Proof. 1. We have ${\left(N^{\ast }\right)}^{\ast }=3\cdot 2^{n+1}-N^{\ast }=3\cdot 2^{n+1}-(3\cdot 2^{n+1}-N)=N$.

2. We have

$N+N^{\ast }=(2^{n+1}+2^n)+(2^{n+1}+2^n)$

or equivalently

$N+N^{\ast }=3\cdot 2^{n+1}$.

3. If $N$ and $N^{\ast }$ have a common factor $a\ne 1$, then this is also a factor of the second part of the equation $N+N^{\ast }=3\cdot 2^{n+1}$.

4. From (14) we obtain

$N-3\cdot 2^n=3\cdot 2^n-N^{\ast }$

and thus

$|N-3\cdot 2^n|=|3\cdot 2^n-N^{\ast }|$.

Now we define the complementary of an odd number. ■

Definition 2. [1] We define as complementary the odd numbers $N$ and $N^{\prime }$ of the interval ${\mathrm{\Omega }}_n$, for which $|N^{\prime }-N|=2^n$ holds.

The following gives the basic properties of complementarity and its relationship to the conjugate.

Proposition 2. [1]

1. $(N^{\prime }{)}^{\prime }=N$.

2. $N$ and $N^{\prime }$ are relatively prime numbers, $(N,N^{\ast })=1$.

3. $(N^{\ast }{)}^{\prime }=(N^{\prime }{)}^{\ast }$.

4. The complementary of conjugates odd numbers are also conjugates.

5. The conjugate of complementary odd numbers are also complementary.

Proof. 1. It results from the equation $|N^{\prime }-N|=2^n$.

2. Let $N=ab,N^{\prime }=ac$ where $a,b,c$ be odd numbers different from one. From Definition 2 we obtain $|N^{\prime }-N|=2^n$ or equivalently $|ac-ab|=2^n$ or equivalently $a|c-b|=2^n$, which is impossible because the odd number $a\ne 1$ cannot be a factor in $2^n$.

3. If $N\in [2^{n+1},3\cdot 2^n]\subseteq {\mathrm{\Omega }}_n$ we have $N<N^{\prime }$ and from Equation $|N^{\prime }-N|=2^n$ we get $N^{\prime }-N=2^n$ or equivalent $N^{\prime }=N+2^n$ and ${\left(N^{\prime }\right)}^{\ast }=3\cdot 2^{n+1}-N^{\prime }=3\cdot 2^{n+1}-(N+2^n)=3\cdot 2^{n+1}-N-2^n=2^{n+2}-N$.

Now, we have $N\in [2^{n+1},3\cdot 2^n]\subseteq {\mathrm{\Omega }}_n$ so $N^{\ast }\in [3\cdot 2^n,2^{n+2}]\subseteq {\mathrm{\Omega }}_n$. Thus, we obtain

$(N^{\ast }{)}^{\prime }=N^{\ast }-2^n=3\cdot 2^{n+1}-N-2^n=2^{n+2}-N$.

From the two previous equations we obtain $(N^{\ast }{)}^{\prime }=(N^{\prime }{)}^{\ast }$. Proof in the case where $N\in [3\cdot 2^n,2^{n+2}]\subseteq {\mathrm{\Omega }}_n$ is similar.

4. Taking into account Proposition 3 we get

$(N^{\ast }{)}^{\prime }+N^{\prime }=(N^{\prime }{)}^{\ast }+N^{\prime }=3\cdot 2^{n+1}$,

so $(N^{\ast }{)}^{\prime }$ and $N^{\prime }$ are conjugate.

5. If $N$ belongs to the interval $[2^{n+1},3\cdot 2^n]$ then $N^{\ast }$ belongs to the interval $[3\cdot 2^n,2^{n+2}]$. Thus, taking into account also 3 of Proposition, we get

$N^{\ast }-(N^{\prime }{)}^{\ast }=N^{\ast }-(N^{\ast }{)}^{\prime }$

or equivalently

$N^{\ast }-(N^{\prime }{)}^{\ast }=3\cdot 2^{n+1}-N-(N^{\ast }-2^n)$

or equivalently

$N^{\ast }-(N^{\prime }{)}^{\ast }=3\cdot 2^{n+1}-(N+N^{\ast })+2^n=3\cdot 2^{n+1}-3\cdot 2^{n+1}+2^n=2^n$.

Therefore $N^{\ast }$ and $(N^{\prime }{)}^{\ast }$ are complementary.

The proof in the case where $N$ belongs to the interval $[3\cdot 2^n,2^{n+2}]$ is similar. ■

For even numbers $E$ we have the following representation,

(16) $$\begin{array}{r}E=2^m\cdot N,\end{array}$$

where $N$ odd number and $m\in {\mathbb{N}}^{\ast }$. Writing $N$ in the form of (1) or $N=1$, if $E$ is a power of 2, from (16) we also obtain $E$ in the form of Theorem 1.

Example 2. For 400 we have

$E=400=2^4\cdot 25$

and taking into account that

$25=2^4+2^3+2^2-2^1-1$

we get

$400=2^4\cdot (2^4+2^3+2^2-2^1-1)=2^8+2^7+2^6-2^5-2^4$.

Left and Right Symmetry of an Odd Number

Euclid's division gives four forms for odd numbers,

$A=A_{\rho }=8\rho +1$,

$W=W_{\rho }=8\rho +3$,

$Y=Y_{\rho }=8\rho +5$,

$B=B_{\rho }=8\rho +7$,

where $\rho \in \mathbb{N}$.

From these forms the $l/r$ symmetry of odd numbers is defined.

Proposition 3. [1]

1. $A=A_0=1$ or $A=2^l\cdot M+1$, where $M$ is an odd number and $l=3,4,5,\dots$.

2. $W=W_0=3$ or $W=2^r\cdot M+3$, where $M$ is an odd number and $r=3,4,5,\dots$.

3. $Y=Y_0=5$ or $Y=2^l\cdot M+5$, where $M$ is an odd number and $l=3,4,5,\dots$.

4. $B=B_0=7$ or $B=2^r\cdot M+7$, where $M$ is an odd number and $r=3,4,5,\dots$.

Proof. We prove 1. 2, 3, and 4 are proven to be similar. We have $A=8\rho +1$. If $\rho =0$ we have $A=1$. If $\rho \ne 0$, then $\rho$ is either an odd number $\rho =M$ or an even number $\rho =2^n\cdot M$,$n=1,2,3,\dots$. If $\rho =M$, we have $A=2^3\cdot M+1$. If $\rho =2^n\cdot M$, $n=1,2,3,\dots$ we have $Q=2^{3+n}\cdot M+1=2^l\cdot M+1$ where $l>3$. ■

Definition 3. [1]

1. The numbers $A=2^l\cdot M+1$, where $M$ is an odd number and $l\ge 3$, have left symmetry $l$.

2. The numbers $W=2^r\cdot M+3$, where $M$ is an odd number and $r\ge 3$, have right symmetry $r$.

3. The numbers $Y=2^l\cdot M+5$, where $M$ is an odd number and $l\ge 3$, have left symmetry $l$.

4. The numbers $B=2^r\cdot M+7$, where $M$ is an odd number and $r\ge 3$, have right symmetry $r$.

Example 3. For $1153$ we get the following.

$\begin{array}{l}1153-1=2^7\cdot 9\\ 1153-3=2\cdot 575\\ 1153-5=2^2\cdot 287\\ 1153-7=2\cdot 573\end{array}$

Therefore $1153$ is of form $A$ with symmetry $l=7$ and $M=9$.

The following correlate $A$, $A^{\ast }$ and $M$, $M^{\ast }$.

Corollary 1. [1] We have $A=2^l\cdot M+1$ if and only if $A^{\ast }=2^L\cdot M^{\ast }-1=B$.

From Definition 3 we get the following.

Proposition 4. [1]

1. We have $A_1{A}_2=A$.

2. We have $B_1{B}_2=A$.

3. We have $A_1{B}_1=B$.

4. If $l\left(A_1\right)<l\left(A_2\right)$, then $l\left(A_1{A}_2\right)=l\left(A_1\right)$.

5. If $l\left(A\right)<r\left(B\right)$, then $r\left(AB\right)=l\left(A\right)$.

6. If $r\left(B\right)<l\left(A\right)$, then $r\left(AB\right)=r\left(B\right)$.

7. If $r\left(B_1\right)<r\left(B_2\right)$, then $l\left(B_1{B}_2\right)=r\left(B_1\right)$.

8. $Symmetry\left(N_1\right)=Symmetry\left(N_2\right)⟶Symmetry\left(N_1{N}_2\right)>Symmetry\left(N_1\right)=Symmetry\left(N_2\right)$.

Example 4. We have $l$ (897) = 7 < $l$ (49153) = 14 = >$l$ (897 × 49153) = $l$ (44090241) = 7.

We have $r$ (143) = $r$ (911) = 3 = > $l$ (143 × 911) = $l$(130273) = 5 > 3.

Corollary 2. [1] If $2^{n+1}+1$ is composite, then

$2^{n+1}+1=(2^m\cdot M_1+1)(2^m\cdot M_2+1)$,

where $M_1$, $M_2$ are odd numbers and $m\in {\mathbb{N}}^{\ast }$.

2. If $2^{n+1}-1$ is composite, then

$2^{n+1}-1=(2^s\cdot M_3+1)(2^s\cdot M_4-1)$,

where $M_3$, $M_4$ are odd numbers and $s\in {\mathbb{N}}^{\ast }$.

Corollary 2 predicts a specific structure for the factors of $2^{n+1}+1$ and $2^{n+1}-1$, in the case where they are composite numbers. Fermat numbers [3]–[6] are of the form $2^{n+1}+1$, if $n+1=2^S$, and $S\in \mathbb{N}$. Mersenne numbers [7], [8] are of the form $2^{n+1}-1$, if $n+1$ is a prime number.

Example 5.

For the composite Fermat number $2^{32}+1=4294967297$ we have $2^{32}+1=(2^7\cdot 5+1)(2^7\cdot 52347+1)$.

For the composite Mersenne number $2^{11}-1=2047$ we have $2^{11}-1=(2^3\cdot 11+1)(2^3\cdot 3-1)$.

The following gives the relationship of $l/r$ symmetry to the conjugate.

Corollary 3. [1] The conjugate odd numbers have a different $l/r$ symmetry.

We now present the algorithms for finding the conjugate and complementary of odd numbers.

Algorithm for finding the conjugate odd number $N$. First, we write the $N$ in the form of (1). We then change the signs of $b_k$ and, $k=0,1,2,\dots ,n-1$.

Algorithm for finding the complementary odd number $N$. First, we determin the binary interval ${\mathrm{\Omega }}_n$ to which $N$ belongs. From $N+2^n$ and $N-2^n$, $N$ is complementary to ${\mathrm{\Omega }}_n$.

Definition and Basic Theorems of the Transpose of an Odd Number

The transposition of an odd number is the third mathematical concept that emerges from Theorem 1.

Definition 4. [1] For odd numbers $\mathrm{\Delta }=B$ or $\mathrm{\Delta }=W$ with right symmetry,

(17) $$\begin{array}{r}\mathrm{\Delta }=2^{n+1}+2^n+b_{n-1}{2}^{n-1}+b_{n-2}{2}^{n-2}+\dots +b_1{2}^1+1,\end{array}$$

we define the transpose $T\left(\mathrm{\Delta }\right)$ of $\mathrm{\Delta }$ as

(18) $$\begin{array}{r}T\left(\mathrm{\Delta }\right)=(\frac{1}{2^{n+1}}+\frac{1}{2^n}+\frac{b_{n-1}}{2^{n-1}}+\frac{b_{n-2}}{2^{n-2}}+\dots +\frac{b_1}{2^1}+1)\cdot 2^{n+1}=2^{n+1}+3+\sum _{k=1}^{n-1}{b}_k\cdot 2^{n+1-k}.\end{array}$$

2. For odd numbers $\text{A }=A$ or $\text{A }=Y$ left symmetry,

(19) $$\begin{array}{r}\text{A }=2^{n+1}+2^n+b_{n-1}{2}^{n-1}+b_{n-2}{2}^{n-2}+\dots +b_1{2}^1-1,\end{array}$$

we define the transpose $T\left(\text{A }\right)$ of $\text{A }$ as

(20) $$\begin{array}{r}T\left(\text{A }\right)=-(\frac{1}{2^{n+1}}+\frac{1}{2^n}+\frac{b_{n-1}}{2^{n-1}}+\frac{b_{n-2}}{2^{n-2}}+\dots +\frac{b_1}{2^1}-1)\cdot 2^{n+1}=2^{n+1}-3-\sum _{k=1}^{n-1}{b}_k\cdot 2^{n+1-k}\end{array}$$

3. If $N=1$ we define,

(21) $$\begin{array}{r}T\left(1\right)=1.\end{array}$$

4. Equations (18), (20) and (21) are equivalent to the following equation

(22) $$\begin{array}{r}T\left(N\right)=2^{n+1}+b_0\cdot (3+\sum _{k=1}^{n-1}{b}_k\cdot 2^{n+1-k}),\end{array}$$

where $n+1=\left[\frac{\ln N}{\ln 2}\right]$.

Equation (22) shows the role played by $l/r$ symmetry in transposition. In left symmetry we have $b_0=-1$. In right symmetry we have $b_0=+1$.

[1] To calculate the transpose $T\left(N\right)$ of an odd number $N\ne 1$, we run the following algorithm. We write $N$ in the form of (1) and calculate $b_k=\pm 1,k=1,2,3,\dots ,n-1$. Then, $T\left(N\right)$ is calculated using (22).

Following is a series of Theorems, Propositions, and Corollaries regarding transpose.

Theorem 2. [1] 1. If $N\ne 1$, we have $T\left(N\right)=1$ if and only if

(23) $$\begin{array}{r}N=2^n-3,n\ge 2,n\in \mathbb{N}.\end{array}$$

2. We have

(24) $$\begin{array}{r}\begin{array}{l}T\left(B\right)=B\\ n+1=\left[{\displaystyle \frac{\ln B}{\ln 2}}\right]\end{array}\}⟺\{\begin{array}{l}{b}_1=1\\ b_{n-l}=b_{l+1}\\ l=1,2,3,\dots ,{\displaystyle \frac{n-2}{2}},n=even\\ l=1,2,3,\dots ,{\displaystyle \frac{n-3}{2}},n=odd\end{array},\end{array}$$

(25) $$\begin{array}{r}\begin{array}{l}T\left(A\right)=A\\ n+1=\left[{\displaystyle \frac{\ln A}{\ln 2}}\right]\end{array}\}⟺\{\begin{array}{l}{b}_1=-1\\ b_{n-l}=-b_{l+1}\\ l=1,2,3,\dots ,{\displaystyle \frac{n-2}{2}},n=even\\ l=1,2,3,\dots ,{\displaystyle \frac{n-3}{2}},n=odd\end{array}.\end{array}$$

Proof. 1. If $N=2^n-3$ we get

$N=2^n-3=(2^n-1)-2=(2^{n-1}+2^{n-2}+2^{n-2}+\dots +2^1+1)-2=2^{n-1}+2^{n-2}+2^{n-2}+\dots +2^1-1$.

Therefore $N$ has left symmetry. Thus, from (20), we obtain the $T\left(N\right)=1$.

We now prove the converse by assuming that $T\left(N\right)=1$. First we prove that $N$ cannot have the right symmetry;

$N=2^{n+1}+2^n+b_{n-1}{2}^{n-1}+b_{n-2}{2}^{n-2}+\dots +b_2{2}^2+b_1{2}^1+1$.

In this case we have

$T\left(N\right)=1$

or equivalently

$2^{n+1}\cdot ({\displaystyle \frac{1}{2^{n+1}}}+{\displaystyle \frac{1}{2^n}}+{\displaystyle \frac{b_{n-1}}{2^{n-1}}}+\dots +{\displaystyle \frac{b_2}{2^2}}+{\displaystyle \frac{b_1}{2^1}}+1)=1$

or equivalently

$1+2^1+b_{n-1}{2}^2+\dots +b_2{2}^{n-1}+b_1{2}^n+2^{n+1}=1$

or equivalently

$2^1+b_{n-1}{2}^2+\dots +b_2{2}^{n-1}+b_1{2}^n+2^{n+1}=0$

which is impossible, since $b_k=\pm 1$, $k=1,2,3,\dots ,n-1$.

We now consider the case where $N\ne 1$ has left symmetry. We have

(26) $$\begin{array}{r}N=2^{n+1}+2^n+b_{n-1}{2}^{n-1}+b_{n-2}{2}^{n-2}+\dots +b_2{2}^2+b_1{2}^1-1,\end{array}$$

where $n+1=\left[\frac{\ln N}{\ln 2}\right]$. From this Equation and (20) we obtain,

$T\left(N\right)=-1-2^1-b_{n-1}{2}^2-\dots -b_2{2}^n+2^{n+1}$.

Starting from this Equation we get the equivalent equations

$\begin{array}{l}T\left(N\right)=1\\ -1-2^1-b_{n-1}{2}^2-\dots -b_2{2}^n+2^{n+1}=1\\ -1-2^1-b_{n-1}{2}^2-\dots -b_2{2}^n+2^{n+1}+2^{n+2}=2^{n+2}+1\\ -1-2^1-b_{n-1}{2}^2-\dots -b_2{2}^n+2^{n+1}+2^{n+2}=2^{n+2}+2^n-2^{n-1}-2^{n-2}-\dots -2^2-2^1-1\end{array}$.

As the representation in (1) is unique, we have

$b_2=b_3=b_3=\dots =b_{n-1}=+1$.

Therefore, (26) implies

$N=2^{n+1}+2^n+2^{n-1}\dots +2^1-1=2(2^n+2^{n-1}+2^{n-2}+\dots +2^1+1)-1=2(2^{n+1}-1)-1=2^{n+2}-3$.

By substituting $n+2⟶n$ we get $N=2^n-3$.

2. We have

(27) $$\begin{array}{r}B=2^{n+1}+2^n+b_{n-1}{2}^{n-1}+b_{n-2}{2}^{n-2}+\dots +b_1{2}^1+1,\end{array}$$

where $n+1=\left[\frac{\ln B}{\ln 2}\right]$. From (18) we have

(28) $$\begin{array}{r}T\left(B\right)=1+2+b_{n-1}{2}^2+b_{n-2}{2}^3+\dots +b_2{2}^n+2^{n+1}.\end{array}$$

From (27) and (28) we get $T\left(B\right)=B$ if and only if

$1+2+b_{n-1}{2}^2+b_{n-2}{2}^3+\dots +b_2{2}^n+2^{n+1}=2^{n+1}+2^n+b_{n-1}{2}^{n-1}+b_{n-2}{2}^{n-2}+\dots +b_2{2}^2+b_1{2}^1+1$.

From this equation we obtain (24). (25) is proved to be similar. ■

Theorem 3. [1] 1. For the numbers $B$ with right symmetry we have

(29) $$\begin{array}{r}T\left(B\right)-T\left(B^{\ast }\right)=6.\end{array}$$

2. For the numbers $A$ with left symmetry we have

(30) $$\begin{array}{r}T\left(A\right)-T\left(A^{\ast }\right)=-6.\end{array}$$

Proof. From (27), we get,

(31) $$\begin{array}{r}{B}^{\ast }=2^{n+1}+2^n-b_{n-1}{2}^{n-1}-b_{n-2}{2}^{n-2}-\dots -b_1{2}^1-1=A.\end{array}$$

From (20) we get

(32) $$\begin{array}{r}T\left(B^{\ast }\right)=-1-2+b_{n-1}{2}^2+b_{n-2}{2}^3+\dots +b_1{2}^n+2^{n+1}.\end{array}$$

From (28) and (32) we obtain the $T\left(B\right)-T\left(B^{\ast }\right)=6$. (30) has been proven to be similar. ■

Theorem 4. [1] For the odd number $N$, $n+1=\left[\frac{\ln N}{\ln 2}\right]$, $N\in {\mathrm{\Omega }}_n=[2^{n+1},2^{n+2}]$, we have

(33) $$\begin{array}{r}T\left(N\right)<2^{n+2}.\end{array}$$

Proof. If $N=\mathrm{\Delta }$, from (18), and considering $b_k=\pm 1,k=0,1,2,\dots ,n-1$, we obtain

$T\left(\mathrm{\Delta }\right)=1+2+b_{n-1}{2}^2+b_{n-2}{2}^3+\dots +b_2{2}^n+2^{n+1}\le 1+2+2^2+\dots +2^n+2^{n+1}=2^{n+2}-1$

or equivalent

$T\left(\mathrm{\Delta }\right)\le 2^{n+2}-1<2^{n+2}$.

The proof is similar in the case where $N$ is of form $\text{A }$. ■

Inequality (33) provides an upper bound for the transpose $T\left(N\right)$ of $N\in {\mathrm{\Omega }}_n$. Equation (23) shows that the transpose of an odd number can take values that are much smaller than this upper bound. Below we will see Theorems, Propositions and equations that confirm this property of the transpose.

Theorem 5. [1] For the numbers $B$ and $A$ of the interval ${\mathrm{\Omega }}_n=[2^{n+1},2^{n+2}]$, $n+1=\left[\frac{\ln B}{\ln 2}\right]=\left[\frac{\ln A}{\ln 2}\right]$, and $n\ge 4$, we have

(34) $$\begin{array}{r}T(B-2)+T\left(B\right)=T\left(A\right)+T(A+2)=2^{n+2}.\end{array}$$

Proof. The smallest odd number $B$ with right symmetry in ${\mathrm{\Omega }}_n=[2^{n+1},2^{n+2}]$ is $B_{min}=2^{n+1}+3$. So for $B>3$, if $B\in {\mathrm{\Omega }}_n$ then $(B-2)\in {\mathrm{\Omega }}_n$. In ${\mathrm{\Omega }}_n$ we have $A_{min}=2^n+1$. So for $A>3$, if $A\in {\mathrm{\Omega }}_n$ then $(A+2)\in {\mathrm{\Omega }}_n$. From Definition 3, we have that

$A=2^{n+1}+2^n+b_{n-1}{2}^{n-1}+b_{n-2}{2}^{n-2}+b_{n-3}{2}^{n-3}+\dots +b_2{2}^2-2-1$

and

$A+2=2^{n+1}+2^n+b_{n-1}{2}^{n-1}+b_{n-2}{2}^{n-2}+b_{n-3}{2}^{n-3}+\dots +b_2{2}^2-2+1$.

From these Equations and (22) we obtain

$T\left(A\right)+T(A+2)=2^{n+2}$.

Equation $T(B-2)+T\left(B\right)=2^{n+2}$ proves to be similar. ■

Theorem 6. [1] For the numbers $B$ and $A$ of the interval ${\mathrm{\Omega }}_n=[2^{n+1},2^{n+2}]$, $n+1=\left[\frac{\ln B}{\ln 2}\right]=\left[\frac{\ln A}{\ln 2}\right]$, and $n\ge 4$, we have

(35) $$\begin{array}{r}T\left(B\right)-T(B-4)=T\left(A\right)-T(A+4)=2^{n+1}.\end{array}$$

Proof. We prove equation $T\left(A\right)-T(A+4)=2^{n+1}$. The equation $T\left(B\right)-T(B-4)=2^{n+1}$ proved to be similar. For odd-number $A$ with left symmetry,

$A=2^{n+1}+2^n+b_{n-1}{2}^{n-1}+b_{n-2}{2}^{n-2}+b_{n-3}{2}^{n-3}+\dots +b_2{2}^2-2-1$

we get

$A+4=2^{n+1}+2^n+b_{n-1}{2}^{n-1}+b_{n-2}{2}^{n-2}+b_{n-3}{2}^{n-3}+\dots +b_2{2}^2+2-1$.

From these Equations and (22) we get

$T\left(A\right)-T(A+4)=2^{n+1}$.

In (34) and (35) we can eliminate $T\left(A\right)$ and $T\left(B\right)$, respectively. Thus, we obtain two equations that relate the transposes of the asymmetric numbers $Y_1=B-2$, $W_1=B-4$ and $W_2=A+2$, $Y_2=A+4$. ■

Corollary 4. [1] For the numbers $B$ and $A$ of the interval ${\mathrm{\Omega }}_n=[2^{n+1},2^{n+2}]$, $n+1=\left[\frac{\ln B}{\ln 2}\right]=\left[\frac{\ln A}{\ln 2}\right]$, and $n\ge 4$, we have

(36) $$\begin{array}{r}T(B-2)+T(B-4)=T(A+2)+T(A+4)=2^{n+1}.\end{array}$$

Theorem 7. [1] For even number $2^m\cdot N$, where $N$ is an odd number and $m\in \mathbb{N}$, we have

$T(2^m\cdot N)=T\left(N\right)$.

Proof. If $N=\mathrm{\Delta }$,

$\mathrm{\Delta }=2^{n+1}+2^n+b_{n-1}{2}^{n-1}+b_{n-2}{2}^{n-2}+\dots +b_1{2}^1+1$, where $n+1=\left[\frac{\ln \mathrm{\Delta }}{\ln 2}\right]$,

we have

$2^m\cdot \mathrm{\Delta }=2^{m+n+1}+2^{m+n}+b_{n-1}{2}^{m+n-1}+b_{n-2}{2}^{m+n-2}+\dots +b_1{2}^{m+1}+2^m$

and from (18) we get

$\begin{array}{l}T(2^m\cdot \mathrm{\Delta })=({\displaystyle \frac{1}{2^{m+n+1}}}+{\displaystyle \frac{1}{2^{m+n}}}+{\displaystyle \frac{b_{n-1}}{2^{m+n-1}}}+{\displaystyle \frac{b_{n-2}}{2^{m+n-2}}}+\dots +{\displaystyle \frac{b_1}{2^{m+1}}}+{\displaystyle \frac{1}{2^m}})\cdot 2^{m+n+1}\\ =({\displaystyle \frac{1}{2^{n+1}}}+{\displaystyle \frac{1}{2^n}}+{\displaystyle \frac{b_{n-1}}{2^{n-1}}}+{\displaystyle \frac{b_{n-2}}{2^{n-2}}}+\dots +{\displaystyle \frac{b_1}{2^1}}+{\displaystyle \frac{1}{2^0}})\cdot 2^{n+1}=T\left(\mathrm{\Delta }\right)\end{array}$.

The proof is similar in the case where $N$ is of form $\text{A }$.

The concept of transpose divides odd numbers into symmetric and asymmetric. ■

Definition 5. [1] Odd number $N$ are categorized as symmetric if

(37) $$\begin{array}{r}T\left(T\left(N\right)\right)=N,\end{array}$$

and as asymmetric if

(38) $$\begin{array}{r}T\left(T\left(N\right)\right)\ne N.\end{array}$$

Definition 4 provides the categorization of odd numbers, depending on their form $A$, $W$, $Y$, or $B$.

Corollary 5. [1] Numbers of the form $A$ and $B$ are symmetric.

2. Numbers of the form $W$ and $Y$ are asymmetric.

Corollary 6. [1] The transposes of $A$, $W$, $Y$, and $B$ have the following forms.

1. $T\left(A\right)=A_1$.

2. $T\left(W\right)=B$.

3. $T\left(Y\right)=A$.

4. $T\left(B\right)=B_1$.

Theorem 8. [1] If an asymmetric odd number belongs to the interval ${\mathrm{\Omega }}_n$, then its transpose belongs to the interval ${\mathrm{\Omega }}_m$, $m<n$.

Proof. Theorem 8 follows from the combination of Proposition 3 and (22). ■

Equation (29) is proved for $b_0=+1$, which is true not only for the numbers $B$ but also for the numbers $W$. (30) is proved for $b_0=-1$, which is true not only for the numbers $A$ but also for the numbers $Y$. Thus, the following Corollary completes Theorem 3.

Corollary 7. [1]

1. We have

$T\left(W\right)-T\left(W^{\ast }\right)=6$.

2. We have

$T\left(Y\right)-T\left(Y^{\ast }\right)=-6$.

The following corollaries arise from the theorems and propositions that we have proven. They give possible forms of the product of odd numbers, taking into account their categorization.

Corollary 8. [1] The product of two odd numbers has one of the following forms.

1. $A\cdot Y_1=Y$.

2. $A\cdot W=W$.

3. $B\cdot Y=W$.

4. $B\cdot W=Y$.

5. $Y\cdot W=B$.

6. $W_1\cdot W_2=A$.

7. $Y_1\cdot Y_2=A$.

Corollary 9. [1] The possible forms of the factors of (composite) odd numbers, depending on their form.

1. $A=A_1{A}_2$.

2. $A=B_1{B}_2$.

3. $A=W_1{W}_2$.

4. $A=Y_1{Y}_2$.

5. $W=A_1{W}_1$.

6. $W=B_1{Y}_1$.

7. $Y=A_1{Y}_1$.

8. $Y=B_1{W}_1$.

9. $B=A_1{B}_1$.

10. $B=Y_1{W}_1$.

Corollary 10. [1] The product of an odd number with its transpose has the following forms.

1. $A\cdot T\left(A\right)=A_1$.

2. $W\cdot T\left(W\right)=Y$.

3. $Y\cdot T\left(Y\right)=Y_1$.

4. $B\cdot T\left(B\right)=A$.

The following is a consequence of Proposition 2.

Corollary 11. [1] Complementary numbers $N$ and $N^{\prime }$ of the interval ${\mathrm{\Omega }}_n$, with $n\ge 3$, are of the same form $A$, $W$, $Y$or $B$.

For the transpose of odd numbers the following applies.

Proposition 5. [1]

A. 1. For $A\in {\mathrm{\Omega }}_n$ we have

$T\left(A\right)=T(3\cdot 2^{n+1}-A)-6$.

2. For $Y\in {\mathrm{\Omega }}_n$ we have

$T\left(Y\right)=T(3\cdot 2^{n+1}-Y)-6$.

B. 1. For $B\in {\mathrm{\Omega }}_n$ we have

$T\left(B\right)=T(3\cdot 2^{n+1}-B)+6$.

2. For $W\in {\mathrm{\Omega }}_n$ we have

$T\left(W\right)=T(3\cdot 2^{n+1}-W)+6$.

Proof. This results from a combination of Theorem 3 and Proposition 1. Thus, we proved that A. 1. Combining (30) and (14), we obtain;

$T(3\cdot 2^{n+1}-A)-T\left(A\right)=6$

or equivalently

$T\left(A\right)=T(3\cdot 2^{n+1}-A)-6$. ■

Theorem 9. [1]

1. For numbers $\mathrm{\Delta }=2^{n+1}+2^n+b_{n-1}{2}^{n-1}+b_{n-2}{2}^{n-2}+b_{n-3}{2}^{n-3}+\dots +b_2{2}^2+b_1{2}^1+1$ we have

(39) $$\begin{array}{r}T\left(\mathrm{\Delta }\right)-T(\mathrm{\Delta }-b_k{2}^{k+1})=2^{n+2-k}\cdot b_k,\end{array}$$

where $k=1,2,3,\dots ,n-1$.

2. For numbers $\text{A }=2^{n+1}+2^n+b_{n-1}{2}^{n-1}+b_{n-2}{2}^{n-2}+b_{n-3}{2}^{n-3}+\dots +b_2{2}^2+b_1{2}^1-1$ we have

(40) $$\begin{array}{r}T(\text{A }-b_k{2}^{k+1})-T\left(\text{A }\right)=2^{n+2-k}\cdot b_k\end{array}$$

where $k=1,2,3,\dots ,n-1$.

Proof. From equation

$\mathrm{\Delta }=2^{n+1}+2^n+b_{n-1}{2}^{n-1}+b_{n-2}{2}^{n-2}+b_{n-3}{2}^{n-3}+\dots +b_2{2}^2+b_1{2}^1+1$

we get

$\mathrm{\Delta }-b_k{2}^{k+1}=2^{n+1}+2^n+b_{n-1}{2}^{n-1}+b_{n-2}{2}^{n-2}+b_{n-3}{2}^{n-3}+\dots -b_k{2}^k+\dots +b_2{2}^2+b_1{2}^1+1$.

From these Equations we get respectively

$T\left(\mathrm{\Delta }\right)=2^{n+1}(1+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{b_{n-1}}{2^{n-1}}}+{\displaystyle \frac{b_{n-2}}{2^{n-2}}}+{\displaystyle \frac{b_{n-3}}{2^{n-3}}}+\dots +{\displaystyle \frac{b_k}{2^k}}+\dots +{\displaystyle \frac{b_2}{2^2}}+{\displaystyle \frac{b_1}{2^1}}+1)$,

$T(\mathrm{\Delta }-b_k{2}^{k+1})=2^{n+1}(1+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{b_{n-1}}{2^{n-1}}}+{\displaystyle \frac{b_{n-2}}{2^{n-2}}}+{\displaystyle \frac{b_{n-3}}{2^{n-3}}}+\dots -{\displaystyle \frac{b_k}{2^k}}+\dots +{\displaystyle \frac{b_2}{2^2}}+{\displaystyle \frac{b_1}{2^1}}+1)$

and finally we get $T\left(\mathrm{\Delta }\right)-T(\mathrm{\Delta }-b_k{2}^{k+1})=2^{n+2-k}\cdot b_k$.

2 of Theorem is proven similarly. ■

5. Consequences of Theorem 9

There are “simple” transposes in which both parts are a function of $n=3,4,5,\dots$ and small odd numbers. Here is a list of ten such simple transposes.

[1] List of simple transposes.

$\begin{array}{l}A=2^n+1\stackrel{T}{⟷}{2}^{n+1}-7\\ A=2^n-7\stackrel{T}{⟷}{2}^{n-1}+1\\ W=2^n+3\stackrel{T}{⟶}7\\ W=2^n-5\stackrel{T}{⟶}{2}^{n-1}-1\\ W=3\cdot (2^n+1)\stackrel{T}{⟶}15\\ Y=2^n-3\stackrel{T}{⟶}1\\ Y=2^n+5\stackrel{T}{⟶}{2}^n-7\\ Y=5\cdot (2^n+1)\stackrel{T}{⟶}{2}^{n+2}-23\end{array}$

$\begin{array}{l}B=2^n+7\stackrel{T}{⟷}{2}^n+7\\ B=7\cdot (2^n+1)\stackrel{T}{⟷}{2}^{n+2}+31\end{array}$

The list in [1] shows the relationship between the two transposes, which is not apparent in (22). For example, For the asymmetric $Y=2^n+5$ we have,

$2^n+5-T(2^n+5)=2^n+5-(2^n-7)=5-(-7)=12$.

The value 12 is the minimum difference between the two transposes. In addition, $Y=2^n+5$ and $T\left(Y\right)$ belong to the consecutive intervals ${\mathrm{\Omega }}_n$,

$Y\in {\mathrm{\Omega }}_{n-1}=[2^n,2^{n+1}]$ and $T\left(Y\right)\in {\mathrm{\Omega }}_{n-2}=[2^{n-1},2^n]$.

The smallest asymmetric number of the form $Y$ of the interval ${\mathrm{\Omega }}_{n-1}$ transposes the largest symmetric number of the form $A$ of the immediately preceding interval ${\mathrm{\Omega }}_{n-2}$. In no other case, there are two transposes that are very close to each other.

We prove two transposes in the list. The technique we follow is applicable to computing all simple transposes. First we prove the following.

$2^n-7\stackrel{T}{⟶}{2}^{n-1}+1$

For $2^n-7$ we have the following sequence of equations

$2^n-7=4\cdot 2^{n-2}-7$

$2^n-7=3\cdot 2^{n-2}+2^{n-2}-7$

$2^n-7=2^{n-1}+2^{n-2}+2^{n-2}-1-2-2^2$

$2^n-7=2^{n-1}+2^{n-2}+(2^{n-2}-1)-2-2^2$

$2^n-7=2^{n-1}+2^{n-2}+(2^{n-3}+2^{n-4}+2^{n-5}+\dots +1)-2-2^2$

$2^n-7=2^{n-1}+2^{n-2}+2^{n-3}+2^{n-4}+2^{n-5}+\dots +2^2+2+1-2-2^2$

$2^n-7=2^{n-1}+2^{n-2}+2^{n-3}+2^{n-4}+2^{n-5}+\dots +2^2-2^1-1$

and taking into account the (20) we get

$T(2^n-7)=-(1+2+2^2+\dots +2^{n-3}-2^{n-2}-2^{n-1})=2^{n-1}+1$.

We now prove the following.

$2^n+7\stackrel{T}{⟶}{2}^n+7$

For $2^n+7$ we have the following sequence of equations

$2^n+7=2^n+1+2^2+2$

$2^n+7=2^n+(2^{n-1}-2^{n-2}-2^{n-3}-2^{n-4}-\dots -2^2-2-1)+2^2+2$

$2^n+7=2^n+2^{n-1}-2^{n-2}-2^{n-3}-2^{n-4}-\dots -2^2-2-1+2^2+2$

$2^n+7=2^n+2^{n-1}-2^{n-2}-2^{n-3}-2^{n-4}-\dots -2^2+2+1$

and taking into account the (18) we get

$T(2^n+7)=2^n+7$.

The main utility of simple transposes arises from their combination with Theorem 9. From this combination we obtained a set of structural equations for the transpose. The equations resulting from five such combinations are given below;

For

$\text{A }=2^n-7$, $n\ge 4$,

we have

$\text{A }=2^n-7=2^{n-1}+2^{n-2}+2^{n-3}+2^{n-4}+2^{n-5}+\dots +2^2-2-1$.

Thus from (40) we get

$T(2^n-7-2^{k+1})-T(2^n-7)=2^{(n-2)+2-k}$

and with the equation $T(2^n-7)=2^{n-1}+1$ we get

(41) $$\begin{array}{r}T(2^n-7-2^{k+1})=2^{n-k}+2^{n-1}+1,k=2,3,4,\dots ,n-3,n\ge 4.\end{array}$$

We have $b_1=-1$. Thus from (40) and $T(2^n-7)=2^{n-1}+1$ we get

$T(2^n-7+2^2)-(2^{n-1}+1)=-2^{(n-2)+2-1}$

or equivalently $T(2^n-3)=1$.

Working similarly, from

$\mathrm{\Delta }=2^n+7=2^n+2^{n-1}-2^{n-2}-2^{n-3}-2^{n-4}-\dots -2^2+2+1$, $n\ge 4$,

$\mathrm{\Delta }=2^{n+2}-1=2^{n+1}+2^n+2^{n-1}+\dots +2^2+2+1$, $n\ge 4$,

$\text{A }=2^n-3=2^{n-1}+2^{n-2}+2^{n-3}+\dots +2^2+2-1$, $n\ge 4$,

$\mathrm{\Delta }=2^n-5=2^{n-1}+2^{n-2}+2^{n-3}+\dots +2^2-2+1$, $n\ge 4$,

we obtain the following equations.

(42) $$\begin{array}{r}T(2^n+7+2^{k+1})=2^n+7+2^{n+1-k},k=2,3,4,\dots ,n-2,n\ge 4,\end{array}$$

(43) $$\begin{array}{r}T(2^n-1-2^{k+1})=2^n-1-2^{n-k},k=1,2,3,\dots ,n-3,n\ge 4,\end{array}$$

(44) $$\begin{array}{r}T(2^n-3-2^{k+1})=2^{n-k}+1,k=1,2,3,\dots ,n-3,n\ge 4,\end{array}$$

(45) $$\begin{array}{r}T(2^n-5-2^{k+1})=2^{n-1}-1-2^{n-k},k=2,3,4,\dots ,n-3,n\ge 4,\end{array}$$

From (41)–(45), by providing appropriate values of $n$ and $k$, we obtain an extremely large amount of information about the structure of the transpose of an odd number. Similar equations also arise from the combination of other simple transposes with Theorem 9. We give an example.

Example 6.

By substituting the maximum allowable value of $k$ into (41)–(45) we obtain;

$\begin{array}{l}T(2^n-7-2^{n-2})=2^{n-1}+9\\ T(2^n+7+2^{n-1})=2^n+15\\ T(2^n-1-2^{n-2})=2^n-9\\ T(2^n-3-2^{n-2})=9\\ T(2^n-5-2^{n-2})=2^{n-1}-9\end{array}$

These Equations are valid for any $n\ge 4$.

By substituting the minimum common allowable value $k=2$ into (41)–(45), we obtain the following:

$\begin{array}{l}T(2^n-15)=2^{n-2}+2^{n-1}+1\\ T(2^n+15)=2^n+7+2^{n-1}\\ T(2^n-9)=2^n-1-2^{n-2}\\ T(2^n-11)=2^{n-2}+1\\ T(2^n-13)=2^{n-1}-1-2^{n-2}\end{array}$

These Equations are valid for any $n\ge 4$.

The above equations are confirmed by (22). The difference between (41)–(45) and (22) is that their two parts have a structure, that is not apparent if (22) is used.

Definition and Basic Properties of Octets

The alternation transpose/conjugate or conjugate/transpose provides a mathematical object consisting of eight odd numbers. Thus we arrive at the following definition.

Definition 6. [1] We define the following transpose/conjugate alternation as the octet of odd number $N$,

(46) $$\begin{array}{r}\{N,T\left(N\right),{\left(T\left(N\right)\right)}^{\ast },T\left({\left(T\left(N\right)\right)}^{\ast }\right),N^{\ast },T\left(N^{\ast }\right),{\left(T\left(N^{\ast }\right)\right)}^{\ast },T\left({\left(T\left(N^{\ast }\right)\right)}^{\ast }\right)\}.\end{array}$$

From (11) it follows that the conjugate numbers belong to the same binary interval ${\mathrm{\Omega }}_n$. From Definition 6 if $N$ is a symmetric number, then $N$ and $T\left(N\right)$ belong to the same binary interval ${\mathrm{\Omega }}_n$. From Theorem 8 it follows that if $N$ is an asymmetric number and belongs to the binary interval ${\mathrm{\Omega }}_n$, then $T\left(N\right)$ belongs to the previous binary interval. Thus we arrive at the following definition.

Definition 7. [1] 1. We define the octet whose numbers belong to the same binary interval as symmetric.

2. We define as asymmetric the octet which has at least two numbers belonging to different binary intervals.

We now present the symbolic representation of an octet. For the conjugates $N$ and $N^{\ast }$ we use the symbolism $N\stackrel{\ast }{⟷}{N}^{\ast }$. With $N_1\stackrel{T}{⟷}{N}_2$ we denote that $T\left(N_1\right)=N_2$ and $T\left(N_2\right)=N_1$. This case concerns symmetric numbers $N_1$ and $N_2$. If $T\left(N_1\right)=N_2$ and $T\left(N_2\right)\ne N_1$, we write $N_1\stackrel{T}{⟶}{N}_2$. In this case $N_1$ is an asymmetric number and $N_2$ is symmetric. We apply the representation of an octet to the next example.

Example 7. For $N=649$, from Definition 6 we obtain

{649, 745, 791, 655, 887, 751, 785, 881}.

All octet numbers belong to the binary interval ${\mathrm{\Omega }}_8$. Therefore we obtain a symmetric octet. Here, we provide representation of the octet.

$\begin{array}{clllllllc}& 649& \stackrel{T}{⟷}& 745& \stackrel{\ast }{⟷}& 791& \stackrel{T}{⟷}& 655& \\ & {}^{\ast }\updownarrow & & & & & & {\updownarrow }^{\ast }& \\ & 887& \stackrel{T}{⟷}& 751& \stackrel{\ast }{⟷}& 785& \stackrel{T}{⟷}& 881& \end{array}$

In Fig. 1 the numbers of the octet are placed at the vertices of a regular octagon. The illustration shows the process of transition from one number to another within the symmetrical octet.

Fig. 1. A typical symmetrical octet.

From (14) and (18), (20) we obtain the following representations.

(47) $$\begin{array}{l}{2}^n+1\stackrel{\ast }{⟷}{2}^{n+1}-1\\ 2^n+7\stackrel{\ast }{⟷}{2}^{n+1}-7\\ 2^n+1\stackrel{T}{⟷}{2}^{n+1}-7\\ 2^n+7\stackrel{T}{⟷}{2}^n+7\\ 2^{n+1}-1\stackrel{T}{⟷}{2}^{n+1}-1\end{array}$$

Considering (47) we get the symmetric octets

(48) $$\begin{array}{r}\{2^{n+1}+1,2^{n+2}-7,2^{n+1}+7,2^{n+1}+7,2^{n+2}-1,2^{n+2}-1,2^{n+1}+1,2^{n+2}-7\},\end{array}$$

where $n=3,4,5,\dots$.

From (47), it follows that the octet is symmetric. Putting $n+1=2^S$, $S\in \mathbb{N}$ or $n+2=P$ where $P$ is a prime number, it follows that the Fermat and Mersenne numbers belong to the octet (48). In addition, notice that this octet consisted of four different numbers.

We now investigate the relationship between asymmetric numbers to octets. Two conjugate numbers, symmetric and asymmetric, belong to the same binary interval. The same applies to transposes of symmetric numbers. In contrast, if an asymmetric number belongs to a binary interval, its transpose belongs to a previous binary interval. Furthermore, the transpose was symmetric. Consequently, an asymmetric number produces the octet to which its transpose belongs. We give an example.

Example 8. For $Y=11965$ and $W=11971$, we use $T\left(11965\right)=649$ and $T\left(11971\right)=887$. Furthermore, it is $W=Y+6$; therefore, $W=11971$ and $Y=11965$ produce the same symmetric octet (it is a consequence of Corollary 7). The schematic representation is as follows.

$\begin{array}{lllllll}11965& & & & & & \\ {\downarrow }^T& & & & & & \\ 649& \stackrel{T}{⟷}& 745& \stackrel{\ast }{⟷}& 791& \stackrel{T}{⟷}& 655\\ {}^{\ast }\updownarrow & & & & & & {\updownarrow }^{\ast }\\ 887& \stackrel{T}{⟷}& 751& \stackrel{\ast }{⟷}& 785& \stackrel{T}{⟷}& 881\\ {\uparrow }^T& & & & & & \\ 11971& & & & & & \end{array}$

The different association between two octets results from 4 of Proposition 2. For its representation we have the following notation for the symbolism of complementary $N$ and $N^{\prime }$.

$N\stackrel{\prime }{⟷}{N}^{\prime }$

With this symbolism, 4 of Proposition 2 is written as follows.

$\begin{array}{lll}{N}_1& \stackrel{\prime }{⟷}& N_2\\ {\updownarrow }^{\ast }& & {\updownarrow }^{\ast }\\ N_1^{\ast }& \stackrel{\prime }{⟷}& N_2^{\ast }\end{array}$

Each of the two conjugate pairs belonged to a different symmetric octet. Thus, for symmetric numbers, we obtain the following schematic representation.

$\begin{array}{lllll}\stackrel{T}{⟷}& N_1& \stackrel{\prime }{⟷}& N_2& \stackrel{T}{⟷}\\ & {\updownarrow }^{\ast }& & & {\updownarrow }^{\ast }\\ \stackrel{T}{⟷}& N_1^{\ast }& \stackrel{\prime }{⟷}& N_2^{\ast }& \stackrel{T}{⟷}\end{array}$

We give an example.

Example 9.

If $N_1=145$ we get the following schematic representation.

$\begin{array}{lllllllllllllll}151& \stackrel{T}{⟷}& 167& \stackrel{\ast }{⟷}& 217& \stackrel{T}{⟷}& 145& \stackrel{\prime }{⟷}& 209& \stackrel{T}{⟷}& 209& \stackrel{\ast }{⟷}& 175& \stackrel{T}{⟷}& 215\\ {\updownarrow }^{\ast }& & & & & & {\updownarrow }^{\ast }& & {\updownarrow }^{\ast }& & & & & & {\updownarrow }^{\ast }\\ 233& \stackrel{T}{⟷}& 162& \stackrel{\ast }{⟷}& 223& \stackrel{T}{⟷}& 239& \stackrel{\prime }{⟷}& 175& \stackrel{T}{⟷}& 215& \stackrel{\ast }{⟷}& 169& \stackrel{T}{⟷}& 169\end{array}$

From Proposition 3 we get the following.

Corollary 12.

The complementary numbers of the intervals ${\mathrm{\Omega }}_n$, $n=4,5,6,\dots$ have the same form $A$, $W$, $Y$ or $B$.

From Corollary 12, it follows that for asymmetric numbers of the intervals ${\mathrm{\Omega }}_n$and, $n=4,5,6,\dots$, we have the following representation.

$\begin{array}{lllll}\stackrel{T}{\leftarrow }& N_1& \stackrel{\prime }{⟷}& N_2& \stackrel{T}{⟶}\\ & {\updownarrow }^{\ast }& & {\updownarrow }^{\ast }& \\ \stackrel{T}{\leftarrow }& N_1^{\ast }& \stackrel{\prime }{⟷}& N_2^{\ast }& \stackrel{T}{⟶}\end{array}$

In ${\mathrm{\Omega }}_1$, ${\mathrm{\Omega }}_2$ and ${\mathrm{\Omega }}_3$ complementary numbers do not necessarily have the same form.

Analysis of the Octets into Quadruples and Pairs of Odd Numbers

In this section we address the question of how many different numbers a symmetric octet can have. There are symmetric octets consisting of eight different numbers

$\begin{array}{lllllll}137& \stackrel{T}{⟷}& 185& \stackrel{\ast }{⟷}& 199& \stackrel{T}{⟷}& 143\\ {}^{\ast }\updownarrow & & & & & & {\updownarrow }^{\ast }\\ 247& \stackrel{T}{⟷}& 191& \stackrel{\ast }{⟷}& 193& \stackrel{T}{⟷}& 241\end{array}$,

symmetric octets consisting of 4 different numbers

$\begin{array}{lllllll}41& \stackrel{T}{⟷}& 41& \stackrel{\ast }{⟷}& 55& \stackrel{T}{⟷}& 47\\ {}^{\ast }\updownarrow & & & & & & {\updownarrow }^{\ast }\\ 55& \stackrel{T}{⟷}& 47& \stackrel{\ast }{⟷}& 49& \stackrel{T}{⟷}& 49\end{array}$,

and symmetric octets consisting of 2 different numbers

$\begin{array}{lllllll}9& \stackrel{T}{⟷}& 9& \stackrel{\ast }{⟷}& 15& \stackrel{T}{⟷}& 15\\ {}^{\ast }\updownarrow & & & & & & {\updownarrow }^{\ast }\\ 15& \stackrel{T}{⟷}& 15& \stackrel{\ast }{⟷}& 9& \stackrel{T}{⟷}& 9\end{array}$.

As a consequence of the equations $1^{\ast }=1$, $T\left(1\right)=1$ and $3^{\ast }=3$, $T\left(3\right)=3$, {1, 1, 1, 1, 1, 1, 1, 1} and {3, 3, 3, 3, 3, 3, 3, 3} are the only octets consisting of the same number.

The structure of a symmetric quadruple is determined by Theorem 3,

(49) $$\begin{array}{r}\{A,A+6,A^{\ast }-6,A^{\ast }\}.\end{array}$$

In general, a symmetric octet consists of two symmetric quadruples,

(50) $$\begin{array}{r}\{A_1,A_1+6,A^{\ast }-6,A_1^{\ast },A_2,A_2+6,A_2^{\ast }-6,A_2^{\ast }\}.\end{array}$$

The differences between the two homologous numbers of a quadruple are the same, and we define this as the distance between the two quadruples. Here we give the distance $D$ from the difference in the numbers $A_2$ and $A_1$ of octet (50),

(51) $$\begin{array}{r}D=|A_2-A_1|.\end{array}$$

It is obvious that distance $D$ can also be defined for quadruples that do not belong to the same octet. An octet has eight or four different numbers for $D\ne 0$ or $D=0$ respectively.

There are cases where an octet consists of two different numbers, that is, it consists of a pair of numbers. The pairs are given by the following representation:

$\{Y=3\cdot 2^{n+1}-3,Y+6=3\cdot 2^{n+1}+3,Y^{\ast }-6=3\cdot 2^{n+1}-3,Y^{\ast }=3\cdot 2^{n+1}+3\}$, $n=2,3,4,\dots$,

or equivalently

(52) $$\begin{array}{r}\{3\cdot 2^{n+1}-3,3\cdot 2^{n+1}+3\},n=2,3,4,\dots .\end{array}$$

For pairs we have the following.

Proposition 6. [1] 1. The pair of ${\mathrm{\Omega }}_2$ is

$\{9,15\}$.

2. The pair of ${\mathrm{\Omega }}_{n+1}$, is

$\{Y,W\}=\{3\cdot 2^{n+1}-3,3\cdot 2^{n+1}+3\}$, $n\ge 2$.

3. If $n\ge 2$ the pair of ${\mathrm{\Omega }}_{n+1}$ always produces $\{9,15\}$.

Proof. 1. If $n=1$, from (52) we get

$\{3\cdot 2^2-3,3\cdot 2^2+3\}=\{9,15\}$.

2. We have

$(3\cdot 2^{n+1}-3)-5=3\cdot 2^{n+1}-8=8\cdot (2^{n-2}-1)=8m$,

therefore $3\cdot 2^{n+1}-3$ is of the form $Y$. Similarly we have

$(3\cdot 2^{n+1}+3)-3=3\cdot 2^{n+1}=8\cdot 3\cdot 2^{n-2}=8m$,

so $3\cdot 2^{n+1}+3$ is of the form $W$.

3. We have

$Y=3\cdot 2^{n+1}-3=2^{n+2}+2^{n+1}-2^n+2^{n-1}+2^{n-2}+2^{n-3}+\dots +2^2+2^1+1$,

and from (22) we get $T\left(Y\right)=9$. Similarly we get $T\left(W\right)=15$.

From 2 of Definition 7 it follows that the following quadruples are asymmetric. ■

$\{Y,Y+4,Y^{\ast }-4,Y^{\ast }\}$, $\{W,W+4,W^{\ast }-4,W^{\ast }\}$

Numbers $Y+4$ and $W^{\ast }-4$ are in the form $A$. $Y^{\ast }-4$ and $W+4$ are of form $B$. For the asymmetric numbers $Y$ and $W$ we have,

$Y\stackrel{T}{⟶}A$

and

$W\stackrel{T}{⟶}B$.

We now prove the following.

Proposition 7. [1]

1. We have

(53) $$\begin{array}{r}Y=2^m\cdot (A+3)-3\stackrel{T}{⟶}T\left(A\right)\stackrel{T}{⟷}A.\end{array}$$

if and only if

(54) $$\begin{array}{r}W=2^m\cdot (B-3)+3\stackrel{T}{⟶}T\left(B\right)\stackrel{T}{⟷}B,\end{array}$$

where $m=0,1,2,\dots$.

2. We have

(55) $$\begin{array}{r}{Y}^{\prime }=2^m\cdot (B-3)-3\stackrel{T}{⟶}{\left(T\left(B\right)\right)}^{\ast }\stackrel{\ast }{⟷}T\left(B\right).\end{array}$$

if and only if

(56) $$\begin{array}{r}{W}^{\prime }=2^m\cdot (A+3)+3\stackrel{T}{⟶}{\left(T\left(A\right)\right)}^{\ast }\stackrel{\ast }{⟷}T\left(A\right),\end{array}$$

where $m=0,1,2,\dots$.

Proof. We prove 1. Substituting $B=A^{\ast }$ in (54) we have

$W=2^m\cdot (A^{\ast }-3)+3\stackrel{T}{⟶}T\left(A^{\ast }\right)\stackrel{T}{⟷}{A}^{\ast }$,

or equivalently, changing the symbolism,

(57) $$\begin{array}{r}T(2^m\cdot (A^{\ast }-3)+3)=T\left(A^{\ast }\right).\end{array}$$

Now, considering (6) when $A$ is in the interval ${\mathrm{\Omega }}_n$, $Y$ belongs to the interval ${\mathrm{\Omega }}_{n+m}$. Thus, from Proposition 5, we obtain

$T\left(Y\right)=T(3\cdot 2^{m+n+1}-Y)-6$.

From this Equation we obtain the following successive equivalent equations.

$T\left(Y\right)=T(3\cdot 2^{m+n+1}-2^m\cdot (A+3)+3)-6$,

$T\left(Y\right)=T(2^m\cdot (3\cdot 2^{n+1}-A-3)+3)-6$,

$T\left(Y\right)=T(2^m\cdot (A^{\ast }-3)+3)-6$.

From this Equation and (57) we have

$T\left(Y\right)=T\left(A^{\ast }\right)-6$

and in schematic representation we get,

$Y\stackrel{T}{⟶}T\left(A^{\ast }\right)-6$

and taking into account (30), $T\left(A^{\ast }\right)-6=T\left(A\right)$, we get,

$Y\stackrel{T}{⟶}T\left(A^{\ast }\right)-6\stackrel{T}{⟷}T\left(A\right)\stackrel{T}{⟷}A$.

Similarly, from (53) we obtain (54). The proofs of the equivalence of (55) and (56) are similar. ■

In the applications of Proposition 7, all four equations have been shown to be true. Proposition 7 is a step towards completing the proof. Assuming that the Equations of Proposition 7 are valid, we obtain the following corollary.

Corollary 13. [1] For every symmetric number of the binary interval ${\mathrm{\Omega }}_n=[2^{n+1},2^{n+2}]$, $n\in \mathbb{N}$, there are infinite asymmetric numbers, which belong to the intervals ${\mathrm{\Omega }}_m$, $m=n+1,n+2,n+3,\dots$, whose transpose is equal to the symmetric number.

Example 10.

Considering that $T\left(745\right)=649$ and $T\left(751\right)=887$ we get the following.

$\begin{array}{l}{2}^m\cdot (745+3)-3\stackrel{T}{⟶}T\left(745\right)=649\\ 2^m\cdot (751-3)+3\stackrel{T}{⟶}T\left(751\right)=887\end{array}$

These Equations are valid for all $m\in \mathbb{N}$. In Example 8 we apply these Equations if $m=4$.

Discussion

From Theorem (1), a set of mathematical concepts emerges that provide a new framework for the study of Natural Numbers. Octets play a central role in the framework. Further investigation of their properties is necessary to highlight the contribution of the theory we have presented to the Number Theory.

Conflict of Interest

Author declares no conflict of interest.