## 3 Use of Special Type Moduli
\begin{pmatrix}((x_{3,n-1}\lor x_{3,n})\ldots (x_{3,0}\lor x_{3,n})||\bar{x}_{3,n-1}\ldots\bar{x}_{3,0}+2^{n})+\\ \hskip 14.226378 pt+\big{(}1-2^{n+1}\big{)}\end{pmatrix} \right|_{2^{2n}-1} \tag{14}\]  
or  
\[\left|x_{3}(2^{n}-1)2^{-1}\right|_{2^{2n}-1}=\left|2^{-1}\left( \begin{pmatrix} x_{3,n-1}\ldots x_{3,0}||\bar{x}_{3,n-1}\ldots\bar{x}_{3,1}||(\overline{x _{3,0}\lor x_{3,n}}))+\\ \hskip 14.226378 pt+\left(\underbrace{0\ldots 0}_{n-1}||\bar{x}_{3,n}||\underbrace{0 \ldots 0}_{n-1}||x_{3,n}||0\right)+\\ \hskip 14.226378 pt+\big{(}1-2^{n+1}\big{)}\end{pmatrix} \right|_{2^{2n}-1} \tag{15}\]  
Depending on the chosen representation of the third term (14) or (15), the article [8] forms two variants of the reverse conversion constructed using the formula:  
\[X_{h}=|w_{1}+w_{2}+w_{3}+C_{k}|_{2^{2n}-1} \tag{16}\]In the first case, based on the Formula (14) we have:  
\[w_{1} =\left(\bar{x}_{1,n-1}\ldots\bar{x}_{1,0}||\underbrace{(0\ldots 0)}_{n}\right)\] \[w_{2} =(x_{2,0}||x_{2,n-1}\ldots x_{2,0}||x_{2,n-1}\ldots x_{2,1})\] \[w_{3} =(\bar{x}_{3,0}||(x_{3,n-1}\lor x_{3,n})\ldots(x_{3,0}\lor x_{3,n} )||\bar{x}_{3,n-1}\ldots\bar{x}_{3,1})\] \[C_{k} =\left|\left(2^{-n}-1\right)+2^{-1}\left(1-2^{n+1}\right)+2^{n-1 }\right|_{2^{2n}-1}=2^{2n-1}+2^{n-1}-1=\left(1\underbrace{0\ldots 0}_{n} \underbrace{1\ldots 1}_{n-1}\right)\]  
In the article [8], there is a typo that \(C_{k}\) has \(n+2\) zeros. However, it is obvious that \(2^{2n-1}\) contains one and \(2n-1\) zeros, and the number (\(2^{n-1}-1\)) contains \(n-1\) one. Then adding the numbers will result in the lowest number \(n-1\) being a one, and the remaining \(2n-1-(n-1)=n\) zeros will remain zeros. For example, for \(n=2\): \(2^{3}+2-1=9=1001_{2}\). For the second case, the formula (15) gives:  
\[w_{1} =\left(\bar{x}_{1,n-1}\ldots\bar{x}_{1,0}||\bar{x}_{3,n}|| \underbrace{(0\ldots 0)}_{n-2}||x_{3,n}\right)\] \[w_{2} =(x_{2,0}||x_{2,n-1}\ldots x_{2,0}||x_{2,n-1}\ldots x_{2,1})\] \[w_{3} =\left((\overline{x}_{3,0}\lor x_{3,n})||x_{3,n-1}\ldots x_{3,0} ||\bar{x}_{3,n-1}\ldots\bar{x}_{3,1}\right)\] \[C_{k} =\left|\left(2^{-n}-1\right)+2^{-1}\left(1-2^{n+1}\right)\right| _{2^{2n}-1}=2^{2n-1}-1=\left(0\underbrace{1\ldots 1}_{2n-1}\right)\]  
The addition of \(C_{k}\) in the calculation of \(X_{h}\) by the Formula (16) is equivalent to the addition of \(2^{n}C_{k}\) to \(X_{23}\). Moreover, the latter can also be obtained by simultaneous addition of \(c_{2}=|2^{n}C_{k}|_{2^{n}-1}=|C_{k}|_{2^{n}-1}\) to \(x_{2}\) modulo \(2^{n}-1\) and \(c_{3}=|2^{n}C_{k}|_{2^{n}+1}=|-C_{k}|_{2^{n}+1}\) to \(x_{3}\) modulo \(2^{n}+1\) (leaving the remainder \(x_{1}\) of even modulo unchanged, since \(|2^{n}C_{k}|_{2^{n}}=0\)). In the first case we obtain:  
\[c_{2} =\left|2^{n}\left(2^{2n-1}+2^{n-1}-1\right)\right|_{2^{n}-1}=0,\] \[w_{1} =\left(\bar{x}_{1,n-1}\ldots\bar{x}_{1,0}||\underbrace{(0\ldots 0)}_{n}\right)\]0  
In the second case:  
\[w_{1} =\left(\bar{x}_{1,n-1}\ldots\bar{x}_{1,0}||\underbrace{(0\ldots 0)}_{n}\right)\]1 \[w_{1} =\left(\bar{x}_{1,n-1}\ldots\bar{x}_{1,0}||\underbrace{(0\ldots 0)}_{n}\right)\]2  
Then the Formula (16) will be reduced to:  
\[w_{1} =\left(\bar{x}_{1,n-1}\ldots\bar{x}_{1,0}||\underbrace{(0\ldots 0)}_{n}\right)\]3  
with further use of the Formula (12). Consider a similar example with RNS {8,7,9}, \(n=3\), for the number \(X=10=\) (2,3,1). In the first case, the coefficients \(c_{2}=0\) and \(c_{3}=1\) do not depend on the value of \(n\). Then the offset value of \(X^{\prime}=(2,3,|1+1|_{9})\). From this follows  
\[w_{1} =\left(\bar{x}_{1,n-1}\ldots\bar{x}_{1,0}||\underbrace{(0\ldots 0)}_{n}\right)\]4 \[w_{1} =\left(\bar{x}_{1,n-1}\ldots\bar{x}_{1,0}||\underbrace{(0\ldots 0)}_{n}\right)\]5 \[w_{1} =\left(\bar{x}_{1,n-1}\ldots\bar{x}_{1,0}||\underbrace{(0\ldots 0)}_{n}\right)\]6  
Then \(X_{h}=|40+45+42|_{63}=1\), and \(X=2+2^{3}\cdot 1=10\). For the second case for \(n=3\)\(c_{2}=2^{2}-1=3\) and \(c_{3}=2^{3}+1=5\). Then the offset value of \(X^{\prime}=(2,|3+3|_{7},|1+5|_{9})\).