# Radix-2\({}^{2}\) Algorithm for the Odd New Mersenne Number Transform (ONMNT)
## 4 Derivation of Radix-\(2^{2}\) Algorithm for IONMNT
 \tag{48}\]  
Let us define the two following terms:  
\[\sum_{k=0}^{\frac{N}{4}-1}X_{o}\bigg{(}k+\lambda\frac{N}{4}\bigg{)} \beta_{1}\bigg{(}\frac{2k+1}{2}l\bigg{)}\beta\bigg{(}-4n\frac{2k+1}{2}\bigg{)} \\ =\sum_{k=0}^{\frac{N}{4}-1}X_{o}\bigg{(}\lambda\frac{N}{4}-k-1 \bigg{)}\beta_{1}\bigg{(}\frac{2k+1}{2}l\bigg{)}\beta\bigg{(}4n\frac{2k+1}{2} \bigg{)}, \tag{49}\]  
\[\sum_{k=0}^{\frac{N}{4}-1}X_{o}\bigg{(}k+\lambda\frac{N}{4}\bigg{)} \beta_{2}\bigg{(}\frac{2k+1}{2}l\bigg{)}\beta\bigg{(}-4n\frac{2k+1}{2}\bigg{)} \\ =-\sum_{k=0}^{\frac{N}{4}-1}X_{o}\bigg{(}\lambda\frac{N}{4}-k-1 \bigg{)}\beta_{2}\bigg{(}\frac{2k+1}{2}l\bigg{)}\beta\bigg{(}4n\frac{2k+1}{2} \bigg{)}. \tag{50}\]Substituting the relations (49) and (50) into (48) leads to  
\[ \begin{split} x(4n+l)=\sum_{\lambda=0}^{3}&\bigg{[} \bigg{\{}\sum_{k=0}^{\frac{N}{4}-1}X_{o}\bigg{(}\lambda\frac{N}{4}+k\bigg{)} \beta_{1}\bigg{(}\frac{2k+1}{2}l\bigg{)}\beta\bigg{(}4n\frac{2k+1}{2}\bigg{)} \beta_{1}\bigg{(}\lambda l\frac{N}{4}\bigg{)}\bigg{\}}\\ &+\bigg{\{}\sum_{k=0}^{\frac{N}{4}-1}X_{o}\bigg{(}\lambda\frac{N}{4 }-k-1\bigg{)}\beta_{1}\bigg{(}\frac{2k+1}{2}l\bigg{)}\beta\bigg{(}4n\frac{2k+1}{ 2}\bigg{)}\beta_{2}\bigg{(}\lambda l\frac{N}{4}\bigg{)}\bigg{\}}\\ &-\bigg{\{}\sum_{k=0}^{\frac{N}{4}-1}X_{o}\bigg{(}\lambda\frac{N}{4 }-k-1\bigg{)}\beta_{2}\bigg{(}\frac{2k+1}{2}l\bigg{)}\beta\bigg{(}4n\frac{2k+1}{ 2}\bigg{)}\beta_{1}\bigg{(}\lambda l\frac{N}{4}\bigg{)}\bigg{\}}\\ &-\bigg{\{}\sum_{k=0}^{\frac{N}{4}-1}X_{o}\bigg{(}\lambda\frac{N} {4}+k\bigg{)}\beta_{2}\bigg{(}\frac{2k+1}{2}l\bigg{)}\beta\bigg{(}4n\frac{2k+1 }{2}\bigg{)}\beta_{2}\bigg{(}\lambda l\frac{N}{4}\bigg{)}\bigg{\}}\bigg{]}. \end{split}  \tag{51}\]  
Upon reorganizing the terms and isolating the common summation term \(\sum_{k=0}^{\frac{N}{4}-1}(\cdot)\) and \(\beta\bigg{(}4n\frac{2k+1}{2}\bigg{)}\) from Equation (51), the result is as follows:  
\[ \begin{split} x(4n+l)=\sum_{k=0}^{\frac{N}{4}-1}\bigg{[}\sum_{ \lambda=0}^{3}&\bigg{[}\bigg{\{}X_{o}\bigg{(}k+\lambda\frac{N}{4 }\bigg{)}\beta_{1}\bigg{(}\lambda l\frac{N}{4}\bigg{)}\\ &+X_{o}\bigg{(}\lambda\frac{N}{4}-k-1\bigg{)}\beta_{2}\bigg{(} \lambda l\frac{N}{4}\bigg{)}\bigg{\}}\beta_{1}\bigg{(}l\frac{2k+1}{2}\bigg{)} \\ &-\bigg{\{}X_{o}\bigg{(}\lambda\frac{N}{4}-k-1\bigg{)}\beta_{1} \bigg{(}\lambda l\frac{N}{4}\bigg{)}\\ &+X_{o}\bigg{(}k+\lambda\frac{N}{4}\bigg{)}\beta_{2}\bigg{(} \lambda l\frac{N}{4}\bigg{)}\beta_{2}\bigg{(}l\frac{2k+1}{2}\bigg{)}\bigg{\}} \bigg{]}\bigg{]}\beta\bigg{(}4n\frac{2k+1}{2}\bigg{)}.\end{split}  \tag{52}\]  
Equation (52) can be rewritten as follows:  
\[x(4n+l)=\sum_{k=0}^{\frac{N}{4}-1}y(l,k)\beta\bigg{(}4n\frac{2k+1}{2}\bigg{)} \tag{53}\]  
where  
\[ \begin{split} y(l,k)=\sum_{\lambda=0}^{3}&\bigg{[} \bigg{\{}X_{o}\bigg{(}k+\lambda\frac{N}{4}\bigg{)}\beta_{1}\bigg{(}\lambda l \frac{N}{4}\bigg{)}+X_{o}\bigg{(}\lambda\frac{N}{4}-k-1\bigg{)}\beta_{2}\bigg{(} \lambda l\frac{N}{4}\bigg{)}\bigg{\}}\beta_{1}\bigg{(}l\frac{2k+1}{2}\bigg{)} \\ &-\bigg{\{}X_{o}\bigg{(}\lambda\frac{N}{4}-k-1\bigg{)}\beta_{1} \bigg{(}\lambda l\frac{N}{4}\bigg{)}+X_{o}\bigg{(}k+\lambda\frac{N}{4}\bigg{)} \beta_{2}\bigg{(}\lambda l\frac{N}{4}\bigg{)}\bigg{\}}\beta_{2}\bigg{(}l\frac{ 2k+1}{2}\bigg{)}\bigg{]}.\end{split}