# Radix-2\({}^{2}\) Algorithm for the Odd New Mersenne Number Transform (ONMNT)
## 4 Derivation of Radix-\(2^{2}\) Algorithm for IONMNT
)\):  
\[\beta\bigg{\{}\bigg{(}\frac{2k+1}{2}l+\lambda l\frac{N}{4}\bigg{)} \bigg{(}4n\frac{2k+1}{2}+\lambda nN\bigg{)}\bigg{\}}\\ =\] \[+\bigg{\{}\beta_{1}\bigg{(}\frac{2k+1}{2}l\bigg{)}\beta_{2}\bigg{(} \lambda l\frac{N}{4}\bigg{)}+\beta_{2}\bigg{(}\frac{2k+1}{2}l\bigg{)}\beta_{1 }\bigg{(}\lambda l\frac{N}{4}\bigg{)}\bigg{\}}\beta\bigg{(}-4n\frac{2k+1}{2} \bigg{)}\] \[= \beta_{1}\bigg{(}\frac{2k+1}{2}l\bigg{)}\beta_{1}\bigg{(}\lambda l \frac{N}{4}\bigg{)}\beta\bigg{(}4n\frac{2k+1}{2}\bigg{)} \tag{45}\] \[-\beta_{2}\bigg{(}\frac{2k+1}{2}l\bigg{)}\beta_{2}\bigg{(}\lambda l \frac{N}{4}\bigg{)}\beta\bigg{(}4n\frac{2k+1}{2}\bigg{)}\] \[\beta\bigg{\{}\bigg{(}\frac{2k+1}{2}l+\lambda l\frac{N}{4}\bigg{)} \bigg{(}4n\frac{2k+1}{2}+\lambda nN\bigg{)}\bigg{\}}\\ =\beta_{1}\bigg{(}\frac{2k+1}{2}l+\lambda l\frac{N}{4}\bigg{)} \beta\bigg{(}4n\frac{2k+1}{2}\bigg{)}+\beta_{2}\bigg{(}\frac{2k+1}{2}l+\lambda l \frac{N}{4}\bigg{)}\beta\bigg{(}-4n\frac{2k+1}{2}\bigg{)} \tag{40}\]0 \[\beta\bigg{\{}\bigg{(}\frac{2k+1}{2}l+\lambda l\frac{N}{4}\bigg{)} \bigg{(}4n\frac{2k+1}{2}+\lambda nN\bigg{)}\bigg{\}}\\ =\beta_{1}\bigg{(}\frac{2k+1}{2}l+\lambda l\frac{N}{4}\bigg{)} \beta\bigg{(}4n\frac{2k+1}{2}\bigg{)}+\beta_{2}\bigg{(}\frac{2k+1}{2}l+\lambda l \frac{N}{4}\bigg{)}\beta\bigg{(}-4n\frac{2k+1}{2}\bigg{)} \tag{40}\]1  
Rearranging and simplifying (45) results in the following:  
\[\beta\bigg{\{}\bigg{(}\frac{2k+1}{2}l+\lambda l\frac{N}{4}\bigg{)} \bigg{(}4n\frac{2k+1}{2}+\lambda nN\bigg{)}\bigg{\}}\\ =\beta_{1}\bigg{(}\frac{2k+1}{2}l+\lambda l\frac{N}{4}\bigg{)} \beta\bigg{(}4n\frac{2k+1}{2}\bigg{)}+\beta_{2}\bigg{(}\frac{2k+1}{2}l+\lambda l \frac{N}{4}\bigg{)}\beta\bigg{(}-4n\frac{2k+1}{2}\bigg{)} \tag{40}\]2Substituting the value from (46) into (38) and ignoring \(\langle\rangle_{M_{p}}\) for the derivation,  
\[ \begin{split} x(4n+l)=&\sum_{k=0}^{\frac{N}{4}-1}\sum_ {\lambda=0}^{3}X_{o}\Big{(}k+\lambda\frac{N}{4}\Big{)}\\ &\bigg{[}\bigg{\{}\beta\bigg{(}4n\frac{2k+1}{2}\bigg{)}\beta_{1} \bigg{(}\lambda l\frac{N}{4}\bigg{)}\\ &+\beta\bigg{(}-4n\frac{2k+1}{2}\bigg{)}\beta_{2}\bigg{(}\lambda l \frac{N}{4}\bigg{)}\bigg{\}}\beta_{1}\bigg{(}\frac{2k+1}{2}l\bigg{)}\\ &+\bigg{\{}\beta\bigg{(}-4n\frac{2k+1}{2}\bigg{)}\beta_{1}\bigg{(} \lambda l\frac{N}{4}\bigg{)}\\ &-\beta\bigg{(}4n\frac{2k+1}{2}\bigg{)}\beta_{2}\bigg{(}\lambda l \frac{N}{4}\bigg{)}\bigg{\}}\beta_{2}\bigg{(}\frac{2k+1}{2}l\bigg{)}\bigg{]}. \end{split}  \tag{47}\]  
Reorganizing the equation and factoring out the shared summation term \(\sum_{\lambda=0}^{3}\) from (47) yields the following result:  
\[ \begin{split} x(4n+l)=&\sum_{\lambda=0}^{3}\biggr{[} \bigg{\{}\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{)} \beta_{1}\bigg{(}\lambda l\frac{N}{4}\bigg{)}\bigg{\}}\\ &+\bigg{\{}\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{)}\beta_{2}\bigg{(}\lambda l\frac{N}{4}\bigg{)}\bigg{\}}\\ &+\bigg{\{}\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{)}\beta_{1}\bigg{(}\lambda l\frac{N}{4}\bigg{)}\bigg{\}}\\ &\bigg{\{}\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{)}\beta_{2}\bigg{(}\lambda l\frac{N}{4}\bigg{)}\bigg{\}}\bigg{]}. \end{split}