ArafuraOpal
Registered Member
$$\sum_{n=1}^{40} \left( \sum_{m=1}^n \left( \frac {a \sin \left( {3^{( m+(n^2-n)/2)}} \theta / b + l(\frac {2 \pi}{3}) \right) }{3^{(m+(n^2-n)/2)}} \right) \right) , l=-1,0,1 $$
$$\sum_{n=1}^8 \left( \sum_{m=1}^n \left( \frac {a \left( \sin \left( {3^{( m+(n^2-n)/2)}} \theta / b + k(\frac { \pi}{3} ) + \cos \left( {3^{( m+(n^2-n)/2)}} \theta / b + k(\frac { \pi}{3} ) \right) \right) }{2 \times 3^{(m+(n^2-n)/2)}} \right) \right) , k=-1,1$$
$$\sum_{n=1}^8 \left( \sum_{m=1}^n \left( \frac {a \left( \sin \left( {3^{( m+(n^2-n)/2)}} \theta / b + k(\frac { \pi}{3} ) + \cos \left( {3^{( m+(n^2-n)/2)}} \theta / b + k(\frac { \pi}{3} ) \right) \right) }{2 \times 3^{(m+(n^2-n)/2)}} \right) \right) , k=-1,1$$