User:Irismus/SandboxPage

abc

 * Full formula: $$\tfrac{1}{3}(6 \times \lfloor \textbf{Quas} \times \tfrac{20}{3} \rfloor) + \tfrac{2}{3}\big( (6 \times \lfloor \textbf{Quas} \times \tfrac{20}{3} \rfloor) \times \sum_{n=1}^{i}(0.98^n) \big), \quad i = \lfloor \textbf{Quas} \times \tfrac{20}{9} \rfloor$$


 * Knockback distance from previous edit: $$(\lfloor Quas \times \tfrac{20}{9} \rfloor \times 6) + \big( 6 \times \sum_{n=1}^{i}(0.98^n) \big), i = \lfloor Quas \times \tfrac{40}{9} - \lfloor Quas \times \tfrac{20}{9} \rfloor \rfloor$$ [?]


 * C(3+3-1,3)=C(5,3)=10