I am new to math stack exchange and it is my first question. I know how to solve discrete double summation without floor function.
Main problem : $$\sum_{x=1}^{\lfloor(n-1)/7\rfloor} \sum_{y=1}^{\lfloor(n-x)/(6x-1)\rfloor}(6x-1)(6y+1) \qquad, \{n \in \mathbb{Z^+} \}$$ After simplification to single summation, it comes out to be: $$\sum_{x=1}^{\lfloor(n-1)/7\rfloor} \sum_{y=1}^{\lfloor(n-x)/(6x-1)\rfloor}(6x-1)(6y+1)$$ $$=\sum_{x=1}^{\lfloor(n-1)/7\rfloor}(6x-1) \sum_{y=1}^{\lfloor(n-x)/(6x-1)\rfloor}(6y+1)$$ $$=\sum_{x=1}^{\lfloor(n-1)/7\rfloor}(6x-1)\left(3\left\lfloor\frac{(n-x)}{(6x-1)}\right\rfloor\left(\left\lfloor\frac{(n-x)}{(6x-1)}\right\rfloor+1\right)+\left\lfloor\frac{(n-x)}{(6x-1)}\right\rfloor\right)$$ $$=\sum_{x=1}^{\lfloor(n-1)/7\rfloor}(6x-1)\left(3\left\lfloor\frac{n-x}{6x-1}\right\rfloor^2\ +\ 4\left\lfloor\frac{n-x}{6x-1}\right\rfloor\right)$$ But after this I am stuck. Discrete Math is college level math and I am a high school student.(not brilliant as other students) So i don't know where to find materials for discrete sums with floor functions.