## Floor function experimenting

Results from exploring the floor function and presented for general interest. Be good.

When $k = 1, 2, 3, ...$ :

(1) $\left\lfloor \frac{1}{n}(k-1) + 1 \right\rfloor = \underbrace{1, ..., 1}_{n}, \underbrace{2, ..., 2}_{n}, \underbrace{3, ..., 3}_{n}, ...$

(2) $\left\lfloor \sqrt{2k} + \frac{1}{2} \right\rfloor = 1, \underbrace{2, 2}_{}, \underbrace{3, 3, 3}_{}, \underbrace{4, 4, 4, 4}_{} ...$
(3) $\left\lfloor \sqrt{2k} + \frac{1}{2} \right\rfloor - \left\lfloor \sqrt{2k} - \frac{1}{2} \right\rfloor = 1, \underbrace{1, 0}_{}, \underbrace{1, 0, 0}_{}, \underbrace{1, 0, 0, 0}_{}, ...$

(4) $\left\lfloor \sqrt{k-1} + \frac{3}{2} \right\rfloor = 1, \underbrace{2, 2}_{}, \underbrace{3, 3, 3, 3}_{}, \underbrace{4, 4, 4, 4, 4, 4}_{}, ...$
(5) $\left\lfloor \sqrt{k-1} + 1 \right\rfloor = 1, \underbrace{2, 2, 2}_{}, \underbrace{3, 3, 3, 3, 3}_{}, \underbrace{4, 4, 4, 4, 4, 4, 4}_{}, ...$

(6) $k - \left\lfloor k-1 \right\rfloor ^2 = 1, \underbrace{1, 2, 3}_{}, \underbrace{1, 2, 3, 4, 5}_{}, \underbrace{1, 2, 3, 4, 5, 6, 7}_{}, ...$

(7) $k - n \left\lfloor \frac{1}{n} (k-1) \right\rfloor = \underbrace{1, 2, 3, ..., n}_{}, \underbrace{1, 2, 3, ..., n}_{}, \underbrace{1, 2, 3, ..., n}_{}, ...$

(8) $k - \frac{1}{2} \left\lfloor \sqrt{2k} - \frac{1}{2} \right\rfloor \left\lfloor \sqrt{2k} + \frac{1}{2} \right\rfloor = 1, \underbrace{1, 2}_{}, \underbrace{1, 2, 3}_{}, \underbrace{1, 2, 3, 4}_{}, ...$
(9) $1 - k + \left\lfloor \sqrt{2k} + \frac{1}{2} \right\rfloor + \frac{1}{2} \left\lfloor \sqrt{2k} + \frac{1}{2} \right\rfloor \left\lfloor \sqrt{2k} - \frac{1}{2} \right\rfloor= 1, \underbrace{2, 1}_{}, \underbrace{3, 2, 1}_{}, \underbrace{4, 3, 2, 1}_{}, ...$

In particular when presenting the series product $(\sum_{j=1}^n a_j)(\sum_{j=1}^n b_j)$ as the sum of the terms of the nxn matrix $\begin{bmatrix} a_1 b_1 & a_1 b_2 & ... & a_1 b_n \\ a_2 b_1 & a_2 b_2 & ... & ... \\ ... & ... & ... & ... \\ a_n b_1 & ... & ... & a_n b_n \end{bmatrix}$ it can be seen that the horizontal a's follow (1) while the respective b's accord to (7) above.

Judging from this we get the result $(\sum_{j=1}^n a_j)(\sum_{j=1}^n b_j) = (\sum_{j=1}^{n^2} a_{\left\lfloor \frac{1}{n}(j-1) + 1 \right\rfloor} b_{j - n \left\lfloor \frac{1}{n} (j-1) \right\rfloor})$.
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