Floor function experimenting

Ferramentarius

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

When [itex] k = 1, 2, 3, ... [/itex] :

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

(2) [itex] \left\lfloor \sqrt{2k} + \frac{1}{2} \right\rfloor = 1, \underbrace{2, 2}_{}, \underbrace{3, 3, 3}_{}, \underbrace{4, 4, 4, 4}_{} ... [/itex]
(3) [itex] \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}_{}, ... [/itex]

(4) [itex] \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}_{}, ... [/itex]
(5) [itex] \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}_{}, ... [/itex]

(6) [itex] 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}_{}, ... [/itex]

(7) [itex] 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}_{}, ... [/itex]

(8) [itex] 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}_{}, ... [/itex]
(9) [itex] 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}_{}, ... [/itex]

In particular when presenting the series product [itex](\sum_{j=1}^n a_j)(\sum_{j=1}^n b_j)[/itex] as the sum of the terms of the nxn matrix [itex] \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} [/itex] 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 [itex] (\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}) [/itex].