It is the sum of known sequence

Number Theory Level pending

Let $f(n) = \left \lfloor \frac{n}{\lfloor \sqrt{n} \rfloor } \right \rfloor \text { for } n = \{ 1,2,3,\ldots, 1000 \}$

Determine the sum of all $i$ 's which satisfy $f(i+1) < f(i)$ .

Notation: $\lfloor \cdot \rfloor$ denotes the floor function .

1 solution

Ossama Ismail
Mar 8, 2017

$f(i+1) < f(i)$ , when $(i+1)$ is a perfect square.

or when $i = ( 3,8,15, \cdots , 960 )$

Answer $= 3 +8 + 15 + \cdots + 960 = \sum_{n=1}^{31} n^2 - 32 = ((31)(31+1)(2*31 +1)/6) -32 = 10385$

Note :: $\sum n^2 = n(n+1)(2n+1)/6$