Floor function

Let $x=m+k$ , with $m$ a positive integer and $0 \le k < 1$ .

Then, $\left \lfloor{x^2 \cdot \left \lfloor{x}\right \rfloor}\right \rfloor =\left \lfloor{(m^2+k^2+2mk)(m)} \right \rfloor = m^3+ \left \lfloor{ 2km^2+mk^2 } \right \rfloor$

Let $f(m) =2km^2+mk^2$ be a parabula with $k$ a parameter. $k=0$ is $y=0$ . Reminding that the coefficient $2k$ fixes the concavity of the conic, let $k$ increase such that the parabula becomes slowly "tighter". Now, focus our attention on $x=2$ , for instance, and define $A (2,y)$ the point of intersection between the parabula and $x=2$ . Then, increasing $k$ , the ordinate of $A$ assumes all values among 0 and a "limit" value given from $k=1$ .

If $k=1$ we get $f(m) =2m^2+m$ : this is the function that gives us a "limit" interval for every $m$ . In fact, we are able to get all possible value $\epsilon$ from the set: $\epsilon \in \{m^3+0, m^3+1, ..., m^3 + 2m^2+m-1 \}$ exploiting the fact that $k$ is a real number ( $m^3 + 2m^2+m$ can't be obtained since $k<1$ from hypothesis).

Every set is made of $2m^2+m$ elements. Since $12^3 < 2017 < 13^3$ , the answer to our problem is: $\sum_{m=1} ^{12} (2m^2+m)-10 = 2 \cdot \frac{12 \cdot 13 \cdot 25}{6}+\frac{12 \cdot 13}{2}-10 = \boxed{1368}$

$-10$ since the last 10 number are greater than 2017.