Is there a closed-form expression for $\sum_{k=1}^{n}\lfloor k^{q} \rfloor$ for $q \in \mathbb{Q}_{> 0}$? There are well-known closed-form expressions for summations such as $\sum_{k=1}^{n}\lfloor k^{\frac{1}{2}} \rfloor$, $\sum_{k=1}^{n}\lfloor k^{\frac{1}{3}} \rfloor$, $\sum_{k=1}^{n}\lfloor k^{\frac{1}{4}} \rfloor$, etc. For example, we have that $$\sum_{k=1}^{n}\lfloor k^{\frac{1}{3}} \rfloor = -\frac{1}{4} \left\lfloor\sqrt[3]{n}\right\rfloor \left( \left\lfloor\sqrt[3]{n}\right\rfloor^{3} + 2 \left\lfloor\sqrt[3]{n}\right\rfloor^{2} + \left\lfloor\sqrt[3]{n}\right\rfloor - 4(n+1)  \right)$$ for all $n \in \mathbb{N}$. 
However, Mathematica is unable to evaluate the sum $\sum_{k=1}^{n}\lfloor k^{\frac{2}{3}} \rfloor$. Furthermore, there is no closed-form expression for this summation given in the OEIS sequence http://oeis.org/A032514 corresponding to this sum. 
More generally, Mathematica is not able to evaluate summations such as $\sum_{k=1}^{n}\lfloor k^{\frac{4}{3}} \rfloor$, $\sum_{k=1}^{n}\lfloor k^{\frac{3}{4}} \rfloor$, $\sum_{k=1}^{n}\lfloor k^{\frac{3}{7}} \rfloor$, etc. Letting $q \in \mathbb{Q}$ be positive, it appears that there is a known closed-form expression for $\sum_{k=1}^{n}\lfloor k^{q} \rfloor$ if and only if $q \in \mathbb{N}$ or $q$ is of the form $q = \frac{1}{r}$ where $r \in \mathbb{N}$. So it is natural to ask:
(1) Is there a closed-form expression for $\sum_{k=1}^{n}\lfloor k^{\frac{2}{3}} \rfloor$?
(2) More generally, is there a closed-form expression for $\sum_{k=1}^{n}\lfloor k^{q} \rfloor$ for $q \in \mathbb{Q}_{> 0} \setminus \mathbb{N} \setminus \left\{ \frac{1}{2}, \frac{1}{3}, \ldots \right\}$?
 A: It's unlikely that there is a closed form expression for what you want.  Consider the expression
$$
\sum_{k=1}^{n}\left\lfloor f(k)\right\rfloor,
$$
where $f(k)$ is a monotonically increasing function with $f(1)=1$.  Let $g(m)$ be the smallest integer value of $k$ such that $f(k)\ge m$; i.e., $g(m)=\left\lceil f^{-1}(m)\right\rceil$. Then the terms in the sum are at least as large as $1$ starting at $k = g(1)$, and at least as large as $2$ starting at $k= g(2)$, and so on.  As long as $g(m)\le n$, each term contributes $n-g(m)+1$ to the sum.  So 
$$
\sum_{k=1}^{n}\left\lfloor f(k)\right\rfloor = \sum_{m=1}^{\lfloor f(n)\rfloor} \left(n + 1 - \left\lceil f^{-1}(m)\right\rceil\right)=(n+1)\lfloor f(n)\rfloor - \sum_{m=1}^{\lfloor f(n)\rfloor} \left\lceil f^{-1}(m)\right\rceil.
$$
This will simplify in certain cases where the inverse is integer-valued (so the ceiling function goes away) and "nice" (so the resulting sum can be evaluated).  Two standard examples are when the inverse is a polynomial (e.g., $f(k)=k^{1/q}$, so $f^{-1}(m)=m^q$) and when the inverse is an exponential function (e.g., $f(k)=1+\log_b k$, so $f^{-1}(m)=b^{m-1}$).  In your case you have
$$
\sum_{k=1}^{n}\left\lfloor k^{2/3} \right\rfloor=(n+1)\left\lfloor n^{2/3}\right\rfloor
 - \sum_{m=1}^{\lfloor n^{2/3}\rfloor} \left\lceil m^{3/2}\right\rceil,
$$
but this sum doesn't seem any easier to evaluate.
