<s>[INST] Sum of $\lfloor k^{1/3} \rfloor$ I am faced with the following sum:
$$\sum_{k=0}^m \lfloor k^{1/3} \rfloor$$
Where $m$ is a positive integer. I have determined a formula for the last couple of terms such that $\lfloor n^{1/3} \rfloor^3 = \lfloor m^{1/3} \rfloor^3$. For example if the sum is from 0 to 11 I can find the sum of the terms 8 through 11. I am stuck on what formula can be applied to find the sum of the terms before the final stretch however. 
 [/INST] Note: This answer provides a solution for the general case $\lfloor \sqrt[p]{k}\rfloor$ with $p\geq 1$.

The following is valid for $m\geq 1, p\geq 1$
\begin{align*}
\sum_{k=1}^{m}\lfloor \sqrt[p]{k}\rfloor
=m\lfloor \sqrt[p]{m}\rfloor^{p+1}
-\frac{1}{p+1}\sum_{j=0}^{p}\binom{p+1}{j}B_j\lfloor \sqrt[p]{m}\rfloor^{p+1-j}+m\lfloor \sqrt[p]{m}\rfloor\tag{1}
\end{align*}
with $B_j$ the $j$-th Bernoulli Number.
The special cases $p=1,2,3$ give
\begin{align*}
\sum_{k=1}^m\lfloor k\rfloor&=\frac{1}{2}m^2+\frac{1}{2}m\\
\sum_{k=1}^m\lfloor \sqrt{k}\rfloor&=
m\lfloor\sqrt{m}\rfloor
-\frac{1}{3}\lfloor\sqrt{m}\rfloor^3-\frac{1}{2}\lfloor\sqrt{m}\rfloor^2+\frac{5}{6}\lfloor\sqrt{m}\rfloor\\
\sum_{k=1}^m\lfloor \sqrt[3]{k}\rfloor&=m\lfloor\sqrt[3]{m}\rfloor-\frac{1}{4}\lfloor\sqrt[3]{m}\rfloor^4
-\frac{1}{2}\lfloor\sqrt[3]{m}\rfloor^3-\frac{1}{4}\lfloor\sqrt[3]{m}\rfloor^2+\lfloor\sqrt[3]{m}\rfloor\tag{2}\\
\end{align*}

For convenient calculations we consider two aspects:


*

*We introduce a variable $a=\lfloor\sqrt[p]{m}\rfloor$. So, we have
$$a\leq \sqrt[p]{m} < a+1$$

*We use the Iverson Bracket notation, so we can replace the expression $\lfloor x\rfloor$ by
$$\lfloor x\rfloor=\sum_{j\geq 0}[1\leq j \leq x]$$

Special case: $m=a^p,a=\lfloor\sqrt[p]{m}\rfloor$
We start the calculation by conveniently assuming $m=a^p$. We obtain
  \begin{align*}
\sum_{k=1}^{m}\lfloor\sqrt[p]{k}\rfloor&=\sum_{k=1}^m\sum_{j\geq 0}[1\leq j \leq \sqrt[p]{k}][0\leq k \leq a^p]\\
&=\sum_{j=1}^{a}\sum_{k=1}^{m}[j^p\leq k \leq a^p]\\
&=\sum_{j=1}^{a}(a^p-j^p+1)\\
&=a^{p+1}-\sum_{j=1}^{a}j^p+a\tag{3}\\
&=a^{p+1}-\frac{1}{p+1}\sum_{j=0}^{p}\binom{p+1}{j}B_j\,a^{p+1-j}+a\tag{4}
\end{align*}

Comment:


*

*In (3) we use the known representation of the sum of $p$-th powers of natural numbers by the Bernoulli Numbers
$$\sum_{k=1}^{n}k^{p}=\frac{1}{p+1}\sum_{j=0}^{p}\binom{p+1}{j}B_j\,n^{p+1-j}$$

General case: $m\geq a^p,a=\lfloor\sqrt[p]{m}\rfloor$
In the general case we let again $a=\lfloor\sqrt[p]{m}\rfloor$ and have additionally to consider the terms for 
  $a^p< k \leq m$.
  They are all equal to $a$, so they sum up to $$(m-a^p)a.$$ Adding this to (4) we get the general formula with $a=\lfloor \sqrt[p]{m}\rfloor$
\begin{align*}
\color{blue}{\sum_{k=1}^{m}\lfloor\sqrt[p]{k}\rfloor}&\color{blue}{=ma-\frac{1}{p+1}\sum_{j=0}^{p}\binom{p+1}{j}B_j\,a^{p+1-j}+a}\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Box
\end{align*}



Let's finally compute the special case $p=3$. We need the Bernoulli numbers $B_0=1, B_1=\frac{1}{2}, B_2=\frac{1}{6}$ and $B_3=0$.
\begin{align*}
\sum_{k=1}^{m}\lfloor\sqrt[3]{k}\rfloor&=na-\frac{1}{4}\sum_{j=0}^{3}\binom{4}{j}B_j\,a^{4-j}+a\\
&=na-\frac{1}{4}\left[\binom{4}{0}B_0a^4+\binom{4}{1}B_1a^3+\binom{4}{2}B_2a^2\right]+a\\
&=na-\frac{1}{4}a^4-\frac{1}{2}a^3-\frac{1}{4}a^2+a\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Box
\end{align*}
The other cases ($p=1,2$) from the claim (2) follow similarly.

Note: This approach ($p=2$) can be found in  Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik.
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