I need to find a clear formula (without summation) for the following sum:

$$\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$$

Well, the first few elements look like this:


In general, we have $(2^2-1^2)$ $1$'s, $(3^2-2^2)$ $2$'s etc.

Still I have absolutely no idea how to generalize it for $n$ first terms...

  • 2
    $\begingroup$ Split the sum. You have one regular part for which you get a nice closed form formula, and the remaining sum from $\lfloor\sqrt{n}\rfloor^2$ to $n$. $\endgroup$ – Daniel Fischer Feb 9 '14 at 12:13
  • $\begingroup$ This is a nice problem ! $\endgroup$ – Claude Leibovici Feb 9 '14 at 14:19
  • $\begingroup$ Is this problem homework ? $\endgroup$ – Claude Leibovici Feb 9 '14 at 14:30
  • $\begingroup$ I'm preparing for my exams this week and doing old ones. Some are easy, but some I need help with. $\endgroup$ – Arek Krawczyk Feb 9 '14 at 14:31
  • $\begingroup$ Did you finish with this one ? What about your formula ? $\endgroup$ – Claude Leibovici Feb 9 '14 at 18:16


We have $$p\le\sqrt k< p+1\iff p^2\le k<(p+1)^2\Rightarrow \lfloor \sqrt{k} \rfloor=p$$ so

$$\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor=\sum\limits_{p=1}^{\lfloor \sqrt{n+1}\rfloor-1} \sum_{k=p^2}^{(p+1)^2-1}\lfloor \sqrt{k} \rfloor=\sum\limits_{p=1}^{\lfloor \sqrt{n+1}\rfloor-1}p(2p+1)$$ Now use the fact $$\sum_{k=1}^n k=n(n+1)/2$$ and $$\sum_{k=1}^n k^2=n(n+1)(2n+1)/6$$ to get the desired closed form.


The following is valid for $n\geq 1$

\begin{align*} \sum_{k=1}^{n}\lfloor\sqrt{k}\rfloor =n\lfloor\sqrt{n}\rfloor -\frac{1}{3}\lfloor\sqrt{n}\rfloor^3-\frac{1}{2}\lfloor\sqrt{n}\rfloor^2+\frac{5}{6}\lfloor\sqrt{n}\rfloor \end{align*}

For convenient calculations we consider two aspects:

  • We introduce a variable $a=\lfloor\sqrt{n}\rfloor$. So, we have $$a\leq \sqrt{n} < a+1$$
  • 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: $n=a^2,a=\lfloor\sqrt{n}\rfloor$

We start the calculation by conveniently assuming $n=a^2$. We obtain \begin{align*} \sum_{k=1}^{n}\lfloor\sqrt{k}\rfloor&=\sum_{k=1}^n\sum_{j\geq 0}[1\leq j \leq \sqrt{k}][0\leq k \leq a^2]\\ &=\sum_{j=1}^{a}\sum_{k=1}^{n}[j^2\leq k \leq a^2]\\ &=\sum_{j=1}^{a}(a^2-j^2+1)\\ &=a^3-\frac{1}{6}a(a+1)(2a+1)+a\\ &=\frac{2}{3}a^3-\frac{1}{2}a^2+\frac{5}{6}a\tag{1} \end{align*}

General case: $n\geq a^2,a=\lfloor\sqrt{n}\rfloor$

In the general case we let again $a=\lfloor\sqrt{n}\rfloor$ and have additionally to consider the terms for $a^2< k \leq n$. They are all equal to $a$, so they sum up to $$(n-a^2)a.$$ Adding this to (1) we get the general formula with $a=\lfloor \sqrt{n}\rfloor$

\begin{align*} \sum_{k=1}^{n}\lfloor\sqrt{k}\rfloor&=na-\frac{1}{3}a^3-\frac{1}{2}a^2+\frac{5}{6}a\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Box \end{align*}

Note: This approach can be found in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik

Note: I've also added here an answer for the general case $\sum_{k=1}^n\lfloor\sqrt[p]{k}\rfloor$ with $p\geq 1$.

  • 2
    $\begingroup$ .This is a very interesting approach, for sure. Thank you ! $\endgroup$ – Claude Leibovici May 10 '15 at 2:26

I am very bad in the area of discrete mathematics (as well in other) but I have been fascinated by the problem set in your post.

I am sure that Daniel Fisher's comment and Sami Ben Romdhane's answer are very useful; however, I have not been able to finish the work.

So, what I used is computer simulation and data regression in order to establish some relations. Later, RIES was used to identify the rational values of the obtained coefficients. As you see, this is a very empirical process but I hope it could help you.

I set the problem in the most general manner, loking for $$\sum\limits_{k=1}^n \lfloor k^{1/p} \rfloor$$.
What I found is that the sum starts with a first term which is $$(n+1) \left\lfloor \sqrt[p]{n}\right\rfloor$$ to which is added a polynomial (no constant term) of degree $(p+1)$ of a variable which is $$1+\left\lfloor \sqrt[p]{n}\right\rfloor$$ So, for the first successive values of $p$, I obtained after some simplifications (I am sure that more simplifications could be done) $$ (n+1) \left\lfloor \sqrt{n}\right\rfloor +\frac{1}{3} \left(-\left(\left\lfloor \sqrt{n}\right\rfloor +1\right)^3+\frac{3}{2} \left(\left\lfloor \sqrt{n}\right\rfloor +1\right)^2-\frac{1}{2} \left(\left\lfloor \sqrt{n}\right\rfloor +1\right)\right) $$ $$ (n+1) \left\lfloor \sqrt[3]{n}\right\rfloor +\frac{1}{4} \left(-\left(\left\lfloor \sqrt[3]{n}\right\rfloor +1\right)^4+2 \left(\left\lfloor \sqrt[3]{n}\right\rfloor +1\right)^3-\left(\left\lfloor \sqrt[3]{n}\right\rfloor +1\right)^2\right) $$ $$ (n+1) \left\lfloor \sqrt[4]{n}\right\rfloor +\frac{1}{5} \left(-\left(\left\lfloor \sqrt[4]{n}\right\rfloor +1\right)^5+\frac{5}{2} \left(\left\lfloor \sqrt[4]{n}\right\rfloor +1\right)^4-\frac{5}{3} \left(\left\lfloor \sqrt[4]{n}\right\rfloor +1\right)^3+\frac{1}{6} \left(\left\lfloor \sqrt[4]{n}\right\rfloor +1\right)\right) $$ $$ (n+1) \left\lfloor \sqrt[5]{n}\right\rfloor +\frac{1}{6} \left(-\left(\left\lfloor \sqrt[5]{n}\right\rfloor +1\right)^6+3 \left(\left\lfloor \sqrt[5]{n}\right\rfloor +1\right)^5-\frac{5}{2} \left(\left\lfloor \sqrt[5]{n}\right\rfloor +1\right)^4+\frac{1}{2} \left(\left\lfloor \sqrt[5]{n}\right\rfloor +1\right)^2\right) $$

I do not know how this will be of any use to you; however, I must confess that I had a great time with this problem.

  • 1
    $\begingroup$ The formula you have can be simplified as $(n+1)\lfloor \sqrt[p]{n}\rfloor - \frac{1}{p+1}\left( B_{p+1}(\sqrt[p]{n}+1) - B_{p+1}(0)\right)$ where $B_k(x)$ are Bernoulli polynomials. $\endgroup$ – achille hui May 9 '15 at 21:42
  • $\begingroup$ @ClaudeLeibovici: I've added here an answer for the general case, which could be of interest. $\endgroup$ – Markus Scheuer May 9 '15 at 22:03

Consider evaluation of sums of following form:


where $p$ is a positive integer $\ge 2$. In the special case $p = 2$, this reduces to the sum we want to calculate.

Let $a = \lfloor \sqrt[p]{n} \rfloor$ and take a sufficiently small $\epsilon > 0$ such that $$\lfloor \sqrt[p]{k} + \epsilon \rfloor = \lfloor \sqrt[p]{k} \rfloor\quad\text{ for all } k = 1, 2, \ldots, n$$

The introduction of the $\epsilon$ piece make the function $\lfloor \sqrt[p]{x} + \epsilon\rfloor$ continuous at the jumps of $\lfloor x \rfloor$. This allows us to rewrite $\mathcal{S}_p$ as a Riemann-Stieljes integral. $$ \mathcal{S}_p = \sum_{k=1}^n \lfloor \sqrt[p]{k} \rfloor = \sum_{k=1}^n \lfloor \sqrt[p]{k} + \epsilon \rfloor = \int_{1^-}^{n^+} \lfloor \sqrt[p]{x} + \epsilon \rfloor d \lfloor x \rfloor $$

Integrate by part, we get $$ \mathcal{S}_p = \bigg[ \lfloor \sqrt[p]{x} + \epsilon \rfloor \lfloor x \rfloor \bigg]_{x=1-}^{n^+} - \int_{1^-}^{n^+} \lfloor x \rfloor d \lfloor \sqrt[p]{x} + \epsilon \rfloor = na - \int_{1+\epsilon}^{\sqrt[p]{n}+\epsilon} \lfloor (s-\epsilon)^p \rfloor d \lfloor s \rfloor $$ Convert the last integral back into a sum, we find $$ \mathcal{S}_p = na - \sum_{s=2}^a (s^p - 1) = (n+1)a - \sum_{s=1}^a s^p = (n+1)a - \frac{B_{p+1}(a+1)-B_{p+1}(0)}{p+1} $$ where $B_k(x)$ is the Bernoulli polynomial of order $k$

For the special case $p = 2$, the sum we want is $$ \bbox[8pt,border: 1px solid blue;]{ \sum_{k=1}^n \lfloor\sqrt{k}\rfloor = \mathcal{S}_2 = (n+1) a - \frac16 a(a+1)(2a+1),\quad a = \lfloor\sqrt{n}\rfloor } $$

This is equivalent to the formula in Markus Scheuer's answer.

For the case $p = 3, 4, 5$. this leads to

$$\begin{align} \sum_{k=1}^n \lfloor\sqrt[3]{k}\rfloor = \mathcal{S}_3 & = (n+1) a - \frac14\left((a+1)^4 - 2(a+1)^3 + (a+1)^2\right)\\ \sum_{k=1}^n \lfloor\sqrt[4]{k}\rfloor = \mathcal{S}_4 & = (n+1) a - \frac15\left( (a+1)^5 - \frac52(a+1)^4 + \frac53 (a+1)^3 - \frac16(a+1) \right)\\ \sum_{k=1}^n \lfloor\sqrt[5]{k}\rfloor = \mathcal{S}_5 & = (n+1) a - \frac16\left( (a+1)^6-3(a+1)^5 + \frac52 (a+1)^4 - \frac12 (a+1)^2 \right)\\ \end{align} $$ reproducing the formulas in Claude Leibovici's answer.

  • $\begingroup$ I've also added here an answer for the general case a few minutes ago! :-) (+1) $\endgroup$ – Markus Scheuer May 9 '15 at 22:06
  • 1
    $\begingroup$ @MarkusScheuer thanks, I just saw that (and upvote it like your answer here). $\endgroup$ – achille hui May 9 '15 at 22:11
  • $\begingroup$ @achillehui. This is a very interesting approach, for sure. Thank you. $\endgroup$ – Claude Leibovici May 10 '15 at 2:25

Hint in the form of an example: If you start the sum at $0$, you can write

$$\begin{align} \lfloor\sqrt0 \rfloor+\lfloor\sqrt1 \rfloor+\lfloor\sqrt2 \rfloor+\cdots+\lfloor\sqrt{11} \rfloor&=0+1+1+1+2+2+2+2+2+3+3+3\\ &=3+3+3+3+3+3+3+3+3+3+3+3\\ &-1-1-1-1-1-1-1-1-1\\ &-1-1-1-1\\ &-1\\ &=12\cdot3-(1+4+9)\\ &=(11+1)\lfloor\sqrt{11}\rfloor-(1^2+2^2+\cdots+\lfloor\sqrt{11}\rfloor^2) \end{align}$$

Remark: The main reason for including the unnecessary term $0$ is that it makes it easier to describe the pattern of what's being subtracted. (Note, there are not $11$ terms in $1^2+2^2+\cdots+\lfloor\sqrt{11}\rfloor^2$, but only $\lfloor\sqrt{11}\rfloor=3$ terms.) Properly understood, the pattern explains the generalization to $\lfloor\sqrt[p]n\rfloor$ in Claude Leibovici's answer.


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