Find a formula for $\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$

Question

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:

$1,1,1,2,2,2,2,2,3,3,3,...$

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...

Answer 1

Hint

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.

Answer 2

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.

Answer 3

The way I approached this problem is dividing the sum into two halves. Where the first half contains all the numbers starting from $1$ and reaching upto $m$ such that $\sqrt{m}$ is greatest integer $<\lfloor\sqrt{n}\rfloor$, and second half contains the leftover numbers. Intuitively $m$ can be calculated by using simple formula $\lfloor\sqrt{n}\rfloor - 1$Mathematically:

$\displaystyle\sum_{k=1}^n\lfloor\sqrt{k}\rfloor = \underbrace{\displaystyle\sum_{p=1}^{\lfloor\sqrt{n}\rfloor-1}p\times(2p + 1)}^{\text{first part}} + \underbrace{(n - {\lfloor\sqrt{n}\rfloor}^2 + 1)\times{\lfloor\sqrt{n}\rfloor}}^{\text{second part}}$

The same can be resolved using the following identities.

$\displaystyle\sum_{x = 1}^n x = \frac{x\times(x+1)}{2}$

and

$\displaystyle\sum_{x = 1}^n x^2 = \frac{x\times(x+1)\times(2 \times x+1)}{6}$

I didn't tested this solution with every value, but the solution seems convincing.

I am taking an example to demonstrate it, taking $n = 5$
The answer should contain $\underbrace{1+1+1+2+2}+2+2+2+3+... = 7$

The $\lfloor\sqrt{5}\rfloor-1 = 1$(upper limit for the first part)
So, our first part is calculated as $1\times (2+1) = 3$
Similarly second part is calculated as
$(5-{\lfloor\sqrt{5}\rfloor}^2)\times \lfloor\sqrt{5}\rfloor = (5 - 2^2+1)\times 2 = (2)\times 2 = 4$
So we get $4 + 3 = 7$ as our answer, which is correct.

The same intuition can be used to derive formula for $\displaystyle\sum_{k=1}^n\lfloor\sqrt[x]{k}\rfloor$.
i.e.

$\displaystyle\sum_{k=1}^n\lfloor\sqrt[x]{k}\rfloor = \displaystyle\sum_{p=1}^{\lfloor\sqrt[x]{p}\rfloor}p\times ((p+1)^x-p^x)+(n - (\rfloor\sqrt[x]{n}\rfloor)^x + 1)\times\lfloor\sqrt[x]{n}\rfloor$

P.S. This is just an intuition, if anything wrong, feel free to comment.