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

Solution 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.

Solution 2:

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 $p\geq 1$.

Solution 3:

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.