Let's look at the following sequence:


I'm trying to calculate:

$$\sum_{n=1}^{k} a_n$$


I have a Closed Form for this sequence.

$$a_n=n- 3 \bigg\lfloor \frac{n-1}{3} \bigg\rfloor$$

The problem is, I'm looking for a closed form for this summation:

$$\sum_{n=1}^{k}\left( n- 3 \bigg\lfloor \frac{n-1}{3} \bigg\rfloor\right)$$

Is it possible?

  • $\begingroup$ There is a closed form with a bunch of floor functions. $\endgroup$ Jan 8, 2019 at 9:52
  • $\begingroup$ You recieved 5 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you. $\endgroup$
    – 5xum
    Jan 15, 2019 at 8:49
  • $\begingroup$ @5xum I've been upvote before. Now, I accept. By the way, If you add the most recent result to your answer, it will be perfect. Regards. $\endgroup$
    – user548054
    Jan 15, 2019 at 9:03

5 Answers 5


Writing down a couple of the sums:


and comparing that to the sequence$$1,3,5,7,9\cdots$$

gives you a clue that the difference between the arithmetic sequence and the sequence you want to describe is simply $$0,0,1,0,0,1,0,0,1,0,0,1\dots$$ which is a sequence you can describe in closed form in a way similar to $a_n$.

That is, you can see that $$\sum_{i=n}^k a_n = 2k-1 + b_k$$

where $b_k$ is equal to $1$ if $k$ is divisible by $3$ and $0$ otherwise.

You can express $b_n$ algebraically by taking $a_n$ and any function for which $f(1)=f(2)=0$ and $f(3)=1$, and you have $b_n=f(a_n)$.

I can't think of any "elegant" function $f$ at the moment, but a quadratic polynomial can surely do it, since we only have a restriction on three points. The quadratic polynomial that satisfies $f(1)=f(2)=0$ and $f(3)=1$ is $$f(x)=\frac12x^2-\frac32 x + 1.$$


Thanks to BarryCipra, a nicer function (more in the spirit of your solution) for $b_k$ is

$$b_k = \left\lfloor 1 + \left\lfloor\frac k3\right\rfloor - \frac k3\right\rfloor$$

  • 2
    $\begingroup$ How about $b_k=\lfloor1+\lfloor{k\over3}\rfloor-{k\over3}\rfloor$? $\endgroup$ Jan 8, 2019 at 10:10
  • $\begingroup$ @BarryCipra That works, yeah. I added it to my answer (hope you don't mind) $\endgroup$
    – 5xum
    Jan 8, 2019 at 10:22
  • $\begingroup$ My pleasure. But you mean $b_k$, not $b_n$. $\endgroup$ Jan 8, 2019 at 10:25
  • $\begingroup$ And What exactly is the result? $\sum_{i=1}^{n}a_n$ $\endgroup$
    – user548054
    Jan 8, 2019 at 11:24
  • $\begingroup$ @Beginner, a complete answer is $$\sum_{n=1}^ka_n=2k-1+\left\lfloor 1 + \left\lfloor\frac k3\right\rfloor - \frac k3\right\rfloor$$ Is that what your comment is asking for? If you like, it can be simplified to $$\sum_{n=1}^ka_n=2k+\left\lfloor \left\lfloor\frac k3\right\rfloor - \frac k3\right\rfloor$$ $\endgroup$ Jan 8, 2019 at 12:10

Subtracting the "average" sequence

$$2,2,2,2,2,2,2,2,2,\cdots$$ you get


which sums as a periodic one


The latter can be expressed as

$$\frac{(n-1)\bmod3-n\bmod 3-2}3.$$

So globally,

$$2n+\frac{(n-1)\bmod3-n\bmod 3-2}3.$$

  • $\begingroup$ (+1) , It would be great if we found a closed form that could be expressed by the floor function $\endgroup$
    – user548054
    Jan 8, 2019 at 9:43
  • $\begingroup$ @Beginner: modulo and floor are interchangeable, with simple arithmetic. $\endgroup$
    – user65203
    Jan 8, 2019 at 9:46

If $n \equiv 0 \pmod{3}$, i.e. say $n=3s$ (where $s \geq 1$), then $a_n=3s-3\lfloor s-\frac{1}{3}\rfloor=3s-3(s-1)=3$.

If $n \equiv 1 \pmod{3}$, i.e. say $n=3s+1$ (where $s \geq 0$), then $a_n=3s+1-3\lfloor s\rfloor=1$.

If $n \equiv 2 \pmod{3}$, i.e. say $n=3s+2$ (where $s \geq 0$), then $a_n=3s+2-3\lfloor s+\frac{1}{3}\rfloor=2$.

So depending on what $k$ is we can count the number of terms which are $1's$, $2's$ and $3's$.

For $k=1,2$, the sum will be $\color{red}{1}$ and $\color{red}{3}$, respectively. So for $k \geq 3$, we do the following:

If $k=3t$ for $t \geq 1$, then $$\sum_{n=1}^k\left(n-3\left\lfloor n-\frac{1}{3}\right\rfloor\right)=\sum_{n=1}^{3t}\left(n-3\left\lfloor n-\frac{1}{3}\right\rfloor\right)=t(1+2+3)=6t=\color{blue}{2k}.$$

If $k=3t+1$ for $t \geq 1$, then $$\sum_{n=1}^k\left(n-3\left\lfloor n-\frac{1}{3}\right\rfloor\right)=\sum_{n=1}^{3t+1}\left(n-3\left\lfloor n-\frac{1}{3}\right\rfloor\right)=t(1+2+3)+1=6t+1=\color{blue}{2k-1}.$$

Likewise we can get the expressions for $k=3t+2$ as $$\sum_{n=1}^k\left(n-3\left\lfloor n-\frac{1}{3}\right\rfloor\right)=\sum_{n=1}^{3t+2}\left(n-3\left\lfloor n-\frac{1}{3}\right\rfloor\right)=t(1+2+3)+(1+2)=6t+3=\color{blue}{2k-1}.$$

  • $\begingroup$ I think, if there is an exist a closed form, it should be expressed by floor function. $\endgroup$
    – user548054
    Jan 8, 2019 at 9:26
  • $\begingroup$ @Beginner you are summing up to $k$ terms, so the answer is a function of $k$. I am not sure what you mean by the closed form has to be in terms of floor function. $\endgroup$
    – Anurag A
    Jan 8, 2019 at 9:29
  • $\begingroup$ No, I didn't do this. $\endgroup$
    – user548054
    Jan 8, 2019 at 9:32

If you consider$$b_k=\sum_{n=1}^{k}\left( n- 3 \bigg\lfloor \frac{n-1}{3} \bigg\rfloor\right)$$ you could see, after some simple manipulationd that it corresponds to the recurrence relation $$b_k=b_{k-1}+b_{k-3}-b_{k-4}\qquad \text{with}\qquad b_1=1, b_2=3, b_3=6, b_4=7$$ Using the conventional method you should end with $$b_k=2k-\frac{2}{3} \left(1-\cos \left(\frac{2 \pi k}{3}\right)\right)$$


If you enjoy the floor function, using a CAS, I got for $b_k$ the small monster $$\frac{1}{2} \left(k^2+k-3 \left\lfloor \frac{k}{3}\right\rfloor ^2-3 \left\lfloor \frac{k-2}{3}\right\rfloor \left(\left\lfloor \frac{k-2}{3}\right\rfloor +1\right)-3 \left\lfloor \frac{k-1}{3}\right\rfloor \left(\left\lfloor \frac{k-1}{3}\right\rfloor +1\right)+3 \left\lfloor \frac{k}{3}\right\rfloor \right)$$

Which one do you prefer ?

  • $\begingroup$ Of course, Floor function :) $\endgroup$
    – user548054
    Jan 8, 2019 at 10:06
  • $\begingroup$ @Beginner. Joke or serious ? $\endgroup$ Jan 8, 2019 at 10:09
  • $\begingroup$ Both are beautiful. :) Thank you very much (+) $\endgroup$
    – user548054
    Jan 8, 2019 at 10:34

Sorry can't comment on @5xum's answer. Building on the quadratic function we can derive: $S_k=\frac{1}{2}((k-3\lfloor\frac{k}{3}\rfloor)^2+k+9\lfloor\frac{k}{3}\rfloor)$, which is pretty neat.

To derive this, note that from @5xum's answer, $S_k=2k-1+\frac{1}{2}(x^2 -3x+2)$ where $x=k-3\lfloor\frac{k}{3}\rfloor$. We can write $k=x+3\lfloor\frac{k}{3}\rfloor$, so that: $$S_k=\frac{1}{2}(x^2 -3x+2+4x+12\lfloor\frac{k}{3}\rfloor-2)=\frac{1}{2}(x^2+x+12\lfloor\frac{k}{3}\rfloor). $$ Substituting $x$ back yields the result.


You must log in to answer this question.