Ytukyg

Let's look at the following sequence:

$a_n=left{1,2,3,1,2,3,1,2,3,1,2,3,...right}$

I'm trying to calculate:

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

Attempts:

I have a Closed Form for this sequence.

$$a_n=n- 3 bigglfloor frac{n-1}{3} biggrfloor$$

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

$$sum_{n=1}^{k}left( n- 3 bigglfloor frac{n-1}{3} biggrfloorright)$$

Is it possible?

Writing down a couple of the sums:

$$1,3,6,7,9,12,13,15,18,dots$$

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

gives you a clue that the difference between the arithmetic sequence and the sequence you want to describe is simply $$0,0,1,0,0,1,0,0,1,0,0,1dots$$ 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 $$f(x)=frac12x^2-frac32 x + 1.$$

Edit:

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

$$b_k = leftlfloor 1 + leftlfloorfrac k3rightrfloor - frac k3rightrfloor$$

Subtracting the "average" sequence

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

$$-1,0,1,-1,0,1,-1,0,1,cdots$$

which sums as a periodic one

$$-1,-1,0,-1,-1,0,-1,-1,0,cdots$$

The latter can be expressed as

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

So globally,

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

If you consider$$b_k=sum_{n=1}^{k}left( n- 3 bigglfloor frac{n-1}{3} biggrfloorright)$$ 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
$$b_k=2k-frac{2}{3} left(1-cos left(frac{2 pi k}{3}right)right)$$

Edit

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

Which one do you prefer ?

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

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

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

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
$$sum_{n=1}^kleft(n-3leftlfloor n-frac{1}{3}rightrfloorright)=sum_{n=1}^{3t}left(n-3leftlfloor n-frac{1}{3}rightrfloorright)=t(1+2+3)=6t=color{blue}{2k}.$$

If $k=3t+1$ for $t geq 1$, then
$$sum_{n=1}^kleft(n-3leftlfloor n-frac{1}{3}rightrfloorright)=sum_{n=1}^{3t+1}left(n-3leftlfloor n-frac{1}{3}rightrfloorright)=t(1+2+3)+1=6t+1=color{blue}{2k-1}.$$

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