Let $k$ be a fixed integer and consider the sequence $1, k, k+1, 2k, 2k+1, 3k, 3k+1, 4k, 4k+1, \dotsc$. Is there a closed form for this sequence?

Question

Let $k$ be a fixed integer and consider the sequence $$1, k, k+1, 2k, 2k+1, 3k, 3k+1, 4k, 4k+1, \dotsc.$$

These are all the integers that are congruent to either 0 or 1 mod $k$.

It is immediate that for $k=2$, the sequence in closed form is $a_n = n$ if we use $n$ to index. However, for larger fixed $k$ I am not able to work out a closed form. Initially I tried using floor/ceilings but I could not find anything to account for the alternating sized gaps between terms. Does such a form even exist?

Answer 1

The following gives $a_0=1, a_1=k, a_2=k+1, a_3=2k, \,\dots\,$:

$$ a_n = \left\lfloor \frac{n+1}{2} \right\rfloor \, k + (n+1) \bmod 2 $$

Answer 2

We can try something of the form $$a_n=\alpha n+\beta (-1)^nn+\gamma.$$ From $$\begin{align}\alpha(2n+1)+2\gamma&=a_n+a_{n+1}\\&=(k+1, 2k+1,3k+1,4k+1, 5k+1,\ldots )\\&=nk+1,\end{align}$$ we find $\alpha=\frac k2$, $\gamma=\frac12-\frac k4$. From this we infer $\beta=\frac k4-\frac12$, and - hooray! - that works.

Answer 3

If you are willing to prepend a zero then there is a straightforward way:

If $n=0,1,2,3,4,5,...$ then $\lfloor {n \over 2}\rfloor =0,0,1,1,2,2,...$ and $n-2\lfloor {n \over 2}\rfloor = 0,1,0,1,0,1,...$ and so the sequence can be written as $k \lfloor {n \over 2}\rfloor + n-2\lfloor {n \over 2}\rfloor$.

Answer 4

From observation, $$\begin{aligned} &\begin{cases}T_{2m}&=mk\\\\ T_{2m+1}&=mk+1\end{cases}\\ \\ \Longrightarrow &\begin{cases}\text{For even $n$:} \;\;\quad T_n&=\displaystyle\frac n2k\\ \text{For odd $n$:}\qquad T_n&=\displaystyle\frac {n-1}2k+1\;\;=\frac n2k+(1-\frac k2)\end{cases}\end{aligned}$$

Summarising both cases, $$T_n =\frac n2k+\left(n-2\bigg\lfloor\frac n2\bigg\rfloor\right)\left(1-\frac k2\right) =\color{red}{n+\left(k-2\right)\bigg\lfloor\frac n2\bigg\rfloor}$$

Answer 5

A variation: \begin{align*} &a_0=1,\ a_2=k+1,\ a_4=2k+1,\ \ \ldots,\qquad\qquad &a_{2n}&=nk+1\qquad\qquad &n\geq 0\\ &a_1=k,\ a_3=2k,\ \ \ \ \ \ a_5=3k,\ \ \ \ \ \ \ \ \ \ldots,\qquad\qquad &a_{2n-1}&=nk\qquad\qquad &n\geq 1\\ \end{align*}

It follows for $n\geq 0$ \begin{align*} \color{blue}{a_n}&=\left(\frac{nk}{2}+1\right)\cdot\frac{1+(-1)^n}{2}+\frac{(n+1)k}{2}\cdot\frac{1-(-1)^n}{2}\tag{1}\\ &\color{blue}{=\frac{1}{2}\left(nk+1+(-1)^n+k\cdot\frac{1-(-1)^n}{2}\right)} \end{align*}

Note:

• The first part in (1) respects even indices: $2l=n\rightarrow l=\frac{n}{2}$.

• The second part in (1) covers odd indices: $2l-1=n\rightarrow l=\frac{n+1}{2}$.