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My specific question is about how I would write a sequence in explicit form, when the sequence has an explicit form for its odd and even terms. So, using notation, if:

$$a_n = \begin{cases} a_{n_1} & n\space \text{even} \\ a_{n_2} & n\space\text{odd} \\ \end{cases}$$

What is an explicit form for $a_n$ in terms of $a_{n_1}$ and $a_{n_2}$? I assume there's an easy answer but I just thought of this on a whim and don't really care enough to spend hours trying to figure out a solution. For that matter, what if we have this case? Suppose $a_n$ goes by a different rule not depending on its remainder mod $2,$ but its remainder mod $k$. So, say it's defined as follows:

$$a_n = \begin{cases} a_{n_1} & n\equiv 0\space(\text{mod}\space k) \\ a_{n_2} & n\equiv 1\space(\text{mod}\space k) \\ .\\ .\\ .\\ a_{n_{k-1}} & n\equiv k-1\space(\text{mod}\space k)\\ \end{cases}$$

Can we get one explicit expression for $a_n$?

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  • $\begingroup$ If you want, you can use roots of unity (or equivalently sine and cosines) to get a single "formula". But that is pointless and \begin{cases}...\end{cases} is already explicit. $\endgroup$ Jun 1, 2021 at 5:50
  • $\begingroup$ Perhaps "explicit" wasn't the right word, as you're correct, it is explicit. I should have said "not piecewise" or something to that effect. $\endgroup$
    – Luna145
    Jun 1, 2021 at 6:00

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One can make use of periodic functions such as trigonometric functions. For example, in the case of $\hbox{mod }2$, write $a_n=|\sin(n\pi)|a_{n_2}+|\cos(n\pi)|a_{n_1}$. One, can also make use of roots of unity. For example, the same formula can be written as $a_n=\dfrac{1-(-1)^n}{2}a_{n_2}+\dfrac{1+(-1)^n}{2}a_{n_1}$. Similarly, for $\hbox{mod k}$, try to make use of $k$th roots of unity. For $\hbox{mod } 3$, note that $\dfrac{1+{\omega}^n+{\omega}^{2n}}{3}$ is $0$ for $n \neq 3k, k \in \mathbb{Z}$ and is $1$ for $n=3k, k \in \mathbb{Z}$.

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