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This question is a generalized form of the problem I asked before:

Algebraic Closed Form for $\sum_{n=1}^{k}\left( n- 3 \lfloor \frac{n-1}{3} \rfloor\right)$

Let, look at this periodic sequence:

$$a_n=\left\{a_1,a_2,a_3,a_4,a_5,\cdots a_k ; a_1,a_2,a_3,a_4,a_5, \cdots a_k; a_1,a_2,a_3,a_4,a_5,\cdots a_k;\cdots \right\}$$, where $\left\{a_1, a_2, a_3, \cdots a_k \right\} \in \mathbb{Z^{+}}$ and $a_{k+1}=a_1, a_{k+2}=a_2, a_{k+3}=a_3, \cdots a_{2k}=a_k, \cdots$

For the sequence $a_k=\left\{ a_1,a_2,a_3,...,a_k \right\}$ , $k$ is a finite number. $a_1,a_2,...a_k$ are arbitary numbers.

Verbally, the series $a_n$ consists of an infinite number of periodic repetitions of the finite series $a_k.$

Finally my question is:

a) If there is an exist a algebraic closed form, for finite series $a_k$, in this case, does the $ a_n $ series always have a algebraic closed form?

b) If there is not an exist a algebraic closed form, for finite series $a_k$, in this case, does the $ a_n $ series always have a algebraic closed form?

I mean ,for example:

a)

$a_n=\left\{ 1,3,5,7,1,3,5,7,1,3,5,7,1,3,5,7,1,3,5,7\cdots\right\}$

b)

$a_n=\left\{1,8,2,6,5,9,1,8,2,6,5,9,1,8,2,6,5,9,1,8,2,6,5,9\cdots\right\}$

Thank you very much.

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  • $\begingroup$ A finite sequence always does have a closed form, and so does a periodic sequence. $\endgroup$ – Robert Israel Jan 8 at 12:39
  • $\begingroup$ @RobertIsrael for any arbitary numbers for $a_1, a_2,a_3,...a_k$, There's an always closed form for $a_n$. Do I understand correct? $\endgroup$ – Elementary Jan 8 at 12:46
  • $\begingroup$ @Beginner, it depends on what you consider a "closed form". That term is not really well-defined. In a sense you have given a "closed form" in your definition of the sequence. $\endgroup$ – Mees de Vries Jan 8 at 12:53
  • $\begingroup$ @MeesdeVries I tried to give 2 example for a good understanding of the question.. $\endgroup$ – Elementary Jan 8 at 12:56
  • $\begingroup$ This might be a closed form for your first sequence: $$ a_k = \begin{cases} 1&\text{if $k = 1 \mod4$,}\\ 3&\text{if $k = 2 \mod4$,}\\ 5&\text{if $k = 3 \mod4$,}\\ 7&\text{else.} \end{cases} $$ Does that satisfy what you're looking for in a "closed form"? $\endgroup$ – Mees de Vries Jan 8 at 13:08
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A periodic sequence with period $P$ can always be written as a trigonometric polynomial $$a_n = \sum_{j=0}^{P-1} b_j \cos(2 \pi j n/P) + \sum_{j=1}^{P-1} c_j \sin(2 \pi j n/P)$$ for some coefficients $b_j$ and $c_j$ (look up Finite Fourier Transform).

Thus your first example can be written as $$a_n = 4-\cos \left( \pi\,j/2 \right) -\cos \left( \pi\,j \right) -\cos \left( 3\,\pi\,j/2 \right) -\sin \left( \pi\,j/2 \right) +\sin \left( 3\,\pi\,j/2 \right) $$

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Let $f(x)$ be a polynomial interpolating $a_1,a_2,a_3,\dots,a_k$.

Then the sequence is given by $a_n = f(1+((n-1) \bmod k))$.

If you can't use mod directly, but can use floor, then note that $$ a \bmod b = a - b \left\lfloor \dfrac{a}{b} \right\rfloor $$ for $a,b \in \mathbb N$.

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