# How can I get sequence $4,4,2,4,4,2,4,4,2\ldots$ into equation?

How can I write an equation that expresses the nth term of the sequence:

$$4, 4, 2, 4, 4, 2, 4, 4, 2, 4, 4, 2,\ldots$$

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nth term of decimal expansion 442/999 ;) –  wim Sep 11 '12 at 13:25
Why isn't this posted to tex stackexchange? –  g33kz0r Sep 12 '12 at 3:09
@g33kzor Why should it be? I don't think the question was how to literally write (in TeX) the equation. –  Austin Mohr Sep 12 '12 at 12:04

$$\frac{14}{3} - \frac{8}{3}\cos^2 (\frac{2 \pi n}{3})$$

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Added: The original formula was typed late at night, and sufered from a couple of computational blunders; hopefuly the present formula is correct.

Of course, the square on the cosine is unnecessary (I only put it there because I thought, due to miscalculation, that it simplified the coefficients).

In some sense the more natural formula is the one without the squared cosine, namely

$$\frac{10}{3} - \frac{4}{3}\cos(\frac{2 \pi n}{3})$$

(as noted by the OP below).

Note that the existence of such a formula is not accidental or without interest. It is an illustration of finite Fourier theory (or, if you prefer, character theory of the finite abelian group $\mathbb Z/3\mathbb Z$). In general, any function of $n$ that depends only on $n \bmod N$ can be written as a linear combination of the functions $e^{2 \pi i n /N}$.

The most familiar example is probably the formula $(-1)^n$ for the sequence $-1,1,-1,1,\ldots$.

Whether such a formula is ever computationally useful is outside my area of expertise, but there is no doubt about the theoretical utility of finite Fourier theory.

[See Lubin's answer for an answer more explictly in keeping with this remark.]

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@MattE: Exactly what I need, BUT, this seems to work better: $$4(\frac{5}{6}-\frac{1}{3}\cos(\frac{2\pi n}{3} ))$$ –  ben Sep 11 '12 at 5:14
This seems to give 5, 5, 2, 5, 5, 2... wolframalpha.com/input/… –  Deebster Sep 11 '12 at 14:04
Amazing how the most computationally inefficient formula is the most up voted. –  asmeurer Sep 11 '12 at 18:38
@asmeurer: Dear asmuerer, You might find the formula dubious, but in fact it is an example of finite Fourier theory: a function depending on the congruence class of $n$ mod $N$ admits a Fourier expansion. It is analogous to writing $-1,1,-1,1, \ldots$ as $(-1)^n$, and is useful in some contexts (at least theoretical ones, which is what I am more familiar with). Regards, –  Matt E Sep 12 '12 at 0:30
It certainty is mathematically interesting. But if the user wanted it for a program (possible, but it wasn't stated), then @AlexBecker's answer would be better. I've seen very computationally inefficient answers to sequence problems on math SE for problems where the question specifically stated it was for a computer program. See for example math.stackexchange.com/questions/162495/…. –  asmeurer Sep 13 '12 at 17:11

$$f(n) = \begin{cases} 4 \text{ if } n \equiv 0 \text{ or } 1 \text{ (mod 3)}\\ 2 \text{ if } n \equiv 2 \text{ (mod 3)} \end{cases}$$

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(+1) You were first; that has to count for something :-) –  robjohn Jul 29 '13 at 19:45
@robjohn Indeed... by 12 seconds. I did not check the timestamps and decided that Alex must have been first to get that many upvotes. Oh well, at least my meta post brought some justice here. :) –  40 votes Jul 29 '13 at 20:14

How about $$x_n=\begin{cases} 4 &\text{if }n\equiv 0,1\:(\bmod 3)\\ 2 &\text{if }n\equiv 2\:(\bmod 3)\\ \end{cases}$$ assuming you start indexing from $0$.

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Assuming you only want to determine the nth entry of a sequence, this would be the most efficient way of calculating it. –  dstibbe Sep 11 '12 at 9:18
Seriously? My highest upvoted non CW answer is this? –  Alex Becker Sep 11 '12 at 17:41
ROFL. I'm guessing this is merely caused by people wanting to top the "approved" answer, since that answer is computationally very inefficient. –  dstibbe Sep 11 '12 at 20:37
At least it is a better answer than the one offered by that lunkhead Austin Mohr. –  MJD Sep 11 '12 at 22:29
@Matt E I'm quite familier with the Fourier theory. However, if someone is not going to use the formula for anything else besides determing the value of the nth position, then it is overly complex (and thus inefficient) when above formula provides the same. –  dstibbe Sep 12 '12 at 8:39

$$a_n:=\left\{\begin{array}{}4\,,&\text{if}\,\;\;n\neq 0\pmod 3\\2\,,&\text{if}\,\;\;n=0\pmod 3\end{array}\right....?$$

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Or 2+ 2*(n %% 3 != 0) . (Not sure of the local convention for modulo remainder so used the R operator) –  DWin Sep 11 '12 at 16:43

Try $a_n=2^{1+\lceil n/3 \rceil - \lfloor n/3 \rfloor}$, where $\lceil n/3 \rceil$ is the least integer $\geq n/3$ and where $\lfloor n/3 \rfloor$ is the greatest integer $\leq n/3$. Then, if 3 divides $n$, you get $2^1$; if it doesn't, you get $2^2$.

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$$x_n= 2+ 3 \left\{ \frac{n}{3} \right\} + 3\left\{ \frac{n}{3}\right\}\left(2- 3\left\{ \frac{n}{3} \right\}\right) \,.$$

where $\{ \}$ denotes the fractional part.

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The "quotients" $a_j$ of the simple continued fraction for $$\frac{17 + \sqrt {442}}{9}.$$

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$x_n = 4(n^2 \bmod 3)+2(1-(n^2 \bmod 3))=2+2(n^2 \bmod 3)$, assuming you start indexing from $1$.

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$a_{n+2} = |a_{n+1} - a_n| + 2$, where $a_1 = a_2 = 4$.

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$2n^2+4n+4 \pmod 6$ for $n \geq 0$.

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$$4-2\cdot\mathbf 1_{3\mid n}\qquad\text{or}\qquad 2+2\cdot\mathbf 1_{\gcd(3,n)=1}$$

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Which has exactly enough characters to be accepted. –  Did Sep 11 '12 at 5:52
And even a downvote... Hallelujah! –  Did Sep 11 '12 at 18:47
Expanded version with many more characters. –  Did Sep 15 '12 at 4:35

$$\Large 2^{2-0^{(n \text{ mod } 3)}}$$

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The On-Line Encyclopedia of Integer Sequences, is always a good place to start looking for them. One often needs to search for one excluding constant multiplication factor and/or drop a few initial terms. And/or add a constant factor (as Theóphile points out below)

For this one we can use 1,2,2,1,2,2,1,2,2,... (http://oeis.org/A130196), drop the initial "1" and multiply by 2

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Alternatively, $0,1,1,0,1,1, ...$ . –  Théophile Sep 12 '12 at 13:39

$$x_n = 3 + (-1)^{((n+2) mod 3)}$$

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$x_n= \begin{cases} 4,&n=0,1\\ (x_{n-2} + x_{n-1})\,\bmod 4 + 2,&n\ge2 \end{cases}$
One can avoid using 'mod' by writing the second clause as $10-x_{n-2}-x_{n-1}$ –  Marc van Leeuwen Oct 7 '12 at 12:36
Mimicking the cosine answer of @Matt E, I suggest setting $\omega=(1+\sqrt{-3})/2$ and taking $a_n=2+2|\omega^n-\omega^{2n}|/\sqrt3$.