25
$\begingroup$

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

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

$\endgroup$
2
  • 36
    $\begingroup$ nth term of decimal expansion 442/999 ;) $\endgroup$
    – wim
    Sep 11, 2012 at 13:25
  • 2
    $\begingroup$ Will people kindly please stop up voting this question. Thank you. $\endgroup$
    – ben
    Jun 12, 2014 at 2:09

17 Answers 17

59
$\begingroup$

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 $0$.

$\endgroup$
9
  • $\begingroup$ Assuming you only want to determine the nth entry of a sequence, this would be the most efficient way of calculating it. $\endgroup$
    – dstibbe
    Sep 11, 2012 at 9:18
  • 33
    $\begingroup$ Seriously? My highest upvoted non CW answer is this? $\endgroup$ Sep 11, 2012 at 17:41
  • 3
    $\begingroup$ ROFL. I'm guessing this is merely caused by people wanting to top the "approved" answer, since that answer is computationally very inefficient. $\endgroup$
    – dstibbe
    Sep 11, 2012 at 20:37
  • 4
    $\begingroup$ At least it is a better answer than the one offered by that lunkhead Austin Mohr. $\endgroup$
    – MJD
    Sep 11, 2012 at 22:29
  • 1
    $\begingroup$ @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. $\endgroup$
    – dstibbe
    Sep 12, 2012 at 8:39
40
$\begingroup$

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

--

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.]

$\endgroup$
17
  • 3
    $\begingroup$ @MattE: Exactly what I need, BUT, this seems to work better: $$4(\frac{5}{6}-\frac{1}{3}\cos(\frac{2\pi n}{3} ))$$ $\endgroup$
    – ben
    Sep 11, 2012 at 5:14
  • 1
    $\begingroup$ This seems to give 5, 5, 2, 5, 5, 2... wolframalpha.com/input/… $\endgroup$
    – Deebster
    Sep 11, 2012 at 14:04
  • 7
    $\begingroup$ Amazing how the most computationally inefficient formula is the most up voted. $\endgroup$
    – asmeurer
    Sep 11, 2012 at 18:38
  • 4
    $\begingroup$ @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, $\endgroup$
    – Matt E
    Sep 12, 2012 at 0:30
  • 1
    $\begingroup$ 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/…. $\endgroup$
    – asmeurer
    Sep 13, 2012 at 17:11
29
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ (+1) You were first; that has to count for something :-) $\endgroup$
    – robjohn
    Jul 29, 2013 at 19:45
  • 4
    $\begingroup$ @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. :) $\endgroup$
    – 40 votes
    Jul 29, 2013 at 20:14
14
$\begingroup$

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

$\endgroup$
3
  • 9
    $\begingroup$ Which has exactly enough characters to be accepted. $\endgroup$
    – Did
    Sep 11, 2012 at 5:52
  • 2
    $\begingroup$ And even a downvote... Hallelujah! $\endgroup$
    – Did
    Sep 11, 2012 at 18:47
  • $\begingroup$ Expanded version with many more characters. $\endgroup$
    – Did
    Sep 15, 2012 at 4:35
12
$\begingroup$

What about

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

$\endgroup$
1
  • $\begingroup$ Or 2+ 2*(n %% 3 != 0) . (Not sure of the local convention for modulo remainder so used the R operator) $\endgroup$
    – DWin
    Sep 11, 2012 at 16:43
10
$\begingroup$

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

$\endgroup$
8
$\begingroup$

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

See PURELY PERIODIC

$\endgroup$
7
$\begingroup$

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

$\endgroup$
6
$\begingroup$

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 $2^2$.

$\endgroup$
6
$\begingroup$

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

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

$\endgroup$
5
$\begingroup$

Mimicking the cosine answer of @Matt E, I suggest setting $\omega=(1+\sqrt{-3})/2$ and taking $a_n=2+2|\omega^n-\omega^{2n}|/\sqrt3$.

$\endgroup$
1
  • $\begingroup$ Dear Lubin, Indeed this is what I had in mind (as I indicated in an edit to my answer, which also seemed to be rife with typos; hopefully they are now fixed and agree with your answer). Regards, $\endgroup$
    – Matt E
    Sep 12, 2012 at 1:05
5
$\begingroup$

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

$\endgroup$
1
  • 1
    $\begingroup$ Alternatively, $0,1,1,0,1,1, ...$ . $\endgroup$
    – Théophile
    Sep 12, 2012 at 13:39
4
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

$$3+ { \left( -1 \right) }^{ (n+1)\mod 3 } ,\; \mathbb{for} \; n = 1,2,...$$

$\endgroup$
0
$\begingroup$

$x_n= \begin{cases} 4,&n=0,1\\ (x_{n-2} + x_{n-1})\,\bmod 4 + 2,&n\ge2 \end{cases} $

$\endgroup$
1
  • $\begingroup$ One can avoid using 'mod' by writing the second clause as $10-x_{n-2}-x_{n-1}$ $\endgroup$ Oct 7, 2012 at 12:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .