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

Question

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

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

Answer 1

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 $0$.

Answer 2

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

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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.]

Answer 3

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

Answer 4

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

Answer 5

What about

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

Answer 6

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

Answer 7

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

See PURELY PERIODIC

Answer 8

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

Answer 9

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 $2^2$.

Answer 10

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

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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