What Rule generates the sequence $8,8,10,12,12,14,16,16$?

Question

What is the rule that generates the following sequence? I can not solve it.

                8 8 10 12 12 14 16 16

For example: 1 3 5 7 9 11 Rule: $n+2$

Answer 1

It looks to me like repeated (add 0, add 2, add 2)

Answer 2

Let $a_{n+3} = 4+a_n$ and $a_1 = 8$, $a_2 = 8$, and $a_3 = 10$.

Answer 3

Let $\alpha=n\pmod 3$ and $\beta=(n-1)\mod 3$ with $0\leq\alpha,\beta\lt 3$ then,

$a_n=a_{n-1}+2(1-(1-\delta(\alpha))(1-\delta(\beta)))$ where $\delta $ is dirac-delta function.

Answer 4

The simplest answer is $$a_n = 8 + 2 \left\lfloor \frac{2n}{3} \right\rfloor$$ with $n$ start at $0$. By trial and error, a more complicated one is $$a_n = \frac13(4n+22)+ \frac{4}{\sqrt{27}}\cos(\frac{4 n + 1}{6}\pi)\tag{*1}$$ If one inspect carefully about this expression, one can see that it is a linear combination of 4 terms:

$$n,\;\;1,\;\;e^{i\frac{2\pi}{3} n}\;\text{ and }\;e^{-i\frac{2\pi}{3} n}$$ The last 3 terms are powers of the cubic roots of unity. This means these parts satisfy a recurrence relation of the from $p_{n+3} - p_{n} = 0$. The $n$ term satisfies a recurrence relation of the form $q_{n+2} - 2 q_{n+1} + q_{n} = 0$. Combine these, we see $a_n$ satisfies a $4^{th}$ order recurrence relation:

$$a_{n+4} - a_{n+3} - a_{n+1} + a_{n} = 0$$

We can use these and the initial values: $a_0 = 8, a_1 = 8, a_2 = 10$ and $a_3 = 12$ as a recursive "definition" of all $a_n$. Alternatively, one can skip the part on $n$ term and use the inhomogeneous $3^{rd}$ order recurrence relation:

$$a_{n+3} - a_{n} = 4\tag{*2}$$

with the initial values for $a_0, a_1, a_2$ as definition of the recurrence relation.

Of course, this is going the tortuous route. One can start with the obvious relation $(*2)$, working backwards and get the explicit answer $(*1)$.

Answer 5

You could also try to find and get any ideas of the general term. Here

Answer 6

Clearly the $n$th term is $$-48 + \frac{3034}{21} n - \frac{2853}{20} n^2 + \frac{8477}{120}n^3 - \frac{153}{8}n^4 + \frac{23}{8}n^5 - \frac{9}{40}n^6 + \frac{1}{140}n^7.$$

To be less facetious: there are infinitely many rules that could generate this sequence. Unless you have more information about the process that generated it, you have no way of knowing which rule (if any!) it came from.

Answer 7

Assuming the $n$th term is $\frac43n+\frac{16}3+\frac43\,\mathbf 1_{n=1\bmod{3}}+\frac23\,\mathbf 1_{n=0\bmod{3}}$...

Keep the three last values $x$, $y$ and $z$ of the sequence:

• If $z=x+2$, set $t=z+2$.
• If $z=x+4$, set $t=z$.
• Replace $(x,y,z)$ by $(y,z,t)$.
• Iterate.

Initialize suitably.

Or, equivalently:

Keep the three last values $x$, $y$ and $z$ of the sequence:

• Set $t=x+4$.
• Replace $(x,y,z)$ by $(y,z,t)$.
• Iterate.

Initialize suitably.

Answer 8

Along the lines of Did's answer: Keep the last two terms $x,y$. If $x=y$ then the next term is $z=x+2$. If $x \neq y,$ then if $x$ is divisible by 4, put $z=x+4$, and if $x$ is not divisible by $4$ put $z=x+2.$ Update $(x,y)$ to $(y,z)$ and iterate. [Initial values $8,8$]