What's the mathematical formula for the sequence 3 3 6 6 12 12 24 24 etc?

Question

The closest I could find is $y=2^{\operatorname{ceil}\left(0.5x\right)}\frac{3}{2}x$ but its not quite right.

I possible, I'd like to not use sums (∑).

What is the next term in this sequence is the typical ill-posed question. For what I know, $3,3,6,6,12,12,24,24$ could continue with $1728,153,81,0,0,0,21,\pi$. See this related question: https://math.stackexchange.com/questions/1790642/general-formula-for-the-1-5-19-65-211-sequence/1790666#1790666 — Jack D'Aurizio, Apr 07 '18 at 20:09

score 3 · Answer 1 · answered Apr 07 '18 at 19:08

3

You can use, for $n\geq 1$, $$a_n = 3\cdot 2^{\lfloor\frac{n-1}{2}\rfloor}$$ where $\lfloor x\rfloor$ is the floor of $x$.

answered Apr 07 '18 at 19:08

Clement C.

score 0 · Accepted Answer · answered Apr 07 '18 at 19:09

0

$3 \cdot2^{ceil(0.5*x)-1}$, for $x=1,2,\ldots$

answered Apr 07 '18 at 19:09

user52227

score 0 · Answer 3 · answered Apr 07 '18 at 19:23

0

For $n\ge 1,$

$$a_{2n}=a_{2n-1}=3.2^{n-1}$$

answered Apr 07 '18 at 19:23

hamam_Abdallah

3 Answers3