General formula for nth element of the sequence 1, 1, 0, 1, 1, 0, 1, 1, 0, ...

Question

Give a general formula for nth element of the sequence 1, 1, 0, 1, 1, 0, 1, 1, 0, ...

I have been trying any combination, but I can't find a solution.

score 2 · Answer 1 · answered Dec 09 '20 at 16:57

2

How about $a_n=\dfrac43\sin^2\left(\dfrac{n\pi}3\right)$?

answered Dec 09 '20 at 16:57

J. W. Tanner

using this formula in excel =(4/3)POWER(SIN((nPI())/3),2) but i get a wrong answer – M. Carlo Bramini Dec 09 '20 at 17:04
1

when I use that formula in Excel, it looks right – J. W. Tanner Dec 09 '20 at 17:06
1

@Luther Double-check the values of $n$. – J.G. Dec 09 '20 at 17:07
1

yep all good thanks – M. Carlo Bramini Dec 09 '20 at 17:12
I realized that we needed something with period $3$; that suggested sin or cos of multiples of $\frac\pi3$; if squared, $\sin\left(\frac{n\pi}3\right)$ gives $0$ when $n$ is a multiple of $3$ and $\frac34$ otherwise, so I scaled by $\frac43$ – J. W. Tanner Dec 09 '20 at 17:17

score 2 · Accepted Answer · answered Dec 09 '20 at 17:02

2

I think $a_n =1-\frac{1}{3}\left(1+e^{\frac{2i\pi}{3}n}+e^{\frac{4i\pi}{3}n}\right)$ works.

answered Dec 09 '20 at 17:02

2

I think that works too; $+1$ – J. W. Tanner Dec 09 '20 at 17:08
@Jet Chung trying it on excel but it gives me an error
=1-(1/3)(1 + EXP(((COMPLEX(0,2)PI())J8/3)J8)+EXP(((COMPLEX(0,4)PI())J8/3)*J8))

I'm probably writing the formula incorrectly
– M. Carlo Bramini Dec 09 '20 at 18:00
I'll mark your correct if the answer is more accurate – M. Carlo Bramini Dec 09 '20 at 18:03
@Luther I don't really know how to use Excel, but a quick Google search tells me COMPLEX(COS(2PI()J8/3), SIN(2PI()J8/3)) and COMPLEX(COS(4PI()J8/3), SIN(4PI()J8/3)) should work. – Dec 09 '20 at 20:08
Where I'm using $e^{i\theta} = \cos{\theta} + i \sin(\theta)$. – Dec 09 '20 at 20:10
This seems to work in Google sheets (I assume it would work the same in Excel): =COMPLEX(ROUND(2/3-(COS(2PI()/3A1)+COS(4PI()/3A1))/3, 5), ROUND((SIN(2PI()/3A1)+SIN(4PI()/3A1))/3,5)) – Dec 09 '20 at 20:26

score 0 · Answer 3 · answered Dec 09 '20 at 17:21

0

If $n$ is divisible by $3$, then $a_n=0$. Otherwise, $a_n=1$.

So, the formula $a_n=\min(n \bmod 3,1)$ should work.

answered Dec 09 '20 at 17:21

Geoffrey Trang

3 Answers3