Need to figure out a formula that calculates -1,-1,+1,+1,-1,-1,+1,+1

[mesa]

So I need a formula that calculates -1,-1,+1,+1,-1,-1,+1,+1,... for a value of n that is not a piece wise.

So far I have come up with (-1)^[2n/3] but I don't like using greatest integer

I also did:
[cos(n × ∏)]!
The notation is not setup correctly and anyone that knows how to do it proper let me know. It is supposed to function like this:

cos(n∏)cos((n-1)∏)cos((n-2)∏)cos((n-3)∏)... until cos((n-n)∏)

I have dug in algebraically but don't think there is a solution, anyone have any thoughts on this?

 

[jedishrfu]

-cos(n*pi/2) --> -1, 0, 1, 0 ...
-sin(n*pi/2) --> 0, -1, 0, 1 ...

 

[Dickfore]

The period for your sequence is 4, so it is represented as a discrete Fourier series:
<br /> x_n = \sum_{m = 0}^{3}{a_{m} \, \exp \left(\frac{2 \pi i \, m \, n}{4} \right)},<br />
where the coefficients are found from:
<br /> a_{m} = \frac{1}{4} \, \sum_{n = 0}^{3}{x_{n} \, \exp \left(-\frac{2 \pi i \, m \, n}{4} \right)}<br />

Do the calculation by using x_0 = x_1 = -1, x_2 = x_3 = +1 for a_{0,1,2,3}, and express the complex exponentials via trigonometric functions through the Euler identity:
<br /> e^{i \alpha} = \cos \alpha + i \sin \alpha<br />

 

[mesa]

The period for your sequence is 4, so it is represented as a discrete Fourier series:
<br /> x_n = \sum_{m = 0}^{3}{a_{m} \, \exp \left(\frac{2 \pi i \, m \, n}{4} \right)},<br />
where the coefficients are found from:
<br /> a_{m} = \frac{1}{4} \, \sum_{n = 0}^{3}{x_{n} \, \exp \left(-\frac{2 \pi i \, m \, n}{4} \right)}<br />

Do the calculation by using x_0 = x_1 = -1, x_2 = x_3 = +1 for a_{0,1,2,3}, and express the complex exponentials via trigonometric functions through the Euler identity:
<br /> e^{i \alpha} = \cos \alpha + i \sin \alpha<br />

I am not familiar with this. What does it do?
It looks complex.

 

[coolul007]

<br /> i^{(n)(n+1)}<br />

 

[Curious3141]

<br /> i^{(n)(n+1)}<br />

Nice and simple.

 

[jedishrfu]

nice solution, need to mention for all n>0.

 

[mesa]

<br /> i^{(n)(n+1)}<br />

That's awesome :)
I played with i but gave up on it (apparently too quickly!)
Very simple solution

 

[Bohrok]

You can usually get sin or cos to get you a periodic sequence nicely, especially if you don't want to use i, so here's one more way:\sqrt{2}\cos\left(\frac{\pi}{4} + \frac{n\pi}{2}\right)

 

[jedishrfu]

or my solution: cos(n*pi/2 + pi) + sin(n*pi/2 + pi)