# Bounded recursive sequence

I would like to know if there are known bounded recursive sequence (non monotonic):

It shouldn't be a constant, neither a convergent sequence, nor a periodic one.

(I am not asking for a true random generator) but the aim is to get a different bounded number in every step, I would like for example a sequence for digits of an irrational number defined recursively, for example $\pi$ or $e$ digits.

(if where possible non periodical but I could be fine a 'long' period)

Does anybody know a sequence with these characteristics?

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why shall it be recursive ? – Dominic Michaelis Mar 13 '13 at 18:23

The sequence $a_n=\sin n$ is bounded, not monotonic, not constant, doesn't converge, and isn't periodic.
I'm not certain whether you are using "recursive" in the math-logic sense of the term, or if you just want it to satisfy a recurrence relation. Assuming the latter, $a_n$ satisfies a three-term constant coefficient linear recurrence. To find it, eliminate $\cos n$ from the formulas, $\sin(n+1)=\sin n\cos1+\cos n\sin1$ and $\sin(n+2)=\sin n\cos2+\cos n\sin2$.
Alternatively, consider (almost) any solution of the recurrence $$2a_n=3a_{n-1}-2a_{n-2}$$ The solutions are of the form $a_n=b_+(r_+)^n+b_-(r_-)^n$ where $$r_{\pm}={3\pm\sqrt{-7}\over4}$$ and $b_{\pm}$ are constant depending on the initial conditions $a_0$ and $a_1$. Since $r_{\pm}$ have modulus $1$ but are not roots of unity, the sequence satisfies all the criteria (so long as you stay away from $a_0=a_1=0$). It can be expressed in terms of sines and cosines so it's really a second cousin to the first example.
Yes is a recurrence relation, I didn't knew about $2a_n=3a_{n-1}-2a_{n-2}$, I would have thought that it was a periodic function, is amazing, thanks – Hernán Eche Mar 14 '13 at 17:47