Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

I've read about Spigot algorithms, but they are not defined recursively.

share|cite|improve this question
why shall it be recursive ? – Dominic Michaelis Mar 13 '13 at 18:23
up vote 1 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.