Find all sequences that satisfy the recurrence relation

$$u_n\cdot (u_{n+1})^2-u_{n+1}-u_n+1=0, \text{with }u_0=1$$

My try

First, we find $u_1$, which follows $u_0=1$.

$u_0\cdot (u_{1})^2-u_{1}-u_0+1=0$



$u_1=0 \vee u_1=1$

Clearly, there are two possible sequences $a$ and $b$, which respectively begin with $1,0\ldots$ and $1,1,\ldots$.

Then, we find the $u_2$ for sequence $a$:

$u_1\cdot (u_{2})^2-u_{2}-u_1+1=0$




I think the answer is:

$1,0,1,0,1,\ldots$ and $1,1,1,1,1,\ldots$


Am I right?

  • $\begingroup$ You may think of $1,0,1,0,1,1,1,0,1,1,...$ or $1,0,1,1,1,0,1,0,1,1,...$ ;) $\endgroup$ – Leaning Sep 28 '14 at 12:49
  • $\begingroup$ Oh, shit, forgotten. Thanks! How should I write this down formally? $\endgroup$ – rae306 Sep 28 '14 at 12:57
  • $\begingroup$ As suggested by Dom, I think we may write the set of all sequences as $\{\{u_n\}|\forall n: u_n\in\{0;1\},u_n+u_{n+1}\ne 0\}$ $\endgroup$ – Leaning Sep 28 '14 at 13:01

You're basically there, but you seem to have forgotten that after a 1 you could have either a 0 or a 1. Basically it is all sequences of 1's and 0's where a one always follows a 0.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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