# Closed-form representation of a recursively defined sequence

I am trying to find a closed-form representation of the following recursively defined sequence

$$π_0 = 0, π_1 = 0, π_2 = β2, π_3 = 0$$ $$π_{π+4} = β2π_{π+2} β π_n$$

I've been working with a lot of homework problems similar to this, but most of the time they are setup in the opposite way, meaning I generally have to find $a_0$ , $a_1$, $a_2$ etc. I am looking for a direction or hint to go in, even a website that may be useful for learning this type of problem would be greatly appreciated. Unfortunately, I am in an online class where it is difficult to get outside help.

I need to go to sleep but will update this in the morning after I attempt the problem. Hopefully I figure it out before I even log back onto my computer!

Thank you all in advance, I've learned so much from using this website in the past week.

• Have a look at the "characteristic polynomial" method. en.wikipedia.org/wiki/…. You will face two little technical difficulties: roots are imaginary and double. For an example of how the method works, you may look at en.wikipedia.org/wiki/Fibonacci_number#Closed-form_expression. – user65203 Sep 26 '14 at 10:48
• This is less general than what @Yves suggests, but in this special case, it just might work: calculate $a_4$, $a_5$, and so on, until you see a pattern; write down a closed form formula for this pattern; then prove by induction that this pattern is for real. – Gerry Myerson Sep 26 '14 at 10:54
• I spent half of the day today (after a physics exam, which went very well) watching lectures on this material for the question. This is the only question I cannot get and I really appreciate the answer/response from you Macavity @Macavity but I do not understand it. I am at a loss. Never taking 4 accelerated math courses at the same time again. – mar10 Sep 27 '14 at 6:09
• The question you just closed was a duplicate, which is not nice, and also a clear-cut case of plagiarism, which is far more serious. – Did Sep 27 '14 at 6:26

Let $e_n = a_{2n}$ and $o_n = a_{2n+1}$. Then both $e_n$ and $o_n$ satisfy the recurrence $f_{n+2} +2f_{n+1}+f_{n} = 0$, we just have initial values different.
Now $o_0=o_1=0 \implies o_n = 0$ for all $n$. Similarly the pattern $e_n=(-1)^n(2n)$ emerges quickly, and can be proven by induction easily.
• The GF is also easily written out as $\dfrac{-2 x^2}{(1 + x^2)^2}$, which is even (all odd coefficients vanish) and is readily expanded to get the terms found above. Nothing surprising. – Semiclassical Sep 26 '14 at 13:33