What is closed form of f when $f(n)=\frac{1}{f(n-1)}+f(n-2), f(2)=1, f(1)=0$?

Question

I've tried to solve this problem this way with characteristic equation:

What is the closed form of the $f$ with $f(1)=1$, $f(2)=7$ and $f(n)=7f(n-1)-12f(n-2)$ ($n\ge 3$)?

$F_n=x^n$

$x^{n-2}(x^2-\frac{1}{x}-1) = 0$

$x^3-x-1 = 0$

but this equation have only one real solution and i'm confused if i can further use this method or not. Any ideas?

Answer 1

Following @Cesareo's answer and checking the initial values, we find that $$ \tag1f(n)f(n+1)=n-1$$ for all $n$. With the double factorial $$ n!!=\prod_{0\le k<\frac n2}(n-2k)=\begin{cases} m!\cdot 2^m,&n=2m\\ \frac{n!}{m!^2\cdot 4^m},&n=2m+1\\ \end{cases},$$ we can see that (at least for $n\ge 3$) $$ f(n)=\frac{(n-2)!!}{(n-3)!!}$$ fits the recursion because it correctly produces $$ f(3)=\frac{1!!}{0!!}=1$$ and makes $$ f(n+1)f(n)=\frac{(n-1)!!(n-2)!!}{(n-2)!!(n-3)!!}=n-1.$$

Answer 2

Welcome to MSE.

If you try the MATLAB code below

clc 
clear 
a(1)=0; 
a(2)=1; 
for i=3:30  
    a(i)=1/a(i-1)+a(i-2); 
   end;
plot(a)  

then you will see the oscillatory behavior of the sequence. Like this:

Are you sure that there is not a typo?

Answer 3

Hint.

Making $$ G(n) = f(n)f(n-1) $$

we have the recurrence

$$ G(n) -G(n-1) = 1 $$

with solution

$$ G(n) = c_0 + n $$

and now from $G(2) = f(2)f(1) = 0 = c_0+2$ we have

$$ f(n)f(n-1) = n-2 $$

now as

$$ \cases{ f(n+1)f(n) = n-1\\ f(n)f(n-1) = n-2 }\Rightarrow f(n+1) = \frac{n-1}{n-2}f(n-1) $$

with initial conditions $f(2)=1,\ f(3) = 1$ and the recurrence solution is

$$ f(n) = \frac{\left((\pi -2) (-1)^n+2+\pi \right) \Gamma \left(\frac{n}{2}\right)}{2 \sqrt{\pi } \Gamma \left(\frac{n-1}{2}\right)} $$