TITLE: Number of ways to represent any N as sum of odd numbers?
QUESTION [13 upvotes]: I was solving some basic Math Coding Problem and found that For any number $N$, the number of ways to express $N$ as sum of Odd Numbers is $Fib[N]$ where $Fib$ is Fibonnaci , I don't have a valid proof for this and didnot understand that how this can be solved using recurrences Can someone provide with it ?
If you are not getting it Suppose for N=4 number of ways to write it as sum of Odd Numbers is 3 which is Fibonnaci at $3$ 
$4=> 1+1+1+1$ 
$4=> 1+3$ 
$4=> 3+1$ 
NOTE-> the composition is ordered $( 1+3)$ and $(3+1)$ are different .
UPD -> I do not claim that I observed it myself but in the problem solution I found it , I asked to just find some valid proof / reason to it

REPLY [4 votes]: We have from first principles that the number of compositions into odd parts is given by
$$[z^N] \sum_{q\ge 1} \left(\frac{z}{1-z^2}\right)^q.$$
This simplifies to
$$[z^N] \frac{z/(1-z^2)}{1-z/(1-z^2)}
= [z^N] \frac{z}{1-z-z^2}.$$
Now $$F(z) = \frac{z}{1-z-z^2}$$ is the OGF of the Fibonacci numbers and we have the claim.