Fountains of Coins and Fibonacci Numbers I recently came across a problem in Herbert Wilf's Generatingfunctionology book that I can't come up with an elegant solution to and can't find any solutions online. The problem statement begins on page 38
https://www.math.upenn.edu/~wilf/gfologyLinked2.pdf
and concerns 'fountains of coins', basically what you get when you stack rows of coins in 2 dimensions. The book goes through a proof that the generating function for the number of block fountains, basically fountains with no gaps, that can be with a base that is exactly $k$ coins wide. The generating function ends up being 
$$F(x) = \frac{1-2x}{1-3x+x^2}$$
It turns out that if you look at the terms in the sequence, they are the odd terms of the Fibonacci numbers, $F_{2N-1}$. The book states this and leaves it as an exercise, but I can't seem to make any progress on any sort of nice combinatorial argument to show this. If anyone has any insight, it would be greatly appreciated!

Let us call : $Fib(x)$ the generating function associated to the Fibonacci number $(F_k)$. We know that :
$$Fib(x)=\frac{x}{1-x-x^2} $$
Now if we want to make appear something like a serie with only the $F_{2k+1}$ it seems like a good idea to compute :
$$Fib(x)-Fib(-x)=2\sum_{k=0}^{\infty}F_{2k+1}x^{2k+1} $$
$$Fib(x)-Fib(-x)=2x\sum_{k=0}^{\infty}F_{2k+1}(x^2)^k $$
Hence, we see that if $G(y)$ is the generating function for the $(F_{2k+1})$ we must have that :
$$G(x^2)=\frac{Fib(x)-Fib(-x)}{2x} $$
Now it suffices to compute :
$$Fib(x)+Fib(-x)=\frac{x}{1-x-x^2}+\frac{x}{1+x-x^2} $$
$$Fib(x)+Fib(-x)=\frac{2x-2x^3}{1-3x^2+x^4}$$
Hence we have that :
$$G(x^2)=\frac{1-x^2}{1-3x^2+x^4} $$
Hence :
$$G(y)=\frac{1-y}{1-3y+y^2} $$
Now the generating function for the $F_{2k-1}$ is exactly $yG(y)$ hence :
$$\sum_{k\geq 0}F_{2k-1}y^k=\frac{y-y^2}{1-3y+y^2} $$
On the other hand compute :
$$F(y)-1=\frac{1-2y}{1-3y+y^2}-1=\frac{y-y^2}{1-3y+y^2} $$
Then you see that $F(y)-1=yG(y)$ this is exactly saying that the $k$-coefficient of $F(y)$ for $k\geq 1$ is $F_{2k-1}$. 
