I'm new to symbolic method, and I'd like to clarify a concept through an example on Fibonacci numbers.
Let $A_n := \{ \text{words in the alphabet} \ \{1,2\} \text{which sum up to} \ n \}$. It is known that $|A_n| = F_{n+1}$. I'd like to find an expression for $$\sum\limits_{n \geq 0}F_{n+1}^2x^n.$$ The idea is to consider $B_n := \{ (\alpha,\beta) : \alpha \wedge \beta \in A_n\}$ (since we know that $B_n = F_{n+1}^2$) and a statistic giving the length(identifying an element of $B_n$ as a $2 \times n$ strip made of squares $1\times 1$ and dominoes $1\times2$) on $\bigcup\limits_n B_n$. In other words we have the following:
What I don't get is how for example there are only $1$ "prime" tile of length $2$, since I don't see how the element $(11,2)$ or $(2,11)$, could be made out the prime given in the text. What am I missing?
Any help or hint would be appreciated.
Ok so this took me a really long time to figure out and I think the solution writer could have phrased this much better.
They are not saying there is only $1$ prime factor of length $2$. Rather there is $1$ of the "obvious type". There are then $2$ more that fall under the type of even length $2n\geq 2$ (in this case $n=1$). So in total there are $\boxed{3}$ prime factors of length $2$.
And when you go expand $1-x-x^2-\sum_{n\geq 2} 2x^n$, you will see that the coefficient of $x^2$ is $-3$ as we expected since the sum starts at $n=2$.
Does that answer all your questions?