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Let a sequence $(a_n)$ by $a_1=1,\ a_2=4,\ a_{n}=a_{n-1}+a_{n-2}$

Denote $f(x)=\sum_{i=1}^\infty a_i x^i$. For which $0

I tried tackling this problem (I found online) yet I only thought of expressing $a_n$ with only $n$. I think it's not going to help much and the closed formula is rather ugly (I think I could do it using the same trick done for Fibbonaci with diagonalization of matrices).

I'm not too much of an exprert when it comes to this kind of questions (as I know only some Linear Algebra and Calculus) so I probably won't understand too complicated solutions, which I ask you to refrain from. If you have a doubt whether I know something you're welcome to ask me in the comments.

Other than that I do thank for any help in advance!

EDIT: I now see that $x$ much be rational, as otherwise $f(x)$ isn't rational. But how do I know which rationals satisfy this?

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    Hint: $\mathbb{Z}$ is closed under addition and multiplication. – 2017-01-30
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    Yes, but given a sum is whole doesn't indicate anything about the numbers it's made of being whole. – 2017-01-30
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    You may explicitly compute $f$ in terms of a rational function. To be precise, $(1-x-x^2)f(x) = ax + bx^2$ for some $a, b$ and it is not hard to figure out what they are. At the end, your question boils down to solving certain quadratic equation. – 2017-01-30
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    Okay, I understand how you got that equation with $f$, However, finding $a$ and $b$ isnt that easy for me. Assuming I have them, then I can see what whole numbers $f$ get (which is obviously finite, as $f$ is bounded in $(0,1)$) – 2017-01-30
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    I got $a=a_1, b=a_2-a_1$ is that correct or close? – 2017-01-30

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Observe that for all $a,b,x,y\in\Bbb{C},$ $$ax^{n+1}+by^{n+1}=(x+y)(ax^n+by^n)-xy(ax^{n-1}+by^{n-1}).$$ By choose $x+y=-xy=1,$ we can have $a_n=ax^n+by^n,$ where $a$ and $b$ determined by the initial values $a_1=1, a_2=4.$
Now note that $$f(z)=\sum_{k=1}^{\infty}a_kz^k=a\sum_{k=1}^{\infty}(xz)^k+b\sum_{k=1}^{\infty}(yz)^k.$$ Two geometric series in the right hand side are convergent if and onlf if $|z|\lt\min\{|x|^{-1},|y|^{-1}\}.$ When they are convergent we have $$f(z)=\dfrac{axz}{1-xz}+\dfrac{byz}{1-yz}.$$ Since you can compute all parameters $a,b,x,y$ it is not difficult to plot the graph of the rational function $f$ over the domain $\left(0,\min\{1, |x|^{-1},|y|^{-1}\}\right).$