Consider the infinite polynomial series $A_F(x)=x{F}_1+x^2{F}_2+x^3{F}_3+\dots$, where $F_k$ is the $k$th term in the Fibonacci sequence: $1,1,2,3,5,8,\dots$; that is, $F_k=F_{k-1}+F_{k-2}$, $F_1=1$ and $F_2=1$.

For this problem we shall be interested in values of $x$ for which $A_F(x)$ is a positive integer.

Surprisingly

$\begin{array}{rl}{A}_F(\frac{1}{2})& =(\frac{1}{2})\times 1+(\frac{1}{2}{)}^2\times 1+(\frac{1}{2}{)}^3\times 2+(\frac{1}{2}{)}^4\times 3+(\frac{1}{2}{)}^5\times 5+\cdots \\ & =\frac{1}{2}+\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{5}{32}+\cdots \\ & =2\end{array}$

The corresponding values of $x$ for the first five natural numbers are shown below.

$x$	$A_F(x)$
$\sqrt{2}-1$	$1$
$\frac{1}{2}$	$2$
$\frac{\sqrt{13}-2}{3}$	$3$
$\frac{\sqrt{89}-5}{8}$	$4$
$\frac{\sqrt{34}-3}{5}$	$5$

We shall call $A_F(x)$ a golden nugget if $x$ is rational, because they become increasingly rarer; for example, the $10$th golden nugget is $74049690$.

Find the $15$th golden nugget.