Analyzing the Coefficient of a generating function and its asymptotic The generating function
$\sum_{n\geq 0} D(n) x^n = \frac{1}{\sqrt{1-6x+x^2}}$ is the gf of the Delannoy number.
See last paragraph in https://en.wikipedia.org/wiki/Delannoy_number
In this link, they mentioned that the coefficient of this gf  is $D(n) = \sum_{k=0}^{n} {n \choose k} { n+k \choose k}$.
My question is how can this coefficient be calculated/extracted from its gf.And how they conclude that it behaves asymptotically as
${\displaystyle D(n)={\frac {c\,\alpha ^{n}}{\sqrt {n}}}\,(1+O(n^{-1}))}$
where
$\alpha =3+2{\sqrt {2}}\approx 5.828$
and
$c=(4\pi (3{\sqrt {2}}-4))^{-1/2}\approx 0.5727$.
How to calculate/extract $D(n)$ from its gf and how to analyze it asymptotically then?
Is that possible to do that with a direct calculations?
Edit
I tried to find the asymptotic of the codfficidnt in the above gf this way:
The gf has two poles: $r_1=3+\sqrt{8}$, $r_2=3-\sqrt{8}$. So the smallest/dominant pole is $r_2=3-\sqrt{8}$.
We can write the gf as:
$A(x)=1/ \sqrt{r_2-x} * 1/ \sqrt{r_1-x}$
Then A(x) behaves as:
$1/ \sqrt {r_1-r_2} * 1/ \sqrt{r_2-x}=
1/ \sqrt {2 * 8^{1/4}} * \sum_{n\geq 0} {-1/2 \choose n} (-1)^n (r_1)^{n+1}x^n$
Here I used the expansion's formula  of $(1+x)^a$ for $a=-1/2$ and $x=-x/r_2$.
So the coefficient behaves asymptotically as:
$1/ \sqrt 2 * 1/ 8^{1/4} * (3+\sqrt 8)^{n+1/2} {-1/2 \choose n} (-1)^n$.
We can simplify it more, by using the asymptotic of ${-1/2 \choose n}$.
Is that approach fine?

We use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a series. This way we can write for instance
\begin{align*}
\binom{n}{k}=[x^k](1+x)^n\tag{1}
\end{align*}

We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^n}&\color{blue}{\binom{n}{k}\binom{n+k}{k}}\\
&=\sum_{k=0}^n\binom{n}{k}[x^k](1+x)^{n+k}\tag{2}\\
&=[x^0](1+x)^n\sum_{k=0}^n\binom{n}{k}\left(\frac{1+x}{x}\right)^k\tag{3}\\
&=[x^0](1+x)^n\left(1+\frac{1+x}{x}\right)^n\tag{4}\\
&=[x^{-1}](1+x)^n(1+2x)^n\frac{1}{x^{n+1}}\tag{5}\\
&=[y^{-1}]\left.\frac{1}{(1+x)(1+2x)\,\frac{1-2x^2}{(1+x)^2(1+2x)^2}}\right|_{x=g(y)}\cdot\frac{1}{y^{n+1}}\tag{6}\\
&=[y^n]\left.\frac{(1+x)(1+2x)}{1-2x^2}\right|_{x=g(y)}\tag{7}\\
&\,\,\color{blue}{=[y^n]\frac{1}{\sqrt{1-6y+y^2}}}\tag{8}\\
\end{align*}
and the claim follows.

Comment:

*

*In (2) we use the coefficient of operator according to (1).


*In (3) we use the linearity of the coefficient of operator and apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$.


*In (4) we apply the binomial theorem.


*In (5) we prepare for the substitution rule which is the essence of this derivation.


*In (6) we use the substitution rule (Rule 5, one-dimensional case) from section 1.2.2 of G. P. Egorychev's Classic: Integral Representation and the Computation of Combinatorial Sums. Here we have
\begin{align*}
f(x)=(1+x)(1+2x)\qquad\qquad y(x)&=\frac{x}{f(x)}=\frac{x}{(1+x)(1+2x)}\tag{9}\\
y^{\prime}(x)&=\frac{1-2x^2}{(1+x)^2(1+2x)^2}
\end{align*}
and apply the substitution rule:
\begin{align*}
[x^{-1}]f(x)^n\cdot \frac{1}{x^{n+1}}=[y^{-1}]\left.\frac{1}{\left(f(x)\cdot y^{\prime}(x)\right)}\right|_{x=g(y)}\cdot\frac{1}{y^{n+1}}
\end{align*}
where $x=g(y)$ is the inverted function given by (9).


*In (7) we do a simplification and apply the rule as we did in (3).


*In (8) we note that with $y(x)=\frac{x}{(1+x)(1+2x)}$ we obtain:
\begin{align*}
1-6y+y^2&=1-\frac{6x}{(1+x)(1+2x)}+\frac{x^2}{(1+x)^2(1+2x)^2}\\
&=\frac{(1+x)^2(1+2x)^2-6x(1+x)(1+2x)+x^2}{(1+x)^2(1+2x)^2}\\
&=\frac{(2x^2-1)^2}{(1+x)^2(1+2x)^2}\\
\frac{1}{\sqrt{1-6y+y^2}}&=\frac{(1+x)(1+2x)}{1-2x^2}
\end{align*}

Notes:

*

*Another approach of the asymptotic coefficient expansion of $\frac{1}{\sqrt{1-6x+x^2}}$ can be found in this answer.


*A somewhat more demanding application of the substitution rule (two-dimensional case) is given in this answer.
