Once again my troubles are with $$[z^n]\frac{f(z)}{(1-z/p)^\alpha}\sim \frac{f(p)}{\Gamma(\alpha)}n^{\alpha-1}p^n$$
In Sedgewick and Flajolet's book, they claim that this formula allows us to go, ignoring the constant from:
$$ (1- \sqrt{1-4x})/2$$ to $$ 4^n/(n\sqrt{n\pi })$$
I didn't quite get this:
1) Isn't $p$ here 1/4, so we would end up with $4^{-n}$ instead?
2) Shouldn't the catalan recurrence be $ (1- \sqrt{1-4x})/2x$ ? Does this x make a difference to the application of the formula?
Cheers
The transfer theorem should be $$[z^n]\frac{f(z)}{(1-z/p)^\alpha}\sim \frac{f(p)}{\Gamma(\alpha)}n^{\alpha-1}p^{\color\red {-n}}$$ and the generating function for the Catalan numbers should be $$(1- \sqrt{1-4x})/(\color\red{2x})$$ which does make a difference in the application of the theorem, as you surmised.