Generalization of some infinite series containing binomial coefficients

Question

On a page here. There are some infinite series in the form $$\sum_{k=1}^{\infty}\frac{k^n}{\binom{2k}{k}}=\frac{a}{b}+\frac{c \pi}{d}$$ Where $n \in {[0,1,2,3,4...]}$ and for some natural numbers $a,b,c,d$. the question is that- Is there any formula to find out $a , b , c , d$ with respect to every $n$?

Answer 1

I believe these come from $$ \sum_{k=1}^\infty \dfrac{t^k}{{2k}\choose k} = \dfrac{t}{4-t} + 4 \dfrac{\sqrt{t}}{(4-t)^{3/2}} \arcsin(\sqrt{t}/2)$$ If this function is $F(t)$, then $$ \sum_{k=1}^\infty \dfrac{k^n}{{2k} \choose k} = \left. \left(t \dfrac{d}{dt}\right)^n F(t)\right|_{t=1}$$ For example, for $n=13$ I get (with Maple's help) $$ \sum_{k=1}^\infty \dfrac{k^{13}}{{2k} \choose k} = {\frac {727348814}{3}}+{\frac {1315508114654}{59049}}\,\sqrt {3}\pi $$