Recall the definition of Catalan numbers: $$C_n=\frac1{n+1}\binom{2n}n=\frac{2^n(2n-1)!!}{(n+1)!}.\tag1$$ Now consider the following series with a parameter $n\in\mathbb N^+$: $$S_n=\frac{2\cdot18^n}{3\sqrt5}\cdot\sum_{k=1}^\infty\frac{C_{2k}\cdot k^n}{5^{2k}}.\tag2$$ It appears that $S_n$ is always integer (and odd), with the following values for $n=1,2,3,...:$ $$1, 43, 3009, 318507, 46065921, 8482079403, 1899432317889, 501282878789547,...$$ (click here to see more terms)
Can we prove that $S_n$ is always integer (odd)? Can we find an explicit formula for $S_n$ not involving an infinite summation?
Let
$$f(x) = \sum_{k=1}^{\infty} \frac1{2 k+1} \binom{4 k}{2 k} x^k $$
Then differentiating, multiplying, integrating, etc., we find that
$$f(x) = \frac1{4 \sqrt{x}} \left [(1+4 \sqrt{x})^{1/2} - (1-4 \sqrt{x})^{1/2} \right ] - 1$$
It should be plain that, if we define the operator
$$\mathcal{D} = x \frac{d}{dx} $$
Then
$$S_n = \frac{2 \cdot 18^n}{3 \sqrt{5}} \mathcal{D}^n f\left (\frac1{25} \right ) $$