While doing some computational experiments, I found the following (conjectured) identity:
$$2^{2n-1}\sum_{k=0}^{\infty}(-1)^{k}\frac{a_{n,k}}{2^k}=\binom{2n}{n}$$
where $a_{k}$ is defined as
$$a_{n,k+1}=\sum_{i=1}^{n-1}\frac{a_{i,k}}{i+1}$$
with $a_{n,0}=1$.
If we truncate the sum to the third term, for example, we obtain
$$2^{2n-1}\left(1-\frac{1}{2}(H_{n}-1)+\frac{1}{4\cdot2}\left(H_{n}^2-H_{n}^{(2)}-2(H_{n}-1)\right)\right)$$
where $H_{n}^{(q)}$ is the $q$-th generalized harmonic number, $\sum_{k=1}^{n}\frac{1}{k^{q}}$.
The values for $n=1$, $n=2$ and $n=3$ are $2$, $6$ and $20$ like in the central binomial coefficient, and then it starts to grow bigger.
How can it be proved? I'm trying with generating functions, but I don't know how to simplify $\sum_{m=0}^{\infty}2^{2m-1}a_{m,k}x^{m}$
Best Answer
Consider the function $$A_k(x)=\sum _{n=0}^{\infty}a_{n,k}x^n,$$ your recursion can be state like $A_0(x)=1/(1-x)$ and $$A_k(x)=\frac{1}{1-x}\int A_{k-1}(x)dx$$ Show that this implies that $A_k(x)=\frac{1}{k!(1-x)}ln^k(1/(1-x))$.
Hint: Convert your main equation to the form $$\sum _{k=0}^{\infty}\frac{(-1)^kA_k(4x)}{2^k}=\frac{1}{\sqrt{1-x}},$$ at the end you are like composing an exponential with the logarithm you get from the recursion.