Some background.
I was asked to find an arithmetic function $f$ such that $f*f=\mathbf 1$ where $\mathbf 1$ is the constant function 1 and $*$ denotes Dirichlet convolution. I was able to prove that there are two solutions $\pm f$ and that $f$ is multiplicative. Next, I would have to evaluate $f$ at prime powers. I constructed a few values and my conjecture is that $$f(p^n)=\frac{2n-1\choose n}{2^{2n-1}}$$ for $p$ prime and $n>0$. To prove this, I only need to show that
$$\sum_{k=1}^n{2k-1\choose k}{2n-2k+1\choose n-k+1}=4^n-{2n+1\choose n+1}\qquad\text{for }\;n\geq0.$$ (This is simply expressing $(f*f)(p^{n+1})=1$ explicitly, plugging in the conjecture.)
For readers who don't really understand what I'm talking about and who are merely interested in the proof of the identity, you can just start reading from here.
Hoping for a combinatorial proof, I interpreted the summation as follows. Given a set of $n+1$ indistinguishable marbles and $n+1$ distinguishable bags (say $b_1,\ldots,b_{n+1}$), the term ${2k-1\choose k}{2n-2k+1\choose n-k+1}$ counts the number of ways to put the marbles in the bags such that there are exactly $k$ marbles in the first $k$ bags $b_1,\ldots,b_k$.
Equivalently, if we identify a configuration of the marbles with a monotonic path in a $n+1\times n+1$ grid such that the path starts in the bottom left corner and ends in the upper right corner, the sum $${2n+1\choose n+1}+\sum_{k=1}^n{2k-1\choose k}{2n-2k+1\choose n-k+1}$$ counts the number of times a path 'crosses' or 'touches' the main diagonal in a point that is not the 'origin', if we summate over all possible paths. (There are ${2n+2\choose n+1}$ such paths in total.) For example, the following path touches the main diagonal $4$ times: At $(2,2)$, $(3,3)$, $(4,4)$ and $(7,7)$.
(We do not count $(0,0)$ because the summation doesn't.) However, interpreting the summation like this I can't get any further. Any other ideas or suggestions on how to approach this problem?
Edit: There are some errors in my reasoning above, let's try again.
Using the identity ${2n-1\choose n}=\frac12{2n\choose n}$ it can be rewritten as $$\sum_{k=0}^n{2k\choose k}{2n-2k\choose n-k}=4^n$$ which looks much better and holds for all $n\geq0$, making it more naturally. This form may give some ideas for combinatorial proofs but I don't really see any.
The term ${2k\choose k}{2n-2k\choose n-k}$ counts the number of $n\times n$ monotonic paths intersecting the diagonal at $(k,k)$. So the summation counts the number of intersection points with the diagonal (all of them this time, including the origin) summing over all paths.
As in Arthur's comment, it would suffice to find a bijection between all $2n$-monotonic paths (no matter their width or height) and the pairs $(p,s)$ where $p$ is a $n\times n$ path and $s$ an intersection point with the diagonal. Perhaps there is a weird bijection which would then solve the question.
For the sake of a proof with induction, I considered all paths that intersect the diagonal for the first time at $(k,k)$ and all possible continuations and their intersection points, summed for $k$ from $1$ to $n$ using the induction hypothesis and a trick with Catalan numbers. (Writing out the details would be tedious, I think the reasoning becomes clear when you see the sum.)
It turns out to be sufficient to prove
$$\sum_{k=1}^n2C_{k-1}\left(4^{n-k}+{2n-2k\choose n-k}\right)=4^n$$ where $C_n=\frac1{n+1}{2n\choose n}$ denotes the $n$th Catalan number. However this doesn't seem to be a simplification.