The expected number of returns of a simple random walk in $1d$ after $2n$ steps is a well known problem, with answer $\sum_{m=1}^n \frac{\binom{2m}{m}}{2^{2m}}$.
A question that requires more information is the probability distribution of the number of returns of a simple random walk of $2n$ steps in $1d$. The mean of this distribution would be the quantity above.
According to OEIS A276418, the number of paths of length $2n$ that hit the origin exactly $k$ times is merely $$N_k = 2^k \binom{2n-k}{n-k}$$ which would give a probability distribution, for $k$ an integer between $0$ and $n$ inclusive, $$P_k = \frac{2^k \binom{2n-k}{n-k}}{2^{2n}}$$
A quick numerical check verifies that the mean of this distribution equals $\sum_{m=1}^n \frac{\binom{2m}{m}}{2^{2m}}$ for several values of $n$.
How can I derive OEIS A276418; that is, how can I derive $N_k$, or, equivalently, $P_k$? The probability to hit the origin at step $m$ and $m'$ are not independent, which is providing something of a stumbling block. I hope to be able to generalize the results here to higher dimensions, but I'll restrict this question to the $1d$ case.