Counting particular Dyck paths

Question

The OEIS entry for Pascal’s triangle contains the following intriguing remark:

$C(n,k)$ = the number of Dyck paths of semilength $n$, with $k$ "u"'s in odd numbered positions and $k$ returns to the x-axis.

Example: {U = u in odd position and _ = return to x axis}

$\mathrm{binomial}(4,1)=1$ (Uudududd_)
$\mathrm{binomial}(4,2)=3$ (Uududd_Ud_, Ud_Uududd_, Uudd_Uudd_)
$\mathrm{binomial}(4,3)=3$ (Ud_Ud_Uudd_, Uudd_Ud_Ud_, Ud_Uudd_Ud_)
$\mathrm{binomial}(4,4)=1$ (Ud_Ud_Ud_Ud_)

– Roger Ford, Nov 05 2014

It isn’t obvious to me why this should be true. I found the edit where this remark was added, and it doesn’t include any links or references.

Can anyone give a proof – or supply a reference, if this is a known result?

Added: As Anthony Quas points out in the comments, the example doesn’t match the description. Based on the example the claim should presumably be that $n \choose k$ is the number of paths of semilength $n+1$ that have $k+1$ u’s in odd positions and $k+1$ returns. But this error makes me wonder if the whole thing is nonsense. I’ll check some more examples.

Answer 1

The answer is pretty simple. Consider any of $k$ parts between consecutive touches of $x$-axis. Note that we can assign a unique U character to each of these parts, namely, the first character, which meets our quota of U characters. The last character of each part obviously has to be d. Between the first and the last character of a part, no character at odd position can be U (since the quota is met), so they have to be d. All the other characters in this part have to be u to compensate the balance, so the only way a part can look like is Uudud...udd for some number of repetitions of ud. So the answer is equal to the number of ways to cut a sequence of length $n$ into $k$ non-empty parts, which is ${n - 1 \choose k - 1}$.