Speaking only for myself, the reason that I don't publish all my results is that I write very slowly. (It has nothing to do with the nature of combinatorics.) It takes a long time for me to arrange my work into something publishable; it's much quicker to put some results into a slide presentation for a talk.

But I'd like to thank you for the references. I do intend to publish this result some day and it's good to know that there are other references to it.

For the benefit of readers who, like myself, think in terms of counting lattice paths rather than sums of random variables associating with random walks, it may be hepful to restate these formulas in terms of lattice paths. We consider paths made up of up steps $(1,1)$ and down steps $(1,-1)$, starting at the origin. (The horizontal components are irrelevant and are included only for convenience in visualization.) The result of Chung and Feller is that among the $4^n$ paths of length $2n$, the number with $2k$ steps above the $x$-axis is $\binom{2k}{k}\binom{2n-2k}{n-k}$. The analogous result for paths of odd length is that  the number of paths of length $2j + 2k − 1$ with $2j − 1$ steps above the $x$-axis and $2k$ steps below is 
$\frac{j}{2(j+k)}\binom{2j}{j}\binom{2k}{k}$. (A path of odd length cannot end on the $x$-axis; if it ends above the $x$-axis it must have an odd number of steps above the $x$-axis. Switching $j$ and $k$ allows us to count paths with an even number of steps above the $x$-axis.)