N is not going to be that simple to calculate. Any given adjacency you find could have multiple possible cycles that could be found within it (So it may incorrectly be counted multiple times). Adjacencies don't need to be as neatly organized as curr3nt's answer was.
I think the bit debt idea will always have situations that produce contradictions, so I'm thinking it should be discarded.
I looked over plainglazed's 38, and I think it works. Nice.
Edit: I got another 38. It's slightly more complicated (1 more branch needed), but is based on yours, so I won't bother posting it. It uses 6of8's and 5of8's in combinations instead of groups of 5's in a bigger group of 20.
I didn't really explain it too well. Here's a (hopefully) better attempt to show that with 3 bits of information, you can get 4 of the next 5 cards right and still end up with 2 bits after
A couple examples
Example 2 shows that this method does not work as I thought it did, since I reached a contradiction. So back to the drawing board... hopefully something from this can be salvaged.
I think it may still work, but that I can't 'borrow' quite as many bits as I thought. I think there needs to be a 1 bit buffer at all times. Hopefully that solves the issue, but I'll need to look into it.