Midlevels Conjecture

[Edwin Clark]

> a(n) = number of ways to cycle through all the
> subsets of size n and n+1 of a set of size 2n+1 by adding or
> deleting one element at a time
> 
> a(1) is 2, because the two cycles are
> 
> 1 13 3 23 2 12 1 ...
> 
> and its reverse

a(n) =2b(n) where b(n) is the number of Hamiltonian cycles 
in the so-called bipartite Kneser graph H(2n+1,n) whose vertices 
are the subsets of size n and n+1 of a set of size 2n+1.  
A is adjacent to B iff one set is contained in the other.

Using the Hamiltonian cycle finder in the program Groups & Graphs, 
I got  b(2) = 24. 

Then when I tried the case n = 3, I got the reply: "Too much
output. Logging of Hamiltonian cycles is being terminated. 
10382 cycles were found so far."

So all I can say is b(3) >=10382. 

Maybe someone else has a program that will not give up so soon.

--Edwin