Fix one edge $e$ of the graph (1-skeleton) of an icosahedron.
By a computer search, I found that there are 1024 Hamiltonian cycles that include $e$.
[But see **edit** below re directed vs. undirected!]
With the two endpoints of $e$ fixed, there are 10 "free" vertices in the cycle.
Because $1024=2^{10}$, it makes me wonder if there might be a combinatorial viewpoint that
makes it evident that there are 1024 cycles including a fixed edge.
It could just be a numerical coincidence, but if anyone
sees an idea for an argument, I'd appreciate hearing it. Thanks!

Incidentally, this MathWorld page says there are 2560 Hamiltonian cycles all together (without
the fixed edge condition). (Thanks to Kristal Cantwell for pointing me to this page.)

**Edit.** I apologize for misleading! :-/ When I looked at the full output of paths more carefully,
I realize I inadvertently computed *directed* cycles, so each is represented twice, i.e., both
$$ \lbrace 2, 7, 6, 11, 8, 9, 4, 10, 12, 5, 3, 1 \rbrace $$
$$ \lbrace 1, 3, 5, 12, 10, 4, 9, 8, 11, 6, 7, 2 \rbrace $$
are included, etc. So there are 512 undirected cycles, 1024 directed cycles.
The paths are listed here: hpaths.html.

triangleson each side of the cycle. This may incite to count pairs of triangulated 12-gons (without interior vertices) which "match up" to an icosahedron. Triangulated 12-gons are counted by the 10-th Catalan number, but there are far less triangulations with at most 4 triangles at each vertex, which is obviously necessary (and maybe sufficient) to have an icosahedral "complement". Translated in terms of binary trees, this amounts to a limitation to depth at most 4 trees, or up/right paths between the diagonal and some (3rd?) subdiagonal in a 10x10 square... – BS. Sep 5 '10 at 14:06