4 added 27 characters in body

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! 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.

3 paths->cycles; added 4 characters in body; added 23 characters in body; added 7 characters in body

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!] 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 they are 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 pathscycles, 1024 directed pathscycles. The paths are listed here: hpaths.html.

2 directed vs. undirected; added 2 characters in body

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 $e$. 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 they are directed, 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 paths, 1024 directed paths. The paths listed here: hpaths.html.

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