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a(n) = number of faces in a concertina n-cube.
1

%I #24 Apr 03 2018 10:17:07

%S 1,3,13,87,805,9303,128533

%N a(n) = number of faces in a concertina n-cube.

%H Tilman Piesk, <a href="https://en.wikiversity.org/wiki/Formulas_in_predicate_logic">Formulas in predicate logic</a> (Wikiversity)

%H Tilman Piesk, <a href="https://commons.wikimedia.org/wiki/File:Concertina_cube_Hasse_diagram.png">Skeleton</a> and <a href="https://commons.wikimedia.org/wiki/File:Concertina_cube_with_direction_colors;_ortho_rhomb.png">solid</a> representation of a concertina cube.

%e A concertina 3-cube has 26 0-faces (vertices), 42 1-faces (edges), 18 2-faces and 1 3-face (the polyhedron itself). Together this makes a(3) = 87 faces.

%Y Row sums of A300700.

%K nonn,more

%O 0,2

%A _Tilman Piesk_, Mar 11 2018