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a(n) = number of edges in a concertina n-cube.
2

%I #9 Apr 09 2018 10:59:58

%S 0,1,6,42,344,3230,34452

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

%C n-place formulas in first-order logic like Ax Ey P(x, y) can be ordered by implication. This Hasse diagram has A000629(n) vertices and a(n) edges.

%C This is the second diagonal on the right in A300700, the triangle of faces in the concertina n-cube.

%C The corresponding sequence for cocoon concertina n-cubes, which have more internal vertices and edges, is A300694.

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

%H Tilman Piesk, Image of a concertina <a href="https://commons.wikimedia.org/wiki/File:Concertina_square_graph.svg">square with 6</a> and <a href="https://commons.wikimedia.org/wiki/File:Concertina_cube_Hasse_diagram.png">cube with 42</a> edges

%H Tilman Piesk, <a href="https://github.com/watchduck/concertina_hypercubes/tree/master/computed_results/hasse">Lists of edges</a> for n=2..5

%H Tilman Piesk, <a href="https://github.com/watchduck/concertina_hypercubes/blob/master/coordinates_hasse.py">Python code used to generate the sequence</a>

%F a(n) = A300700(n, n-1).

%Y Cf. A300700, A000629, A300694.

%K nonn,more

%O 0,3

%A _Tilman Piesk_, Apr 03 2018