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Number of different n-dimensional convex regular polytopes that can tile n-dimensional space.
0

%I #13 Nov 01 2024 12:44:54

%S 1,3,1,3,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,

%T 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,

%U 1,1,1,1,1,1,1,1,1,1,1

%N Number of different n-dimensional convex regular polytopes that can tile n-dimensional space.

%C The only n-dimensional convex regular polytope that can tile n-dimensional space for all n>4 is the n-hypercube

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Space-FillingPolyhedron.html">Space-Filling Polyhedron</a>.

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/List_of_regular_polytopes">Regular Polytopes</a>.

%H Dominika Závacká, Cristina Dalfó, and Miquel Angel Fiol, <a href="https://ceur-ws.org/Vol-3792/paper19.pdf">Integer sequences from k-iterated line digraphs</a>, CEUR: Proc. 24th Conf. Info. Tech. - Appl. and Theory (ITAT 2024) Vol 3792, 156-161. See p. 161, Table 2.

%F a(n)=3 for n = 2 & 4. a(n)=1 for all other n.

%e a(2)=3 because the plane can be tiled by equilateral triangles, squares or regular hexagons. a(3)=1 since the only platonic solid that can tile 3-dimensional space is the cube. a(4)=3 because the 4-dimensional space can be tiled by hypercubes (tesseracts), hyperoctahedra or 24-cell polytopes.

%Y Cf. A053016, A060296.

%K nonn

%O 1,2

%A _Sergio Pimentel_, Aug 23 2006