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%I #8 Jul 07 2024 13:14:42
%S 16,80,106,132
%N a(n) is the minimum number of hypercubes needed to admit a hole of size n in the 4D tesseractic honeycomb.
%e 16 hypercubes surround a single vertex so a(0) = 16.
%e A 3 X 3 X 3 X 3 hypercube will admit a single hole in its center, so a(1) = 3*3*3*3 - 1 = 80.
%Y Cf. A345205 (3D equivalent), A235382 (2D equivalent).
%K nonn,more
%O 0,1
%A _Abraham Maxfield_, Jun 22 2024