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%I #21 Feb 03 2022 14:02:47
%S 8,27,125,729,4913,35937,274625,2146689,16974593,135005697,1076890625,
%T 8602523649,68769820673,549957165057,4398851866625,35187593412609,
%U 281487861809153,2251851353686017,18014604668698625,144116012711149569,1152924803144876033,9223385231000600577
%N a(n) = (2^n + 1)^3.
%C The number of vertices when starting with a cube (n=0) and iterating by dividing every cube into 8 equal cubes.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (15,-70,120,-64).
%F G.f.: (8 - 93*x + 280*x^2 - 216*x^3)/(1 - 15*x + 70*x^2 - 120*x^3 + 64*x^4). - _Stefano Spezia_, Apr 12 2021
%e a(2) = 125; since after iterating twice on the original cube, there are now 64 cubes inside. This 4 X 4 X 4 arrangement of cubes has 5*5=25 vertices on each face across the 5 slices, thus 125 vertices.
%t (2^Range[0, 25] + 1)^3 (* _Wesley Ivan Hurt_, Apr 11 2021 *)
%o (Python)
%o def A343318(n): return (2**n+1)**3 # _Chai Wah Wu_, Feb 03 2022
%Y a(n) = A000578(A000051(n)).
%K nonn,easy
%O 0,1
%A _Steve Bosze_, Apr 11 2021