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Minimum number of unit faces required to construct n unit cubes.
1

%I #13 Oct 05 2021 22:27:29

%S 6,11,16,20,25,29,33,36,41,45,49,52,57,61,65,68,72,75,80,84,88,91,95,

%T 98,102,105,108,113,117,121,124,128,131,135,138,141,146,150,154,157,

%U 161,164,168,171,174,178,181,184,189,193,197,200,204,207,211,214,217,221,224

%N Minimum number of unit faces required to construct n unit cubes.

%F a(n^3) = 3*(n^2)*(n+1) = A270205(n+1). - _Mohammed Yaseen_, Aug 22 2021

%e a(2)=11 because 6 unit faces are required to construct each cube but 1 face is shared. I.e., a(2)=6+5=11.

%Y Cf. A117227, A270205.

%K easy,nonn

%O 1,1

%A _Ron R. King_, May 18 2006

%E Missing term a(56)=214 inserted by _Mohammed Yaseen_, Aug 22 2021