A166083 - OEIS

%I #17 Jul 11 2023 15:29:38

%S 8,12,16,18,20,24,27,28,30,32,36,40,45,48,54,60,64,72,75,80,84,90,96,

%T 100,105,112,120,125,128,140,144,150,160,168,175,180,192,200,210,216,

%U 225,240,245,252,256,270,280,288,294,300,315,324,336,343,350,360,378

%N Maximal volume of a closed box created by using at most n voxels as the boundary, skipping values of n for which the volume is the same as for n-1.

%C For example, a 3 X 3 X 3 box can be created by using top and bottom plates of 3 X 3 X 1 voxels, and using 8 voxels to connect them, totaling 26 voxels.

%H Li-yao Xia, <a href="/A166083/a166083.txt">Examples of (N, Volume(N)), N<1000</a>

%F For each N (starting at 8), calculate the max Volume(N)=w*h*d such that (N <= (w*h*d - (w-1)*(h-1)*(d-1)). Keep only those N for which Volume(N)>Volume(N-1). The minimum box is 2 X 2 X 2 voxels to prevent overlapping voxels (multiple voxels occupying the same location in space) or degenerate cases.

%e N Volume

%e 8 8

%e 12 12

%e 16 16

%e 18 18

%e 20 20

%e 24 24

%e 26 27

%e 28 28

%e 30 30

%e 32 32

%o (Java)

%o int lastMax = 0;

%o for (int voxels = 8; voxels <= 1000; voxels++) {

%o   int max = 0;

%o   for (int depth = voxels / 4; depth >= 2; depth--) {

%o     for (int width = voxels / (2 * depth); width >= 2; width--) {

%o       int remaining = voxels - 2 * width * depth;

%o       int height = 2 + remaining / (2 * (width - 1 + depth - 1));

%o       int volume = width * depth * height;

%o       if (max < volume) {

%o         max = volume;

%o       }

%o     }

%o   }

%o   if (lastMax < max) {

%o    lastMax = max;

%o    System.out.println(voxels + " " + max);

%o   }

%o }

%Y Cf. A166082 (full sequence without conditions), A166084 (sequence where the enclosed empty space must increase).

%K nonn

%O 1,1

%A _Mark Jeronimus_, Oct 06 2009, Dec 01 2009

%E Minor edits by _N. J. A. Sloane_, Dec 05 2009