A166083 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.

8, 12, 16, 18, 20, 24, 27, 28, 30, 32, 36, 40, 45, 48, 54, 60, 64, 72, 75, 80, 84, 90, 96, 100, 105, 112, 120, 125, 128, 140, 144, 150, 160, 168, 175, 180, 192, 200, 210, 216, 225, 240, 245, 252, 256, 270, 280, 288, 294, 300, 315, 324, 336, 343, 350, 360, 378

Offset

1,1

Comments

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.

Links

Table of n, a(n) for n=1..57.

Li-yao Xia, Examples of (N, Volume(N)), N<1000

Formula

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.

Example

N Volume
8 8
12 12
16 16
18 18
20 20
24 24
26 27
28 28
30 30
32 32

Prog

(Java)
int lastMax = 0;
for (int voxels = 8; voxels <= 1000; voxels++) {
  int max = 0;
  for (int depth = voxels / 4; depth >= 2; depth--) {
    for (int width = voxels / (2 * depth); width >= 2; width--) {
      int remaining = voxels - 2 * width * depth;
      int height = 2 + remaining / (2 * (width - 1 + depth - 1));
      int volume = width * depth * height;
      if (max < volume) {
        max = volume;
      }
    }
  }
  if (lastMax < max) {
   lastMax = max;
   System.out.println(voxels + " " + max);
  }
}

Crossrefs

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

Sequence in context: A111087 A341610 A337373 * A326111 A091375 A282671

Adjacent sequences: A166080 A166081 A166082 * A166084 A166085 A166086

Keyword

nonn

Author

Mark Jeronimus, Oct 06 2009, Dec 01 2009

Extensions

Minor edits by N. J. A. Sloane, Dec 05 2009

Status

approved