

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 n1.


3



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
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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.
Liyao 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  (w1)*(h1)*(d1)). Keep only those N for which Volume(N)>Volume(N1). 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: A071280 A033942 A111087 * A326111 A091375 A282671
Adjacent sequences: A166080 A166081 A166082 * A166084 A166085 A166086


KEYWORD

nonn


AUTHOR

Mark Jeronimus (mark.jeronimus(AT)gmail.com), Oct 06 2009, Dec 01 2009


EXTENSIONS

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


STATUS

approved



