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A166082 Maximal volume of a closed box created by using at most n voxels as the boundary. 2
8, 8, 8, 8, 12, 12, 12, 12, 16, 16, 18, 18, 20, 20, 20, 20, 24, 24, 27, 27, 28, 28, 30, 30, 32, 32, 36, 36, 36, 36, 36, 36, 40, 40, 45, 45, 48, 48, 48, 48, 48, 48, 54, 54, 54, 54, 60, 60, 64, 64, 64, 64, 64, 64, 64, 64, 72, 72, 75, 75, 80, 80, 80, 80, 80, 80, 84, 84, 84, 84, 90 (list; graph; refs; listen; history; text; internal format)
OFFSET

8,1

COMMENTS

For example, a 3x3x3 box can be created by using top and bottom plates of 3x3x1 voxels, and using 8 voxels to connect them, totaling 26 voxels.

LINKS

Table of n, a(n) for n=8..78.

FORMULA

For a given value of N (at least 8), calculate the max Volume(N)=w*h*d such that (N <= (w*h*d - (w-1)*(h-1)*(d-1)). The minimum box is 2x2x2 voxels to prevent overlapping voxels (multiple voxels occupying the same location in space) or degenerate cases.

PROG

(Java) // input: int 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;

        }

    }

}

CROSSREFS

Cf. A166083 (Sequence with only the Volume(N)>Volume(N-1) condition), A166084 (Sequence where the enclosed empty space must increase)

Sequence in context: A165926 A216865 A216412 * A145446 A248762 A186985

Adjacent sequences:  A166079 A166080 A166081 * A166083 A166084 A166085

KEYWORD

base,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

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Last modified January 18 23:05 EST 2019. Contains 319282 sequences. (Running on oeis4.)