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Minimum number of curves of unitlength flexible1 lengthsrequired to form a convex perimeter around n non-overlapping unit circles.

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Jon E. Schoenfield: Name: “… unit flexible lengths …” -> “… flexible unit lengths …”, maybe?

Eckard Specht, <a href="http://hydra.nat.uni-magdeburg.de/packing/cci/cci.html">The best known packings of equal circles in a circle (complete up to N = 2600)</a>

Eckard Specht, <a href="http://hydra.nat.uni-magdeburg.de/packing/cci/cci.html">The best known packings of equal circles in a circle (complete up to N = 2600)</a>.

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Robert G. Wilson v: Jon, Most of the time, a disc on a hex grid is optimal, but not always. For n=4, they can be on a hex grid or a normal lattice.

Jon E. Schoenfield: Agreed! (That was why I wrote “1. There exist optimal layouts in which each circle’s center is placed at a lattice point on a hexagonal grid.” rather than something more restrictive, like “For every optimal layout, each circle’s center can be placed at a lattice point on a hexagonal grid.” The n=4 case where you had mentioned “four circles arranged in a square or parallelogram” made it clear that such a statement would be false.)

0, 7, 11, 13, 15, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 33

a(9) = 22 = ceiling(2*piPi+12+4*cos(Pi/6));

a(11) = 24 = ceiling(2*piPi+14+4*cos(Pi/6));

a(12) = 25 = ceiling(2*piPi+18);

a(13) = 26 = ceiling(2*piPi+16+4*cos(Pi/6));

a(14) = 27 = ceiling(2*piPi+20);

a(15) = 28 = ceiling(2*piPi+18+4*cos(Pi/6));

a(16) = 29 = ceiling(2*Pi+22);

a(17) = 30 = ceiling(2*Pi+20+4*cos(Pi/6));

a(18) = 31 = ceiling(2*Pi+24);

a(19) = 33 = ceiling(2*Pi+26);

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Otto von Guericke, UniversityEckard MagdeburgSpecht, <a href="http://hydra.nat.uni-magdeburg.de/packing/cci/cci.html">The best known packings of equal circles in a circle (complete up to N = 2600)</a>

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Jon E. Schoenfield: I’m too cheap to spend money to access the Kallrath/Frey article … is the layout of the circles for n=6 the same thing you’d get if you took the “2,3,2” layout for n=7 and removed from it any one circle other than the one at the center?

Jon E. Schoenfield: Some observations I’ve made (unencumbered by any knowledge of the content at any of the links, since reading any of them might spoil the fun :-) ) about optimal (i.e., minimum-convex-perimeter) layouts for each n up to 15 (unless I’ve made a mistake, which is entirely possible!):

1. There exist optimal layouts in which each circle’s center is placed at a lattice point on a hexagonal grid.

2. Given any such optimal layout (i.e., an optimal layout in which each circle’s center is at a lattice point on a hex grid) with n-1 circles, an optimal layout with n disks can be obtained simply by adding the n-th circle to an appropriate empty lattice point (i.e., without moving any of the other n-1 circles).

3. Any empty lattice point that lies at the minimum distance from the centroid of the (optimal) (n-1)-circle layout above is an appropriate lattice point at which the n-th circle can be added.

Some questions that may be of interest: Are the above observations correct? If they are, up to how high a value of n do they hold? Is it possible that they hold for all n? If so, is there some simple, fast algorithm for constructing an optimal layout for any n? If so, can it be used to obtain a fairly simple formula for the optimal perimeter as a function of n (hence a formula for a(n))?

Jon E. Schoenfield: (’Sorry … where I said “disks”, I should’ve said “circles”.)

Otto von Guericke, University Magdeburg, <a href="http://hydra.nat.uni-magdeburg.de/packing/cci/cci.html"> ">The best known packings of equal circles in a circle (complete up to N = 2600)</a>

Mathematics, Stack Exchange, <a href="https://math.stackexchange.com/questions/1591973/circle-packing-in-an-elastic-container"> ">Circle Packing in an Elastic Container</a>.

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