A343850 Integer part of the area of an irregular hexagon, formed by the regular overlap of two regular pentagons each of side length n.

2, 9, 22, 39, 62, 89, 122, 159, 201, 248, 301, 358, 420, 488, 560, 637, 719, 806, 898, 995, 1098, 1205, 1317, 1434, 1556, 1683, 1815, 1952, 2094, 2240, 2392, 2549, 2711, 2878, 3050, 3226, 3408, 3595, 3787, 3983, 4185, 4392, 4603, 4820, 5042, 5268, 5500, 5736

Offset: 1

Author

Torlach Rush, May 01 2021

Keywords

• nonn

Comments

To overlap "regularly", the two pentagons must line up on their vertices. This means that the overlapping region will also be an equilateral polygon and that it will have an even number of sides (half from each of the original pentagons), and will have all but two of its internal angles equal to (3*Pi)/5, as all but two of the angles will be angles from the original regular pentagons.

The sum of the internal angles of the irregular hexagon is 4 * (3*Pi/5) + 2 * (4*Pi/5) = 4*Pi.

The symmetries of the irregular hexagon are that of a rectangle.

A255605(n) < a(n) <= A255606(n).

This object tiles the plane in at least two ways. The first is obvious:

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Another way to tile the plane with this object is to copy an instance of the object, rotate it through +- (2 * Pi)/5 and stack the copy on the original object. By translation symmetries the new object also tiles the plane.

Examples

			a(1) = floor(2 * cos(Pi/10) * (1 + sin(Pi/10))) = 2.
a(2) = floor(8 * cos(Pi/10) * (1 + sin(Pi/10))) = 9.
a(5) = floor(50 * cos(Pi/10) * (1 + sin(Pi/10))) = 62.
		

Links

• Mathrecreation, Regular Polygons, Intersecting Regularly, includes image of the irregular hexagon.

Crossrefs

Cf. A255605, A255606.

Programs

Mathematica

Table[Floor[2 n^2Cos[Pi/10](1+Sin[Pi/10])],{n,48}] (* Stefano Spezia, May 11 2021 *)

PARI

a(n) = floor(2 * n^2 * cos(Pi/10) * (1 + sin(Pi/10)))

PARI

my(c=50+22*quadgen(20)); a(n) = sqrtint(floor(c*n^4))>>2; \\ Kevin Ryde, May 11 2021

Formula

a(n) = floor(2 * n^2 * cos(Pi/10) * (1 + sin(Pi/10))).