This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A343850 #58 Jun 24 2021 02:56:05
%S A343850 2,9,22,39,62,89,122,159,201,248,301,358,420,488,560,637,719,806,898,
%T A343850 995,1098,1205,1317,1434,1556,1683,1815,1952,2094,2240,2392,2549,2711,
%U A343850 2878,3050,3226,3408,3595,3787,3983,4185,4392,4603,4820,5042,5268,5500,5736
%N A343850 Integer part of the area of an irregular hexagon, formed by the regular overlap of two regular pentagons each of side length n.
%C A343850 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.
%C A343850 The sum of the internal angles of the irregular hexagon is 4 * (3*Pi/5) + 2 * (4*Pi/5) = 4*Pi.
%C A343850 The symmetries of the irregular hexagon are that of a rectangle.
%C A343850 A255605(n) < a(n) <= A255606(n).
%C A343850 This object tiles the plane in at least two ways. The first is obvious:
%C A343850 \_/ \_/ \_/ \_/
%C A343850 / \_/ \_/ \_/ \
%C A343850 \_/ \_/ \_/ \_/
%C A343850 / \_/ \_/ \_/ \
%C A343850 \_/ \_/ \_/ \_/
%C A343850 / \_/ \_/ \_/ \
%C A343850 \_/ \_/ \_/ \_/
%C A343850 / \_/ \_/ \_/ \
%C A343850 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.
%H A343850 Mathrecreation, <a href="http://www.mathrecreation.com/2015/07/regular-polygons-intersecting-regularly.html">Regular Polygons, Intersecting Regularly</a>, includes image of the irregular hexagon.
%F A343850 a(n) = floor(2 * n^2 * cos(Pi/10) * (1 + sin(Pi/10))).
%e A343850 a(1) = floor(2 * cos(Pi/10) * (1 + sin(Pi/10))) = 2.
%e A343850 a(2) = floor(8 * cos(Pi/10) * (1 + sin(Pi/10))) = 9.
%e A343850 a(5) = floor(50 * cos(Pi/10) * (1 + sin(Pi/10))) = 62.
%t A343850 Table[Floor[2 n^2Cos[Pi/10](1+Sin[Pi/10])],{n,48}] (* _Stefano Spezia_, May 11 2021 *)
%o A343850 (PARI) a(n) = floor(2 * n^2 * cos(Pi/10) * (1 + sin(Pi/10)))
%o A343850 (PARI) my(c=50+22*quadgen(20)); a(n) = sqrtint(floor(c*n^4))>>2; \\ _Kevin Ryde_, May 11 2021
%Y A343850 Cf. A255605, A255606.
%K A343850 nonn
%O A343850 1,1
%A A343850 _Torlach Rush_, May 01 2021