cp's OEIS Frontend

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.

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A255604 Table read by antidiagonals, T(n,k) is the integer part of the area of a regular k-gon with side length n.

Original entry on oeis.org

0, 1, 1, 3, 4, 1, 6, 9, 6, 2, 10, 16, 15, 10, 3, 15, 25, 27, 23, 14, 4, 21, 36, 43, 41, 32, 19, 6, 27, 49, 61, 64, 58, 43, 24, 7, 35, 64, 84, 93, 90, 77, 55, 30, 9, 43, 81, 110, 127, 130, 120, 98, 69, 37, 11, 52, 100, 139, 166, 178, 173, 154, 123, 84, 44, 13, 62, 121, 172
Offset: 1

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Author

Kival Ngaokrajang, Feb 27 2015

Keywords

Comments

See illustration in the links.

Examples

			See table in the links.
		

Crossrefs

Row 1: A011865.
Columns 1,2,3,4,10: A171971, A000290, A255605, A255606, A172526.

Programs

  • Mathematica
    t[n_, k_] := Floor[k*n^2/(4 Tan[Pi/k])]; Table[t[n - k + 1, k], {n, 3, 14}, {k, 3 , n}] // Flatten
    (* to view table: Table[t[n, k], {k, 3, 6}, {n, 6}] // TableForm *) (* Robert G. Wilson v, Feb 28 2015 *)
  • PARI
    {for(i=1,20,for(n=3,i-1,a=floor(n*(i-n)^2/(4*tan(Pi/n)));print1(a,", ")))}

Formula

T(n,k) = floor(k*n^2/(4*tan(Pi/k))), n >=1, k >=3.

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

Original entry on oeis.org

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

Views

Author

Torlach Rush, May 01 2021

Keywords

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:
\/ \/ \/ \/
/ \/ \/ \_/ \
\/ \/ \/ \/
/ \/ \/ \_/ \
\/ \/ \/ \/
/ \/ \/ \_/ \
\/ \/ \/ \/
/ \/ \/ \_/ \
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.
		

Crossrefs

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))).
Showing 1-2 of 2 results.