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.

A255575 a(n) = floor(((sqrt(sqrt(3))^3)/sqrt(Pi))^n).

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 15, 20, 26, 33, 43, 56, 72, 92, 119, 153, 197, 253, 325, 419, 539, 693, 891, 1146, 1474, 1896, 2439, 3136, 4034, 5188, 6672, 8581, 11036, 14194, 18254, 23476, 30192, 38830, 49938, 64225, 82598, 106227, 136616, 175698, 225961, 290603, 373737
Offset: 0

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Author

Kival Ngaokrajang, Feb 25 2015

Keywords

Comments

Inspired by A255405, but starting with a unit circle and an equilateral triangle whose area is equal to Pi.
a(n) is the curvature (rounded down) of the n-th nested circle. See illustration in the links.

Crossrefs

Cf. A255405.

Programs

  • Maple
    A255575:=n->floor(((sqrt(sqrt(3))^3)/sqrt(Pi))^n): seq(A255575(n), n=0..70); # Wesley Ivan Hurt, Apr 28 2017
  • Mathematica
    Table[Floor[(Sqrt[Sqrt[3]]^3/Sqrt[Pi])^n], {n, 51}] (* Michael De Vlieger, Feb 25 2015 *)
  • PARI
    {for(n=0,100,a=floor(((sqrt(sqrt(3))^3)/sqrt(Pi))^n);print1(a,", "))}

Formula

a(n) = floor(((sqrt(sqrt(3))^3)/sqrt(Pi))^n).