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.

Showing 1-2 of 2 results.

A255162 Rational part of circle radii in nested circles and hexagons (see comment).

Original entry on oeis.org

2, 0, 24, -288, 3744, -48384, 625536, -8087040, 104550912, -1351655424, 17474476032, -225913577472, 2920656642048, -37758842634240, 488153991315456, -6310954007396352, 81589295984541696, -1054802999903256576, 13636707550653579264
Offset: 0

Views

Author

Kival Ngaokrajang, Feb 15 2015

Keywords

Comments

Inspired by Vitruvian Man, but using hexagons instead of squares, starting with a hexagon whose sides are of length 4 (in some units). The radius of the circle is an integer in the real quadratic number field Q(sqrt(3)), namely R(n) = A(n) + B(n)*sqrt(3) with A(0)=2, A(n) = a(n), and B(0) = 1, B(n) = A255163(n). See illustrations in the links.

Crossrefs

Programs

  • PARI
    {a=2;b=1;print1(a,", ");for(n=1,30,c=12*b-6*a;d=4*a-6*b;print1(c,", ");a=c;b=d)}

Formula

Conjectures from Colin Barker, Feb 15 2015: (Start)
a(n) = -12*a(n-1) + 12*a(n-2).
G.f.: -2*(12*x+1) / (12*x^2 - 12*x - 1).
(End)

A255405 a(n) = floor((2/sqrt(Pi))^n).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 5, 6, 6, 7, 8, 9, 11, 12, 14, 16, 18, 20, 23, 26, 29, 33, 37, 42, 47, 53, 60, 68, 77, 87, 98, 111, 125, 141, 159, 180, 203, 229, 258, 292, 329, 371, 419, 473, 534, 602, 680, 767, 865, 977, 1102, 1244, 1403, 1584, 1787, 2016, 2275, 2567
Offset: 0

Views

Author

Kival Ngaokrajang, Feb 22 2015

Keywords

Comments

Inspired by squaring the circle and Vitruvian Man, but starting with a unit circle and a square whose sides are of length sqrt(Pi), A002161. a(n) is the curvature (rounded down) of the n-th circle. See illustrations in the links.

Crossrefs

Programs

  • Mathematica
    Table[Floor[(2/Sqrt[Pi])^n], {n,0,50}] (* G. C. Greubel, Jan 09 2017 *)
  • PARI
    {for(n=1,100,a=floor(2^n/sqrt(Pi)^n);print1(a,", "))}

Formula

a(n) = floor((2/sqrt(Pi))^n).
Showing 1-2 of 2 results.