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 1-5 of 5 results.

A042944 Curvatures in diagram constructed by inscribing 2 circles of curvature 2 inside circle of curvature -1, continuing indefinitely to inscribe circles wherever possible.

Original entry on oeis.org

-1, 2, 3, 6, 11, 14, 15, 18, 23, 26, 27, 30, 35, 38, 39, 42, 47, 50, 51, 54, 59, 62, 63, 66, 71, 74, 75, 83, 86, 87, 90, 95, 98, 99, 102, 107, 110, 111, 114, 119, 122, 123, 126, 131, 134, 135, 138, 143, 146, 147, 150, 155, 158, 162, 167, 170, 171, 174, 179, 182, 183
Offset: 1

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Author

Keywords

Comments

The sequence seems to follow a pattern where differences between consecutive terms are 3,1,3,5,3,1,3,5,..., which would give A218155. However, some curvatures (starting with 78, listed in A042945) are in that sequence, but missing from the circle diagram.

References

  • Clifford A. Pickover, The Mathematics of OZ, Mental Gymnastics From Beyond The Edge, Cambridge University Press, Chapter 104 'Circle Mathematics,' figure courtesy of Allan Wilks, Cambridge, UK, 2002, pages 219-220.

Crossrefs

A189226 Curvatures in the nickel-dime-quarter Apollonian circle packing, ordered first by generation and then by size.

Original entry on oeis.org

-11, 21, 24, 28, 40, 52, 61, 157, 76, 85, 96, 117, 120, 132, 181, 213, 237, 376, 388, 397, 132, 156, 160, 189, 204, 205, 216, 237, 253, 285, 288, 309, 316, 336, 349, 405, 412, 421, 453, 460, 469, 472, 517, 544, 565, 616, 628, 685, 717, 741, 1084, 1093, 1104, 1125, 1128, 1140
Offset: 1

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Author

Jonathan Sondow, Apr 18 2011

Keywords

Comments

For a circle, curvature = 1/radius. The curvatures of a quarter, nickel, and dime are approximately proportional to 21, 24, and 28, respectively. Three mutually tangent circles with curvatures 21, 24, 28 can be inscribed in a circle of curvature 11.
Apollonius's and Descartes's Theorems say that, given three mutually tangent circles of curvatures a, b, c, there are exactly two circles tangent to all three, and their curvatures are a + b + c +- 2*sqrt(ab + ac + bc). (Here negative curvature of one of the two circles means that the three circles are inscribed in it.)
Fuchs (2009) says "An Apollonian circle packing ... is made by repeatedly inscribing circles into the triangular interstices in a Descartes configuration of four mutually tangent circles. Remarkably, if the original four circles have integer curvature, all of the circles in the packing will have integer curvature as well." That is because if a + b + c - 2s*qrt(ab + ac + bc) is an integer, then so is a + b + c + 2*sqrt(ab + ac + bc).
For n > 1, the n-th generation of the packing has 4*3^(n-2) circles.
Infinitely many of the curvatures are prime numbers A189227. In fact, in any integral Apollonian circle packing that is primitive (i.e., the curvatures have no common factor), the prime curvatures constitute a positive fraction of all primes (Bourgain 2012) and there are infinitely many pairs of tangent circles both of whose curvatures are prime (Sarnak 2007, 2011).
Fuchs and Sanden (2012) report on experiments with the nickel-dime-quarter Apollonian circle packing, which they call the coins packing P_C.

Examples

			The 1st-generation curvatures are -11, 21, 24, 28, the 2nd are 40, 52, 61, 157, and the 3rd are 76, 85, 96, 117, 120, 132, 181, 213, 237, 376, 388, 397. The 4th generation begins 132, 156, 160, 189, 204, 205, 216, ....
As 21 + 24 + 28 +- 2*sqrt(21*24 + 21*28 + 24*28) = 157 or -11, the sequence begins -11, 21, 24, 28, ... and 157 is in it.
The primes 157 and 397 are the curvatures of two circles that are tangent.
		

Crossrefs

Programs

  • Mathematica
    root = {-11, 21, 24, 28};
    triples = Subsets[root, {3}];
    a = {root};
    Do[
      ng = Table[Total@t + 2 Sqrt@Total[Times @@@ Subsets[t, {2}]], {t, triples}];
      AppendTo[a, Sort@ng];
      triples = Join @@ Table[{t, r} = tr; Table[Append[p, r], {p, Subsets[t, {2}]}], {tr, Transpose@{triples, ng}}]
      , {k, 3}];
    Flatten@a (* Andrey Zabolotskiy, May 29 2022 *)

Formula

a(n) == 0, 4, 12, 13, 16, or 21 (mod 24).

Extensions

Terms a(28) and beyond from Andrey Zabolotskiy, May 29 2022

A218155 Numbers congruent to 2, 3, 6, 11 mod 12.

Original entry on oeis.org

2, 3, 6, 11, 14, 15, 18, 23, 26, 27, 30, 35, 38, 39, 42, 47, 50, 51, 54, 59, 62, 63, 66, 71, 74, 75, 78, 83, 86, 87, 90, 95, 98, 99, 102, 107, 110, 111, 114, 119, 122, 123, 126, 131, 134, 135, 138, 143, 146, 147, 150, 155, 158, 159, 162, 167, 170, 171, 174
Offset: 1

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Author

Jean-Claude Babois, Oct 22 2012

Keywords

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{2, -2, 2, -1}, {2, 3, 6, 11}, 100] (* T. D. Noe, Nov 11 2012 *)
  • PARI
    for(m=2,175,if(binomial(m,4)%binomial(m,2)==0,print1(m,", "))) \\ Hugo Pfoertner, Aug 11 2020

Formula

a(n) = 2a(n-1) - 2a(n-2) + 2a(n-3) - a(n-4). - Charles R Greathouse IV, Nov 09 2012
G.f.: x^2*(x^3+4*x^2-x+2) / ((x-1)^2*(x^2+1)). - Colin Barker, Jan 07 2013
{m>1|C(m,4)==0 (mod C(m,2))}. - Gary Detlefs, Jan 11 2014
Sum_{n>=1} (-1)^(n+1)/a(n) = (2*sqrt(3)+1)*Pi/24 - log(2+sqrt(3))/(4*sqrt(3)) - log(2)/6. - Amiram Eldar, Mar 18 2022

Extensions

Edited by Andrey Zabolotskiy, Aug 11 2020

A045679 Numbers congruent to 0,1,4,9 mod 12 missing from A045673 (conjectured to be finite).

Original entry on oeis.org

13, 21, 37, 45, 48, 61, 69, 85, 93, 109, 117, 120, 132, 133, 141, 157, 165, 181, 189, 205, 208, 213, 229, 237, 241, 252, 253, 261, 277, 285, 300, 301, 309, 325, 328, 333, 340, 349, 357, 360, 373, 381, 397, 405, 421, 429, 445, 453, 468, 469, 477
Offset: 1

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Crossrefs

Extensions

Offset 1 and name corrected by Michel Marcus, Sep 13 2019

A294510 Residues modulo 24 of curvatures in the Apollonian circle packing A042944.

Original entry on oeis.org

2, 3, 6, 11, 14, 15, 18, 23
Offset: 1

Views

Author

Jonathan Sondow, Nov 16 2017

Keywords

Comments

Fuchs and Sanden proved that all curvatures in the Apollonian circle packing -1, 2, 3, 6 are congruent mod 24 to either 2, 3, 6, 11, 14, 15, 18, or 23.
Additional comments and links are in A189226.

Crossrefs

Showing 1-5 of 5 results.