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 1-6 of 6 results.

A110956 a(n) = permanent of the n X n matrix M of zeros and ones defined as follows: if we concatenate the rows of M to form a vector v of length n^2, v_i = 0 if i is an entry from the sequence A045506 (inscribe 2 spheres of curvature 2 inside sphere of curvature -1, continue to inscribe spheres where possible; the sequence gives list of curvatures).

Original entry on oeis.org

2, 6, 20, 0, 368, 2016
Offset: 3

Views

Author

Simone Severini, Sep 26 2005

Keywords

Crossrefs

Cf. A045506.

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

Views

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

A294732 Maximal diameter of the connected cubic graphs on 2*n vertices.

Original entry on oeis.org

1, 2, 3, 5, 6, 8, 9, 11, 12, 14, 15, 17, 18, 20, 21, 23, 24, 26, 27, 29, 30, 32, 33, 35, 36, 38, 39, 41, 42, 44, 45, 47, 48, 50, 51, 53, 54, 56, 57, 59, 60, 62, 63, 65, 66, 68, 69, 71, 72, 74, 75, 77, 78, 80, 81, 83, 84, 86, 87, 89, 90, 92, 93, 95, 96, 98, 99
Offset: 2

Views

Author

Hugo Pfoertner, Dec 13 2017

Keywords

Examples

			a(5)=5 because there exists a unique graph (up to permutations) on 2*5=10 nodes, given by the following adjacency matrix
.
      1 2 3 4 5 6 7 8 9 10
   1  . 1 1 1 . . . . . .
   2  1 . 1 1 . . . . . .
   3  1 1 . . 1 . . . . .
   4  1 1 . . 1 . . . . .
   5  . . 1 1 . 1 . . . .
   6  . . . . 1 . 1 1 . .
   7  . . . . . 1 . . 1 1
   8  . . . . . 1 . . 1 1
   9  . . . . . . 1 1 . 1
  10  . . . . . . 1 1 1 .
.
that requires a shortest possible walk using 5 edges to get from node 1 to node 9.
From _Andrew Howroyd_, Dec 15 2017: (Start)
The following constructions are optimal (see theorem 5 of Caccetta et al.).
Pattern for odd n >= 10. Each additional 4 nodes increases diameter by 3.
      o         o         o         o
    / | \     / | \     / | \     / | \
   o--o  o---o  |  o---o  |  o---o  o--o
    \ | /     \ | /     \ | /     \ | /
      o         o         o         o
Pattern for even n >= 12. Each additional 4 nodes increases diameter by 3.
      o--o         o         o         o
    / |  | \     / | \     / | \     / | \
   o--o  |  o---o  |  o---o  |  o---o  o--o
    \ |  | /     \ | /     \ | /     \ | /
      o--o         o         o         o
(End)
		

Crossrefs

Apart from initial term, duplicate of A267528.

Programs

  • Mathematica
    Prepend[LinearRecurrence[{1, 1, -1}, {2, 3, 5}, 100], 1] (* Jean-François Alcover, Dec 27 2017 *)
  • PARI
    Vec(x^2*(1 + x + x^3) / ((1 - x)^2*(1 + x)) + O(x^100)) \\ Colin Barker, Dec 16 2017

Formula

From Andrew Howroyd, Dec 15 2017: (Start)
a(n) = (6*n - 11 - (-1)^n)/4 for n > 2.
a(n) = a(n-2) + 3 for n > 4. (End)
From Colin Barker, Dec 16 2017: (Start)
G.f.: x^2*(1 + x + x^3) / ((1 - x)^2*(1 + x)).
a(n) = a(n-1) + a(n-2) - a(n-3) for n > 5. (End)
a(n) = A267528(n-1). - Hugo Pfoertner, Oct 10 2018
E.g.f.: (6 + 2*x + x^2 + 3*(x - 2)*cosh(x) + (3*x - 5)*sinh(x))/2. - Stefano Spezia, Feb 20 2023

Extensions

Terms a(12) and beyond from Andrew Howroyd, Dec 15 2017

A104566 Triangle read by rows: T(i,j) is the (i,j)-entry (1 <= j <= i) of the product H*R of the infinite lower triangular matrices H = [1; 1,2; 1,2,1; 1 2,1,2; ...] and R = [1; 1,1; 1,1,1; 1,1,1,1; ...].

Original entry on oeis.org

1, 3, 2, 4, 3, 1, 6, 5, 3, 2, 7, 6, 4, 3, 1, 9, 8, 6, 5, 3, 2, 10, 9, 7, 6, 4, 3, 1, 12, 11, 9, 8, 6, 5, 3, 2, 13, 12, 10, 9, 7, 6, 4, 3, 1, 15, 14, 12, 11, 9, 8, 6, 5, 3, 2, 16, 15, 13, 12, 10, 9, 7, 6, 4, 3, 1, 18, 17, 15, 14, 12, 11, 9, 8, 6, 5, 3, 2, 19, 18, 16, 15, 13, 12, 10, 9, 7, 6, 4, 3, 1
Offset: 1

Views

Author

Gary W. Adamson, Mar 15 2005

Keywords

Examples

			The first few rows are
  1;
  3, 2;
  4, 3, 1;
  6, 5, 3, 2;
  ...
		

Crossrefs

Row sums yield A001082.
Columns 1, 3, 5, ... (starting at the diagonal entry) yield A032766.
Columns 2, 4, 6, ... (starting at the diagonal entry) yield A045506.
Row sums = 1, 5, 8, 16, 21, ... (generalized octagonal numbers, A001082). A006578(2n-1) = A001082(2n).

Programs

  • Maple
    T:=proc(i,j) if j>i then 0 elif i mod 2 = 1 and j mod 2 = 1 then 3*(i-j)/2+1 elif i mod 2 = 0 and j mod 2 = 0 then 3*(i-j)/2+2 elif i+j mod 2 = 1 then 3*(i-j+1)/2 else fi end: for i from 1 to 14 do seq(T(i,j),j=1..i) od; # yields sequence in triangular form # Emeric Deutsch, Mar 24 2005

Formula

For 1 <= j <= i: T(i,j) = 3(i-j+1)/2 if i and j are of opposite parity; T(i,j) = 3(i-j)/2 + 1 if both i and j are odd; T(i,j) = 3(i-j)/2 + 2 if both i and j are even. - Emeric Deutsch, Mar 24 2005

Extensions

More terms from Emeric Deutsch, Mar 24 2005

A114119 Row sums of triangle A114118.

Original entry on oeis.org

1, 3, 5, 6, 8, 9, 11, 12, 14, 15, 17, 18, 20, 21, 23, 24, 26, 27, 29, 30, 32, 33, 35, 36, 38, 39, 41, 42, 44, 45, 47, 48, 50, 51, 53, 54, 56, 57, 59, 60, 62, 63, 65, 66, 68, 69, 71, 72, 74, 75, 77, 78, 80, 81, 83, 84, 86, 87, 89, 90, 92, 93, 95, 96, 98, 99, 101, 102, 104, 105
Offset: 0

Views

Author

Paul Barry, Nov 13 2005

Keywords

Comments

Taken modulo 3 yields 1,0,2,0,2,0,2,0,2,...; a(n) is congruent to 0 or 2 (mod 3) for n > 0.

Crossrefs

Programs

Formula

a(n) = 3*floor((n + 1)/2) + 2*((n+1) mod 2) - 0^n.
a(n) = Sum_{k=0..n} Sum_{j=0..n} binomial(floor((n + k + j)/3), k)*binomial(k, floor((n + k + j)/3)).
G.f.: 1 - x*(-3 - 2*x + 2*x^2)/((1 + x)*(x - 1)^2). - R. J. Mathar, Oct 25 2011
E.g.f.: ((4 + 3*x)*cosh(x) + 3*(1 + x)*sinh(x) - 2)/2. - Stefano Spezia, Feb 20 2023
Showing 1-6 of 6 results.