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 1-8 of 8 results.

A248930 Decimal expansion of c = 2*Product_{prime p == 3 (mod 4)} (1 - 2/(p*(p-1)^2)), a constant related to the problem of integral Apollonian circle packings.

Original entry on oeis.org

1, 6, 4, 9, 3, 3, 7, 6, 8, 9, 0, 9, 8, 0, 3, 0, 7, 0, 1, 0, 2, 5, 9, 4, 2, 9, 3, 3, 3, 6, 0, 1, 7, 8, 9, 6, 3, 6, 6, 9, 2, 3, 5, 7, 6, 6, 2, 5, 6, 6, 1, 1, 4, 4, 9, 0, 5, 7, 7, 2, 4, 8, 8, 3, 8, 4, 2, 5, 6, 4, 5, 1, 8, 9, 4, 8, 0, 7, 7, 2, 5, 2, 0, 6, 9, 0, 2, 0, 4, 2, 4, 8, 5, 2, 5, 3, 6, 0, 1, 0, 2, 7, 0, 1, 7
Offset: 1

Views

Author

Jean-François Alcover, Oct 17 2014

Keywords

Examples

			1.64933768909803...
		

Crossrefs

Programs

  • Mathematica
    kmax = 25; Do[ P[k] = Product[p = Prime[n]; If[Mod[p, 4] == 3, 1 - 2/(p*(p - 1)^2) // N[#, 40]&, 1], {n, 1, 2^k}]; Print["P(", k, ") = ", P[k]], {k, 10, kmax}]; c = 2*P[kmax]; RealDigits[c, 10, 15] // First
    (* -------------------------------------------------------------------------- *)
    $MaxExtraPrecision = 1000; digits = 121;
    f[p_] := (1 - 2/(p*(p - 1)^2));
    coefs = Rest[CoefficientList[Series[Log[f[1/x]], {x, 0, 1000}], x]];
    S[m_, n_, s_] := (t = 1; sums = 0; difs = 1; While[Abs[difs] > 10^(-digits - 5) || difs == 0, difs = (MoebiusMu[t]/t) * Log[If[s*t == 1, DirichletL[m, n, s*t], Sum[Zeta[s*t, j/m]*DirichletCharacter[m, n, j]^t, {j, 1, m}]/m^(s*t)]]; sums = sums + difs; t++]; sums);
    P[m_, n_, s_] := 1/EulerPhi[m] * Sum[Conjugate[DirichletCharacter[m, r, n]] * S[m, r, s], {r, 1, EulerPhi[m]}] + Sum[If[GCD[p, m] > 1 && Mod[p, m] == n, 1/p^s, 0], {p, 1, m}];
    m = 2; sump = 0; difp = 1; While[Abs[difp] > 10^(-digits - 5) || difp == 0, difp = coefs[[m]]*P[4, 3, m]; sump = sump + difp; m++];
    RealDigits[Chop[N[2*Exp[sump], digits]], 10, digits - 1][[1]] (* Vaclav Kotesovec, Jan 16 2021 *)

Extensions

More digits from Vaclav Kotesovec, Jun 27 2020

A248938 Decimal expansion of beta = G^2*(2/3)*Product_{prime p == 3 (mod 4)} (1 - 2/(p*(p-1)^2)) (where G is Catalan's constant), a constant related to the problem of integral Apollonian circle packings.

Original entry on oeis.org

4, 6, 1, 2, 6, 0, 9, 0, 8, 6, 1, 3, 8, 6, 1, 3, 0, 3, 3, 2, 8, 5, 2, 9, 8, 4, 6, 4, 2, 4, 6, 0, 7, 5, 1, 5, 8, 0, 1, 3, 8, 3, 4, 4, 3, 7, 6, 5, 8, 8, 2, 0, 6, 3, 0, 0, 7, 0, 3, 9, 7, 7, 5, 1, 9, 0, 7, 1, 2, 8, 1, 6, 0, 7, 2, 2, 0, 7, 4, 9, 8, 3, 7, 9, 1, 0, 4, 2, 6, 0, 7, 2, 6, 2, 1, 4, 8, 0, 7, 2, 3, 1, 6, 3, 1, 6
Offset: 0

Views

Author

Jean-François Alcover, Oct 17 2014

Keywords

Examples

			0.4612609086138613...
		

Crossrefs

Programs

  • Mathematica
    kmax = 25; Clear[P]; Do[P[k] = Product[p = Prime[n]; If[Mod[p, 4] == 3 , 1 - 2/(p*(p - 1)^2) // N[#, 40]&, 1], {n, 1, 2^k}]; Print["P(", k, ") = ", P[k]], {k, 10, kmax}]; beta = Catalan^2*(2/3)*P[kmax]; RealDigits[beta, 10, 16] // First
    (* -------------------------------------------------------------------------- *)
    $MaxExtraPrecision = 1000; digits = 121;
    f[p_] := (1 - 2/(p*(p - 1)^2));
    coefs = Rest[CoefficientList[Series[Log[f[1/x]], {x, 0, 1000}], x]];
    S[m_, n_, s_] := (t = 1; sums = 0; difs = 1; While[Abs[difs] > 10^(-digits - 5) || difs == 0, difs = (MoebiusMu[t]/t) * Log[If[s*t == 1, DirichletL[m, n, s*t], Sum[Zeta[s*t, j/m]*DirichletCharacter[m, n, j]^t, {j, 1, m}]/m^(s*t)]]; sums = sums + difs; t++]; sums);
    P[m_, n_, s_] := 1/EulerPhi[m] * Sum[Conjugate[DirichletCharacter[m, r, n]] * S[m, r, s], {r, 1, EulerPhi[m]}] + Sum[If[GCD[p, m] > 1 && Mod[p, m] == n, 1/p^s, 0], {p, 1, m}];
    m = 2; sump = 0; difp = 1; While[Abs[difp] > 10^(-digits - 5) || difp == 0, difp = coefs[[m]]*P[4, 3, m]; sump = sump + difp; m++];
    RealDigits[Chop[N[2*Catalan^2/3 * Exp[sump], digits]], 10, digits - 1][[1]] (* Vaclav Kotesovec, Jan 16 2021 *)

Formula

beta = (G^2/3)*A248930, where G is Catalan's constant A006752.

Extensions

More digits from Vaclav Kotesovec, Jun 27 2020

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

A135849 a(n) is the ratio of the sum of the bends (curvatures) of the circles in the n-th generation of an Apollonian packing to the sum of the bends in the initial four-circle configuration.

Original entry on oeis.org

1, 5, 39, 297, 2259, 17181, 130671, 993825, 7558587, 57487221, 437222007, 3325314393, 25290849123, 192350849805, 1462934251071, 11126421459153, 84622568920011, 643601286982629, 4894942589100999, 37228736851860105, 283145067047577843, 2153474325825042429
Offset: 1

Views

Author

Colin Mallows, Mar 06 2008

Keywords

Comments

These ratios are independent of the starting configuration.
For more comments, references and links, see A189226.

Examples

			Starting with the configuration with bends (-1,2,2,3) with sum(bends) = 6, the next generation contains four circles with bends 3,6,6,15. The sum is 30 = 6*a(2). The third generation has 12 circles with sum(bends) = 234 = 6*a(3).
		

Crossrefs

Programs

  • Magma
    I:=[1, 5, 39]; [n le 3 select I[n] else  8*Self(n-1) - 3*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Dec 25 2012
  • Mathematica
    CoefficientList[Series[(2 z^2 - 3 z + 1)/(3 z^2 - 8 z + 1), {z, 0, 100}], z] (* and *) LinearRecurrence[{8, -3}, {1, 5, 39}, 100] (* Vladimir Joseph Stephan Orlovsky, Jul 03 2011 *)
  • PARI
    Vec((2*x^3 - 3*x^2 + x)/(3*x^2 - 8*x + 1)+O(x^99)) \\ Charles R Greathouse IV, Jul 03 2011
    

Formula

For n >= 4, a(n) = 8*a(n-1) - 3*a(n-2).
For n>2, [a(n+2), a(n+3)] = the 2 X 2 matrix [0,1; -3,8]^n * [5,39]. Example: [0,1; -3,8]^3 * [5,39] = [a(5), a(6)] = [2259, 17181]. - Gary W. Adamson, Mar 09 2008 (typo corrected by Jonathan Sondow, Dec 24 2012)
a(n) = floor(C * A138264(n)), where C = 1.057097576... = (1/2)*((1/9) + sqrt((1/81) + 4)). Example: a(7) = 130671 = floor(C * A138264(7)) = floor(C * 123613). A135849(n)/A138264(n) tends to C. - Gary W. Adamson, Mar 09 2008
O.g.f.: 2*x/3 +7/9 +(59*x-7)/(9*(1-8*x+3*x^2)). - R. J. Mathar, Apr 24 2008
a(n) = 31*sqrt(13)*(A^n - B^n)/234 - 7*(A^n + B^n)/18 for n>1 where A=3/(4-sqrt(13)) and B=3/(4+sqrt(13)). - R. J. Mathar, Apr 24 2008

A137246 a(n) is the ratio of the sum of the squares of the bends (curvatures) of the n-th generation of an Apollonian packing to the sum of the squares of the bends of the initial four-circle configuration.

Original entry on oeis.org

1, 17, 339, 6729, 133563, 2651073, 52620771, 1044462201, 20731381707, 411494247537, 8167690805619, 162119333369769, 3217883594978523, 63871313899461153, 1267772627204287491, 25163838602387366361, 499473454166134464747, 9913977567515527195857
Offset: 1

Views

Author

Colin Mallows, Mar 09 2008

Keywords

Comments

These ratios are independent of the starting configuration. Similar ratios of third and higher moments are not so independent.
See A189226 for additional comments, references and links.

Examples

			Starting with the configuration with bends (-1,2,2,3) with sum(bends^2) = 18, the next generation contains four circles with bends 3,6,6,15. The sum of their squares is 306 = 18*a(2). The third generation has 12 circles with sum(bends^2) = 6102 = 18*a(3).
		

Crossrefs

Programs

  • GAP
    a:=[1,17,339];; for n in [4..30] do a[n]:=20*a[n-1]-3*a[n-2]; od; a; # G. C. Greubel, May 24 2019
  • Magma
    R:=PowerSeriesRing(Integers(), 30); Coefficients(R!(x*(1-x)*(1-2*x)/(1-20*x+3*x^2)));  // Bruno Berselli, Jul 04 2011
    
  • Mathematica
    CoefficientList[Series[(2z^2-3z+1)/(3z^2-20z+1), {z, 0, 30}], z] (* and *) LinearRecurrence[{20, -3}, {1, 17, 339}, 30] (* Vladimir Joseph Stephan Orlovsky, Jul 03 2011 *)
  • PARI
    Vec(x*(1-2*x)*(1-x)/(1-20*x+3*x^2)+O(x^30)) \\ Charles R Greathouse IV, Jul 03 2011
    
  • Sage
    a=(x*(1-x)*(1-2*x)/(1-20*x+3*x^2)).series(x, 30).coefficients(x, sparse=False); a[1:] # G. C. Greubel, May 24 2019
    

Formula

For n >= 4, a(n) = 20*a(n-1) - 3*a(n-2).
O.g.f.: x*(1-x)*(1-2*x)/(1-20*x+3*x^2). - R. J. Mathar, Mar 31 2008
a(n) = ((41+sqrt(97))*(10+sqrt(97))^(n-1) - (41-sqrt(97))*(10-sqrt(97))^(n-1))/(6*sqrt(97)) for n>1. - Bruno Berselli, Jul 04 2011

A154636 a(n) is the ratio of the sum of the bends of the circles that are drawn in the n-th generation of Apollonian packing to the sum of the bends of the circles in the initial configuration of 3 circles.

Original entry on oeis.org

1, 2, 18, 138, 1050, 7986, 60738, 461946, 3513354, 26720994, 203227890, 1545660138, 11755597434, 89407799058, 679995600162, 5171741404122, 39333944432490, 299156331247554, 2275248816682962, 17304521539721034, 131610425867719386, 1000969842322591986
Offset: 0

Views

Author

Colin Mallows, Jan 13 2009

Keywords

Comments

For comments and more references and links, see A189226.

Examples

			Starting from three circles with bends -1,2,2 summing to 3, the first derived generation consists of two circles, each with bend 3. So a(1) is (3+3)/3 = 2.
		

Crossrefs

Other sequences relating to the two-dimensional case are A135849, A137246, A154637. For the three-dim. case see A154638 - A154645. Five dimensions: A154635.
Cf. also A189226, A189227.

Programs

  • Mathematica
    CoefficientList[Series[(5 z^2 - 6 z + 1)/(3 z^2 - 8 z + 1), {z, 0, 100}], z] (* and *) LinearRecurrence[{8, -3}, {1, 2, 18}, 100] (* Vladimir Joseph Stephan Orlovsky, Jul 03 2011 *)
  • PARI
    Vec((1 - x)*(1 - 5*x) / (1 - 8*x + 3*x^2) + O(x^30)) \\ Colin Barker, Jul 15 2017

Formula

G.f.: (1 - x)*(1 - 5*x) / (1 - 8*x + 3*x^2).
From Colin Barker, Jul 15 2017: (Start)
a(n) = ((-(-7+sqrt(13))*(4+sqrt(13))^n - (4-sqrt(13))^n*(7+sqrt(13)))) / (3*sqrt(13)) for n>0.
a(n) = 8*a(n-1) - 3*a(n-2) for n>2.
(End)

Extensions

More terms from N. J. A. Sloane, Nov 22 2009

A060790 Inscribe two circles of curvature 2 inside a circle of curvature -1. Sequence gives curvatures of the smallest circles that can be sequentially inscribed in such a diagram.

Original entry on oeis.org

-1, 2, 2, 3, 15, 38, 110, 323, 927, 2682, 7754, 22403, 64751, 187134, 540822, 1563011, 4517183, 13054898, 37729362, 109039875, 315131087, 910745750, 2632104062, 7606921923, 21984412383, 63536130986, 183622826522, 530679817859, 1533693138351, 4432455434478
Offset: 0

Views

Author

Brian Galebach, Apr 26 2001

Keywords

Comments

The ratio of successive terms approaches the constant phi+sqrt(phi) ~= 2.89005363826396..., where phi is the golden ratio (sqrt(5)+1)/2. The ratio between the curvatures of two successively smaller circles approaches this constant in any apollonian packing as the curvatures increase.
For more comments, references and links, see A189226.

Examples

			After circles of 2, 2, 3, 15 have been inscribed in the diagram, the next smallest circle that can be inscribed has a curvature of 38.
		

Crossrefs

Programs

  • GAP
    a:=[-1,2,2,3];; for n in [5..30] do a[n]:=2*a[n-1]+2*a[n-2]+2*a[n-3]-a[n-4]; od; a; # Muniru A Asiru, Sep 12 2018
  • Maple
    seq(coeff(series((x-1)*(1-3*x-3*x^2)/(1-2*x-2*x^2-2*x^3+x^4),x,n+1), x, n), n = 0 .. 30); # Muniru A Asiru, Sep 12 2018
  • Mathematica
    CoefficientList[Series[(-3 z^3 + 4 z - 1)/(z^4 - 2 z^3 - 2 z^2 - 2 z + 1), {z, 0, 100}], z] (* and *) LinearRecurrence[{2, 2, 2, -1}, {-1, 2, 2, 3}, 100] (* Vladimir Joseph Stephan Orlovsky, Jul 03 2011 *)
  • PARI
    { for (n=0, 200, if (n>3, a=2*a1 + 2*a2 + 2*a3 - a4; a4=a3; a3=a2; a2=a1; a1=a, if (n==0, a=a4=-1, if (n==1, a=a3=2, if (n==2, a=a2=2, a=a1=3)))); write("b060790.txt", n, " ", a); ) } \\ Harry J. Smith, Jul 12 2009
    

Formula

a(n) = 2a(n-1) + 2a(n-2) + 2a(n-3) - a(n-4).
G.f.: -(1-x)*(1 - 3*x - 3*x^2)/(1 - 2*x - 2*x^2 - 2*x^3 + x^4). - Colin Barker, Apr 22 2012
Lim_{n -> inf} a(n)/a(n-1) = A318605. - A.H.M. Smeets, Sep 12 2018

Extensions

Corrected by T. D. Noe, Nov 08 2006

A154637 a(n) is the ratio of the sum of squares of the bends of the circles that are added in the n-th generation of Apollonian packing, to the sum of squares of the bends of the initial three circles.

Original entry on oeis.org

1, 2, 66, 1314, 26082, 517698, 10275714, 203961186, 4048396578, 80356048002, 1594975770306, 31658447262114, 628384017931362, 12472705016840898, 247568948283023874, 4913960850609954786, 97536510167350024098, 1935988320795170617602, 38427156885401362279746, 762735172745641733742114
Offset: 0

Views

Author

Colin Mallows, Jan 13 2009

Keywords

Comments

For more references and links, see A189226.

Examples

			Starting with three circles with bends -1,2,2, the ssq is 9. The first derived generation has two circles, each with bend 3. So a(1) = (9+9)/9 = 2.
		

Crossrefs

For starting with four circles, see A137246. For sums of bends, see A135849 and A154636. For three dimensions, see A154638 - A154645.
Cf. also A189226, A189227.

Programs

  • Mathematica
    CoefficientList[Series[(29 z^2 - 18 z + 1)/(3 z^2 - 20 z + 1), {z, 0, 100}], z] (* and *) LinearRecurrence[{20, -3}, {1, 2, 66}, 100] (* Vladimir Joseph Stephan Orlovsky, Jul 03 2011 *)
  • PARI
    Vec((1-18*x+29*x^2)/(1-20*x+3*x^2) + O(x^30)) \\ Colin Barker, Nov 16 2016

Formula

G.f.: (1-18*x+29*x^2) / (1-20*x+3*x^2).
From Colin Barker, Nov 16 2016: (Start)
a(n) = ((133-13*sqrt(97))*(10+sqrt(97))^n - (10-sqrt(97))^n*(133+13*sqrt(97))) / (3*sqrt(97)) for n>0.
a(n) = 20*a(n-1) - 3*a(n-2) for n>2.
(End)

Extensions

More terms from N. J. A. Sloane, Nov 22 2009
Showing 1-8 of 8 results.