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 1-3 of 3 results.

A249457 The numerator of curvatures of touching circles inscribed in a special way in the larger segment of a unit circle divided by a chord of length sqrt(84)/5.

Original entry on oeis.org

10, 100, 2890, 96100, 3237610, 109202500, 3683712490, 124263300100, 4191798484810, 141402777864100, 4769968258260490, 160906295771812900, 5427884341892493610, 183099910962324064900, 6176546013641762558890, 208354665265158340802500, 7028469704892605715408010
Offset: 0

Views

Author

Kival Ngaokrajang, Oct 29 2014

Keywords

Comments

The denominators are conjectured to be A005032.
Refer to comments and links of A240926. Consider a unit circle with a chord of length sqrt(84)/5. This has been chosen such that the larger sagitta has length 7/5. The input, besides the unit circle C, is the circle C_0 with radius R_0 = 7/10, touching the chord and circle C. The following sequence of circles C_n with radii R_n, n >= 1, is obtained from the conditions that C_n touches (i) the circle C, (ii) the chord and (iii) the circle C_(n-1). The curvature of the n-th circle is C_n = 1/R_n, n >= 0, and its numerator is conjectured to be a(n).
If one considers the curvature of touching circles inscribed in the smaller segment (sagitta length 3/5), the rational sequence would be A249458/A169634. See an illustration given in the link.
For the proof and the formula for the rational curvatures of the circles in the larger segment see a comment under A249862. C_n = (5/7)*(S(n, 34/3) - (17/3)*S(n-1, 34/3) + 1), n >= 0, with Chebyshev's S polynomials (A049310). - Wolfdieter Lang, Nov 07 2014

Crossrefs

Programs

  • Magma
    I:=[10,100,2890]; [n le 3 select I[n] else 37*Self(n-1) - 111*Self(n-2) + 27*Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 20 2017
  • Mathematica
    LinearRecurrence[{37, -111, 27},{10, 100, 2890},16] (* Ray Chandler, Aug 11 2015 *)
    CoefficientList[Series[10*(1 - 27*x + 30*x^2)/((1 - 34*x + 9*x^2)*(1 - 3*x)), {x, 0, 50}], x] (* G. C. Greubel, Dec 20 2017 *)
  • PARI
    {
    r=0.7;dn=7;print1(round(dn/r),", ");r1=r;
    for (n=1,40,
         if (n<=1,ab=2-r,ab=sqrt(ac^2+r^2));
         ac=sqrt(ab^2-r^2);
         if (n<=1,z=0,z=(Pi/2)-atan(ac/r)+asin((r1-r)/(r1+r));r1=r);
         b=acos(r/ab)-z;
         r=r*(1-cos(b))/(1+cos(b)); dn=dn*3;
         print1(round(dn/r),", ");
    )
    }
    
  • PARI
    x='x+O('x^30); Vec(10*(1 - 27*x + 30*x^2)/((1 - 34*x + 9*x^2)*(1 - 3*x))) \\ G. C. Greubel, Dec 20 2017
    

Formula

Empirical g.f.: -10*(30*x^2-27*x+1) /((3*x - 1)*(9*x^2-34*x+1)). - Colin Barker, Oct 29 2014
From Wolfdieter Lang, Nov 07 2014: (Start)
a(n) = 5*(A249862(n) + 3^n) = 5*3^n*(S(n, 34/3) - (17/3)*S(n-1, 34/3) + 1), n >= 0, with Chebyshev's S polynomials (A049310). See the comments on A249862 for the proof.
O.g.f.: 5*((1 - 17*x)/(1 - 34*x + 9*x^2) + 1/(1-3*x)) = 10*(1 - 27*x + 30*x^2)/((1 - 34*x + 9*x^2)*(1 - 3*x)) proving the conjecture of Colin Barker above. (End)
E.g.f.: 5*exp(3*x)*(1 + exp(14*x)*cosh(2*sqrt(70)*x)). - Stefano Spezia, Mar 24 2023

Extensions

Edited. Name and comment small changes, keyword easy added. - Wolfdieter Lang, Nov 07 2014
a(16) from Stefano Spezia, Mar 24 2023

A249458 The numerators of curvatures of touching circles inscribed in a special way in the smaller segment of unit circle divided by a chord of length sqrt(84)/5.

Original entry on oeis.org

10, 100, 1690, 36100, 835210, 19802500, 472931290, 11318832100, 271066588810, 6492762648100, 155527144782490, 3725543446072900, 89243180863948810, 2137770243127864900, 51209104645650371290, 1226685938180259902500
Offset: 0

Views

Author

Kival Ngaokrajang, Oct 29 2014

Keywords

Comments

The denominators are conjectured to be A169634.
Refer to comments and links of A240926. Consider a unit circle with a chord of length sqrt(84)/5. This has been chosen such that the smaller sagitta has length 3/5. The input, besides the circle C, is the circle C_0 with radius R_0 = 3/10, touching the chord and circle C. The following sequence of circles C_n with radii R_n, n >= 1, is obtained from the conditions that C_n touches (i) the circle C, (ii) the chord and (iii) the circle C_(n-1). The curvature of the n-th circle is C_n = 1/R_n, n >= 0, and its numerator is conjectured to be a(n). If one considers the curvature of touching circles inscribed in the larger segment (sagitta length 7/5), the sequence would be A249457/A005032. See an illustration given in the link.
For the proof and the formula for the rational curvatures of the circles in the smaller segment see a comment under A249864. C_n = (5/(3*7))*(7*S(n, 26/7) - 13*S(n-1, 26/7) + 7), n >= 0, with Chebyshev's S polynomials (A049310). - Wolfdieter Lang, Nov 08 2014

Crossrefs

Programs

  • Magma
    I:=[10, 100, 1690]; [n le 3 select I[n] else 33*Self(n-1) - 231*Self(n-2) + 343*Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 20 2017
  • Mathematica
    LinearRecurrence[{33, -231, 343},{10, 100, 1690},16] (* Ray Chandler, Aug 11 2015 *)
    CoefficientList[Series[10*(1 - 23*x + 70*x^2)/((1 - 26*x + (7*x)^2)*(1 - 7*x)), {x, 0, 50}], x] (* G. C. Greubel, Dec 20 2017 *)
  • PARI
    {
    r=0.3;dn=3;print1(round(dn/r),", ");r1=r;
    for (n=1,40,
         if (n<=1,ab=2-r,ab=sqrt(ac^2+r^2));
         ac=sqrt(ab^2-r^2);
         if (n<=1,z=0,z=(Pi/2)-atan(ac/r)+asin((r1-r)/(r1+r));r1=r);
         b=acos(r/ab)-z;
         r=r*(1-cos(b))/(1+cos(b)); dn=dn*7;
         print1(round(dn/r),", ");
    )
    }
    
  • PARI
    x='x+O('x^30); Vec(10*(1 - 23*x + 70*x^2)/((1 - 26*x + (7*x)^2)*(1 - 7*x))) \\ G. C. Greubel, Dec 20 2017
    

Formula

Empirical g.f.: -10*(70*x^2-23*x+1) / ((7*x-1)*(49*x^2-26*x+1)). - Colin Barker, Oct 29 2014
From Wolfdieter Lang, Nov 09 2014 (Start)
a(n) = 5*(A249864(n) + 7^n) = (5*7^n)*(S(n, 26/7) - (13/7)*S(n-1, 26/7) + 1), n >= 0, with Chebyshev's S polynomials (A049310). See the comments on A249864 for the proof.
O.g.f.: 5*((1 - 13*x)/(1 - 26*x + (7*x)^2) + 1/(1-7*x)) = 10*(1 - 23*x + 70*x^2)/((1 - 26*x + (7*x)^2)*(1 - 7*x)) proving the conjecture of Colin Barker above. (End)

Extensions

Edited. In name and comment small changes, keyword easy and crossrefs added. - Wolfdieter Lang, Nov 08 2014

A248833 The curvature of touching circles inscribed in a special way in the larger segment of circle of radius 1/6 divided by a chord of length sqrt(8/75).

Original entry on oeis.org

10, 25, 160, 1225, 9610, 75625, 595360, 4687225, 36902410, 290532025, 2287353760, 18008298025, 141779030410, 1116223945225, 8788012531360, 69187876305625, 544714997913610, 4288532107003225, 33763541858112160, 265819802757894025, 2092794880205040010, 16476539238882426025
Offset: 0

Views

Author

Kival Ngaokrajang, Oct 15 2014

Keywords

Comments

Refer to comment of A240926. Consider a circle C of radius 1/6 (in some length units) with a chord of length sqrt(8/75). This has been chosen such that the larger sagitta has length 1/5. The input, besides the circle C, is the circle C_0 with radius R_0 = 1/10, touching the chord and circle C. The following sequence of circles C_n with radii R_n, n >= 1, is obtained from the conditions that C_n touches (i) the circle C, (ii) the chord and (iii) the circle C_(n-1). The curvature of the n-th circle, C_n = 1/R_n, n >= 0, is conjectured to be a(n). If one considers the curvature of touching circles inscribed in the smaller segment (sagitta length 2/15), the sequence would be A248834. See an illustration given in the link.

Crossrefs

Programs

  • Magma
    I:=[10,25,160]; [n le 3 select I[n] else 9*Self(n-1)-9*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Oct 29 2014
  • Mathematica
    CoefficientList[Series[- 5 (5 x^2 - 13 x + 2)/((x - 1) (x^2 - 8 x + 1)), {x, 0, 30}], x] (* Vincenzo Librandi, Oct 29 2014 *)
    LinearRecurrence[{9,-9,1}, {10,25,160}, 30] (* G. C. Greubel, Dec 20 2017 *)
  • PARI
    {
    r=0.6;print1(round(6/r),", ");r1=r;
    for (n=1,40,
         if (n<=1,ab=2-r,ab=sqrt(ac^2+r^2));
         ac=sqrt(ab^2-r^2);
         if (n<=1,z=0,z=(Pi/2)-atan(ac/r)+asin((r1-r)/(r1+r));r1=r);
         b=acos(r/ab)-z;
         r=r*(1-cos(b))/(1+cos(b));
         print1(round(6/r),", ");
    )
    }
    
  • PARI
    Vec(-5*(5*x^2-13*x+2)/((x-1)*(x^2-8*x+1)) + O(x^100)) \\ Colin Barker, Oct 15 2014
    

Formula

From Colin Barker, Oct 15 2014: (Start)
a(n) = 9*a(n-1) - 9*a(n-2) + a(n-3).
G.f.: -5*(5*x^2-13*x+2) / ((x-1)*(x^2-8*x+1)). (End)
a(n) = 5*(2+(4-sqrt(15))^n+(4+sqrt(15))^n)/2. - Colin Barker, Mar 03 2016
E.g.f.: 5*exp(x)*(1 + exp(3*x)*cosh(sqrt(15)*x)). - Stefano Spezia, Aug 27 2025
Showing 1-3 of 3 results.