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 1-7 of 7 results.

A246643 A sequence used in the touching circle problem described in A247512.

Original entry on oeis.org

1, 10, 121, 1690, 25921, 420250, 7027801, 119508490, 2050368961, 35341836010, 610665665401, 10564982353210, 182902930753921, 3167536046903290, 54865571909148121, 950426408617182250, 16464857882672822401, 285238628280432626890, 4941562979309619843961
Offset: 0

Views

Author

Wolfdieter Lang, Sep 30 2014

Keywords

Comments

This sequence appears in the touching circle problem considered in A247512. There the rational curvatures are b'(n) = a(n)/9^(n-1), and A247512(n) = floor(b'(n)).
See the W. Lang link, part II) with the details where B'(n) plays the role of a(n).

Crossrefs

Programs

  • Magma
    I:=[1, 10, 121]; [n le 3 select I[n] else 31*Self(n-1) - 279*Self(n-2) + 729*Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 20 2017
  • Mathematica
    CoefficientList[Series[(1 - 21*x + 90*x^2)/((1 - 9*x)*(1 - 22*x + 81*x^2)), {x, 0, 50}], x] (* or *) LinearRecurrence[{31,-279,729}, {1, 10, 121}, 50] (* G. C. Greubel, Dec 20 2017 *)
  • PARI
    Vec(-(6*x-1)*(15*x-1)/((9*x-1)*(81*x^2-22*x+1)) + O(x^100)) \\ Colin Barker, Sep 30 2014
    

Formula

One step recurrence: a(n) = 11*a(n-1) - 9^(n-1) + 20*sqrt((a(n-1) - 9^(n-1))*a(n-1)/10), a(0) = 1, n >= 1.
a(n) = (9^n)*(1 + S(n, 22/9) - (11/9)*S(n-1, 22/9))/2, with Chebyshev's S-polynomials (see A049310). For 9^n*S(n, 22/9) see A246645(n). The positive integer sequence sqrt((a(n) - 9^n)*a(n)/10) = A246645(n-1).
O.g.f.: (1 - 21*x + 90*x^2)/((1 - 9*x)*(1 - 22*x + 81*x^2)) = (1/2)*((1 -11*x)/(1 - 22*x + 81*x^2 ) - 1/(1 - 9*x)).
For the proofs see the W. Lang link with a(n) = B'(n).
a(n) = 31*a(n-1)-279*a(n-2)+729*a(n-3). - Colin Barker, Sep 30 2014

A247335 The curvature of touching circles inscribed in a special way in the larger segment of circle of radius 10/9 divided by a chord of length 4/3.

Original entry on oeis.org

1, 10, 361, 13690, 519841, 19740250, 749609641, 28465426090, 1080936581761, 41047124680810, 1558709801289001, 59189925324301210, 2247658452522156961, 85351831270517663290, 3241121929827149048041, 123077281502161146162250
Offset: 0

Views

Author

Kival Ngaokrajang, Sep 18 2014

Keywords

Comments

Refer to comment of A240926. Consider a circle C of radius 10/9 (in some length units) with a chord of length 4/3. This has been chosen such that the larger sagitta has length 2. The smaller sagitta has length 2/9. The input, besides the circle C is the circle C_0 with radius R_0 = 1, 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, the sequence would be A247512. See an illustration given in the link.
a(n) also seems to be A078986(i)^2 and 10*A097315(j)^2 interleaved; where i = n/2 for n even, j = n/2 - 1/2 for n odd; as following:
1 = 1^2
10 = 10*1^2
361 = 19^2
13690 = 10*37^2
519841 = 721^2
19740250 = 10*1405^2
749609641 = 27379^2
...
A078986; Chebyshev... polynomial: 1, 19, 721, 27379, ...
A097315; Pell equation... : 1, 37, 1405, 53353, ...

Crossrefs

Programs

  • Magma
    I:=[39,-39,1]; [n le 3 select I[n] else Self(n-1) - 10*Self(n-2) + 361*Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 20 2017
  • Mathematica
    LinearRecurrence[{39,-39,1}, {1, 10, 361}, 50] (* or *) Table[Round[((19 + 6*Sqrt[10])^(-n)*(1 + (19 + 6*Sqrt[10])^n)^2)]/4, {n, 0, 30}] (* G. C. Greubel, Dec 20 2017 *)
  • PARI
    {
    r=0.9;print1(1,", ");r1=r;
    for (n=1,50,
    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));
    an=floor(9/(10*r));
    print1(if(an>9,an,10),", ")
    )
    }
    
  • PARI
    Vec(-(10*x^2-29*x+1)/((x-1)*(x^2-38*x+1)) + O(x^20)) \\ Colin Barker, Mar 03 2016
    

Formula

Conjectures from Colin Barker, Sep 18 2014: (Start)
a(n) = 39*a(n-1) - 39*a(n-2) + a(n-3).
G.f.: -(10*x^2-29*x+1) / ((x-1)*(x^2-38*x+1)). (End)
From Wolfdieter Lang, Sep 30 2014 (Start)
See the W. Lang link for proofs of the following statements.
One step nonlinear recurrence: a(n) = -9 + 19*a(n-1) + 60*sqrt(a(n-1)*(a(n-1) - 1)/10), n>=1, with a(0) = 1.
a(n) = (1 + A078986(n))/2 = (2 + S(n, 38) - S(n-2, 38))/4 =
(1 + S(n, 38) -19*S(n-1, 38))/2 for n>=0, with Chebyshev's S-polynomials (see A049310). S(n, 38) = A078987(n).
The G.f. conjectured by Colin Barker above follows from the one for Chebyshev's T(n, 19) = A078986(n): (1/(1-x) + (1-19*x)/(1-38*x+x^2))/2 = (1-29*x+10*x^2)/((1-x)* (1-38*x+x^2)).
The four term recurrence conjectured by Colin Barker above follows from the expanded g.f. denominator: (1-x)* (1-38*x+x^2) = 1- 39*x + 39*x^2 - x^3.
(End)
a(n) = ((19+6*sqrt(10))^(-n)*(1+(19+6*sqrt(10))^n)^2)/4. - Colin Barker, Mar 03 2016

A246645 Expansion of 1/(1 - 22*x + 81*x^2), used in A246643.

Original entry on oeis.org

1, 22, 403, 7084, 123205, 2136706, 37027927, 641541208, 11114644489, 192557340910, 3335975296411, 57794311907332, 1001260862952013, 17346399720450394, 300518663950795615, 5206352229561021616, 90197737270328030737, 1562635689352773925318, 27071968446864455867299
Offset: 0

Views

Author

Wolfdieter Lang, Sep 30 2014

Keywords

Comments

This sequence is used in the formula for the curvature in a touching circle problem considered in A247512 and A246643.

Crossrefs

Programs

  • Magma
    I:=[1, 22]; [n le 2 select I[n] else 22*Self(n-1) - 81*Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 20 2017
  • Mathematica
    CoefficientList[Series[1/(1 - 22*x + 81*x^2), {x, 0, 50}], x] (* or *) LinearRecurrence[{22,-81}, {1,22}, 50] (* G. C. Greubel, Dec 20 2017 *)
  • PARI
    Vec(1/(1 - 22*x + 81*x^2) + O(x^40)) \\ Michel Marcus, Sep 30 2014
    

Formula

O.g.f.: 1/(1 - 22*x + 81*x^2).
a(n) = 9^n*S(n, 22/9) with Chebyshev's S-polynomials (see A049310).
a(n) = 22*a(n-1) - 81*a(n-2), n >= 1, a(-1) = 0 and a(0) = 1.
a(n) = 9^n*(ap^(n+1) - am^(n+1))/(ap - am), n >= 1, with ap := (11 + 2*sqrt(10))/9 and am = 1/ap = (11 - 2*sqrt(10))/9 (Binet - de Moivre formula). a(0) = 1 (via L'Hopital's rule).
a(n) = 9^(n+1)*sinh(2*(n + 1)*arccsch(3))/(2*sqrt(10)). - Federico Provvedi, Feb 02 2021

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

A248834 The numerator of curvature of touching circles inscribed in a special way in the smaller segment of circle of radius 1/6 divided by a chord of length sqrt(8/75).

Original entry on oeis.org

15, 25, 245, 3025, 39605, 525625, 6997445, 93219025, 1242045605, 16549536025, 220514700245, 2938258798225, 39150987330005, 521669482807225, 6951013841444645, 92619168339300625, 1234109231890228805, 16443956730548563225, 219108411138085022645, 2919522145350504838225
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 smaller sagitta has length 2/15. The input, besides the circle C, is the circle C_0 with radius R_0 = 1/15, 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). The denominator is A000244 for n > 0. If one considers the curvature of touching circles inscribed in the larger segment (sagitta length 1/5), the sequence would be A248833. See an illustration given in the link.

Crossrefs

Programs

  • PARI
    {
    r=0.4;print1(round(6/r),", ");r1=r;dn=1;
    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)*dn),", ");
    dn=dn*3
    )
    }

Formula

Conjecture: a(n) = 17*a(n-1) - 51*a(n-2) + 27*a(n-3) for n > 3. - Colin Barker, Oct 15 2014
Empirical g.f.: 5*(54*x^3-117*x^2+46*x-3) / ((3*x-1)*(9*x^2-14*x+1)). - Colin Barker, Oct 15 2014

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 1-7 of 7 results.