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.

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

Links

Crossrefs

Cf. A240926, A078986, A097315, A247512, A247335, A247512, A248834, A169634, A249457, A049310, A249863, A249864.

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

A249862 A special solution of X(n)^2 - 280*Y(n)^2 = 3^(2*n), n >= 0; here the X member.

Original entry on oeis.org

1, 17, 569, 19193, 647441, 21840257, 736741769, 24852657833, 838359690401, 28280555553137, 953993651593049, 32181259154185433, 1085576868377967281, 36619982192463218657, 1235309202728347728809, 41670933053031653811593, 1405693940978521100034881
Offset: 0

Views

Author

Wolfdieter Lang, Nov 07 2014

Keywords

Comments

The member Y(n) = A248163(n-1) with A248163(-1) = 0.
This pair of sequences (X(n), Y(n)) appears in the solution of the touching circles and chord problem proposed by Kival Ngaokrajang in A249457. The curvatures (inverse radii) b(n) (for bend) of the circles in the lower section (on the left hand side) are here considered.
The derivation of the solution follows the lines given in the Wolfdieter Lang link in A240926, part I. Now the original radius of the large circle is R = 1 l.u. (length units) and the larger sagitta is 7/5 l.u. The circle radii are R(n), n >= 0, starting with R(0) = 7/10. Then a rescaling is done by a factor of 10/7 in the lengths: r = (10/7)*R = 10/7 l.u., such that the larger sagitta has 2 l.u. and r(n) = (10/7)*R(n).
The (nonlinear) recurrence for the curvature b(n) = 1/r(n) is written for bhat(n) := 3^n*b(n) and found to be: bhat(n) = 17*bhat(n-1) - 7*3^(n-1) + 140*sqrt((bhat(n-1) - 3^(n-1))*bhat(n-1)/(7*10)), n >= 1 with input bhat(0) = 1. This looks like A249457(n)/10. We search therefore for a positive integer solution which will then be the unique solution.
Define Y(n) := sqrt((bhat(n)-3^(n))*bhat(n)/(7*10)). This means (for positive bhat(n)) bhat(n) = (3^n + sqrt(9^n + 280*Y(n)^2))/2. Isolating the root and squaring yields X(n)^2 - 289*Y(n)^2 = 9^n, with X(n) := 2*bhat(n) - 3^n for n >= 0. For fixed n there are infinitely many solutions of this Diophantine equation. Here we only need a special solution for each n, which has to be an odd positive integer X(n) and X(0) = 1. Such a solution is X(n) = 3^(n-1)*(3*S(n, 34/3) - 17*S(n-1, 34/3)) and Y(n) = 3^(n-1)*S(n-1, 34/3) given in A249863(n-1), with Chebyshev's S-polynomial S(n, x). The proof is easy (once the o.g.f. for one of the sequences X or Y has been guessed by superseeker). Inserting the given formulas one has to prove S(n, 34/3)^2 + S(n-1, 34/3)^2 = 1 + (34/3)*S(n, 34/3)*S(n-1,34/3) which reduces after use of the recurrence relation for S to the well known Cassini-Simson identity S(n-1, x)^2 = 1 + S(n, x)*S(n-2, x), with S(-2) = -1, n >= 0, here with x = 34/3.
The solution for bhat(n) is then (X(n) + 3^n)/2. This satisfies indeed the original recurrence with input due to the recurrence relation of S(n, 34/3). Therefore, A249457(n)/10 = bhat(n).

Links

Crossrefs

Cf. A249457, A240926, A248163.

Programs

  • Magma
    I:=[1,17]; [n le 2 select I[n] else 34*Self(n-1) - 9*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 08 2014
  • Mathematica
    CoefficientList[Series[(1 - 17 x) / (1 - 34 x + 9 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Nov 08 2014 *)
LinearRecurrence[{34,-9},{1,17},30] (* Harvey P. Dale, Dec 13 2016 *)
  • PARI
    x='x+O('x^30); Vec((1 - 17*x)/(1 - 34*x + 9*x^2)) \\ G. C. Greubel, Dec 20 2017

Formula

a(n) = 3^(n-1)*(3*S(n, 34/3) - 17*S(n-1, 34/3)), n >= 0, with the scaled Chebyshev S sequence 3^n*S(n, 34/3) given in A248163.
O.g.f.: (1 - 17*x)/(1 - 34*x + 9*x^2).
a(n) = 34*a(n-1) - 9*a(n-2), a(0) = 0, a(1) = 17.
E.g.f.: exp(17*x)*cosh(2*sqrt(70)*x). - Stefano Spezia, Mar 24 2023

A249864 A special solution of X(n)^2 - 120*Y(n)^2 = 7^(2*n), n >= 0. The present sequence gives the X values.

Original entry on oeis.org

1, 13, 289, 6877, 164641, 3943693, 94468609, 2262942877, 54207552961, 1298512176013, 31105146481249, 745106711887837, 17848622331502561, 427553951736562573, 10241820250907001409, 245337182888490470557
Offset: 0

Views

Author

Wolfdieter Lang, Nov 09 2014

Keywords

Comments

The member Y(n) = A249863(n-1) with A249863(-1) = 0.
This pair of sequences (X(n), Y(n)) appears in the solution of the touching circles and chord problem proposed by Kival Ngaokrajang in A249458. The curvatures (inverse radii) b(n) (for bend) of the circles in the smaller section (on the left hand side) are here considered.
The derivation of the solution follows the lines given in the Wolfdieter Lang link in A240926, part II. Now the original radius of the large circle is R = 1 l.u. (length units) and the larger sagitta h is 7/5 l.u. The circle radii are R(n), n >= 0, starting with R(0) = 3/10. Then a rescaling is done by r = (10/7)*R = 10/7 l.u., such that the larger sagitta has 2 l.u. and r(n) = (10/7)*R(n).
The (nonlinear) recurrence for the curvature b(n) = 1/r(n) is written for bhat(n) := 3*7^(n-1)*b(n) and found to be: bhat(n) = (1/7)*(91*bhat(n-1) - 3*7^n + 2*sqrt(210)*sqrt((7*bhat(n-1) - 7^n)*bhat(n-1))), n >= 1, with input bhat(0) = 1. This looks like A249458(n)/10. We search therefore for a positive integer solution which will then be the unique solution.
Define Y(n) := sqrt((bhat(n)-7^(n))*bhat(n)/(30)). This means (for positive bhat(n)) bhat(n) = (7^n + sqrt(7^(2*n) + 120*Y(n)^2))/2. Isolating the root and squaring yields X(n)^2 - 120*Y(n)^2 = 7^(2*n), with X(n) := 2*bhat(n) - 7^n for n >= 0. For fixed n there are infinitely many solutions of this Diophantine equation. Here we only need a special solution for each n, which has to be an odd positive integer X(n) and X(0) = 1. This solution is X(n) = 7^(n-1)*(7*S(n, 26/7) - 13*S(n-1, 26/7)) and Y(n) = 7^(n-1)*S(n-1, 26/7) given in A249863(n-1), with Chebyshev's S-polynomial S(n, x). The proof is easy (once the o.g.f. for one of the sequences X or Y has been guessed, e.g., by superseeker). Inserting the given formulas one has to prove S(n, 26/7)^2 + S(n-1, 26/7)^2 = 1 + (26/7)*S(n, 26/7)*S(n-1, 26/7) which reduces after use of the recurrence relation for S to the well known Cassini-Simson identity S(n-1, x)^2 = 1 + S(n, x)*S(n-2, x), with S(-2) = -1, n >= 0, here with x = 26/7.
The solution for bhat(n) is then (X(n) + 7^n)/2. This satisfies indeed the original recurrence with input due to the recurrence relation of S(n, 26/7). Therefore, A249458(n)/10 = bhat(n).

Links

Crossrefs

Cf. A249457, A249458, A249863.

Programs

  • Magma
    I:=[1,13]; [n le 2 select I[n] else 26*Self(n-1)-49*Self(n-2): n in [1..40]]; // Vincenzo Librandi, Nov 09 2014
  • Maple
    f:= gfun:-rectoproc({a(n)=26*a(n-1)-49*a(n-2),a(0)=1,a(1)=13},a(n),remember):
seq(f(n),n=0..50); # Robert Israel, Nov 18 2014
  • Mathematica
    LinearRecurrence[{26, -49}, {1, 13} , 30] (* or *) CoefficientList[ Series[(1 - 13*x)/(1 - 26*x + (7*x)^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 20 2017 *)
  • PARI
    x='x+O('x^30); Vec((1 - 13*x)/(1 - 26*x + (7*x)^2)) \\ G. C. Greubel, Dec 20 2017

Formula

a(n) = (7^n)*(S(n, 26/7) - (13/7)*S(n-1, 26/7)), n >= 0, with the scaled Chebyshev S sequence 7^n*S(n, 26/7) given in A249863.
O.g.f.: (1 - 13*x)/(1 - 26*x + (7*x)^2).
a(n) = 26*a(n-1) - 49*a(n-2), a(0) = 1, a(1) = 13.
a(n) = (r^n + s^n)/2 where r,s are the roots of x^2 - 26*x + 49. - Robert Israel, Nov 18 2014
Showing 1-3 of 3 results.

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