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.

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

Offset: 0

Author

Kival Ngaokrajang, Oct 29 2014

Keywords

• nonn
• frac
• easy

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

• Kival Ngaokrajang, Illustration of initial terms.
• Eric Weisstein's World of Mathematics, Sagitta.
• Index entries for linear recurrences with constant coefficients, signature (33,-231,343).
• Index entries for sequences related to Chebyshev polynomials.

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