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.

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

Offset: 0

Author

Kival Ngaokrajang, Sep 18 2014

Keywords

• nonn
• easy

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, ...

Links

• Colin Barker, Table of n, a(n) for n = 0..600
• Kival Ngaokrajang, Illustration of initial terms
• Wolfdieter Lang, Curvature computation for A247335 and A247512.
• Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
• Giovanni Lucca, Circle chains inside the arbelos and integer sequences, Int'l J. Geom. (2023) Vol. 12, No. 1, 71-82.
• Index entries for sequences related to Chebyshev polynomials.
• Index entries for linear recurrences with constant coefficients, signature (39,-39,1).

Crossrefs

Cf. A240926, A078986, A097315, A247512.

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