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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.
7
1, 10, 361, 13690, 519841, 19740250, 749609641, 28465426090, 1080936581761, 41047124680810, 1558709801289001, 59189925324301210, 2247658452522156961, 85351831270517663290, 3241121929827149048041, 123077281502161146162250
OFFSET
0,2
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, ...
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
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 *)
PROG
(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
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Kival Ngaokrajang, Sep 18 2014
STATUS
approved