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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 (list; graph; refs; listen; history; text; internal format)
 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 condition that C_n touches i) the circle C, ii) the chord and iii) the circle C_(n-1). The circle curvatures C_n = 1/R_n, n >= 0, are 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. Index entries for linear recurrences with constant coefficients, signature (39,-39,1). 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)^(-n)*(1 + (19 + 6*Sqrt)^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 Cf. A240926, A078986, A097315, A247512. Sequence in context: A185255 A160523 A301310 * A112694 A079914 A051790 Adjacent sequences:  A247332 A247333 A247334 * A247336 A247337 A247338 KEYWORD nonn,easy AUTHOR Kival Ngaokrajang, Sep 18 2014 STATUS approved

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Last modified July 17 08:40 EDT 2019. Contains 325097 sequences. (Running on oeis4.)