%I
%S 4,5,9,20,49,125,324,845,2209,5780,15129,39605,103684,271445,710649,
%T 1860500,4870849,12752045,33385284,87403805,228826129,599074580,
%U 1568397609,4106118245,10749957124,28143753125,73681302249,192900153620,505019158609
%N a(n) = 2 + L(2*n) = 2 + A005248(n), n >= 0, with the Lucas numbers (A000032).
%C This sequence also gives the curvature of touching circles inscribed in a special way in the smaller segment of a circle of radius 5/4 divided by a chord of length 2.
%C Consider a circle C of radius 5/4 (in some length units) with a chord of length 2. This has been chosen such that the larger sagitta also has length 2. The smaller sagitta has length 1/2. The input, besides the circle C, is the circle C_0 with radius R_0 = 1/4, 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)(n1). The circle curvatures C_n = 1/R_n, n >= 0, are conjectured to be a(n). See an illustration given in the link. As found by _Wolfdieter Lang_ (see part II of the proof given by W. Lang in the link), this circle problem is related to the nonnegative solutions of the Pell equation X^2  5*Y^2 = 4: a(n) = 2 + X(n) = 2 + A005248(n). For the larger segment below the chord (with sagitta length 2) the sequence would be A115032, see W. Lang's proof given in part I of the link.
%C If the circle radius and the sagitta length were both equal to 1, the curvature sequence would be A099938.
%H G. C. Greubel, <a href="/A240926/b240926.txt">Table of n, a(n) for n = 0..2375</a>
%H Kival Ngaokrajang, <a href="/A240926/a240926_5.pdf">Illustration of initial terms</a>.
%H Wolfdieter Lang, <a href="/A240926/a240926_8.pdf"> Proof of the coincidence of a(n) with the touching circle problem (part II)</a>.
%H Wolfdieter Lang, <a href="/A240926/a240926_10.pdf">Figures for various touching circle problems</a>.
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (4,4,1).
%F Conjectures (proved in the next entry) from _Colin Barker_, Aug 25 2014 (and Aug 27 2014): (Start)
%F a(n) = (2 + ((1/2)*(3sqrt(5)))^n + ((1/2)*(3+sqrt(5)))^n).
%F a(n) = 4*a(n1)  4*a(n2) + a(n3).
%F G.f.: (5*x^211*x+4) / ((x1)*(x^23*x+1)). (End)
%F From _Wolfdieter Lang_, Aug 26 2014: (Start)
%F a(n) = 2 + S(n, 3)  S(n2, 3) = 2 + 2*S(n, 3)  3*S(n1, 3).
%F a(n) = 3*a(n1)  a(n2)  2, n >= 1, with a(1)= 5 and a(0) = 4 (from the S(n, 3) recurrence or from A005248).
%F The first of the _Colin Barker_ conjectures above is true because of the Binetde Moivre formula for L(2*n) (see the Jul 24 2003 _Dennis P. Walsh_ comment on A005248). With phi = (1+sqrt(5))/2, use 1/phi = phi1, phi^2 = phi+1, (phi1)^2 = 2  phi.
%F His third conjecture (the g.f.) follows from the g.f. of A005248 by adding 2/(1x).
%F His second conjecture (recurrence) with input a(3) = 20, a(2) = 9 and a(1) = 5 (from the above given recurrence) leads to his g.f. with the expanded denominator. Thus all three conjectures are true. (End)
%F a(n) = A005592(n) + 3, with n > 0.  _Zino Magri_, Feb 16 2015
%F a(n) = (phi^n + phi^(n))^2, where phi = A001622 = (1 + sqrt(5))/2.  _Diego Rattaggi_, Jun 10 2020
%F Sum_{k>=0} 1/a(k) = A338303.  _Amiram Eldar_, Oct 22 2020
%t Table[2 + LucasL[2 n], {n, 0, 50}] (* _Vincenzo Librandi_, Oct 08 2015 *)
%o (MAGMA) [2+Lucas(2*n): n in [0..40]]; // _Vincenzo Librandi_, Oct 08 2015
%o (PARI) vector(100, n, n; 2 + fibonacci(2*n1) + fibonacci(2*n+1)) \\ _Altug Alkan_, Oct 08 2015
%Y Cf. A000032, A005248, A005592, A099938, A115032, A246638, A246639, A246640, A246641, A246642, A338303.
%K nonn,easy
%O 0,1
%A _Kival Ngaokrajang_, Aug 03 2014
%E Edited: name changed (after proof has been given in part II of the W. Lang link), comments rewritten, cross refs. and link to Chebyshev index added.  _Wolfdieter Lang_, Aug 26 2014
