cp's OEIS Frontend

A240926 a(n) = 2 + L(2*n) = 2 + A005248(n), n >= 0, with the Lucas numbers (A000032).

%I A240926 #102 May 20 2024 16:48:06
%S A240926 4,5,9,20,49,125,324,845,2209,5780,15129,39605,103684,271445,710649,
%T A240926 1860500,4870849,12752045,33385284,87403805,228826129,599074580,
%U A240926 1568397609,4106118245,10749957124,28143753125,73681302249,192900153620,505019158609
%N A240926 a(n) = 2 + L(2*n) = 2 + A005248(n), n >= 0, with the Lucas numbers (A000032).
%C A240926 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 cut by a chord of length 2.
%C A240926 Consider a circle C of radius 5/4 (in some length units) with a chord of length 2. This has been chosen so 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 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). 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 A240926 If the circle radius and the sagitta length were both equal to 1, the curvature sequence would be A099938.
%C A240926 Essentially a duplicate of A092387. - _R. J. Mathar_, Jul 07 2023
%H A240926 G. C. Greubel, <a href="/A240926/b240926.txt">Table of n, a(n) for n = 0..2375</a>
%H A240926 Wolfdieter Lang, <a href="/A240926/a240926_8.pdf"> Proof of the coincidence of a(n) with the touching circle problem (part II)</a>.
%H A240926 Wolfdieter Lang, <a href="/A240926/a240926_10.pdf">Figures for various touching circle problems</a>.
%H A240926 Kival Ngaokrajang, <a href="/A240926/a240926_5.pdf">Illustration of initial terms</a>.
%H A240926 <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%H A240926 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (4,-4,1).
%F A240926 Conjectures (proved in the next entry) from _Colin Barker_, Aug 25 2014 (and Aug 27 2014): (Start)
%F A240926 a(n) = (2 + ((1/2)*(3-sqrt(5)))^n + ((1/2)*(3+sqrt(5)))^n).
%F A240926 a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3).
%F A240926 G.f.: -(5*x^2-11*x+4) / ((x-1)*(x^2-3*x+1)). (End)
%F A240926 From _Wolfdieter Lang_, Aug 26 2014: (Start)
%F A240926 a(n) = 2 + S(n, 3) - S(n-2, 3) = 2 + 2*S(n, 3)  - 3*S(n-1, 3).
%F A240926 a(n) = 3*a(n-1) - a(n-2) - 2, n >= 1, with a(-1)= 5 and a(0) = 4 (from the S(n, 3) recurrence or from A005248).
%F A240926 The first of the _Colin Barker_ conjectures above is true because of the Binet-de 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 = phi-1, phi^2 = phi+1, (phi-1)^2 = 2 - phi.
%F A240926 His third conjecture (the g.f.) follows from the g.f. of A005248 by adding 2/(1-x).
%F A240926 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 A240926 a(n) = A005592(n) + 3, with n > 0. - _Zino Magri_, Feb 16 2015
%F A240926 a(n) = (phi^n + phi^(-n))^2, where phi = A001622 = (1 + sqrt(5))/2. - _Diego Rattaggi_, Jun 10 2020
%F A240926 Sum_{k>=0} 1/a(k) = A338303. - _Amiram Eldar_, Oct 22 2020
%t A240926 Table[2 + LucasL[2 n], {n, 0, 50}] (* _Vincenzo Librandi_, Oct 08 2015 *)
%o A240926 (Magma) [2+Lucas(2*n): n in [0..40]]; // _Vincenzo Librandi_, Oct 08 2015
%o A240926 (PARI) vector(100, n, n--; 2 + fibonacci(2*n-1) + fibonacci(2*n+1)) \\ _Altug Alkan_, Oct 08 2015
%Y A240926 Cf. A000032, A005248, A005592, A099938, A115032, A246638, A246639, A246640, A246641, A246642, A338303.
%K A240926 nonn,easy
%O A240926 0,1
%A A240926 _Kival Ngaokrajang_, Aug 03 2014
%E A240926 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