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

4, 5, 9, 20, 49, 125, 324, 845, 2209, 5780, 15129, 39605, 103684, 271445, 710649, 1860500, 4870849, 12752045, 33385284, 87403805, 228826129, 599074580, 1568397609, 4106118245, 10749957124, 28143753125, 73681302249, 192900153620, 505019158609

Offset: 0

Author

Kival Ngaokrajang, Aug 03 2014

Keywords

• nonn
• easy

Comments

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.

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.

If the circle radius and the sagitta length were both equal to 1, the curvature sequence would be A099938.

Essentially a duplicate of A092387. - R. J. Mathar, Jul 07 2023

Links

• G. C. Greubel, Table of n, a(n) for n = 0..2375
• Wolfdieter Lang, Proof of the coincidence of a(n) with the touching circle problem (part II).
• Wolfdieter Lang, Figures for various touching circle problems.
• Kival Ngaokrajang, Illustration of initial terms.
• Index entries for sequences related to Chebyshev polynomials.
• Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Crossrefs

Cf. A000032, A005248, A005592, A099938, A115032, A246638, A246639, A246640, A246641, A246642, A338303.

Programs

Magma

[2+Lucas(2*n): n in [0..40]]; // Vincenzo Librandi, Oct 08 2015

Mathematica

Table[2 + LucasL[2 n], {n, 0, 50}] (* Vincenzo Librandi, Oct 08 2015 *)

PARI

vector(100, n, n--; 2 + fibonacci(2*n-1) + fibonacci(2*n+1)) \\ Altug Alkan, Oct 08 2015

Formula

Conjectures (proved in the next entry) from Colin Barker, Aug 25 2014 (and Aug 27 2014): (Start)

a(n) = (2 + ((1/2)*(3-sqrt(5)))^n + ((1/2)*(3+sqrt(5)))^n).

a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3).

G.f.: -(5*x^2-11*x+4) / ((x-1)*(x^2-3*x+1)). (End)

From Wolfdieter Lang, Aug 26 2014: (Start)

a(n) = 2 + S(n, 3) - S(n-2, 3) = 2 + 2*S(n, 3) - 3*S(n-1, 3).

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).

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.

His third conjecture (the g.f.) follows from the g.f. of A005248 by adding 2/(1-x).

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)

a(n) = A005592(n) + 3, with n > 0. - Zino Magri, Feb 16 2015

a(n) = (phi^n + phi^(-n))^2, where phi = A001622 = (1 + sqrt(5))/2. - Diego Rattaggi, Jun 10 2020

Sum_{k>=0} 1/a(k) = A338303. - Amiram Eldar, Oct 22 2020

Extensions

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