OFFSET
0,1
COMMENTS
Let b(k) = Sum_{i=1..k} F(2*n*i) * binomial(k,i) where F(k) denotes the k-th Fibonacci number. The (b(k)) sequence satisfies the recursion: b(k) = a(n) * (b(k-1)-b(k-2)).
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.
Same as the number of Kekulé structures in polyphenanthrene in terms of the number of hexagons. - Parthasarathy Nambi, Aug 22 2006
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Wolfdieter Lang, Figures for various touching circle problems.
I. Lukovits and D. Janezic, Enumeration of conjugated circuits in nanotubes, J. Chem. Inf. Comput. Sci. 44 (2004), 410-414. See Table 1 column 3 on page 411.
Kival Ngaokrajang, Illustration of initial terms.
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).
FORMULA
a(0)=4, a(1)=5, a(2)=9, a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3).
a(n) = 3 + floor((1+phi)^n) where phi = (1+sqrt(5))/2.
a(n) = A005248(n) + 2.
From R. J. Mathar, Mar 18 2009: (Start)
G.f.: -x*(5 - 11*x + 4*x^2)/((x-1)(x^2 - 3*x + 1)).
a(n+1) - a(n) = A002878(n). (End)
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). (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 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 both 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
MATHEMATICA
CoefficientList[Series[-(5-11*x+4*x^2)/((x-1)(x^2-3*x+1)), {x, 0, 30}], x] (* Vincenzo Librandi, May 06 2012 *)
PROG
(Magma) I:=[4, 5, 9]; [n le 3 select I[n] else 4*Self(n-1)-4*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, May 06 2012
(PARI) vector(100, n, n--; 2 + fibonacci(2*n-1) + fibonacci(2*n+1)) \\ Altug Alkan, Oct 08 2015
CROSSREFS
KEYWORD
nonn,easy,changed
AUTHOR
Benoit Cloitre, Mar 20 2004
EXTENSIONS
Better definition from Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 20 2004
a(0)=4 inserted by Sean A. Irvine, Nov 30 2025
Content copied from duplicate A240926 and entry revised by Sean A. Irvine, Nov 30 2025
STATUS
approved