A005592 -id:A005592 - cp's OEIS frontend

A005248 Bisection of Lucas numbers: a(n) = L(2n) = A000032(2n).

2, 3, 7, 18, 47, 123, 322, 843, 2207, 5778, 15127, 39603, 103682, 271443, 710647, 1860498, 4870847, 12752043, 33385282, 87403803, 228826127, 599074578, 1568397607, 4106118243, 10749957122, 28143753123, 73681302247, 192900153618, 505019158607, 1322157322203

Offset: 0

N. J. A. Sloane

Drop initial 2; then iterates of A050411 do not diverge for these starting values. - David W. Wilson
All nonnegative integer solutions of Pell equation a(n)^2 - 5*b(n)^2 = +4 together with b(n)=A001906(n), n>=0. - Wolfdieter Lang, Aug 31 2004
a(n+1) = B^(n)AB(1), n>=0, with compositions of Wythoff's complementary A(n):=A000201(n) and B(n)=A001950(n) sequences. See the W. Lang link under A135817 for the Wythoff representation of numbers (with A as 1 and B as 0 and the argument 1 omitted). E.g., 3=`10`, 7=`010`, 18=`0010`, 47=`00010`, ..., in Wythoff code. a(0) = 2 = B(1) in Wythoff code.
Output of Tesler's formula (as well as that of Lu and Wu) for the number of perfect matchings of an m X n Möbius band where m and n are both even specializes to this sequence for m=2. - Sarah-Marie Belcastro, Jul 04 2009
Numbers having two 1's in their base-phi representation. - Robert G. Wilson v, Sep 13 2010
Pisano period lengths: 1, 3, 4, 3, 2, 12, 8, 6, 12, 6, 5, 12, 14, 24, 4, 12, 18, 12, 9, 6, ... - R. J. Mathar, Aug 10 2012
From Wolfdieter Lang, Feb 18 2013: (Start)
a(n) is also one half of the total number of round trips, each of length 2*n, on the graph P_4 (o-o-o-o) (the simple path with 4 points (vertices) and 3 lines (or edges)). See the array and triangle A198632 for the general case of the graph P_N (there N is n and the length is l=2*k).
O.g.f. for w(4,l) (with zeros for odd l): y*(d/dy)S(4,y)/S(4,y) with y=1/x and Chebyshev S-polynomials (coefficients A049310). See also A198632 for a rewritten form. One half of this o.g.f. for x -> sqrt(x) produces the g.f. (2-3x)/(1-3x+x^2) given below. (End)
Solutions (x, y) = (a(n), a(n+1)) satisfying x^2 + y^2 = 3xy - 5. - Michel Lagneau, Feb 01 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 7xy + y^2 + 45 = 0. - Colin Barker, Feb 16 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 18xy + y^2 + 320 = 0. - Colin Barker, Feb 16 2014
a(n) are the numbers such that a(n)^2-2 are Lucas numbers. - Michel Lagneau, Jul 22 2014
All sequences of this form, b(n+1) = 3*b(n) - b(n-1), regardless of initial values, which includes this sequence, yield this sequence as follows: a(n) = (b(j+n) + b(j-n))/b(j), for any j, except where b(j) = 0. Also note formula below relating this a(n) to all sequences of the form G(n+1) = G(n) + G(n-1).  - Richard R. Forberg, Nov 18 2014
A non-simple continued fraction expansion for F(2n*(k+1))/F(2nk) k>=1 is a(n) + (-1)/(a(n) + (-1)/(a(n) + ... +  (-1)/a(n))) where a(n) appears exactly k times (F(n) denotes the n-th Fibonacci number). E.g., F(16)/F(12) equals 7 + (-1)/(7 + (-1)/7). Furthermore, these a(n) are exactly the positive integers k such that the non-simple infinite continued fraction k + (-1)/(k + (-1)/(k + (-1)/(k + ...))) belongs to Q(sqrt(5)). Compare to Benoit Cloitre and Thomas Baruchel's comments at A002878.  - Greg Dresden, Aug 13 2019
For n >= 1, a(n) is the number of cyclic up-down words of length 2*n over an alphabet of size 3. - Sela Fried, Apr 08 2025

			G.f. = 2 + 3*x + 7*x^2 + 18*x^3 + 47*x^4 + 123*x^5 + 322*x^6 + 843*x^7 + ... - _Michael Somos_, Aug 11 2009

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Richard P. Stanley, Enumerative combinatorics, Vol. 2. Volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Richard André-Jeannin, Summation of Certain Reciprocal Series Related to Fibonacci and Lucas Numbers, The Fibonacci Quarterly, Vol. 29, No. 3 (1991), pp. 200-204.
Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1.
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Pooja Bhadouria, Deepika Jhala and Bijendra Singh, Binomial Transforms of the k-Lucas Sequences and its Properties, The Journal of Mathematics and Computer Science (JMCS), Vol. 8, No. 1 (2014), pp. 81-92; sequences B_1, T_1 and R_1.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 51, 56.
Noureddine Chair, Exact two-point resistance, and the simple random walk on the complete graph minus N edges, Ann. Phys., Vol. 327, No. 12 (2012), pp. 3116-3129, Eq. (18).
Tony Crilly, Double sequences of positive integers, Math. Gaz., Vol. 69 (1985), pp. 263-271.
Pedro P. B. de Oliveira, Enrico Formenti, Kévin Perrot, Sara Riva and Eurico L. P. Ruivo, Non-maximal sensitivity to synchronism in periodic elementary cellular automata: exact asymptotic measures, arXiv:2004.07128 [cs.FL], 2020.
Robert G. Donnelly, Molly W. Dunkum, Murray L. Huber and Lee Knupp, Sign-alternating Gibonacci polynomials, arXiv:2012.14993 [math.CO], 2020.
Sergio Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, Vol. 5, No. 15 (2014), pp. 2226-2234.
Margherita Maria Ferrari and Norma Zagaglia Salvi, Aperiodic Compositions and Classical Integer Sequences, Journal of Integer Sequences, Vol. 20 (2017), Article 17.8.8.
Sela Fried, A formula for the number of up-down words, arXiv:2503.02005 [math.CO], 2025.
Sela Fried, Even-up words and their variants, arXiv:2505.14196 [math.CO], 2025. See p. 8.
Dale Gerdemann, Collision of Digits "Also interesting are the two bisections of the Lucas numbers A005248 (digit minimizer) and A002878 (digit maximizer). I particularly like the multiples of A005248 because I have this image of the two digits piling up on top of each other and then spreading out like waves".
André Gougenheim, About the linear sequence of integers such that each term is the sum of the two preceding Part 1 Part 2, Fib. Quart., Vol. 9, No. 3 (1971), pp. 277-295, 298.
Richard K. Guy, Letter to N. J. A. Sloane, Feb 1986.
Tanya Khovanova, Recursive Sequences.
Emrah Kılıç, Yücel Türker Ulutaş and Neşe Ömür, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq., Vol. 14 (2011), Article 11.5.6, Table 2.
Wentao T. Lu and F. Y. Wu, Close-packed dimers on nonorientable surfaces, Physics Letters A, Vol. 293, No. 5-6 (2002), pp. 235-246. [From Sarah-Marie Belcastro, Jul 04 2009]
Yun-Tak Oh, Hosho Katsura, Hyun-Yong Lee and Jung Hoon Han, Proposal of a spin-one chain model with competing dimer and trimer interactions, arXiv:1709.01344 [cond-mat.str-el], 2017.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992, arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Sara Riva, Factorisation of Discrete Dynamical Systems, Ph.D. Thesis, Univ. Côte d'Azur (France 2023).
Jeffrey Shallit, An interesting continued fraction, Math. Mag., Vol. 48, No. 4 (1975), pp. 207-211.
Jeffrey Shallit, An interesting continued fraction, Math. Mag., Vol. 48, No. 4 (1975), pp. 207-211. [Annotated scanned copy]
Jeffrey Shallit, Proving Properties of phi-Representations with the Walnut Theorem-Prover, arXiv:2305.02672 [math.NT], 2023.
Jeffrey Shallit and N. J. A. Sloane, Correspondence, 1975.
Paweł J. Szabłowski, On moments of Cantor and related distributions, arXiv preprint arXiv:1403.0386 [math.PR], 2014.
Glenn Tesler, Matchings in Graphs on Non-Orientable Surfaces, Journal of Combinatorial Theory, Series B, Vol. 78, No. 2 (2000), pp. 198-231.
Kai Wang, Fibonacci Numbers And Trigonometric Functions Outline, (2019).
Eric Weisstein's World of Mathematics, Phi Number System.
A. V. Zarelua, On Matrix Analogs of Fermat’s Little Theorem, Mathematical Notes, Vol. 79, No. 6 (2006), pp. 783-796. Translated from Matematicheskie Zametki, Vol. 79, No. 6 (2006), pp. 840-855.
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2).
Index entries for sequences related to Chebyshev polynomials.
Index entries for two-way infinite sequences.
Index entries for linear recurrences with constant coefficients, signature (3,-1).

Cf. A000032, A002878 (odd-indexed Lucas numbers), A001906 (Chebyshev S(n-1, 3)), a(n) = sqrt(4+5*A001906(n)^2), A228842.
a(n) = A005592(n)+1 = A004146(n)+2 = A065034(n)-1.
First differences of A002878. Pairwise sums of A001519. First row of array A103997.
Cf. A153415, A201157. Also Lucas(k*n): A000032 (k = 1), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11), A089775 (k = 12).

a005248 n = a005248_list !! n
a005248_list = zipWith (+) (tail a001519_list) a001519_list
-- Reinhard Zumkeller, Jan 11 2012

[Lucas(2*n) : n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

a:= n-> (<<2|3>>. <<3|1>, <-1|0>>^n)[1$2]: seq(a(n), n=0..30); # Alois P. Heinz, Jul 31 2008
with(combinat): seq(5*fibonacci(n)^2+2*(-1)^n, n= 0..26);

a[0] = 2; a[1] = 3; a[n_] := 3a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 27}] (* Robert G. Wilson v, Jan 30 2004 *)
Fibonacci[1 + 2n] + 1/2 (-Fibonacci[2n] + LucasL[2n]) (* Tesler. Sarah-Marie Belcastro, Jul 04 2009 *)
LinearRecurrence[{3, -1}, {2, 3}, 50] (* Sture Sjöstedt, Nov 27 2011 *)
LucasL[Range[0,60,2]] (* Harvey P. Dale, Sep 30 2014 *)

{a(n) = fibonacci(2*n + 1) + fibonacci(2*n - 1)}; /* Michael Somos, Jun 23 2002 */

{a(n) = 2 * subst( poltchebi(n), x, 3/2)}; /* Michael Somos, Jun 28 2003 */

[lucas_number2(n,3,1) for n in range(37)] # Zerinvary Lajos, Jun 25 2008

a(n) = Fibonacci(2*n-1) + Fibonacci(2*n+1).
G.f.: (2-3*x)/(1-3*x+x^2). - Simon Plouffe in his 1992 dissertation.
a(n) = S(n, 3) - S(n-2, 3) = 2*T(n, 3/2) with S(n-1, 3) = A001906(n) and S(-2, x) = -1. U(n, x)=S(n, 2*x) and T(n, x) are Chebyshev's U- and T-polynomials.
a(n) = a(k)*a(n - k) - a(n - 2k) for all k, i.e., a(n) = 2*a(n) - a(n) = 3*a(n - 1) - a(n - 2) = 7*a(n - 2) - a(n - 4) = 18*a(n - 3) - a(n - 6) = 47*a(n - 4) - a(n - 8) etc., a(2n) = a(n)^2 - 2. - Henry Bottomley, May 08 2001
a(n) = A060924(n-1, 0) = 3*A001906(n) - 2*A001906(n-1), n >= 1. - Wolfdieter Lang, Apr 26 2001
a(n) ~ phi^(2*n) where phi=(1+sqrt(5))/2. - Joe Keane (jgk(AT)jgk.org), May 15 2002
a(0)=2, a(1)=3, a(n) = 3*a(n-1) - a(n-2) = a(-n). - Michael Somos, Jun 28 2003
a(n) = phi^(2*n) + phi^(-2*n) where phi=(sqrt(5)+1)/2, the golden ratio. E.g., a(4)=47 because phi^(8) + phi^(-8) = 47. - Dennis P. Walsh, Jul 24 2003
With interpolated zeros, trace(A^n)/4, where A is the adjacency matrix of path graph P_4. Binomial transform is then A049680. - Paul Barry, Apr 24 2004
a(n) = (floor((3+sqrt(5))^n) + 1)/2^n. - Lekraj Beedassy, Oct 22 2004
a(n) = ((3-sqrt(5))^n + (3+sqrt(5))^n)/2^n (Note: substituting the number 1 for 3 in the last equation gives A000204, substituting 5 for 3 gives A020876). - Creighton Dement, Apr 19 2005
a(n) = (1/(n+1/2))*Sum_{k=0..n} B(2k)*L(2n+1-2k)*binomial(2n+1, 2k) where B(2k) is the (2k)-th Bernoulli number. - Benoit Cloitre, Nov 02 2005
a(n) = term (1,1) in the 1 X 2 matrix [2,3] . [3,1; -1,0]^n. - Alois P. Heinz, Jul 31 2008
a(n) = 2*cosh(2*n*psi), where psi=log((1+sqrt(5))/2). - Al Hakanson, Mar 21 2009
From Sarah-Marie Belcastro, Jul 04 2009: (Start)
a(n) - (a(n) - F(2n))/2 - F(2n+1) = 0. (Tesler)
Product_{r=1..n} (1 + 4*(sin((4r-1)*Pi/(4n)))^2). (Lu/Wu) (End)
a(n) = Fibonacci(2n+6) mod Fibonacci(2n+2), n > 1. - Gary Detlefs, Nov 22 2010
a(n) = 5*Fibonacci(n)^2 + 2*(-1)^n. - Gary Detlefs, Nov 22 2010
a(n) = A033888(n)/A001906(n), n > 0. - Gary Detlefs, Dec 26 2010
a(n) = 2^(2*n) * Sum_{k=1..2} (cos(k*Pi/5))^(2*n). - L. Edson Jeffery, Jan 21 2012
From Peter Bala, Jan 04 2013: (Start)
Let F(x) = Product_{n>=0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 1/2*(3 - sqrt(5)). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.31829 56058 81914 31334 ... = 2 + 1/(3 + 1/(7 + 1/(18 + ...))).
Also F(-alpha) = 0.64985 97768 07374 32950 has the continued fraction representation 1 - 1/(3 - 1/(7 - 1/(18 - ...))) and the simple continued fraction expansion 1/(1 + 1/((3-2) + 1/(1 + 1/((7-2) + 1/(1 + 1/((18-2) + 1/(1 + ...))))))).
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((3^2-4) + 1/(1 + 1/((7^2-4) + 1/(1 + 1/((18^2-4) + 1/(1 + ...))))))).
Added Oct 13 2019: 1/2 + 1/2*F(alpha)/F(-alpha) = 1.5142923542... has the simple continued fraction expansion 1 + 1/((3 - 2) + 1/(1 + 1/((18 - 2) + 1/(1 + 1/(123 - 2) + 1/(1 + ...))))). (End)
G.f.: (W(0)+6)/(5*x), where W(k) = 5*x*k + x - 6 + 6*x*(5*k-9)/W(k+1) (continued fraction). - Sergei N. Gladkovskii, Aug 19 2013
Sum_{n >= 1} 1/( a(n) - 5/a(n) ) = 1. Compare with A001906, A002878 and A023039. - Peter Bala, Nov 29 2013
0 = a(n) * a(n+2) - a(n+1)^2 - 5 for all n in Z. - Michael Somos, Aug 24 2014
a(n) = (G(j+2n) + G(j-2n))/G(j), for n >= 0 and any j, positive or negative, except where G(j) = 0, and for any sequence of the form G(n+1) = G(n) + G(n-1) with any initial values for G(0), G(1), including non-integer values. G(n) includes Lucas, Fibonacci. Compare with A081067 for odd number offsets from j. - Richard R. Forberg, Nov 16 2014
a(n) = [x^n] ( (1 + 3*x + sqrt(1 + 6*x + 5*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
From J. M. Bergot, Oct 28 2015: (Start)
For n>0, a(n) = F(n-1) * L(n) + F(2*n+1) - (-1)^n with F(k) = A000045(k).
For n>1, a(n) = F(n+1) * L(n) + F(2*n-1) - (-1)^n.
For n>2, a(n) = 5*F(2*n-3) + 2*L(n-3) * L(n) + 8*(-1)^n. (End)
For n>1, a(n) = L(n-2)*L(n+2) -7*(-1)^n. - J. M. Bergot, Feb 10 2016
a(n) = 6*F(n-1)*L(n-1) - F(2*n-6) with F(n)=A000045(n) and L(n)=A000032(n). - J. M. Bergot, Apr 21 2017
a(n) = F(2*n) + 2*F(n-1)*L(n) with F(n)=A000045(n) and L(n)=A000032(n). - J. M. Bergot, May 01 2017
E.g.f.: exp(4*x/(1+sqrt(5))^2) + exp((1/4)*(1+sqrt(5))^2*x). - Stefano Spezia, Aug 13 2019
From Peter Bala, Oct 14 2019: (Start)
a(n) = F(2*n+2) - F(2*n-2) = A001906(n+1) - A001906(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^2 = [1, 1; 1, 2].
Consequently the Gauss congruences hold: a(n*p^k) = a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k. See Zarelua and also Stanley (Ch. 5, Ex. 5.2(a) and its solution).
Sum_{n >= 1} (-1)^(n+1)/( a(n) + 1/a(n) ) = 1/5.
Sum_{n >= 1} (-1)^(n+1)/( a(n) + 3/(a(n) + 2/(a(n))) ) = 1/6.
Sum_{n >= 1} (-1)^(n+1)/( a(n) + 9/(a(n) + 4/(a(n) + 1/(a(n)))) ) = 1/9.
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 3*x^2 + 8*x^3 + 21*x^4 + ... is the o.g.f. for A001906. (End)
a(n) = n + 2 + Sum_{k=1..n-1} k*a(n-k). - Yu Xiao, May 30 2020
Sum_{n>=1} 1/a(n) = A153415. - Amiram Eldar, Nov 11 2020
Sum_{n>=0} 1/(a(n) + 3) = (2*sqrt(5) + 1)/10 (André-Jeannin, 1991). - Amiram Eldar, Jan 23 2022
a(n) = 2*cosh(2*n*arccsch(2)) = 2*cosh(2*n*asinh(1/2)). - Peter Luschny, May 25 2022
a(n) = (5/2)*(Sum_{k=-n..n} binomial(2*n, n+5*k)) - (1/2)*4^n. - Greg Dresden, Jan 05 2023
a(n) = Sum_{k>=0} Lucas(2*n*k)/(Lucas(2*n)^(k+1)). - Diego Rattaggi, Jan 12 2025

Additional comments from Michael Somos, Jun 23 2001

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

Original entry on oeis.org

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

Kival Ngaokrajang, Aug 03 2014

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

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).

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

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

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

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

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

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

A065034 a(n) = Lucas(2*n) + 1.

Original entry on oeis.org

3, 4, 8, 19, 48, 124, 323, 844, 2208, 5779, 15128, 39604, 103683, 271444, 710648, 1860499, 4870848, 12752044, 33385283, 87403804, 228826128, 599074579, 1568397608, 4106118244, 10749957123, 28143753124, 73681302248, 192900153619, 505019158608, 1322157322204

Offset: 0

N. J. A. Sloane, Nov 04 2001

Harry J. Smith, Table of n, a(n) for n = 0..200
Sela Fried, Even-up words and their variants, arXiv:2505.14196 [math.CO], 2025. See p. 8.
Markus Grassl and Andrew J. Scott, Fibonacci-Lucas SIC-POVMs, arXiv:1707.02944 [quant-ph], 2017.
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Cf. A000032, A000045, A005248, A005592.

Cf. A002878 (first differences). - R. J. Mathar, Jul 18 2009

[ Lucas(2*n) + 1: n in [0..210]]; // Vincenzo Librandi, Apr 15 2011

a:= n-> (<<0|1>, <1|1>>^(2*n). <<2,1>>)[1, 1]+1:
seq(a(n), n=0..30);  # Alois P. Heinz, Nov 01 2016

LucasL[2 Range[30]]+1 (* Harvey P. Dale, Oct 21 2011 *)
LinearRecurrence[{4, -4, 1}, {3, 4, 8}, 30] (* Jean-François Alcover, Jan 08 2019 *)

a(n) = { fibonacci(2*n + 1) + fibonacci(2*n - 1) + 1 } \\ Harry J. Smith, Oct 03 2009

Vec((3-2*x)*(1-2*x)/((1-x)*(1-3*x+x^2)) + O(x^40)) \\ Colin Barker, Nov 01 2016

a(n) = F(2*n+1) + F(2*n-1) + 1 = A005248(n) + 1.
From R. J. Mathar, Jul 18 2009: (Start)
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3).
G.f.: 1/(1-x) + (2-3*x)/(1-3*x+x^2). (End)
a(n) = 1 + ((3-sqrt(5))/2)^n + ((3+sqrt(5))/2)^n. - Colin Barker, Nov 01 2016

a(0)=3 prepended by Joerg Arndt, Nov 01 2016

A210619 Triangle of numbers with n 1's and n 0's in their representation in base of Fibonacci numbers (A014417).

Original entry on oeis.org

2, 6, 7, 17, 19, 20, 46, 51, 53, 54, 122, 135, 140, 142, 143, 321, 355, 368, 373, 375, 376, 842, 931, 965, 978, 983, 985, 986, 2206, 2439, 2528, 2562, 2575, 2580, 2582, 2583, 5777, 6387, 6620, 6709, 6743, 6756, 6761, 6763, 6764, 15126, 16723, 17333, 17566, 17655, 17689, 17702, 17707, 17709, 17710

Offset: 1

Alex Ratushnyak, May 07 2012

There are n such 2n-bit numbers. For example, 17, 19, and 20 all require six bits: 100101, 101001, 101010. The least number in each group is Fib(2n+1) + Fib(2n-1) - 1, which is A005592(n). The greatest number in each group is Fib(2n+2) - 1, which is A035508(n). - T. D. Noe, May 08 2012

			Representation of 20 is 101010, three 1's and three 0's, so 20 is in the sequence.
Representation of 22 is 1000001, two 1's and five 0's, so 22 is not in the sequence.

T. D. Noe, Table of rows 1 to 49, flattened

Cf. A014417, A003714, A000045 (Fibonacci numbers).

Cf. A005592 (column k=1), A035508 (main diagonal), A249450 (second diagonal), A346434 (in Fibonacci base).

nn = 10; f = Join[{0}, Accumulate[Fibonacci[Range[2, 2*nn, 2] - 1]]]; t = Table[hi = f[[n+1]] - 1; Reverse[Table[hi - f[[i]], {i, n - 1}]], {n, 2, nn}]; t = Flatten[t] (* T. D. Noe, May 08 2012 *)

Numbers with equal counts of 1's and 0's in their Zeckendorf representation.
From Kevin Ryde, Jul 24 2021: (Start)
T(n,k) = Fibonacci(2*n+2) - Fibonacci(2*(n-k)) - 1.
G.f.: x*y*(2 - 2*x + x^2 - (1 + x + x^2)*x*y + x^3*y^2) / ( (1-x) * (1 - 3*x + x^2) * (1 - x*y) * (1 - 3*x*y + (x*y)^2) ).
(End)

A277919 Triangle read by rows: CL(n,k) is the number of labeled subgraphs with k edges of the n-cycle C_n.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 7, 6, 3, 1, 15, 16, 10, 4, 1, 31, 40, 30, 15, 5, 1, 63, 96, 84, 50, 21, 6, 1, 127, 224, 224, 154, 77, 28, 7, 1, 255, 512, 576, 448, 258, 112, 36, 8, 1, 511, 1152, 1440, 1248, 810, 405, 156, 45, 9, 1, 1023, 2560, 3520, 3360, 2420, 1362, 605, 210, 55, 10, 1

Offset: 0

John P. McSorley, Nov 03 2016

			For row 3 of the triangle below: there are 7 labeled subgraphs of the triangle C_3 with 0 edges, 6 with 1 edge, 3 with 2 edges, and 1 with 3 edges (C_3 itself).
Triangle begins:
     1;
     1,    1;
     3,    2,    1;
     7,    6,    3,    1;
    15,   16,   10,    4,    1;
    31,   40,   30,   15,    5,    1;
    63,   96,   84,   50,   21,    6,   1;
   127,  224,  224,  154,   77,   28,   7,   1;
   255,  512,  576,  448,  258,  112,  36,   8,  1;
   511, 1152, 1440, 1248,  810,  405, 156,  45,  9,  1;
  1023, 2560, 3520, 3360, 2420, 1362, 605, 210, 55, 10, 1;
  ...

Andrew Howroyd, Table of n, a(n) for n = 0..1274
Thomas Selig, Combinatorial aspects of sandpile models on wheel and fan graphs, arXiv:2202.06487 [math.CO], 2022.

Row sums give A005592.

Middle diagonal gives A110170.

T(n)={[Vecrev(p) | p<-Vec((1 - 2*x + 2*x^2)/((1-x)*(1 - y*x - 2*x + y*x^2)) + O(x*x^n))]}
{ my(A=T(12)); for(n=1, #A, print(A[n])) } \\ Andrew Howroyd, Sep 27 2019

The identity CL(n,k) = 2^(n-2*k) * CL(n,n-k) can be proved combinatorially.

G.f.: (1 - 2*x + 2*x^2)/((1-x)*(1 - y*x - 2*x + y*x^2)). - Andrew Howroyd, Sep 27 2019

More terms from John P. McSorley, Nov 17 2016

A294044 a(0) = 0, a(1) = a(2) = 1; a(2n) = 2a(n) + a(n+1), a(2*n+1) = a(n) + a(n+1).

Original entry on oeis.org

0, 1, 1, 2, 4, 3, 8, 6, 11, 7, 14, 11, 22, 14, 23, 17, 29, 18, 28, 21, 39, 25, 44, 33, 58, 36, 51, 37, 63, 40, 63, 46, 76, 47, 64, 46, 77, 49, 81, 60, 103, 64, 94, 69, 121, 77, 124, 91, 152, 94, 123, 87, 139, 88, 137, 100, 166, 103, 143, 103, 172, 109, 168, 122, 199, 123, 158, 111, 174, 110, 169

Offset: 0

Ilya Gutkovskiy, Oct 22 2017

nonn

			a(0) = 0; a(1) = a(2) = 1;
a(3) = a(2*1+1) = a(1) + a(2) = 2;
a(4) = a(2*2) = 2*a(2) + a(3) = 4;
a(5) = a(2*2+1) = a(2) + a(3) = 3;
a(6) = a(2*3) = 2*a(3) + a(4) = 8, etc.
G.f. = x + x^2 + 2*x^3 + 4*x^4 + 3*x^5 + 8*x^6 + 6*x^7 + 11*x^8 + 7*x^9 + 14*x^10 + ... - _Michael Somos_, Jul 24 2023

Robert Israel, Table of n, a(n) for n = 0..10000
Michael Gilleland, Some Self-Similar Integer Sequences
Ilya Gutkovskiy, Extended graphical example

Cf. A001622, A002487, A002878, A005248, A005592.

f:= proc(n) option remember;
  if n::odd then procname((n+1)/2)+procname((n-1)/2)
  else 2*procname(n/2)+procname(n/2+1)
  fi
end proc:
f(0):= 0: f(1):= 1: f(2):= 1:
map(f, [$0..100]); # Robert Israel, Oct 24 2017

a[0] = 0; a[1] = 1; a[2] = 1; a[n_] := If[EvenQ[n], 2 a[n/2] + a[(n + 2)/2],  a[(n - 1)/2] + a[(n + 1)/2]]; Table[a[n], {n, 0, 70}]

{a(n) = if(n<1, 0, n<3, 1, n%2, a(n\2) + a(n\2+1), 2*a(n\2) + a(n\2+1))}; /* Michael Somos, Jul 24 2023 */

a(n) = a(2*n) - a(2*n+1) for n > 1.
a(n+1) = 2*a(2*n+1) - a(2*n) for n > 1.
a(2^(k+1)) = floor(phi^(2*k+1)) = A002878(k), where phi is the golden ratio (A001622).
a(2^(k+1)+1) = phi^(2*k) + phi^(-2*k) = A005248(k).
a(2^(k+1)-1) = floor(phi^(2*k)) = A005592(k).
G.f. g(x) satisfies g(x) = (x + 2 + 1/x + 1/x^2)*g(x^2) - 1 - 2*x^2. - Robert Israel, Oct 24 2017

A134561 Array T by antidiagonals: T(n,k) = k-th number whose Zeckendorf representation has exactly n terms.

Original entry on oeis.org

1, 2, 4, 3, 6, 12, 5, 7, 17, 33, 8, 9, 19, 46, 88, 13, 10, 20, 51, 122, 232, 21, 11, 25, 53, 135, 321, 609, 34, 14, 27, 54, 140, 355, 842, 1596, 55, 15, 28, 67, 142, 368, 931, 2206, 4180, 89, 16, 30, 72, 143, 373, 965

Offset: 1

Clark Kimberling, Nov 01 2007

A permutation of the natural numbers.

Except for initial terms in some cases, (Row 1) = A000045 (Row 2) = A095096 (Row 3) = A059390 (Row 4) = A111458 (Col 1) = A027941 (Col 2) = A005592.

			19 = 13 + 5 + 1 is the 3rd-largest number (after 12 and 17) that has a 3-term Zeckendorf representation; i.e., the (unique) sum of distinct non-neighboring Fibonacci numbers.
Northwest corner:
1 2 3 5 8 13
4 6 7 9 10 11
12 17 19 20 25 27
33 46 51 53 54 67

C. Kimberling, The Zeckendorf array equals the Wythoff array, Fibonacci Quarterly 33 (1995) 3-8.
Index entries for sequences that are permutations of the natural numbers

Cf. A035513.

cp's OEIS Frontend

A005248 Bisection of Lucas numbers: a(n) = L(2*n) = A000032(2*n).

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

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A065034 a(n) = Lucas(2*n) + 1.

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A210619 Triangle of numbers with n 1's and n 0's in their representation in base of Fibonacci numbers (A014417).

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A277919 Triangle read by rows: CL(n,k) is the number of labeled subgraphs with k edges of the n-cycle C_n.

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A294044 a(0) = 0, a(1) = a(2) = 1; a(2*n) = 2*a(n) + a(n+1), a(2*n+1) = a(n) + a(n+1).

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A005248 Bisection of Lucas numbers: a(n) = L(2n) = A000032(2n).

A294044 a(0) = 0, a(1) = a(2) = 1; a(2n) = 2a(n) + a(n+1), a(2*n+1) = a(n) + a(n+1).