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A005248 Bisection of Lucas numbers: a(n) = L(2*n) = A000032(2*n).
(Formerly M0848 N1067)
145
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,1

COMMENTS

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

REFERENCES

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.

LINKS

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.

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.

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.

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 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).

FORMULA

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

EXAMPLE

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

MAPLE

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);

MATHEMATICA

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 *)

PROG

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

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

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

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

(Haskell)

a005248 n = a005248_list !! n

a005248_list = zipWith (+) (tail a001519_list) a001519_list

-- Reinhard Zumkeller, Jan 11 2012

CROSSREFS

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).

Sequence in context: A131093 A343358 A002864 * A032102 A100388 A186232

Adjacent sequences:  A005245 A005246 A005247 * A005249 A005250 A005251

KEYWORD

nonn,easy

AUTHOR

N. J. A. Sloane

EXTENSIONS

Additional comments from Michael Somos, Jun 23 2001

STATUS

approved

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Last modified September 26 18:11 EDT 2022. Contains 357002 sequences. (Running on oeis4.)