A087215 -id:A087215 - 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

A014448 Even Lucas numbers: L(3n).

Original entry on oeis.org

2, 4, 18, 76, 322, 1364, 5778, 24476, 103682, 439204, 1860498, 7881196, 33385282, 141422324, 599074578, 2537720636, 10749957122, 45537549124, 192900153618, 817138163596, 3461452808002, 14662949395604, 62113250390418

Offset: 0

Mohammad K. Azarian

This is the Lucas sequence V(4,-1). - Bruno Berselli, Jan 08 2013

			a(4) = L(3 * 4) = L(12) = 322. - _Indranil Ghosh_, Feb 05 2017

Indranil Ghosh, Table of n, a(n) for n = 0..1591
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), Volume 8, Issue 1 (2014), pp. 81-92; sequence L_{4,n}.
H. H. Ferns, Problem B-115, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 5, No. 2 (1967), p. 202; Identities for F_{kn} and L{kn}, Solution to Problem B-115 by Stanley Rabinowitz, ibid., Vol. 6, No. 1 (1968), pp. 92-93.
Tanya Khovanova, Recursive Sequences.
Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1.
Roberto Tauraso, Some Congruences for Central Binomial Sums Involving Fibonacci and Lucas Numbers, JIS 19 (2016) # 16.5.4
Wikipedia, Lucas sequence: Specific names.
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2).
Index entries for linear recurrences with constant coefficients, signature (4,1).

Cf. A000032, A000045, A001077.

Cf. Lucas(k*n): A005248 (k = 2), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A089772 (k = 11), A089775 (k = 12).

[Lucas(3*n) : n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

Table[LucasL[3*n], {n,0,100}] (* G. C. Greubel, Nov 07 2018 *)

PARI
```
polsym(x^2-4*x-1,100)
```

a(n)=sum(k=0,n,binomial(n,k)*(fibonacci(n+k-1)+fibonacci(n+k+1))) \\ Paul D. Hanna, Oct 19 2010

[lucas_number2(n,4,-1) for n in range(0, 23)] # Zerinvary Lajos, May 14 2009

G.f.: (2-4*x)/(1-4*x-x^2).
a(n) = 4*a(n-1) +a(n-2) with n>1, a(0)=2, a(1)=4.
a(n) = (2+sqrt(5))^n + (2-sqrt(5))^n.
a(n) = 2*A001077(n).
a(n) = A000032(3*n).
a(n) = Sum_{k=0..n} C(n,k)*Lucas(n+k). - Paul D. Hanna, Oct 19 2010
a(n) = Fibonacci(6*n)/Fibonacci(3*n), n>0. - Gary Detlefs, Dec 26 2010
From Peter Bala, Mar 22 2015: (Start)
a(n) = ( Fibonacci(3*n + 2*k) - F(3*n - 2*k) )/Fibonacci(2*k) for nonzero integer k.
a(n) = ( Fibonacci(3*n + 2*k + 1) + F(3*n - 2*k - 1) )/Fibonacci(2*k + 1) for arbitrary integer k. (End)
a(n) = [x^n] ( (1 + 4*x + sqrt(1 + 8*x + 20*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
a(n) = L(n)*(L(n-1)*L(n+1) + 2*(-1)^n). - J. M. Bergot, Feb 05 2016
From Peter Bala Oct 14 2019: (Start)
Sum_{n >= 1} 1/( a(n) + (-1)^(n+1)*20/a(n) ) = 3/16.
Sum_{n >= 1} (-1)^(n+1)/( a(n) + (-1)^(n+1)*20/a(n) ) = 1/16. (End)
a(n) = (15*Fibonacci(n)^2*Lucas(n) + Lucas(n)^3)/4 (Ferns, 1967). - Amiram Eldar, Feb 06 2022
E.g.f.: 2*exp(2*x)*cosh(sqrt(5)*x). - Stefano Spezia, Jan 18 2025

More terms from Erich Friedman

A056854 a(n) = Lucas(4*n).

Original entry on oeis.org

2, 7, 47, 322, 2207, 15127, 103682, 710647, 4870847, 33385282, 228826127, 1568397607, 10749957122, 73681302247, 505019158607, 3461452808002, 23725150497407, 162614600673847, 1114577054219522, 7639424778862807, 52361396397820127, 358890350005878082, 2459871053643326447

Offset: 0

Barry E. Williams, Aug 29 2000

a(n) and b(n) := A004187(n) are the nonnegative proper and improper solutions of the Pell equation a(n)^2 - 5*(3*b(n))^2 = +4. See the cross-reference to A004187 below. - Wolfdieter Lang, Jun 26 2013

Lucas numbers of the form n^2-2. - Michel Lagneau, Aug 11 2014

			Pell equation: n = 0, 2^2 - 45*0^2 = +4 (improper);  n = 1, 7^2 - 5*(3*1)^2 = +4; n=2, 47^2 - 5*(3*7)^2 = +4. - _Wolfdieter Lang_, Jun 26 2013

R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

G. C. Greubel, Table of n, a(n) for n = 0..1190
Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Dale Gerdemann, Combinatorial proofs of Zeckendorf family identities, Fib. Quart. 46/47 (2009) 249. [_N. J. A. Sloane_, Dec 05 2009]
A. F. Horadam, Special Properties of the Sequence W(n){a,b; p,q}, Fib. Quart., Vol. 5, No. 5 (1967), pp. 424-434.
Tanya Khovanova, Recursive Sequences
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 linear recurrences with constant coefficients, signature (7,-1).

Cf. A004187, A005248.
Cf. quadrisection of A000032: this sequence (first), A056914 (second), A246453 (third, without 11), A288913 (fourth).
Cf. Lucas(k*n): A000032 (k = 1), A005248 (k = 2), A014448 (k = 3), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11), A089775 (k = 12).

[Lucas(4*n): n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

a[0] = 2; a[1] = 7; a[n_] := 7a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 19}] (* Robert G. Wilson v, Jan 30 2004 *)
LinearRecurrence[{7,-1},{2,7},25] (* or *) LucasL[4*Range[0,25]] (* Harvey P. Dale, Aug 08 2011 *)

a(n)=if(n<0,0,polsym(1-7*x+x^2,n)[n+1])

a(n)=if(n<0,0,2*subst(poltchebi(n),x,7/2))

[lucas_number2(n,7,1) for n in range(27)] #Zerinvary Lajos, Jun 25 2008

a(n) = 7*a(n-1) - a(n-2) with a(0)=2, a(1)=7.
a(n) = A000032(4*n), where A000032 = Lucas numbers.
a(n) = 7*S(n-1, 7) - 2*S(n-2, 7) = S(n, 7) - S(n-2, 7) = 2*T(n, 7/2), with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. U(n, x), resp. T(n, x), are Chebyshev's polynomials of the second, resp. first, kind. S(n-1, 7) = A004187(n), n>=0. See A049310 and A053120.
a(n) = ((7+sqrt(45))/2)^n + ((7-sqrt(45))/2)^n.
G.f.: (2-7x)/(1-7x+x^2).
a(n) = A005248(2*n); bisection of A005248.
a(n) = Fibonacci(8*n)/Fibonacci(4*n), n>0. - Gary Detlefs, Dec 26 2010
a(n) = 2 + 5*Fibonacci(2*n)^2 = 2 + 5*A049684(n), n >= 0. This is in Koshy's book (reference under A065563) 15. on p. 88. Compare with the above Chebyshev T formula. - Wolfdieter Lang, Aug 27 2012
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = Product_{n = 0..inf} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 1/2*(7 - 3*sqrt(5)). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.14242 42709 40138 85949 ... = 2 + 1/(7 + 1/(47 + 1/(322 + ...))).
Also F(-alpha) = 0.85670 72882 04563 14901 ... has the continued fraction representation 1 - 1/(7 - 1/(47 - 1/(322 - ...))) and the simple continued fraction expansion 1/(1 + 1/((7-2) + 1/(1 + 1/((47-2) + 1/(1 + 1/((322-2) + 1/(1 + ...))))))). Cf. A005248.
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((7^2-4) + 1/(1 + 1/((47^2-4) + 1/(1 + 1/((322^2-4) + 1/(1 + ...))))))).
Added Oct 13 2019: 1/2 + (1/2)*F(alpha)/F(-alpha) = 1.16675297774947414828... has the simple continued fraction expansion 1 + 1/((7 - 2) + 1/(1 + 1/((322 - 2) + 1/(1 + 1/(15127 - 2) + 1/(1 + ...))))). (End)
a(n) = Fibonacci(4*n+2) - Fibonacci(4*n-2), where Fibonacci(-2) = -1. - Bruno Berselli, May 25 2015
a(n) = sqrt(45*(A004187(n))^2+4).
From Peter Bala, Oct 13 2019: (Start)
a(n) = F(4*n+4)/F(4) - F(4*n-4)/F(4) = A004187(n+1) - A004187(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^4 = [2, 3; 3, 5].
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).
5*Sum_{n >= 1} 1/(a(n) - 9/a(n)) = 1: (9 = Lucas(4)+2 and 5 = Lucas(4)-2)
9*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 5/a(n)) = 1.
Sum_{n >= 1} 1/a(n) = (1/4)*( theta_3((7-3*sqrt(5))/2)^2 - 1 ), where theta_3(q) = 1 + 2*Sum_{n >= 1} q^n^2. Cf. A153415.
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4)*( 1 - theta_3((3*sqrt(5)-7)/2)^2 ).
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 7*x^2 + 48*x^3 + 329*x^4 + ... is the o.g.f. for A004187. (End)
E.g.f.: 2*exp(7*x/2)*cosh(3*sqrt(5)*x/2). - Stefano Spezia, Oct 18 2019
a(2k+1)/7 is the numerator of the continued fraction [3*sqrt(5), 3*sqrt(5), ..., 3*sqrt(5)] with 2k copies of 3*sqrt(5), for k>0. - Greg Dresden and Tracy Z. Wu, Sep 10 2020
a(n) = Sum_{k>=1} Lucas(2*n*k)/(Lucas(2*n)^k). - Diego Rattaggi, Jan 20 2025

More terms from James Sellers, Aug 31 2000

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A023039 a(n) = 18*a(n-1) - a(n-2).

Original entry on oeis.org

1, 9, 161, 2889, 51841, 930249, 16692641, 299537289, 5374978561, 96450076809, 1730726404001, 31056625195209, 557288527109761, 10000136862780489, 179445175002939041, 3220013013190122249, 57780789062419261441

Offset: 0

David W. Wilson

The primitive Heronian triangle 3*a(n) +- 2, 4*a(n) has the latter side cut into 2*a(n) +- 3 by the corresponding altitude and has area 10*a(n)*A060645(n). - Lekraj Beedassy, Jun 25 2002
Chebyshev polynomials T(n,x) evaluated at x=9.
{a(n)} gives all (unsigned, integer) solutions of Pell equation a(n)^2 - 80*b(n)^2 = +1 with b(n) = A049660(n), n >= 0.
{a(n)} gives all possible solutions for x in Pell equation x^2 - D*y^2 = 1 for D=5, D=20 and D=80. The corresponding values for y are A060645 (D=5), A207832 (D=20) and A049660 (D=80). - Herbert Kociemba, Jun 05 2022
Also gives solutions to the equation x^2 - 1 = floor(x*r*floor(x/r)) where r=sqrt(5). - Benoit Cloitre, Feb 14 2004
Appears to give all solutions > 1 to the equation: x^2 = ceiling(x*r*floor(x/r)) where r=sqrt(5). - Benoit Cloitre, Feb 24 2004
For all terms x of the sequence, 5*x^2 - 5 is a square, A004292(n)^2.
The a(n) are the x-values in the nonnegative integer solutions of x^2 - 5y^2 = 1, see A060645(n) for the corresponding y-values. - Sture Sjöstedt, Nov 29 2011
Rightmost digits alternate repeatedly: 1 and 9 in fact, a(2) = 18*9 - 1 == 1 (mod 10); a(3) = 18*1 - 9 == 9 (mod 10) therefore a(2n) == 1 (mod 10), a(2n+1) == 9 (mod 10). - Carmine Suriano, Oct 03 2013

			G.f. = 1 + 9*x + 161*x^2 + 2889*x^3 + 51841*x4 + 930249*x^5 + 16692641*x^6 + ...

Seiichi Manyama, Table of n, a(n) for n = 0..750 (terms 0..200 from Vincenzo Librandi)
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Cf. A001077, A115032.
Row 2 of array A188645.
Row 4 of A322790.

I:=[1, 9]; [n le 2 select I[n] else 18*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 13 2012

a := n -> hypergeom([n, -n], [1/2], -4):
seq(simplify(a(n)), n=0..16); # Peter Luschny, Jul 26 2020

LinearRecurrence[{18, -1}, {1, 9}, 50] (* Sture Sjöstedt, Nov 29 2011 *)
CoefficientList[Series[(1-9*x)/(1-18*x+x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)

{a(n) = fibonacci(6*n) / 2 + fibonacci(6*n - 1)}; /* Michael Somos, Aug 11 2009 */

x='x+O('x^30); Vec((1-9*x)/(1-18*x+x^2)) \\ G. C. Greubel, Dec 19 2017

a(n) ~ (1/2)*(sqrt(5) + 2)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Limit_{n->infinity} a(n)/a(n-1) = phi^6 = 9 + 4*sqrt(5). - Gregory V. Richardson, Oct 13 2002
a(n) = T(n, 9) = (S(n, 18) - S(n-2, 18))/2, with S(n, x) := U(n, x/2) and T(n, x), resp. U(n, x), are Chebyshev's polynomials of the first, resp. second, kind. See A053120 and A049310. S(-2, x) := -1, S(-1, x) := 0, S(n, 18)=A049660(n+1).
a(n) = sqrt(80*A049660(n)^2 + 1) (cf. Richardson comment).
a(n) = ((9 + 4*sqrt(5))^n + (9 - 4*sqrt(5))^n)/2.
G.f.: (1 - 9*x)/(1 - 18*x + x^2).
a(n) = cosh(2*n*arcsinh(2)). - Herbert Kociemba, Apr 24 2008
a(n) = A001077(2*n). - Michael Somos, Aug 11 2009
From Johannes W. Meijer, Jul 01 2010: (Start)
a(n) = 2*A167808(6*n+1) - A167808(6*n+3).
Limit_{k->infinity} a(n+k)/a(k) = a(n) + A060645(n)*sqrt(5).
Limit_{n->infinity} a(n)/A060645(n) = sqrt(5).
(End)
a(n) = (1/2)*A087215(n) = (1/2)*(sqrt(5) + 2)^(2*n) + (1/2)*(sqrt(5) - 2)^(2*n).
Sum_{n >= 1} 1/( a(n) - 5/a(n) ) = 1/8. Compare with A005248, A002878 and A075796. - Peter Bala, Nov 29 2013
a(n) = 2*A115032(n-1) - 1 = S(n, 18) - 9*S(n-1, 18), with A115032(-1) = 1, and see the above formula with S(n, 18) using its recurrence. - Wolfdieter Lang, Aug 22 2014
a(n) = A128052(3n). - A.H.M. Smeets, Oct 02 2017
a(n) = A049660(n+1) - 9*A049660(n). - R. J. Mathar, May 24 2018
a(n) = hypergeom([n, -n], [1/2], -4). - Peter Luschny, Jul 26 2020
a(n) = L(6*n)/2 for L(n) the Lucas sequence A000032(n). - Greg Dresden, Dec 07 2021
a(n) = cosh(6*n*arccsch(2)). - Peter Luschny, May 25 2022

Chebyshev and Pell comments from Wolfdieter Lang, Nov 08 2002

Sture Sjöstedt's comment corrected and reformulated by Wolfdieter Lang, Aug 24 2014

A090307 a(n) = 18*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 18.

Original entry on oeis.org

2, 18, 326, 5886, 106274, 1918818, 34644998, 625528782, 11294163074, 203920464114, 3681862517126, 66477445772382, 1200275886420002, 21671443401332418, 391286257110403526, 7064824071388595886

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 25 2004

Lim_{n-> infinity} a(n)/a(n+1) = 0.0553851... = 1/(9+sqrt(82)) = (sqrt(82)-9).

Lim_{n-> infinity} a(n+1)/a(n) = 18.0553851... = (9+sqrt(82)) = 1/(sqrt(82)-9).

			a(4) = 18*a(3) + a(2) = 18*5886 + 326 = (9+sqrt(82))^4 + (9-sqrt(82))^4 = 106273.9999905903 + 0.000009406 = 106274.

G. C. Greubel, Table of n, a(n) for n = 0..500
Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (18,1).

Cf. A083218, A087215.

Lucas polynomials Lucas(n,m): A000032 (m=1), A002203 (m=2), A006497 (m=3), A014448 (m=4), A087130 (m=5), A085447 (m=6), A086902 (m=7), A086594 (m=8), A087798 (m=9), A086927 (m=10), A001946 (m=11), A086928 (m=12), A088316 (m=13), A090300 (m=14), A090301 (m=15), A090305 (m=16), A090306 (m=17), this sequence (m=18), A090308 (m=19), A090309 (m=20), A090310 (m=21), A090313 (m=22), A090314 (m=23), A090316 (m=24), A330767 (m=25).

m:=18;; a:=[2,m];; for n in [3..20] do a[n]:=m*a[n-1]+a[n-2]; od; a; # G. C. Greubel, Dec 30 2019

m:=18; I:=[2,m]; [n le 2 select I[n] else m*Self(n-1) +Self(n-2): n in [1..20]]; // G. C. Greubel, Dec 30 2019

seq(simplify(2*(-I)^n*ChebyshevT(n, 9*I)), n = 0..20); # G. C. Greubel, Dec 30 2019

LinearRecurrence[{18,1},{2,18},25] (* or *) CoefficientList[ Series[ (2-18x)/(1-18x-x^2),{x,0,25}],x] (* Harvey P. Dale, Apr 22 2011 *)
LucasL[Range[20]-1, 18] (* G. C. Greubel, Dec 30 2019 *)

vector(21, n, 2*(-I)^(n-1)*polchebyshev(n-1, 1, 9*I) ) \\ G. C. Greubel, Dec 30 2019

[2*(-I)^n*chebyshev_T(n, 9*I) for n in (0..20)] # G. C. Greubel, Dec 30 2019

a(n) = 18*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 18.
a(n) = (9+sqrt(82))^n + (9-sqrt(82))^n.
(a(n))^2 = a(2n) - 2 if n=1, 3, 5, ...
(a(n))^2 = a(2n) + 2 if n=2, 4, 6, ...
G.f.: (2-18*x)/(1-18*x-x^2). - Philippe Deléham, Nov 02 2008
a(n) = Lucas(n, 18) = 2*(-i)^n * ChebyshevT(n, 9*i). - G. C. Greubel, Dec 30 2019
E.g.f.: 2*exp(9*x)*cosh(sqrt(82)*x). - Stefano Spezia, Dec 31 2019

More terms from Ray Chandler, Feb 14 2004

A188645 Array of ((k^n)+(k^(-n)))/2 where k=(sqrt(x^2+1)+x)^2 for integers x>=1.

Original entry on oeis.org

1, 3, 1, 17, 9, 1, 99, 161, 19, 1, 577, 2889, 721, 33, 1, 3363, 51841, 27379, 2177, 51, 1, 19601, 930249, 1039681, 143649, 5201, 73, 1, 114243, 16692641, 39480499, 9478657, 530451, 10657, 99, 1, 665857, 299537289, 1499219281, 625447713, 54100801, 1555849, 19601, 129, 1

Offset: 0

Charles L. Hohn, Apr 06 2011

Conjecture: Given function f(x, y)=(sqrt(x^2+y)+x)^2; and constant k=f(x, y); then for all integers x>=1 and y=[+-]1, k may be irrational, but ((k^n)+(k^(-n)))/2 always produces integer sequences; y=1 results shown here; y=-1 results are A188644.

Also square array A(n,k), n >= 1, k >= 0, read by antidiagonals, where A(n,k) is Chebyshev polynomial of the first kind T_{k}(x), evaluated at x=2*n^2+1. - Seiichi Manyama, Jan 01 2019

			Square array begins:
     | 0    1       2          3             4
-----+---------------------------------------------
   1 | 1,   3,     17,        99,          577, ...
   2 | 1,   9,    161,      2889,        51841, ...
   3 | 1,  19,    721,     27379,      1039681, ...
   4 | 1,  33,   2177,    143649,      9478657, ...
   5 | 1,  51,   5201,    530451,     54100801, ...
   6 | 1,  73,  10657,   1555849,    227143297, ...
   7 | 1,  99,  19601,   3880899,    768398401, ...
   8 | 1, 129,  33281,   8586369,   2215249921, ...
   9 | 1, 163,  53137,  17322499,   5647081537, ...
  10 | 1, 201,  80801,  32481801,  13057603201, ...
  11 | 1, 243, 118097,  57394899,  27893802817, ...
  12 | 1, 289, 167041,  96549409,  55805391361, ...
  13 | 1, 339, 229841, 155831859, 105653770561, ...
  14 | 1, 393, 308897, 242792649, 190834713217, ...
  15 | 1, 451, 406801, 366934051, 330974107201, ...
  ...

Row 1 is A001541, row 2 is A023039, row 3 is A078986, row 4 is A099370, row 5 is A099397, row 6 is A174747, row 8 is A176368, (row 1)*2 is A003499, (row 2)*2 is A087215.
Column 1 is A058331, (column 1)*2 is A005899.
A188644 (f(x, y) as above with y=-1).
Diagonal gives A173128.
Cf. A188647.

max = 9; y = 1; t = Table[k = ((x^2 + y)^(1/2) + x)^2; ((k^n) + (k^(-n)))/2 // FullSimplify, {n, 0, max - 1}, {x, 1, max}]; Table[ t[[n - k + 1, k]], {n, 1, max}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jul 17 2013 *)

A(n,k) = (A188647(n,k-1) + A188647(n,k))/2.

A(n,k) = Sum_{j=0..k} binomial(2*k,2*j)*(n^2+1)^(k-j)*n^(2*j). - Seiichi Manyama, Jan 01 2019

Edited and extended by Seiichi Manyama, Jan 01 2019

A087265 Lucas numbers L(8*n).

Original entry on oeis.org

2, 47, 2207, 103682, 4870847, 228826127, 10749957122, 505019158607, 23725150497407, 1114577054219522, 52361396397820127, 2459871053643326447, 115561578124838522882, 5428934300813767249007, 255044350560122222180447

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Oct 19 2003

a(n+1)/a(n) converges to (47+sqrt(2205))/2 = 46.9787137... a(0)/a(1)=2/47; a(1)/a(2)=47/2207; a(2)/a(3)=2207/103682; a(3)/a(4)=103682/4870847; etc. Lim_{n->infinity} a(n)/a(n+1) = 0.02128623625... = 2/(47+sqrt(2205)) = (47-sqrt(2205))/2.
a(n) = a(-n). - Alois P. Heinz, Aug 07 2008
From Peter Bala, Oct 14 2019: (Start)
Let F(x) = Product_{n >= 0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let Phi = 1/2*(sqrt(5) - 1). This sequence gives the partial denominators in the simple continued fraction expansion of the number F(Phi^8) = 1.0212763906... = 1 + 1/(47 + 1/(2207 + 1/(103682 + ...))).
Also F(-Phi^8) = 0.9787231991... has the continued fraction representation 1 - 1/(47 - 1/(2207 - 1/(103682 - ...))) and the simple continued fraction expansion 1/(1 + 1/((47 - 2) + 1/(1 + 1/((2207 - 2) + 1/(1 + 1/((103682 - 2) + 1/(1 + ...))))))).
F(Phi^8)*F(-Phi^8) = 0.9995468962... has the simple continued fraction expansion 1/(1 + 1/((47^2 - 4) + 1/(1 + 1/((2207^2 - 4) + 1/(1 + 1/((103682^2 - 4) + 1/(1 + ...))))))).
1/2 + 1/2*F(Phi^8)/F(-Phi^8) = 1.0217391349... has the simple continued fraction expansion 1 + 1/((47 - 2) + 1/(1 + 1/((103682 - 2) + 1/(1 + 1/(228826127 - 2) + 1/(1 + ...))))). (End)

			a(4) = 4870847 = 47*a(3) - a(2) = 47*103682 - 2207=((47+sqrt(2205))/2)^4 + ( (47-sqrt(2205))/2)^4 =4870846.999999794696 + 0.000000205303 = 4870847.

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.
R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

Indranil Ghosh, Table of n, a(n) for n = 0..596
Tanya Khovanova, Recursive Sequences
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 linear recurrences with constant coefficients, signature (47,-1).

Cf. A000032. Cf. Lucas(k*n): A005248 (k = 2), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11), A089775 (k = 12).

a(n) = A000032(8n).

[ Lucas(8*n) : n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

a:= n-> (Matrix([[2,47]]). Matrix([[47,1],[ -1,0]])^(n))[1,1]:
seq(a(n), n=0..14);  # Alois P. Heinz, Aug 07 2008

LucasL[8*Range[0,20]] (* or *) LinearRecurrence[{47,-1},{2,47},20] (* Harvey P. Dale, Oct 23 2017 *)

a(n) = 47*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 47.
a(n) = ((47+sqrt(2205))/2)^n + ((47-sqrt(2205))/2)^n
(a(n))^2 = a(2n)+2.
G.f.: (2-47*x)/(1-47*x+x^2). - Alois P. Heinz, Aug 07 2008
From Peter Bala, Oct 14 2019: (Start)
a(n) = F(8*n+8)/F(8) - F(8*n-8)/F(8) = A049668(n+1) - A049668(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^8 = [13, 21; 21, 34].
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).
45*Sum_{n >= 1} 1/(a(n) - 49/a(n)) = 1: (49 = Lucas(8) + 2 and 45 = Lucas(8) - 2)
49*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 45/a(n)) = 1.
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 47*x^2 + 2208*x^3 + ... is the o.g.f. for A049668. (End)
E.g.f.: 2*exp(47*x/2)*cosh(21*sqrt(5)*x/2). - Stefano Spezia, Oct 18 2019
From Peter Bala, Apr 16 2025: (Start)
a(n) = Lucas(2*n)^4 - 4*Lucas(2*n)^2 + 2 = 2*T(4, (1/2)*Lucas(2*n)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind; more generally, for k >= 0, Lucas(2*k*n) = 2*T(k, Lucas(2*n)/2).
Sum_{n >= 1} 1/a(n) = (1/4) * (theta_3( (47 - sqrt(2205))/2 )^2 - 1) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4) * (1 - theta_3( (sqrt(2205) - 47)/2 )^2),
where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). See Borwein and Borwein, Proposition 3.5 (i), p. 91. Cf. A153415 and A003499. (End)

Terms a(22)-a(27) from John W. Layman, Jun 14 2004

A065705 a(n) = Lucas(10*n).

Original entry on oeis.org

2, 123, 15127, 1860498, 228826127, 28143753123, 3461452808002, 425730551631123, 52361396397820127, 6440026026380244498, 792070839848372253127, 97418273275323406890123, 11981655542024930675232002, 1473646213395791149646646123, 181246502592140286475862241127

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Oct 25 2003

Lim_{n->infinity} a(n+1)/a(n) = (123 + sqrt(15125))/2 = 122.9918693812...
Lim_{n->infinity} a(n)/a(n+1) = (123 - sqrt(15125))/2 = 0.00813061875578...
From Peter Bala, Oct 14 2019: (Start)
Let F(x) = Product_{n >= 0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let Phi = 1/2*(sqrt(5) - 1). This sequence gives the partial denominators in the simple continued fraction expansion of the number F(Phi^10) = 1.0081300769... = 1 + 1/(123 + 1/(15127 + 1/(1860498 + ...))).
Also F(-Phi^10) = 0.9918699143... has the continued fraction representation 1 - 1/(123 - 1/(15127 - 1/(1860498 - ...))) and the simple continued fraction expansion 1/(1 + 1/((123 - 2) + 1/(1 + 1/((15127 - 2) + 1/(1 + 1/((1860498 - 2) + 1/(1 + ...))))))).
F(Phi^10)*F(-Phi^10) = 0.9999338930... has the simple continued fraction expansion 1/(1 + 1/((123^2 - 4) + 1/(1 + 1/((15127^2 - 4) + 1/(1 + 1/((1860498^2 - 4) + 1/(1 + ...))))))).
1/2 + (1/2)*F(Phi^10)/F(-Phi^10) = 1.0081967213... has the simple continued fraction expansion 1 + 1/((123 - 2) + 1/(1 + 1/((1860498 - 2) + 1/(1 + 1/(28143753123 - 2) + 1/(1 + ...))))). (End)

			a(4) = 228826127 = 123*a(3) - a(2) = 123*1860498 - 15127=((123+sqrt(15125))/2)^4 + ( (123-sqrt(15125))/2)^4 =228826126.99999999562986 + 0.00000000437013 = 228826127.
a(4) = L(10 * 4) = L(40) = 228826127. - _Indranil Ghosh_, Feb 08 2017

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.
R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

Indranil Ghosh, Table of n, a(n) for n = 0..477
Tanya Khovanova, Recursive Sequences
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 linear recurrences with constant coefficients, signature (123,-1).

Cf. A000032: a(n) = A000032(10*n).

Cf. Lucas(k*n): A005248 (k = 2), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A089772 (k = 11), A089775 (k = 12).

[Lucas(10*n): n in [0..90]]; // Vincenzo Librandi, Apr 14 2011

LucasL[10*Range[0, 20]] (* Paolo Xausa, Mar 04 2024 *)

a(n) = 123*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 123.
a(n) = ((123 + sqrt(15125))/2)^n + ((123 - sqrt(15125))/2)^n.
a(n)^2 = a(2*n) + 2.
G.f.: (2 - 123*x)/(1 - 123*x + x^2). -  Philippe Deléham, Nov 18 2008
From Peter Bala, Oct 14 2019: (Start)
a(n) = F(10*n+10)/F(10) - F(10*n-10)/F(10) = A049670(n+1) - A049670(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^10 = [34, 55; 55, 89].
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).
121*Sum_{n >= 1} 1/(a(n) - 125/a(n)) = 1: (125 = Lucas(10) + 2 and 121 = Lucas(10) - 2)
125*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 121/a(n)) = 1.
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 123*x^2 + 15128*x^3 + ... is the o.g.f. for A049670. (End)
E.g.f.: exp((1/2)*(123 - 55*sqrt(5))*x)*(1 + exp(55*sqrt(5)*x)). - Stefano Spezia, Oct 18 2019
From Peter Bala, Apr 16 2025: (Start)
a(n) = Lucas(2*n)^5 - 5*Lucas(2*n)^3 + 5*Lucas(2*n) = 2*T(5, (1/2)*Lucas(2*n)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
Sum_{n >= 1} 1/a(n) = (1/4) * (theta_3( (123 - sqrt(15125))/2 )^2 - 1) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4) * (1 - theta_3( (sqrt(15125) - 123)/2 )^2),
where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). See Borwein and Borwein, Proposition 3.5 (i), p. 91. Cf. A153415 and A003499. (End)

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

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A014448 Even Lucas numbers: L(3n).

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A056854 a(n) = Lucas(4*n).

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A023039 a(n) = 18*a(n-1) - a(n-2).

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A090307 a(n) = 18*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 18.

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