A065705 -id:A065705 - 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

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

A001946 a(n) = 11*a(n-1) + a(n-2).

Original entry on oeis.org

2, 11, 123, 1364, 15127, 167761, 1860498, 20633239, 228826127, 2537720636, 28143753123, 312119004989, 3461452808002, 38388099893011, 425730551631123, 4721424167835364, 52361396397820127, 580696784543856761, 6440026026380244498, 71420983074726546239

Offset: 0

N. J. A. Sloane, Simon Plouffe

For odd n there is the Aurifeuillian factorization a(n) = Lucas[5n] = Lucas[n]*A[n]*B[n] = A000032[n]*A124296[n]*A124297[n], where A[n] = A124296[n] = 5*F(n)^2 - 5*F(n) + 1 and B[n] = A124297[n] = 5*F(n)^2 + 5*F(n) + 1, where F(n) = Fibonacci[n]. The largest prime divisors of a(n) for n>0 are listed in A121171[n] = {11, 41, 31, 2161, 151, 2521, 911, ...}. - Alexander Adamchuk, Oct 25 2006

For more information about this type of recurrence follow the Khovanova link and see A086902 and A054413. - Johannes W. Meijer, Jun 12 2010

J. Riordan, Combinatorial Identities, Wiley, 1968, p. 139.
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).

T. D. Noe, Table of n, a(n) for n = 0..200
Tanya Khovanova, Recursive Sequences
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
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (11, 1).

Cf. A000032, A000045, A121171, A124296, A124297.

[ Lucas(5*n) : n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

A001946:=(-2+11*z)/(-1+11*z+z**2); # Conjectured by Simon Plouffe in his 1992 dissertation

Table[Fibonacci[5n-1]+Fibonacci[5n+1],{n,0,30}] (* Alexander Adamchuk, Oct 25 2006 *)
LinearRecurrence[{11,1},{2,11},20] (* Harvey P. Dale, Jan 25 2024 *)

a(n) = Lucas(5n) = Fibonacci(5n-1) + Fibonacci(5n+1). - Alexander Adamchuk, Oct 25 2006
a(n) = ((11 + 5*sqrt(5))/2)^n + ((11 - 5*sqrt(5))/2)^n. - Tanya Khovanova, Feb 06 2007
Contribution from Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+1) = 11*A097842(n), a(2n) = A065705(n).
a(3n+1) = A041226(5n), a(3n+2) = A041226(5n+3), a(3n+3) = 2* A041226(5n+4).
Limit(a(n+k)/a(k), k=infinity) = (A001946(n) + A049666(n)*sqrt(125))/2.
Limit(A001946(n)/A049666(n), n=infinity) = sqrt(125). (End)
From Peter Bala, Mar 22 2015: (Start)
a(n) = Fibonacci(10*n)/Fibonacci(5*n) for n >= 1.
a(n) = ( Fibonacci(5*n + 2*k) - F(5*n - 2*k) )/Fibonacci(2*k) for nonzero integer k.
a(n) = ( Fibonacci(5*n + 2*k + 1) + F(5*n - 2*k - 1) )/Fibonacci(2*k + 1) for arbitrary integer k.
a(n) = Sum_{k = 0..2*n} binomial(2*n,k)*Lucas(n + k). (End)
a(n) = [x^n] ( (1 + 11*x + sqrt(1 + 22*x + 125*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 26 2015
E.g.f.: 2*cosh(5*sqrt(5)*x/2)*(cosh(11*x/2) + sinh(11*x/2)). - Stefano Spezia, Jan 18 2025

A087215 Lucas(6n): a(n) = 18a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 18.

Original entry on oeis.org

2, 18, 322, 5778, 103682, 1860498, 33385282, 599074578, 10749957122, 192900153618, 3461452808002, 62113250390418, 1114577054219522, 20000273725560978, 358890350005878082, 6440026026380244498, 115561578124838522882, 2073668380220713167378

Offset: 0

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

a(n+1)/a(n) converges to 9 + sqrt(80) = 17.9442719... a(0)/a(1) = 2/18; a(1)/a(2) = 18/322; a(2)/a(3) = 322/5778; a(3)/a(4) = 5778/103682; etc.
Lim_{n -> oo} a(n)/a(n+1) = 0.05572809000084... = 1/(9 + sqrt(80)) = 9 - sqrt(80).
From Peter Bala, Oct 13 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^6) = 1.0555459720... = 1 + 1/(18 + 1/(322 + 1/(5778 + ...))).
Also F(-Phi^6) = 0.9444348576... has the continued fraction representation 1 - 1/(18 - 1/(322 - 1/(5788 - ...))) and the simple continued fraction expansion 1/(1 + 1/((18 - 2) + 1/(1 + 1/((322 - 2) + 1/(1 + 1/((5788 - 2) + 1/(1 + ...))))))).
F(Phi^6)*F(-Phi^6) = 0.9968944099... has the simple continued fraction expansion 1/(1 + 1/((18^2 - 4) + 1/(1 + 1/((322^2 - 4) + 1/(1 + 1/((5788^2 - 4) + 1/(1 + ...))))))).
1/2 + (1/2)*F(Phi^6)/F(-Phi^6) = 1.0588241282... has the simple continued fraction expansion 1 + 1/((18 - 2) + 1/(1 + 1/((5778 - 2) + 1/(1 + 1/(1860498 - 2) + 1/(1 + ...))))). (End)

			a(4) = 103682 = 18*a(3) - a(2) = 18*5778 - 322 = (9 + sqrt(80))^4 + (9 - sqrt(80))^4 = 103681.99999035512... + 0.00000964487... = 103682.

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.

Colin Barker, Table of n, a(n) for n = 0..750
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
P. Bhadouria, D. Jhala, and B. Singh, Binomial Transforms of the k-Lucas Sequences and its Properties, The Journal of Mathematics and Computer Science (JMCS), Volume 8, Issue 1, Pages 81-92; sequence R_4.
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 (18,-1).

Cf. A074919.
Row 2 * 2 of array A188645.
Cf. Lucas(k*n): A000032 (k = 1), A005248 (k = 2), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11), A089775 (k = 12).

[ Lucas(6*n) : n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

a[0] = 2; a[1] = 18; a[n_] := 18a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 15}] (* Robert G. Wilson v, Jan 30 2004 *)
Table[LucasL[6n], {n, 0, 18}]  (* or *) CoefficientList[Series[2*(1 - 9*x)/(1 - 18*x + x^2), {x, 0, 17}], x] (* Indranil Ghosh, Mar 15 2017 *)

Vec(2*(1-9*x)/(1-18*x+x^2) + O(x^20)) \\ Colin Barker, Jan 24 2016

a(n) = if(n<2, 17^n + 1, 18*a(n - 1) - a(n - 2));
for(n=0, 17, print1(a(n),", ")) \\ Indranil Ghosh, Mar 15 2017

a(n) = A000032(6*n).
a(n) = 18*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 18.
a(n) = (9 + sqrt(80))^n + (9 - sqrt(80))^n.
G.f.: 2*(1-9*x)/(1-18*x+x^2). - Philippe Deléham, Nov 17 2008
a(n) = 2*A023039(n). - R. J. Mathar, Oct 22 2010
From Peter Bala, Oct 13 2019: (Start)
a(n) = F(6*n+6)/F(6) - F(6*n-6)/F(6) = A049660(n+1) - A049660(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^6 = [5, 8; 8, 13].
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).
16*Sum_{n >= 1} 1/(a(n) - 20/a(n)) = 1: (20 = Lucas(6) + 2 and 16 = Lucas(6) - 2)
20*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 16/a(n)) = 1.
Series acceleration formulas for sum of reciprocals:
Sum_{n >= 1} 1/a(n) = 1/16 - 20*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 20)).
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/20 + 16*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 16)).
Sum_{n >= 1} 1/a(n) = ( (theta_3(9-4*sqrt(5)))^2 - 1 )/4 and
Sum_{n >= 1} (-1)^(n+1)/a(n) = ( 1 - (theta_3(4*sqrt(5)-9))^2 )/4,
where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). Cf. A153415 and A003499.
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 18*x^2 + 323*x^3 + ... is the o.g.f. for A049660. (End)
E.g.f.: 2*exp(9*x)*cosh(4*sqrt(5)*x). - Stefano Spezia, Oct 18 2019
a(n) = L(2n-1)^2 * F(2n+1) + L(2n+1)^2 * F(2n-1), where F(n) = A000045(n) and L(n) = A000032(n). - Diego Rattaggi, Nov 12 2020
From  Peter Bala, Apr 16 2025: (Start)
a(n) = Lucas(2*n)^3 - 3*Lucas(2*n) = 2*T(3, (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(9 - sqrt(80))^2 - 1) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4) * (1 - theta_3(sqrt(80) - 9)^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)

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

A089775 Lucas numbers L(12n).

Original entry on oeis.org

2, 322, 103682, 33385282, 10749957122, 3461452808002, 1114577054219522, 358890350005878082, 115561578124838522882, 37210469265847998489922, 11981655542024930675232002, 3858055874062761829426214722, 1242282009792667284144565908482, 400010949097364802732720796316482

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 09 2004

a(n+1)/a(n) converges to (322 + sqrt(103680))/2 = 321.996894379... a(0)/a(1) = 2/322; a(1)/a(2) = 322/103682; a(2)/a(3) = 103682/33385282; a(3)/a(4) = 33385282/10749957122; etc. Lim_{n -> inf} a(n)/a(n+1) = 0.00310562... = 2/(322 + sqrt(103680)) = (322 - sqrt(103680))/2.

			a(4) = 10749957122 = 322*a(3) - a(2) = 322*33385282 - 103682 = ((322 + sqrt(103680))/2)^4 + ((322 - sqrt(103680))/2)^4.

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.

Nathaniel Johnston, Table of n, a(n) for n = 0..100
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 (322, -1).

Cf. A000032, A060964.
a(n) = A000032(12n).
Row 9 * 2 of array A188644
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), A065705 (k = 10), A089772 (k = 11).

[ Lucas(12*n) : n in [0..70]]; // Vincenzo Librandi, Apr 15 2011

Table[LucasL[12n], {n, 0, 13}] (* Indranil Ghosh, Mar 15 2017 *)

Vec((2 - 322*x)/(1 - 322*x + x^2) + O(x^14)) \\ Indranil Ghosh, Mar 15 2017

a(n) = 322*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 322
a(n) = ((322 + sqrt(103680))/2)^n + ((322 - sqrt(103680))/2)^n.
(a(n))^2 = a(2n) + 2.
G.f.: (2-322*x)/(1-322*x+x^2). - Philippe Deléham, Nov 02 2008
From Peter Bala, Apr 16 2025: (Start)
a(n) = Lucas(2*n)^6 - 6*Lucas(2*n)^4 + 9*Lucas(2*n)^2 - 2 = 2*T(6, (1/2)*Lucas(2*n)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
320*Sum_{n >= 1} 1/(a(n) - 324/a(n)) = 1: (324 = Lucas(12) + 2 and 320 = Lucas(12) - 2)
324*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 320/a(n)) = 1.
Sum_{n >= 1} 1/a(n) = (1/4) * (theta_3( (322 - sqrt(103680))/2 )^2 - 1) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4) * (1 - theta_3( (sqrt(103680) - 322)/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. (End)

a(11) - a(13) from Vincenzo Librandi, Apr 15 2011

A153179 a(n) = L(11*n)/L(n) where L(n) = A000204(n).

Original entry on oeis.org

199, 13201, 1970299, 224056801, 28374454999, 3450736132801, 426236170575799, 52337681992411201, 6441140796368008699, 792018481913198430001, 97420733208491869044199, 11981539981561372141075201

Offset: 1

Artur Jasinski, Dec 20 2008

nonn

All numbers in this sequence are:
congruent to 99 mod 100 (iff n is odd),
congruent to 1 mod 100 (iff n is even).

G. C. Greubel, Table of n, a(n) for n = 1..475
Index entries for linear recurrences with constant coefficients, signature (89,4895,-83215,-582505,1514513,1514513,-582505,-83215,4895,89,-1).

Cf. A000032, A000204, A110391, A153173, A153175, A153177.

[Lucas(11*n)/Lucas(n): n in [0..30]]; // G. C. Greubel, Dec 21 2017

Table[LucasL[11*n]/LucasL[n], {n, 1, 50}]

{lucas(n) = fibonacci(n+1) + fibonacci(n-1)};
for(n=0,30, print1( lucas(11*n)/lucas(n), ", ")) \\ G. C. Greubel, Dec 21 2017

From R. J. Mathar, Oct 22 2010: (Start)
a(n) = +89*a(n-1) +4895*a(n-2) -83215*a(n-3) -582505*a(n-4) +1514513*a(n-5) +1514513*a(n-6) -582505*a(n-7) -83215*a(n-8) +4895*a(n-9) +89*a(n-10) -a(n-11).
G.f.: -1 -1/(1+x) +(-2-47*x)/(x^2+47*x+1) +(2-3*x)/(x^2-3*x+1) +(-2-7*x)/(x^2+7*x+1) +(2-123*x)/(x^2-123*x+1) +(2-18*x)/(x^2-18*x+1).
a(n) = -(-1)^n -(-1)^n*A087265(n) +A005248(n) -(-1)^n*A056854(n) +A065705(n) +A087215(n). (End)

A299741 Array read by antidiagonals upwards: a(i,0) = 2, i >= 0; a(i,1) = i+2, i >= 0; a(i,j) = (i+2) * a(i,j-1) - a(i,j-2), for i >= 0, j > 1.

Original entry on oeis.org

2, 2, 2, 2, 3, 2, 2, 4, 7, 2, 2, 5, 14, 18, 2, 2, 6, 23, 52, 47, 2, 2, 7, 34, 110, 194, 123, 2, 2, 8, 47, 198, 527, 724, 322, 2, 2, 9, 62, 322, 1154, 2525, 2702, 843, 2, 2, 10, 79, 488, 2207, 6726, 12098, 10084, 2207, 2, 2, 11, 98, 702, 3842, 15127, 39202, 57965, 37634, 5778, 2

Offset: 0

William W. Collier, Feb 18 2018

Note the similarity in form of the recursive steps in the array definition above and the polynomial definition under FORMULA.

			i\j |0  1   2    3      4       5        6          7           8            9
----+-------------------------------------------------------------------------
   0|2  2   2    2      2       2        2          2           2            2
   1|2  3   7   18     47     123      322        843        2207         5778
   2|2  4  14   52    194     724     2702      10084       37634       140452
   3|2  5  23  110    527    2525    12098      57965      277727      1330670
   4|2  6  34  198   1154    6726    39202     228486     1331714      7761798
   5|2  7  47  322   2207   15127   103682     710647     4870847     33385282
   6|2  8  62  488   3842   30248   238142    1874888    14760962    116212808
   7|2  9  79  702   6239   55449   492802    4379769    38925119    345946302
   8|2 10  98  970   9602   95050   940898    9313930    92198402    912670090
   9|2 11 119 1298  14159  154451  1684802   18378371   200477279   2186871698
  10|2 12 142 1692  20162  240252  2862862   34114092   406506242   4843960812
  11|2 13 167 2158  27887  360373  4656962   60180133   777684767  10049721838
  12|2 14 194 2702  37634  524174  7300802  101687054  1416317954  19726764302
  13|2 15 223 3330  49727  742575 11088898  165590895  2472774527  36926027010
  14|2 16 254 4048  64514 1028176 16386302  261152656  4162056194  66331746448
  15|2 17 287 4862  82367 1395377 23639042  400468337  6784322687 114933017342
  16|2 18 322 5778 103682 1860498 33385282  599074578 10749957122 192900153618
  17|2 19 359 6802 128879 2441899 46267202  876634939 16609796639 314709501202
  18|2 20 398 7940 158402 3160100 63043598 1257711860 25091193602 500566160180
  19|2 21 439 9198 192719 4037901 84603202 1772629341 37140612959 778180242798

William W. Collier, a(i,j) = f(i+2,j)
William W. Collier, Experimental Mathematics on Wisteria Tables, Talk to Poughkeepsie ACM Chapter.
OEIS Wiki, The (1,2) Pascal Triangle.

The array first appeared in A298675.
Rows 1 through 29 of the array appear in these OEIS entries: A005248, A003500, A003501, A003499, A056854, A086903, A056918, A087799, A057076, A087800, A078363, A067902, A078365, A090727, A078367, A087215, A078369, A090728, A090729, A090730, A090731, A090732, A090733, A090247, A090248, A090249, A090251. Also entries occur for rows 45, 121, and 320:  A087265, A065705, A089775. Each of these entries asserts that a(i,j)=f(i+2,j) is true for that row.
A few of the columns appear in the OEIS: A008865 (for column 2), A058794 and A007754 (for column 3), and A230586 (for column 5).
Main diagonal gives A343261.

A:= proc(i, j) option remember; `if`(min(i, j)=0, 2,
      `if`(j=1, i+2, (i+2)*A(i, j-1)-A(i, j-2)))
    end:
seq(seq(A(d-k, k), k=0..d), d=0..12);  # Alois P. Heinz, Mar 05 2019

a[, 0] = a[0, ] = 2; a[i_, 1] := i + 2;
a[i_, j_] := a[i, j] =(i + 2) a[i, j - 1] - a[i, j - 2];
Table[a[i - j, j], {i, 0, 10}, {j, 0, i}] // Flatten (* Jean-François Alcover, Dec 07 2019 *)

Let k be an integer, and let r1 and r2 be the roots of x + 1/x = k. Then f(k,n) = r1^n + r2^n is an integer, for integer n >= 0. Theorem: a(i,j) = f(i+2,j), for i,j >= 0. Proof: See the Collier link.
Define polynomials recursively by:
    p[0](n) = 2, for n >= 0 ( [ and ] demark subscripts).
    p[1](n) = n + 2, for n >= 0.
    p[j](n) = p[j-1](n) * p[1](n) - p[j-2](n), for j > 1, n >= 0. The coefficients of these polynomials occur as the even numbered, upward diagonals in the OEIS Wiki link. Conjecture: a(i,j) = p[j](i), i,j >= 0.

Edited by N. J. A. Sloane, Apr 04 2018

cp's OEIS Frontend

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

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

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A001946 a(n) = 11*a(n-1) + a(n-2).

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A087215 Lucas(6*n): a(n) = 18*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 18.

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A087265 Lucas numbers L(8*n).

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

A087215 Lucas(6n): a(n) = 18a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 18.