cp's OEIS Frontend

A002878 Bisection of Lucas sequence: a(n) = L(2*n+1).

%I A002878 M3420 N1384 #431 Jul 25 2025 15:28:58
%S A002878 1,4,11,29,76,199,521,1364,3571,9349,24476,64079,167761,439204,
%T A002878 1149851,3010349,7881196,20633239,54018521,141422324,370248451,
%U A002878 969323029,2537720636,6643838879,17393796001,45537549124,119218851371,312119004989,817138163596,2139295485799
%N A002878 Bisection of Lucas sequence: a(n) = L(2*n+1).
%C A002878 In any generalized Fibonacci sequence {f(i)}, Sum_{i=0..4n+1} f(i) = a(n)*f(2n+2). - _Lekraj Beedassy_, Dec 31 2002
%C A002878 The continued fraction expansion for F((2n+1)*(k+1))/F((2n+1)*k), k>=1 is [a(n),a(n),...,a(n)] where there are exactly k elements (F(n) denotes the n-th Fibonacci number). E.g., continued fraction for F(12)/F(9) is [4, 4,4]. - _Benoit Cloitre_, Apr 10 2003
%C A002878 See A135064 for a possible connection with Galois groups of quintics.
%C A002878 Sequence of all positive integers k such that continued fraction [k,k,k,k,k,k,...] belongs to Q(sqrt(5)). - _Thomas Baruchel_, Sep 15 2003
%C A002878 All positive integer solutions of Pell equation a(n)^2 - 5*b(n)^2 = -4 together with b(n)=A001519(n), n>=0.
%C A002878 a(n) = L(n,-3)*(-1)^n, where L is defined as in A108299; see also A001519 for L(n,+3).
%C A002878 Inverse binomial transform of A030191. - _Philippe Deléham_, Oct 04 2005
%C A002878 General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->infinity} a(n) =  x*(k*x+1)^n, k =(a(1)-3), x=(1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - _Ctibor O. Zizka_, Sep 02 2008
%C A002878 Let r = (2n+1), then a(n), n>0 = Product_{k=1..floor((r-1)/2)} (1 + sin^2 k*Pi/r); e.g., a(3) = 29 = (3.4450418679...)*(4.801937735...)*(1.753020396...). - _Gary W. Adamson_, Nov 26 2008
%C A002878 a(n+1) is the Hankel transform of A001700(n)+A001700(n+1). - _Paul Barry_, Apr 21 2009
%C A002878 a(n) is equal to the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(5)'s along the main diagonal, i's along the superdiagonal and the subdiagonal (i is the imaginary unit), and 0's everywhere else. - _John M. Campbell_, Jun 09 2011
%C A002878 Conjecture: for n > 0, a(n) = sqrt(Fibonacci(4*n+3) + Sum_{k=2..2*n} Fibonacci(2*k)). - _Alex Ratushnyak_, May 06 2012
%C A002878 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
%C A002878 The continued fraction [a(n); a(n), a(n), ...] = phi^(2n+1), where phi is the golden ratio, A001622. - _Thomas Ordowski_, Jun 05 2013
%C A002878 Solutions (x, y) = (a(n), a(n+1)) satisfying  x^2 + y^2 = 3xy + 5. - _Michel Lagneau_, Feb 01 2014
%C A002878 Conjecture: except for the number 3, a(n) are the numbers such that a(n)^2+2 are Lucas numbers. - _Michel Lagneau_, Jul 22 2014
%C A002878 Comment on the preceding conjecture: It is clear that all a(n) satisfy a(n)^2 + 2 = L(2*(2*n+1)) due to the identity (17 c) of Vajda, p. 177: L(2*n) + 2*(-1)^n = L(n)^2 (take n -> 2*n+1). - _Wolfdieter Lang_, Oct 10 2014
%C A002878 Limit_{n->oo} a(n+1)/a(n) = phi^2 = phi + 1 = (3+sqrt(5))/2. - _Derek Orr_, Jun 18 2015
%C A002878 If d[k] denotes the sequence of k-th differences of this sequence, then d[0](0), d[1](1), d[2](2), d[3](3), ... = A048876, cf. message to SeqFan list by P. Curtz on March 2, 2016. - _M. F. Hasler_, Mar 03 2016
%C A002878 a(n-1) and a(n) are the least phi-antipalindromic numbers (A178482) with 2*n and 2*n+1 digits in base phi, respectively. - _Amiram Eldar_, Jul 07 2021
%C A002878 Triangulate (hyperbolic) 2-space such that around every vertex exactly 7 triangles touch. Call any 7 triangles having a common vertex the first layer and let the (n+1)-st layer be all triangles that do not appear in any of the first n layers and have a common vertex with the n-th layer. Then the n-th layer contains 7*a(n-1) triangles. E.g., the first layer (by definition) contains 7 triangles, the second layer (the "ring" of triangles around the first layer) consists of 28 triangles, the third layer (the next "ring") consists of 77 triangles, and so on. - _Nicolas Nagel_, Aug 13 2022
%D A002878 J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.
%D A002878 N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D A002878 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%D A002878 Steven Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.
%H A002878 Vincenzo Librandi, <a href="/A002878/b002878.txt">Table of n, a(n) for n = 0..200</a>
%H A002878 Marco Abrate, Stefano Barbero, Umberto Cerruti and Nadir Murru, <a href="https://www.emis.de/journals/INTEGERS/papers/p38/p38.Abstract.html">Polynomial sequences on quadratic curves</a>, Integers, Vol. 15, 2015, #A38.
%H A002878 Kasper K. S. Andersen, Lisa Carbone and Diego Penta, <a href="https://pdfs.semanticscholar.org/8f0c/c3e68d388185129a56ed73b5d21224659300.pdf">Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields</a>, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
%H A002878 Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL14/Barry3/barry132.html">On the Central Coefficients of Bell Matrices</a>, J. Int. Seq., Vol. 14 (2011), Article 11.4.3, page 9.
%H A002878 Hacène Belbachir, Soumeya Merwa Tebtoub and László Németh, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL23/Nemeth/nemeth7.html">Ellipse Chains and Associated Sequences</a>, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
%H A002878 Joshua P. Bowman, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL27/Bowman/bowman4.html">Compositions with an Odd Number of Parts, and Other Congruences</a>, J. Int. Seq (2024) Vol. 27, Art. 24.3.6. See p. 25.
%H A002878 Nathan D. Cahill, John R. D'Errico and John P. Spence, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/41-1/cahill.pdf">Complex Factorizations of the Fibonacci and Lucas Numbers</a>, Fibonacci Quarterly, 1(41):13-19, 2003.
%H A002878 L. Carlitz, <a href="https://fq.math.ca/Scanned/5-1/elementary5-1.pdf">Problem B-110</a>, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 5, No. 1 (1967), p. 108; <a href="https://fq.math.ca/Scanned/5-5/elementary5-5.pdf">An Infinite Series Equality</a>, Solution to Problem B-110 by the proposer, ibid., Vol. 5, No. 5 (1967), pp. 469-470.
%H A002878 Paul Curtz, <a href="https://web.archive.org/web/*/http://list.seqfan.eu/oldermail/seqfan/2016-March/016180.html">A269500</a>, SeqFan list, March 2, 2016.
%H A002878 Murray Elder and Arkadius Kalka, <a href="https://arxiv.org/abs/1310.0933">Logspace computations for rigid Garside groups</a>, arXiv preprint arXiv:1310.0933 [math.GR], 2013.
%H A002878 Sergio Falcon, <a href="https://doi.org/10.4236/am.2014.515216">Relationships between Some k-Fibonacci Sequences</a>, Applied Mathematics, Vol. 5 (2014), pp. 2226-2234.
%H A002878 Alex Fink, Richard K. Guy and Mark Krusemeyer, <a href="https://cdm.ucalgary.ca/article/view/61940">Partitions with parts occurring at most thrice</a>, Contributions to Discrete Mathematics, Vol. 3, No. 2 (2008), pp. 76-114. See Section 13.
%H A002878 Dale Gerdemann, <a href="https://www.youtube.com/watch?vEQYQ4bEUX34">Collision of Digits</a> "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".
%H A002878 André Gougenheim, About the linear sequence of integers such that each term is the sum of the two preceding <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/9-3/gougenheim-a.pdf">Part 1</a> <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/9-3/gougenheim-b.pdf">Part 2</a>, Fib. Quart., Vol. 9, No. 3 (1971), pp. 277-295, 298.
%H A002878 Tian-Xiao He and Louis W. Shapiro, <a href="https://doi.org/10.1016/j.laa.2017.06.025">Fuss-Catalan matrices, their weighted sums, and stabilizer subgroups of the Riordan group</a>, Lin. Alg. Applic., Vol. 532 (2017) pp. 25-41, example p 34.
%H A002878 Christian Kassel and Christophe Reutenauer, <a href="https://arxiv.org/abs/2507.15780">Pairs of intertwined integer sequences</a>, arXiv:2507.15780 [math.NT], 2025. See p. 13.
%H A002878 Seong Ju Kim, Ryan Stees and Laura Taalman, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL19/Stees/stees4.html">Sequences of Spiral Knot Determinants</a>, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.4.
%H A002878 Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>.
%H A002878 Ioana-Claudia Lazăr, <a href="https://arxiv.org/abs/1904.06555">Lucas sequences in t-uniform simplicial complexes</a>, arXiv:1904.06555 [math.GR], 2019.
%H A002878 D. H. Lehmer, <a href="https://doi.org/10.1090/S0002-9904-1943-07880-9">Recurrence formulas for certain divisor functions</a>, Bull. Amer. Math. Soc., Vol. 49, No. 2 (1943), pp. 150-156.
%H A002878 Giovanni Lucca, <a href="https://forumgeom.fau.edu/FG2019volume19/FG201902index.html">Integer Sequences and Circle Chains Inside a Hyperbola</a>, Forum Geometricorum, Vol. 19 (2019), pp. 11-16.
%H A002878 Donatella Merlini and Renzo Sprugnoli, <a href="https://doi.org/10.1016/j.disc.2016.08.017">Arithmetic into geometric progressions through Riordan arrays</a>, Discrete Mathematics, Vol. 340, No. 2 (2017), pp. 160-174.
%H A002878 Prabha Sivaraman Nair and Rejikumar Karunakaran, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL27/Nair/nair11.html">On k-Fibonacci Brousseau Sums</a>, J. Int. Seq. (2024) Art. No. 24.6.4. See p. 3.
%H A002878 Serge Perrine, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Papers1/54-2/Perrine02242016.pdf">Some properties of the equation x^2=5y^2-4</a>, The Fibonacci Quarterly, Vol. 54, No. 2 (2016) pp. 172-177.
%H A002878 Simon Plouffe, <a href="https://arxiv.org/abs/0911.4975">Approximations de séries génératrices et quelques conjectures</a>, Dissertation, Université du Québec à Montréal, 1992, arXiv:0911.4975 [math.NT], 2009.
%H A002878 Simon Plouffe, <a href="/A000051/a000051_2.pdf">1031 Generating Functions</a>, Appendix to Thesis, Montreal, 1992.
%H A002878 Ryan Stees, <a href="https://commons.lib.jmu.edu/honors201019/84">Sequences of Spiral Knot Determinants</a>, Senior Honors Projects, Paper 84, James Madison Univ., May 2016.
%H A002878 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/FibonacciPolynomial.html">Fibonacci Polynomial</a>.
%H A002878 H. C. Williams and R. K. Guy, <a href="https://doi.org/10.1142/S1793042111004587">Some fourth-order linear divisibility sequences</a>, Intl. J. Number Theory, Vol. 7, No. 5 (2011), pp. 1255-1277.
%H A002878 H. C. Williams and R. K. Guy, <a href="https://www.emis.de/journals/INTEGERS/papers/a17self/a17self.Abstract.html">Some Monoapparitic Fourth Order Linear Divisibility Sequences</a>, Integers, Volume 12A (2012), The John Selfridge Memorial Volume.
%H A002878 <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials</a>.
%H A002878 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (3,-1).
%F A002878 a(n+1) = 3*a(n) - a(n-1).
%F A002878 G.f.: (1+x)/(1-3*x+x^2). - _Simon Plouffe_ in his 1992 dissertation
%F A002878 a(n) = S(2*n, sqrt(5)) = S(n, 3) + S(n-1, 3); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 3) = A001906(n+1) (even-indexed Fibonacci numbers).
%F A002878 a(n) ~ phi^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
%F A002878 Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then (-1)^n*q(n, -1) = a(n). - _Benoit Cloitre_, Nov 10 2002
%F A002878 a(n) = A005248(n+1) - A005248(n) = -1 + Sum_{k=0..n} A005248(k). - _Lekraj Beedassy_, Dec 31 2002
%F A002878 a(n) = 2^(-n)*A082762(n) = 4^(-n)*Sum_{k>=0} binomial(2*n+1, 2*k)*5^k; see A091042. - _Philippe Deléham_, Mar 01 2004
%F A002878 a(n) = (-1)^n*Sum_{k=0..n} (-5)^k*binomial(n+k, n-k). - _Benoit Cloitre_, May 09 2004
%F A002878 From _Paul Barry_, May 27 2004: (Start)
%F A002878 Both bisection and binomial transform of A000204.
%F A002878 a(n) = Fibonacci(2n) + Fibonacci(2n+2). (End)
%F A002878 Sequence lists the numerators of sinh((2*n-1)*psi) where the denominators are 2; psi=log((1+sqrt(5))/2). Offset 1. a(3)=11. - Al Hakanson (hawkuu(AT)gmail.com), Mar 25 2009
%F A002878 a(n) = A001906(n) + A001906(n+1). - _Reinhard Zumkeller_, Jan 11 2012
%F A002878 a(n) = floor(phi^(2n+1)), where phi is the golden ratio, A001622. - _Thomas Ordowski_, Jun 10 2012
%F A002878 a(n) = A014217(2*n+1) = A014217(2*n+2) - A014217(2*n). - _Paul Curtz_, Jun 11 2013
%F A002878 Sum_{n >= 0} 1/(a(n) + 5/a(n)) = 1/2. Compare with A005248, A001906, A075796. - _Peter Bala_, Nov 29 2013
%F A002878 a(n) = lim_{m->infinity} Fibonacci(m)^(4n+1)*Fibonacci(m+2*n+1)/ Sum_{k=0..m} Fibonacci(k)^(4n+2). - _Yalcin Aktar_, Sep 02 2014
%F A002878 From _Peter Bala_, Mar 22 2015: (Start)
%F A002878 The aerated sequence (b(n))n>=1 = [1, 0, 4, 0, 11, 0, 29, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -1, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy.
%F A002878 b(n) = (1/2)*((-1)^n - 1)*F(n) + (1 + (-1)^(n-1))*F(n+1), where F(n) is a Fibonacci number. The o.g.f. is x*(1 + x^2)/(1 - 3*x^2 + x^4).
%F A002878 Exp( Sum_{n >= 1} 2*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*F(n)*x^n.
%F A002878 Exp( Sum_{n >= 1} (-2)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*F(n)*(-x)^n.
%F A002878 Exp( Sum_{n >= 1} 4*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 4*A029907(n)*x^n.
%F A002878 Exp( Sum_{n >= 1} (-4)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 4*A029907(n)*(-x)^n. Cf. A002315, A004146, A113224 and A192425. (End)
%F A002878 a(n) = sqrt(5*F(2*n+1)^2-4), where F(n) = A000045(n). - _Derek Orr_, Jun 18 2015
%F A002878 For n > 1, a(n) = 5*F(2*n-1) + L(2*n-3) with F(n) = A000045(n). - _J. M. Bergot_, Oct 25 2015
%F A002878 For n > 0, a(n) = L(n-1)*L(n+2) + 4*(-1)^n. - _J. M. Bergot_, Oct 25 2015
%F A002878 For n > 2, a(n) = a(n-2) + F(n+2)^2 + F(n-3)^2 = L(2*n-3) + F(n+2)^2 + F(n-3)^2. - _J. M. Bergot_, Feb 05 2016 and Feb 07 2016
%F A002878 E.g.f.: ((sqrt(5) - 5)*exp((3-sqrt(5))*x/2) + (5 + sqrt(5))*exp((3+sqrt(5))*x/2))/(2*sqrt(5)). - _Ilya Gutkovskiy_, Apr 24 2016
%F A002878 a(n) = Sum_{k=0..n} (-1)^floor(k/2)*binomial(n-floor((k+1)/2), floor(k/2))*3^(n-k). - _L. Edson Jeffery_, Feb 26 2018
%F A002878 a(n)*F(m+2n-1) = F(m+4n-2)-F(m), with Fibonacci number F(m), empirical observation. - _Dan Weisz_, Jul 30 2018
%F A002878 a(n) = -a(-1-n) for all n in Z. - _Michael Somos_, Jul 31 2018
%F A002878 Sum_{n>=0} 1/a(n) = A153416. - _Amiram Eldar_, Nov 11 2020
%F A002878 a(n) = Product_{k=1..n} (1 + 4*sin(2*k*Pi/(2*n+1))^2). - _Seiichi Manyama_, Apr 30 2021
%F A002878 Sum_{n>=0} (-1)^n/a(n) = (1/sqrt(5)) * A153387 (Carlitz, 1967). - _Amiram Eldar_, Feb 05 2022
%F A002878 The continued fraction [a(n);a(n),a(n),...] = phi^(2*n+1), with phi = A001622. - _A.H.M. Smeets_, Feb 25 2022
%F A002878 a(n) = 2*sinh((2*n + 1)*arccsch(2)). - _Peter Luschny_, May 25 2022
%F A002878 This gives the sequence with 2 1's prepended: b(1)=b(2)=1 and, for k >= 3, b(k) = Sum_{j=1..k-2} (2^(k-j-1) - 1)*b(j). - _Neal Gersh Tolunsky_, Oct 28 2022 (formula due to Jon E. Schoenfield)
%F A002878 For n > 0, a(n) = 1 + 1/(Sum_{k>=1} F(k)/phi^(2*n*k + k)). - _Diego Rattaggi_, Nov 08 2023
%F A002878 From _Peter Bala_, Apr 16 2025: (Start)
%F A002878 a(3*n+1) = a(n)^3 + 3*a(n).
%F A002878 a(5*n+2) = a(n)^5 + 5*a(n)^3 + 5*a(n).
%F A002878 a(7*n+3) = a(n)^7 + 7*a(n)^5 + 14*a(n)^3 + 7*a(n).
%F A002878 For the coefficients see A034807.
%F A002878 The general result is: for k >= 0, a(k*n + (k-1)/2) = 2 * T(k, a(n)/2), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind and a(n) = ((1 + sqrt(5))/2)^(2*n+1) + ((1 - sqrt(5))/2)^(2*n+1).
%F A002878 Sum_{n >= 0} (-1)^n/a(n) = (1/4)* (theta_3(phi) - theta_3(phi^2)) = 0.815947983588122..., where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122) and phi = (sqrt(5) - 1)/2. See Borwein and Borwein, Exercise 3 a, p. 94 and Carlitz, 1967. (End)
%F A002878 From _Peter Bala_, May 15 2025: (Start)
%F A002878 Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/5 (telescoping series: 5/(a(n) - 1/a(n)) = 1/A001906(n+1) + 1/A001906(n) ).
%F A002878 More generally, for k >= 1, Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - s(k)/a(k*n)) = 1/(1 + a(k)) where s(k) = a(0) + a(1) + ... + a(k-1) = Lucas(2*k) - 2.
%F A002878 For k >= 1, Sum_{n >= 1} (-1)^(n+1)/(a(n) + L(2*k)^2/a(n)) = (1/5) * A064170(k+2).
%F A002878 Sum_{n >= 1} 1/(a(n) + 9/a(n)) = 3/10 (follows from 1/(a(n) + 9/a(n)) = L(2*n)/A081076(n) - L(2*n+2)/A081076(n+1) ).
%F A002878 More generally, it appears that for k >= 1, Sum_{n >= 1} 1/(a(n) + L(2*k)^2/a(n)) is rational.
%F A002878 Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(5) [telescoping product: Product_{k = 1..n} ((a(k) + 1)/(a(k) - 1))^2 = 5*(1 - 4/A240926(n+1)) ]. (End)
%e A002878 G.f. = 1 + 4*x + 11*x^2 + 29*x^3 + 76*x^4 + 199*x^5 + 521*x^6 + ... - _Michael Somos_, Jan 13 2019
%p A002878 A002878 := proc(n)
%p A002878     option remember;
%p A002878     if n <= 1 then
%p A002878         op(n+1,[1,4]);
%p A002878     else
%p A002878         3*procname(n-1)-procname(n-2) ;
%p A002878     end if;
%p A002878 end proc: # _R. J. Mathar_, Apr 30 2017
%t A002878 a[n_]:= FullSimplify[GoldenRatio^n - GoldenRatio^-n]; Table[a[n], {n, 1, 40, 2}]
%t A002878 a[1]=1; a[2]=4; a[n_]:=a[n]= 3a[n-1] -a[n-2]; Array[a, 40]
%t A002878 LinearRecurrence[{3, -1}, {1, 4}, 41] (* _Jean-François Alcover_, Sep 23 2017 *)
%t A002878 Table[Sum[(-1)^Floor[k/2] Binomial[n -Floor[(k+1)/2], Floor[k/2]] 3^(n - k), {k, 0, n}], {n, 0, 40}] (* _L. Edson Jeffery_, Feb 26 2018 *)
%t A002878 a[ n_] := Fibonacci[2n] + Fibonacci[2n+2]; (* _Michael Somos_, Jul 31 2018 *)
%t A002878 a[ n_]:= LucasL[2n+1]; (* _Michael Somos_, Jan 13 2019 *)
%o A002878 (Magma) [Lucas(2*n+1): n in [0..40]]; // _Vincenzo Librandi_, Apr 16 2011
%o A002878 (Haskell)
%o A002878 a002878 n = a002878_list !! n
%o A002878 a002878_list = zipWith (+) (tail a001906_list) a001906_list
%o A002878 -- _Reinhard Zumkeller_, Jan 11 2012
%o A002878 (PARI) a(n)=fibonacci(2*n)+fibonacci(2*n+2) \\ _Charles R Greathouse IV_, Jun 16 2011
%o A002878 (PARI) for(n=1,40,q=((1+sqrt(5))/2)^(2*n-1);print1(contfrac(q)[1],", ")) \\ _Derek Orr_, Jun 18 2015
%o A002878 (PARI) Vec((1+x)/(1-3*x+x^2) + O(x^40)) \\ _Altug Alkan_, Oct 26 2015
%o A002878 (Sage) [lucas_number2(2*n+1,1,-1) for n in (0..40)] # _G. C. Greubel_, Jul 15 2019
%o A002878 (GAP) List([0..40], n-> Lucas(1,-1,2*n+1)[2] ); # _G. C. Greubel_, Jul 15 2019
%o A002878 (Python)
%o A002878 a002878 = [1, 4]
%o A002878 for n in range(30): a002878.append(3*a002878[-1] - a002878[-2])
%o A002878 print(a002878) # _Gennady Eremin_, Feb 05 2022
%Y A002878 Cf. A000204. a(n) = A060923(n, 0), a(n)^2 = A081071(n).
%Y A002878 Cf. A005248 [L(2n) = bisection (even n) of Lucas sequence].
%Y A002878 Cf. A001906 [F(2n) = bisection (even n) of Fibonacci sequence], A000045, A002315, A004146, A029907, A113224, A153387, A153416, A178482, A192425, A285992 (prime subsequence).
%Y A002878 Cf. similar sequences of the type k*F(n)*F(n+1)+(-1)^n listed in A264080.
%K A002878 nonn,easy
%O A002878 0,2
%A A002878 _N. J. A. Sloane_
%E A002878 Chebyshev and Pell comments from _Wolfdieter Lang_, Aug 31 2004