A264080 -id:A264080 - cp's OEIS frontend

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

1, 4, 11, 29, 76, 199, 521, 1364, 3571, 9349, 24476, 64079, 167761, 439204, 1149851, 3010349, 7881196, 20633239, 54018521, 141422324, 370248451, 969323029, 2537720636, 6643838879, 17393796001, 45537549124, 119218851371, 312119004989, 817138163596, 2139295485799

Offset: 0

N. J. A. Sloane

In any generalized Fibonacci sequence {f(i)}, Sum_{i=0..4n+1} f(i) = a(n)*f(2n+2). - Lekraj Beedassy, Dec 31 2002
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
See A135064 for a possible connection with Galois groups of quintics.
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
All positive integer solutions of Pell equation a(n)^2 - 5*b(n)^2 = -4 together with b(n)=A001519(n), n>=0.
a(n) = L(n,-3)*(-1)^n, where L is defined as in A108299; see also A001519 for L(n,+3).
Inverse binomial transform of A030191. - Philippe Deléham, Oct 04 2005
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
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
a(n+1) is the Hankel transform of A001700(n)+A001700(n+1). - Paul Barry, Apr 21 2009
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
Conjecture: for n > 0, a(n) = sqrt(Fibonacci(4*n+3) + Sum_{k=2..2*n} Fibonacci(2*k)). - Alex Ratushnyak, May 06 2012
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
The continued fraction [a(n); a(n), a(n), ...] = phi^(2n+1), where phi is the golden ratio, A001622. - Thomas Ordowski, Jun 05 2013
Solutions (x, y) = (a(n), a(n+1)) satisfying  x^2 + y^2 = 3xy + 5. - Michel Lagneau, Feb 01 2014
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
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
Limit_{n->oo} a(n+1)/a(n) = phi^2 = phi + 1 = (3+sqrt(5))/2. - Derek Orr, Jun 18 2015
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
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
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

			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

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.
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).
Steven Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Marco Abrate, Stefano Barbero, Umberto Cerruti and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
Kasper K. S. Andersen, Lisa Carbone and Diego Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Paul Barry, On the Central Coefficients of Bell Matrices, J. Int. Seq., Vol. 14 (2011), Article 11.4.3, page 9.
Hacène Belbachir, Soumeya Merwa Tebtoub and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Joshua P. Bowman, Compositions with an Odd Number of Parts, and Other Congruences, J. Int. Seq (2024) Vol. 27, Art. 24.3.6. See p. 25.
Nathan D. Cahill, John R. D'Errico and John P. Spence, Complex Factorizations of the Fibonacci and Lucas Numbers, Fibonacci Quarterly, 1(41):13-19, 2003.
L. Carlitz, Problem B-110, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 5, No. 1 (1967), p. 108; An Infinite Series Equality, Solution to Problem B-110 by the proposer, ibid., Vol. 5, No. 5 (1967), pp. 469-470.
Paul Curtz, A269500, SeqFan list, March 2, 2016.
Murray Elder and Arkadius Kalka, Logspace computations for rigid Garside groups, arXiv preprint arXiv:1310.0933 [math.GR], 2013.
Sergio Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, Vol. 5 (2014), pp. 2226-2234.
Alex Fink, Richard K. Guy and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol. 3, No. 2 (2008), pp. 76-114. See Section 13.
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.
Tian-Xiao He and Louis W. Shapiro, Fuss-Catalan matrices, their weighted sums, and stabilizer subgroups of the Riordan group, Lin. Alg. Applic., Vol. 532 (2017) pp. 25-41, example p 34.
Christian Kassel and Christophe Reutenauer, Pairs of intertwined integer sequences, arXiv:2507.15780 [math.NT], 2025. See p. 13.
Seong Ju Kim, Ryan Stees and Laura Taalman, Sequences of Spiral Knot Determinants, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.4.
Tanya Khovanova, Recursive Sequences.
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
D. H. Lehmer, Recurrence formulas for certain divisor functions, Bull. Amer. Math. Soc., Vol. 49, No. 2 (1943), pp. 150-156.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum, Vol. 19 (2019), pp. 11-16.
Donatella Merlini and Renzo Sprugnoli, Arithmetic into geometric progressions through Riordan arrays, Discrete Mathematics, Vol. 340, No. 2 (2017), pp. 160-174.
Prabha Sivaraman Nair and Rejikumar Karunakaran, On k-Fibonacci Brousseau Sums, J. Int. Seq. (2024) Art. No. 24.6.4. See p. 3.
Serge Perrine, Some properties of the equation x^2=5y^2-4, The Fibonacci Quarterly, Vol. 54, No. 2 (2016) pp. 172-177.
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.
Ryan Stees, Sequences of Spiral Knot Determinants, Senior Honors Projects, Paper 84, James Madison Univ., May 2016.
Eric Weisstein's World of Mathematics, Fibonacci Polynomial.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory, Vol. 7, No. 5 (2011), pp. 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012), The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (3,-1).

Cf. A000204. a(n) = A060923(n, 0), a(n)^2 = A081071(n).
Cf. A005248 [L(2n) = bisection (even n) of Lucas sequence].
Cf. A001906 [F(2n) = bisection (even n) of Fibonacci sequence], A000045, A002315, A004146, A029907, A113224, A153387, A153416, A178482, A192425, A285992 (prime subsequence).
Cf. similar sequences of the type k*F(n)*F(n+1)+(-1)^n listed in A264080.

List([0..40], n-> Lucas(1,-1,2*n+1)[2] ); # G. C. Greubel, Jul 15 2019

a002878 n = a002878_list !! n
a002878_list = zipWith (+) (tail a001906_list) a001906_list
-- Reinhard Zumkeller, Jan 11 2012

[Lucas(2*n+1): n in [0..40]]; // Vincenzo Librandi, Apr 16 2011

A002878 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,4]);
    else
        3*procname(n-1)-procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Apr 30 2017

a[n_]:= FullSimplify[GoldenRatio^n - GoldenRatio^-n]; Table[a[n], {n, 1, 40, 2}]
a[1]=1; a[2]=4; a[n_]:=a[n]= 3a[n-1] -a[n-2]; Array[a, 40]
LinearRecurrence[{3, -1}, {1, 4}, 41] (* Jean-François Alcover, Sep 23 2017 *)
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 *)
a[ n_] := Fibonacci[2n] + Fibonacci[2n+2]; (* Michael Somos, Jul 31 2018 *)
a[ n_]:= LucasL[2n+1]; (* Michael Somos, Jan 13 2019 *)

a(n)=fibonacci(2*n)+fibonacci(2*n+2) \\ Charles R Greathouse IV, Jun 16 2011

for(n=1,40,q=((1+sqrt(5))/2)^(2*n-1);print1(contfrac(q)[1],", ")) \\ Derek Orr, Jun 18 2015

Vec((1+x)/(1-3*x+x^2) + O(x^40)) \\ Altug Alkan, Oct 26 2015

a002878 = [1, 4]
for n in range(30): a002878.append(3*a002878[-1] - a002878[-2])
print(a002878) # Gennady Eremin, Feb 05 2022

[lucas_number2(2*n+1,1,-1) for n in (0..40)] # G. C. Greubel, Jul 15 2019

a(n+1) = 3*a(n) - a(n-1).
G.f.: (1+x)/(1-3*x+x^2). - Simon Plouffe in his 1992 dissertation
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).
a(n) ~ phi^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
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
a(n) = A005248(n+1) - A005248(n) = -1 + Sum_{k=0..n} A005248(k). - Lekraj Beedassy, Dec 31 2002
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
a(n) = (-1)^n*Sum_{k=0..n} (-5)^k*binomial(n+k, n-k). - Benoit Cloitre, May 09 2004
From Paul Barry, May 27 2004: (Start)
Both bisection and binomial transform of A000204.
a(n) = Fibonacci(2n) + Fibonacci(2n+2). (End)
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
a(n) = A001906(n) + A001906(n+1). - Reinhard Zumkeller, Jan 11 2012
a(n) = floor(phi^(2n+1)), where phi is the golden ratio, A001622. - Thomas Ordowski, Jun 10 2012
a(n) = A014217(2*n+1) = A014217(2*n+2) - A014217(2*n). - Paul Curtz, Jun 11 2013
Sum_{n >= 0} 1/(a(n) + 5/a(n)) = 1/2. Compare with A005248, A001906, A075796. - Peter Bala, Nov 29 2013
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
From Peter Bala, Mar 22 2015: (Start)
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.
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).
Exp( Sum_{n >= 1} 2*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*F(n)*x^n.
Exp( Sum_{n >= 1} (-2)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*F(n)*(-x)^n.
Exp( Sum_{n >= 1} 4*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 4*A029907(n)*x^n.
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)
a(n) = sqrt(5*F(2*n+1)^2-4), where F(n) = A000045(n). - Derek Orr, Jun 18 2015
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
For n > 0, a(n) = L(n-1)*L(n+2) + 4*(-1)^n. - J. M. Bergot, Oct 25 2015
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
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
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
a(n)*F(m+2n-1) = F(m+4n-2)-F(m), with Fibonacci number F(m), empirical observation. - Dan Weisz, Jul 30 2018
a(n) = -a(-1-n) for all n in Z. - Michael Somos, Jul 31 2018
Sum_{n>=0} 1/a(n) = A153416. - Amiram Eldar, Nov 11 2020
a(n) = Product_{k=1..n} (1 + 4*sin(2*k*Pi/(2*n+1))^2). - Seiichi Manyama, Apr 30 2021
Sum_{n>=0} (-1)^n/a(n) = (1/sqrt(5)) * A153387 (Carlitz, 1967). - Amiram Eldar, Feb 05 2022
The continued fraction [a(n);a(n),a(n),...] = phi^(2*n+1), with phi = A001622. - A.H.M. Smeets, Feb 25 2022
a(n) = 2*sinh((2*n + 1)*arccsch(2)). - Peter Luschny, May 25 2022
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)
For n > 0, a(n) = 1 + 1/(Sum_{k>=1} F(k)/phi^(2*n*k + k)). - Diego Rattaggi, Nov 08 2023
From Peter Bala, Apr 16 2025: (Start)
a(3*n+1) = a(n)^3 + 3*a(n).
a(5*n+2) = a(n)^5 + 5*a(n)^3 + 5*a(n).
a(7*n+3) = a(n)^7 + 7*a(n)^5 + 14*a(n)^3 + 7*a(n).
For the coefficients see A034807.
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).
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)
From Peter Bala, May 15 2025: (Start)
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) ).
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.
For k >= 1, Sum_{n >= 1} (-1)^(n+1)/(a(n) + L(2*k)^2/a(n)) = (1/5) * A064170(k+2).
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) ).
More generally, it appears that for k >= 1, Sum_{n >= 1} 1/(a(n) + L(2*k)^2/a(n)) is rational.
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)

Chebyshev and Pell comments from Wolfdieter Lang, Aug 31 2004

A226205 a(n) = F(n)^2 - F(n-1)^2 or F(n+1) * F(n-2) where F(n) = A000045(n), the Fibonacci numbers.

Original entry on oeis.org

1, 0, 3, 5, 16, 39, 105, 272, 715, 1869, 4896, 12815, 33553, 87840, 229971, 602069, 1576240, 4126647, 10803705, 28284464, 74049691, 193864605, 507544128, 1328767775, 3478759201, 9107509824, 23843770275, 62423800997, 163427632720, 427859097159, 1120149658761

Offset: 1

Michael Somos, Jun 06 2013

A001519(n)^2 = A079472(n)^2 + a(n)^2 and (A001519(n), A079472(n), a(n)) is a Pythagorean triple.
INVERT transform is A052156. PSUM transform is A007598. SUMADJ transform is A088305. BINOMIAL transform is A039717. BINOMIAL transform with 0 prepended is A112091 with 0 prepended. BINOMIAL transform inverse is A084179(n+1).
In general, the difference between squares of two consecutive terms of a second order linear recurrence having a signature of (c,d) will be a third order recurrence with signature (c^2+d,(c^2+d)*d,-d^3). - Gary Detlefs, Mar 13 2025

			G.f. = x + 3*x^3 + 5*x^4 + 16*x^5 + 39*x^6 + 105*x^7 + 272*x^8 + 715*x^9 + ...

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
John P. Bonomo and Montana Ferita, A Small Fib, College Math. J., 2023.
Nurettin Irmak, Product of arbitrary Fibonacci numbers with distance 1 to Fibonomial coefficient, Turk J Math, (2017) 41: 825-828. See p. 828.
C.-A. Laisant, Observations sur les triangles rectangles en nombres entiers et les suites de Fibonacci, Nouvelles Annales de Math. (1919, in French) Série 4, Vol. 19, 391-397.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A001519, A001622, A001654, A007598, A033999, A039717, A052156, A079472, A084179, A088305, A112091, A121646, A290565.
Cf. similar sequences of the type k*F(n)*F(n+1)+(-1)^n listed in A264080.
Cf. A260259: numbers of the form F(n)*F(n+1)-(-1)^n. - Bruno Berselli, Nov 02 2015

[Fibonacci(n)^2-Fibonacci(n-1)^2: n in [1..40]]; // Vincenzo Librandi, Jun 18 2014

a:= n-> (<<0|1|0>, <0|0|1>, <-1|2|2>>^n. <<1,0,3>>)[1, 1]:
seq(a(n), n=0..30);  # Alois P. Heinz, Sep 28 2016

a[ n_] := Fibonacci[n + 1] Fibonacci[n - 2]; (* Michael Somos, Jun 17 2014 *)
CoefficientList[Series[(1 - x)^2/((1 + x) (1 - 3 x + x^2)), {x, 0, 30}], x] (* Vincenzo Librandi, Jun 17 2014 *)

{a(n) = fibonacci( n + 1) * fibonacci( n - 2)};

a(n) = round(2^(-1-n)*(-(-1)^n*2^(3+n)-(3-sqrt(5))^n*(1+sqrt(5))+(-1+sqrt(5))*(3+sqrt(5))^n)/5) \\ Colin Barker, Sep 28 2016

lista(nn) = {my(p = (3*x-1)/(x^3-2*x^2-2*x+1)); for (n=1, nn, p = deriv(p, x); print1(subst(p, x, 0)/n!, ", "); ); } \\ Michel Marcus, May 22 2018

G.f.: x * (1 - x)^2 / ((1 + x) * (1 -3*x + x^2)).
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
a(n) = - A121646(n).
a(n) = -a(1-n) for all n in Z.
a(n) = A121801(n+1) / 2. - Michael Somos, Jun 17 2014
a(n) = a(n-1) + A000045(n-1)^2 - 2*(-1)^n, for n>1. - Alexander Samokrutov, Sep 07 2015
a(n) = F(n-1)*F(n) - (-1)^n. - Bruno Berselli, Oct 30 2015
a(n) = 2^(-1-n)*(-(-1)^n*2^(3+n)-(3-sqrt(5))^n*(1+sqrt(5))+(-1+sqrt(5))*(3+sqrt(5))^n)/5. - Colin Barker, Sep 28 2016
From Amiram Eldar, Oct 06 2020: (Start)
Sum_{n>=3} 1/a(n) = (1/2) * A290565 - 1/4.
Sum_{n>=3} (-1)^(n+1)/a(n) = (3/2) * (1/phi - 1/2), where phi is the golden ratio (A001622). (End)

A236428 a(n) = F(n+1)^2 + F(n+1)*F(n) - F(n)^2, where F = A000045.

Original entry on oeis.org

1, 1, 5, 11, 31, 79, 209, 545, 1429, 3739, 9791, 25631, 67105, 175681, 459941, 1204139, 3152479, 8253295, 21607409, 56568929, 148099381, 387729211, 1015088255, 2657535551, 6957518401, 18215019649, 47687540549, 124847601995, 326855265439, 855718194319

Offset: 0

Richard R. Forberg, Jan 25 2014

(a(n) + a(n+1))/2 = F(2n+2).
a(n) = -a(-n-1), using the negative Fibonacci values.
First differences equal 2*A059929.
Partial sums equal A192873.
Unlike Fibonacci, the divisibility of a(n) by the primes is quite limited, specifically to p = 5, 11, 19, 31, 59, 71, 79, 109, ... where those after 5 are only a subset of primes congruent to {1,4} mod 5.
Values of a(n) mod p, for all primes p exhibit repeating pattern cycles of length k = (p-1)/m or (p+1)/m (except p = 5), based on whether p is congruent to {1,4} mod 5 or {2,3} mod 5. For p = 5, k = 2p = 10. Only the slightest similarity exists here with Fibonacci: there are formulas like this for a cycle length k, but for Fibonacci those are "divisibility cycles" for prime p, not the "pattern cycles" on mod p, and the m values differ for many primes, creating different cycle lengths for the same p.
a(n) has the property: a(k/2 + i) mod p + a(k/2 - 1 - i) mod p = p or 0, for all primes p, and all i 0 <= i <= k/2, in every cycle of length k. Thus, when plotted, the lower and upper halves of a every cycle have an inverted (i.e., flipped) symmetry.
For some primes (e.g., 13, 17, 37, 53, 61, 89, 97) each half-cycle (of length k/2) is internally symmetric (i.e., the second quarter-cycle is a mirror image of the first quarter cycle, and the fourth is a mirror image of the third, on each side of some value at k/4), while the flipped symmetry still holds for the upper and lower halves. See example for p = 61, with k = 30 in pdf file below.
No such symmetries on mod p, of either type, exist for Fibonacci.
a(n) is also (apart from sign) the determinant of a 2 X 2 matrix of squares of successive Fibonacci numbers: a(n) = (-1)^(n)*(F(n+2)^2*F(n-1)^2 -F(n)^2*F(n+1)^2). - R. M. Welukar, Aug 30 2014
For n>1 a(n) is the ceiling of the maximum area of a quadrilateral having sides of length in increasing order F(n), F(n+1), L(n), and L(n+1) with L(n)=A000032(n). - J. M. Bergot, Jan 19 2016
For n>1 a(n) is the numerator of the continued fraction [1, 1, ... 1, 2, 1, 1, ... 1, 2] with n-2 1's before each 2. - Greg Dresden and Kevin Zhanming Zheng, Aug 16 2020
a(n) is the number of edge covers in the rocket graph R_{3,n+1,n}. A rocket graph R_{m,i,j} is a m-cycle with two paths attached to adjacent vertices of the cycle, which have lengths i and j respectively. This is similar to a tadpole graph but with two tails. - Bridget Rozema, Oct 09 2024

Colin Barker, Table of n, a(n) for n = 0..1000
R. C. Alperin, A family of nonlinear recurrences and their linear solutions, Fib. Q., 57:4 (2019), 318-321.
R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., Vol. 58, No. 2 (2020), 140-142.
Richard R. Forberg, Plot of a(n) mod 61
Bridget Rozema and Maisie Smith, Edge Covers of Unions of Path and Cycle Graphs, 2024.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000032, A000045, A059929, A192873.

Cf. similar sequences of the type k*F(n)*F(n+1)+(-1)^n listed in A264080.

[Fibonacci(n+1)^2+Fibonacci(n+1)*Fibonacci(n)- Fibonacci(n)^2: n in [0..30]]; // Vincenzo Librandi, Jan 20 2016

F:=Fibonacci; [F(n+1)^2+F(n)*F(n-1): n in [0..30]]; // Bruno Berselli, Feb 15 2017

a[n_] := Fibonacci[n+1]^2 + Fibonacci[n+1]*Fibonacci[n] - Fibonacci[n]^2; Table[a[n], {n, 0, 26}] (* Jean-François Alcover, Feb 27 2014 *)
LinearRecurrence[{2, 2, -1}, {1, 1, 5}, 40] (* Vincenzo Librandi, Jan 20 2016 *)

F=fibonacci;
a(n)=F(n+1)^2 + F(n+1)*F(n) - F(n)^2;
vector(33,n,a(n-1)) \\ Joerg Arndt, Feb 23 2014

Vec((x^2-x+1)/((x+1)*(x^2-3*x+1)) + O(x^100)) \\ Colin Barker, Dec 20 2014

a(n) = round((2^(-n)*(3*(-2)^n-(3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n))/5) \\ Colin Barker, Sep 28 2016

a(n) = A001654(n) + A226205(n+1).
G.f.: (x^2 - x + 1)/((x + 1)*(x^2 - 3*x + 1)). - Joerg Arndt, Feb 23 2014
a(n) = (2*Lucas(2*n+1) + 3*(-1)^n)/5. - Ralf Stephan, Feb 27 2014
a(n) = 2*a(n-1) + 2*a(n-2)-a(n-3). - Colin Barker, Dec 20 2014
a(n) = F(n-1)*F(n+2) + F(n)*F(n+1). - J. M. Bergot, Dec 20 2014
a(n) = 2*F(n)*F(n+1) + (-1)^n. - Bruno Berselli, Oct 30 2015
a(n) = F(2*n+1) - F(n-1)^2 +(-1)^n for n>0. - J. M. Bergot, Jan 19 2016
a(n) = (2^(-n)*(3*(-2)^n-(3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n))/5. - Colin Barker, Sep 28 2016
a(n) = F(n+1)^2 + F(n)*F(n-1). See also A099016, tenth formula. - Bruno Berselli, Feb 15 2017
2*a(n) = L(n)*L(n+1) - F(n)*F(n+1), where L = A000032. - Bruno Berselli, Sep 27 2017

A014742 Expansion of (1+x^2)/(1 - 2x - 2x^2 + x^3).

Original entry on oeis.org

1, 2, 7, 17, 46, 119, 313, 818, 2143, 5609, 14686, 38447, 100657, 263522, 689911, 1806209, 4728718, 12379943, 32411113, 84853394, 222149071, 581593817, 1522632382, 3986303327, 10436277601, 27322529474, 71531310823, 187271402993, 490282898158, 1283577291479

Offset: 0

N. J. A. Sloane

Let M = a triangle with (1,1,1,3,3,5,5,7,7,...) as the left border and (0,1,2,3,4,5,...) as all other columns. A014742 = lim_{n->infinity} M^n, the left-shifted vector considered as a sequence. - Gary W. Adamson, Jul 26 2010
For n >= 1, a(n) is the ratio of L/h (rounded down), where L = length of short sides of parallelogram appearing in dissection fallacy of square F(n+2) X F(n+2), F(n) is Fibonacci number, and h = perpendicular distance between the long sides LL. The first differences of A069921 give L^2. See illustration. - Kival Ngaokrajang, Jun 27 2014
From Wolfdieter Lang, Jul 15 2014: (Start)
The preceding comment is a conjecture using a(n) = floor(LL(n)*L(n)) with LL(n) = sqrt(F(n+2)^2 + F(n)^2) and L(n) = LL(n-1), n >= 1 (its author agreed with this in an email). See also, e.g., the Koshy reference for the dissection fallacy, sect. 6, 100 - 108.
The proof of the conjecture uses first the identity (LL(n)*LL(n-1))^2 - a(n)^2 = + 1 with a(n) = F(n-1)*F(n) + F(n+1)*F(n+2) (see the formula section for a(n)). This identity is due to the factorization of the left-hand side which is A(n)^2 with A(n) = F(n)*F(n+1) - F(n-1)*F(n+2). But A(n) = (-1)^(n+1) is a special instance of a well known Fibonacci identity (Koshy, p. 88, Nr. 19 for h=-1, k=2, F(-1) = 1). Now one has (LL(n)*LL(n-1))^2 = 1 + a(n)^2, that is LL(n)*LL(n-1) = sqrt(1 + a(n)^2). Because a(n) < sqrt(1 + a(n)^2) <  a(n) + 1 (just square both inequalities using a(n) > 0) one has now proved that floor(LL(n)*LL(n-1)) = a(n), n >= 1. (End)
a(n) = numerator(Re(C(n))), with the complex sequence C(n) defined in the name of A069921. - Wolfdieter Lang, Jul 16 2014

			a(2) = F(1)*F(2) + F(3)*F(4) = 1*1 + 2*3 = 7. - _James R. Buddenhagen_, Jan 06 2009

T. Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley & Sons, 2001.

Colin Barker, Table of n, a(n) for n = 0..1000
Kival Ngaokrajang, Illustration of initial terms
Eric Weisstein's World of Mathematics, Dissection Fallacy
Wikipedia, Missing square puzzle
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A069921. - Kival Ngaokrajang, Jun 27 2014
Cf. similar sequences of the type k*F(n)*F(n+1) + (-1)^n listed in A264080.
Cf. A000032, A000045.

seq(combinat[fibonacci](n-1)*combinat[fibonacci](n)+combinat[fibonacci](n+1)*combinat[fibonacci](n+2), n=0..50); # will give first 50 terms - James R. Buddenhagen, Jan 06 2009

CoefficientList[Series[(1 + x^2)/(1 - 2*x - 2*x^2 + x^3), {x, 0, 30}], x] (* Wesley Ivan Hurt, Jun 27 2014 *)
LinearRecurrence[{2, 2, -1},{1, 2, 7},30] (* Ray Chandler, Sep 23 2015 *)

Vec((1+x^2)/(1-2*x-2*x^2+x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

a(n) = round((2^(-1-n)*((-1)^n*2^(2+n)-3*(3-sqrt(5))^n*(-1+sqrt(5))+3*(1+sqrt(5))*(3+sqrt(5))^n))/5) \\ Colin Barker, Sep 29 2016

a(n) = F(n-1)*F(n) + F(n+1)*F(n+2), where F = A000045. - James R. Buddenhagen, Jan 06 2009
From Wolfdieter Lang, Jul 15 2014: (Start)
G.f.: (1+x^2)/(1 - 2*x - 2*x^2 + x^3) = (1+x^2)/((1+x)*(1 - 3*x + x^2)) = (2/(1+x) + 3*(1+x)/(1 - 3*x + x^2))/5 (see the name).
a(n) = (2*(-1)^n + 3*(F(2*n) + F(2*(n+1))))/5,
a(n) = (2*(-1)^n + L(2*n-1) + L(2*n+3))/5 with L(n) = A000032(n) and L(-1) = -1. (End)
a(n) = 3*F(n)*F(n+1) + (-1)^n. - Bruno Berselli, Oct 30 2015
a(n) = (2^(-1-n)*((-1)^n*2^(2+n) - 3*(3-sqrt(5))^n*(-1+sqrt(5)) + 3*(1+sqrt(5))*(3+sqrt(5))^n))/5. - Colin Barker, Sep 29 2016
a(n) = Fibonacci(n + 2)^2 - 2*Fibonacci(n)^2. - Detlef Meya, Jun 29 2024
E.g.f.: exp(-x)*(2 + 3*exp(5*x/2)*(cosh(sqrt(5)*x/2) + sqrt(5)*sinh(sqrt(5)*x/2)))/5. - Stefano Spezia, Jun 30 2024

Buddenhagen's Jan 06 2009 entries adjusted for offset 0 by Wolfdieter Lang, Jul 15 2014

A061647 Beginning at the well for the topograph of a positive definite quadratic form with values 1, 1, 1 at a superbase (i.e., 1, 1 and 1 are the vonorms of the superbase), these numbers indicate the labels of the edges of the topograph on a path of greatest ascent.

Original entry on oeis.org

1, 3, 9, 23, 61, 159, 417, 1091, 2857, 7479, 19581, 51263, 134209, 351363, 919881, 2408279, 6304957, 16506591, 43214817, 113137859, 296198761, 775458423, 2030176509, 5315071103, 13915036801, 36430039299, 95375081097, 249695203991, 653710530877, 1711436388639

Offset: 1

Darrin Frey (freyd(AT)cedarville.edu), Jun 14 2001

Form the matrix A=[1,1,1;2,1,0;1,0,0]. a(n) is the sum of the second row elements of A^n. - Paul Barry, Sep 22 2004

			a(7) = 417 since a(7) = 2*a(6) + 2*a(5) - a(4) = 2*159 + 2*6 - 23.

J. H. Conway, The Sensual (Quadratic) Form, MAA.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000032, A000045, A001654.

Cf. similar sequences of the type k*F(n)*F(n+1) + (-1)^n listed in A264080.

LinearRecurrence[{2,2,-1},{1,3,9},40] (* Harvey P. Dale, May 31 2015 *)

x='x+O('x^99); Vec(x*(1+x+x^2)/((1+x)*(x^2-3*x+1))) \\ Altug Alkan, Mar 26 2016

a(n) = round(2^(-n)*(-(-2)^n-2*(3-sqrt(5))^n*(1+sqrt(5))+2*(-1+sqrt(5))*(3+sqrt(5))^n)/5) \\ Colin Barker, Sep 30 2016

a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3) for n > 3, with a(1)=1, a(2)=3, a(3)=9.
a(n) = Fibonacci(n)^2 + Fibonacci(n-1)^2 + 2*Fibonacci(n)*Fibonacci(n+1) + 3*Fibonacci(n-1)*Fibonacci(n), with offset 0. - Paul Barry, Sep 22 2004
a(n) = Fibonacci(n+1)^2 - Fibonacci(n-2)^2 - (-1)^n. - Thomas Baruchel, Jul 29 2005
From J. M. Bergot, Aug 26 2013: (Start)
a(n) = F(n-2)*F(n-1) + F(n-1)*F(n) + F(n)*F(n+1), where F=A000045.
a(n) = (L(2*n+1) + L(2*n-1) + L(2*n-3) - (-1)^n)/5, where L=A000032. [Corrected by Ehren Metcalfe, Mar 26 2016]
Starting at n=2 create Pythagorean triangles by using side x = b^2 - a^2, side y = 2*a*b, and side y = a^2 + b^2. For three successive triangles, let a=F(n) and b=F(n+1), a=F(n+1) and b=F(n+2), and a=F(n+2) and b=F(n+3); then a(n+3)=one-half the sum of the three perimeters.
(End)
G.f.: x*(1+x+x^2) / ( (1+x)*(x^2-3*x+1) ). - R. J. Mathar, Aug 28 2013
a(n) = A001654(n) + A001654(n-1) + A001654(n-2). - R. J. Mathar, Aug 28 2013
a(n) = 4*Fibonacci(n-1)*Fibonacci(n) - (-1)^n. - Bruno Berselli, Oct 30 2015
a(n) = Lucas(2*n-1) - Fibonacci(n-1)*Fibonacci(n). See similar identity for A264080. - Bruno Berselli, Nov 04 2015
a(n) = (1/5)*(4*Lucas(2*n-1) - (-1)^n). - Ehren Metcalfe, Mar 26 2016
a(n) = 2^(-n)*(-(-2)^n - 2*(3-sqrt(5))^n*(1+sqrt(5)) + 2*(-1+sqrt(5))*(3+sqrt(5))^n)/5. - Colin Barker, Sep 30 2016
a(n) = (-1)^(n-1) + 3*a(n-1) - a(n-2) with a(1) = 1 and a(2) = 3. - Peter Bala, Nov 12 2017

cp's OEIS Frontend

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

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A226205 a(n) = F(n)^2 - F(n-1)^2 or F(n+1) * F(n-2) where F(n) = A000045(n), the Fibonacci numbers.

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A236428 a(n) = F(n+1)^2 + F(n+1)*F(n) - F(n)^2, where F = A000045.

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A014742 Expansion of (1+x^2)/(1 - 2*x - 2*x^2 + x^3).

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A061647 Beginning at the well for the topograph of a positive definite quadratic form with values 1, 1, 1 at a superbase (i.e., 1, 1 and 1 are the vonorms of the superbase), these numbers indicate the labels of the edges of the topograph on a path of greatest ascent.

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A014742 Expansion of (1+x^2)/(1 - 2x - 2x^2 + x^3).