A226205 -id:A226205 - cp's OEIS frontend

A001654 Golden rectangle numbers: F(n) * F(n+1), where F(n) = A000045(n) (Fibonacci numbers).

0, 1, 2, 6, 15, 40, 104, 273, 714, 1870, 4895, 12816, 33552, 87841, 229970, 602070, 1576239, 4126648, 10803704, 28284465, 74049690, 193864606, 507544127, 1328767776, 3478759200, 9107509825, 23843770274, 62423800998, 163427632719, 427859097160, 1120149658760

Offset: 0

N. J. A. Sloane

a(n)/A007598(n) ~= golden ratio, especially for larger n. - Robert Happelberg (roberthappelberg(AT)yahoo.com), Jul 25 2005
Let phi be the golden ratio (cf. A001622). Then 1/phi = phi - 1 = Sum_{n>=1} (-1)^(n-1)/a(n), an alternating infinite series consisting solely of unit fractions. - Franz Vrabec, Sep 14 2005
a(n+2) is the Hankel transform of A005807 aerated. - Paul Barry, Nov 04 2008
A more exact name would be: Golden convergents to rectangle numbers. These rectangles are not actually golden (ratio of sides is not phi) but are golden convergents (sides are numerator and denominator of convergents in the continued fraction expansion of phi, whence ratio of sides converges to phi). - Daniel Forgues, Nov 29 2009
The Kn4 sums (see A180662 for definition) of the "Races with Ties" triangle A035317 lead to this sequence. - Johannes W. Meijer, Jul 20 2011
Numbers m such that m(5m+2)+1 or m(5m-2)+1 is a square. - Bruno Berselli, Oct 22 2012
In pairs, these numbers are important in finding binomial coefficients that appear in at least six places in Pascal's triangle. For instance, the pair (m,n) = (40, 104) finds the numbers binomial(n-1,m) = binomial(n,m-1). Two additional numbers are found on the other side of the triangle. The final two numbers appear in row binomial(n-1,m). See A003015. - T. D. Noe, Mar 13 2013
For n>1, a(n) is one-half the area of the trapezoid created by the four points (F(n),L(n)), (L(n),F(n)), (F(n+1), L(n+1)), (L(n+1), F(n+1)) where F(n) = A000045(n) and L(n) = A000032(n). - J. M. Bergot, May 14 2014
[Note on how to calculate: take the two points (a,b) and (c,d) with aa(n) = A067962(n-1) / A067962(n-2), n > 1. - Reinhard Zumkeller, Sep 24 2015
Can be obtained (up to signs) by setting x = F(n)/F(n+1) in g.f. for Fibonacci numbers - see Pongsriiam. - N. J. A. Sloane, Mar 23 2017

			G.f. = x + 2*x^2 + 6*x^3 + 15*x^4 + 40*x^5 + 104*x^6 + 273*x^7 + 714*x^8 + ...

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 9.
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).

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from T. D. Noe)
R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., Vol. 58, No. 2 (2020), 140-142.
Paul Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays, JIS 12 (2009) 09.8.6.
A. Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83.
Alfred Brousseau, Fibonacci and Related Number Theoretic Tables, Fibonacci Association, San Jose, CA, 1972. See p. 17.
Ömer Egecioglu, Elif Saygi, and Zülfükar Saygi, The Mostar index of Fibonacci and Lucas cubes, arXiv:2101.04740 [math.CO], 2021. Mentions this sequence.
Shalosh B. Ekhad and Doron Zeilberger, Automatic Counting of Tilings of Skinny Plane Regions, arXiv preprint arXiv:1206.4864 [math.CO], 2012.
Sergio Falcon, On the Sequences of Products of Two k-Fibonacci Numbers, American Review of Mathematics and Statistics, March 2014, Vol. 2, No. 1, pp. 111-120.
Dale Gerdemann, Golden Ratio Base Digit Patterns for Columns of the Fibonomial Triangle, "Another interesting pattern is for Golden Rectangle Numbers A001654. I made a short video illustrating this pattern, along with other columns of the Fibonomial Triangle A010048".
Jonny Griffiths and Martin Griffiths, Fibonacci-related sequences via iterated QRT maps, Fib. Q., 51 (2013), 218-227.
James P. Jones and Péter Kiss, Representation of integers as terms of a linear recurrence with maximal index, Acta Academiae Paedagogicae Agriensis, Sectio Mathematicae, 25. (1998) pp. 21-37. See Lemma 4.1 p. 34.
Ronald Orozco López, Generating Functions of Generalized Simplicial Polytopic Numbers and (s,t)-Derivatives of Partial Theta Function, arXiv:2408.08943 [math.CO], 2024. See p. 11.
Ronald Orozco López, Simplicial d-Polytopic Numbers Defined on Generalized Fibonacci Polynomials, arXiv:2501.11490 [math.CO], 2025. See p. 6.
C. Pita, On s-Fibonomials, J. Int. Seq. 14 (2011) # 11.3.7.
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
Prapanpong Pongsriiam, Integral Values of the Generating Functions of Fibonacci and Lucas Numbers, College Math. J., 48 (No. 2 2017), pp 97ff.
M. Renault, Dissertation
Wikipedia, Illustration of 273 as a golden rectangle number.
Robert G. Wilson v, Letter to N. J. A. Sloane, circa 1993
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A001655, A001656, A001657, A001658, A010048, A067962.
Cf. A005968, A005969, A098531, A098532, A098533, A119283, A128697.
Cf. A000071, A079472, A080145, A239798, A290565.
Bisection of A006498, A070550, A080239.
First differences of A064831.
Partial sums of A007598.

a001654 n = a001654_list !! n
a001654_list = zipWith (*) (tail a000045_list) a000045_list
-- Reinhard Zumkeller, Jun 08 2013

I:=[0,1,2]; [n le 3 select I[n] else 2*Self(n-1) + 2*Self(n-2) - Self(n-3): n in [1..30]]; // G. C. Greubel, Jan 17 2018

with(combinat): A001654:=n->fibonacci(n)*fibonacci(n+1):
seq(A001654(n), n=0..28); # Zerinvary Lajos, Oct 07 2007

LinearRecurrence[{2,2,-1}, {0,1,2}, 100] (* Vladimir Joseph Stephan Orlovsky, Jul 03 2011 *)
Times@@@Partition[Fibonacci[Range[0,30]],2,1] (* Harvey P. Dale, Aug 18 2011 *)
Accumulate[Fibonacci[Range[0, 30]]^2] (* Paolo Xausa, May 31 2024 *)

A001654(n)=fibonacci(n)*fibonacci(n+1);

b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j));
vector(30, n, b(n-1, 2))  \\ Joerg Arndt, May 08 2016

from sympy import fibonacci as F
def a(n): return F(n)*F(n + 1)
[a(n) for n in range(101)] # Indranil Ghosh, Aug 03 2017

from math import prod
from gmpy2 import fib2
def A001654(n): return prod(fib2(n+1)) # Chai Wah Wu, May 19 2022

a(n) = A010048(n+1, 2) = Fibonomial(n+1, 2).
a(n) = A006498(2*n-1).
a(n) = a(n - 1) + A007598(n) = a(n - 1) + A000045(n)^2 = Sum_{j <= n} Fibonacci(j)^2. - Henry Bottomley, Feb 09 2001 [corrected by Ridouane Oudra, Apr 12 2025]
For n > 0, 1 - 1/a(n+1) = Sum_{k=1..n} 1/(F(k)*F(k+2)) where F(k) is the k-th Fibonacci number. - Benoit Cloitre, Aug 31 2002.
G.f.: x/(1-2*x-2*x^2+x^3) = x/((1+x)*(1-3*x+x^2)). (Simon Plouffe in his 1992 dissertation; see Comments to A055870),
a(n) = 3*a(n-1) - a(n-2) - (-1)^n = -a(-1-n).
Let M = the 3 X 3 matrix [1 2 1 / 1 1 0 / 1 0 0]; then a(n) = the center term in M^n *[1 0 0]. E.g., a(5) = 40 since M^5 * [1 0 0] = [64 40 25]. - Gary W. Adamson, Oct 10 2004
a(n) = Sum{k=0..n} Fibonacci(k)^2. The proof is easy. Start from a square (1*1). On the right side, draw another square (1*1). On the above side draw a square ((1+1)*(1+1)). On the left side, draw a square ((1+2)*(1+2)) and so on. You get a rectangle (F(n)*F(1+n)) which contains all the squares of side F(1), F(2), ..., F(n). - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 19 2007
With phi = (1+sqrt(5))/2, a(n) = round((phi^(2*n+1))/5) = floor((1/2) + (phi^(2*n+1))/5), n >= 0. - Daniel Forgues, Nov 29 2009
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3), a(1)=1, a(2)=2, a(3)=6. - Sture Sjöstedt, Feb 06 2010
a(n) = (A002878(n) - (-1)^n)/5. - R. J. Mathar, Jul 22 2010
a(n) = 1/|F(n+1)/F(n) - F(n)/F(n-1)| where F(n) = Fibonacci numbers A000045. b(n) = F(n+1)/F(n) - F(n)/F(n-1): 1/1, -1/2, 1/6, -1/15, 1/40, -1/104, ...; c(n) = 1/b(n) = a(n)*(-1)^(n+1): 1, -2, 6, -15, 40, -104, ... (n=1,2,...). - Thomas Ordowski, Nov 04 2010
a(n) = (Fibonacci(n+2)^2 - Fibonacci(n-1)^2)/4. - Gary Detlefs, Dec 03 2010
Let d(n) = n mod 2, a(0)=0 and a(1)=1. For n > 1, a(n) = d(n) + 2*a(n-1) + Sum_{k=0..n-2} a(k). - L. Edson Jeffery, Mar 20 2011
From Tim Monahan, Jul 11 2011: (Start)
a(n+1) = ((2+sqrt(5))*((3+sqrt(5))/2)^n+(2-sqrt(5))*((3-sqrt(5))/2)^n+(-1)^n)/5.
a(n) = ((1+sqrt(5))*((3+sqrt(5))/2)^n+(1-sqrt(5))*((3-sqrt(5))/2)^n-2*(-1)^n)/10. (End)
From Wolfdieter Lang, Jul 21 2012: (Start)
a(n) = (2*A059840(n+2) - A027941(n))/3, n >= 0, with A059840(n+2) = Sum_{k=0..n} F(k)*F(k+2) and A027941(n) = A001519(n+1) - 1, n >= 0, where A001519(n+1) = F(2*n+1). (End)
a(n) = (-1)^n * Sum_{k=0..n} (-1)^k*F(2*k), n >= 0. - Wolfdieter Lang, Aug 11 2012
a(-1-n) = -a(n) for all n in Z. - Michael Somos, Sep 19 2014
0 = a(n)*(+a(n+1) - a(n+2)) + a(n+1)*(-2*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 19 2014
a(n) = (L(2*n+1) - (-1)^n)/5 with L(k) = A000032(k). - J. M. Bergot, Apr 15 2016
E.g.f.: ((3 + sqrt(5))*exp((5+sqrt(5))*x/2) - 2*exp((2*x)/(3+sqrt(5))+x) - 1 - sqrt(5))*exp(-x)/(5*(1 + sqrt(5))). - Ilya Gutkovskiy, Apr 15 2016
From Klaus Purath, Apr 24 2019: (Start)
a(n) = A061646(n) - Fibonacci(n-1)^2.
a(n) = (A061646(n+1) - A061646(n))/2. (End)
a(n) = A226205(n+1) + (-1)^(n+1). - Flávio V. Fernandes, Apr 23 2020
Sum_{n>=1} 1/a(n) = A290565. - Amiram Eldar, Oct 06 2020
Product_{n>=2} (1 + (-1)^n/a(n)) = phi^2/2 (A239798). - Amiram Eldar, Dec 02 2024
G.f.: x * exp( Sum_{k>=1} F(3*k)/F(k) * x^k/k ), where F(n) = A000045(n). - Seiichi Manyama, May 07 2025

Extended by Wolfdieter Lang, Jun 27 2000

A195971 Number of n X 1 0..4 arrays with each element x equal to the number its horizontal and vertical neighbors equal to 2,0,1,3,4 for x=0,1,2,3,4.

Original entry on oeis.org

0, 1, 3, 4, 5, 9, 16, 25, 39, 64, 105, 169, 272, 441, 715, 1156, 1869, 3025, 4896, 7921, 12815, 20736, 33553, 54289, 87840, 142129, 229971, 372100, 602069, 974169, 1576240, 2550409, 4126647, 6677056, 10803705, 17480761, 28284464, 45765225

Offset: 0

R. H. Hardin, Sep 25 2011

Every 0 is next to 0 2's, every 1 is next to 1 0's, every 2 is next to 2 1's, every 3 is next to 3 3's, every 4 is next to 4 4's.
Column 1 of A195978.
a(n) is the number of total dominating sets in the (n+1)-path graph. - Eric W. Weisstein, Apr 10 2018
Equivalently, a(n) is the number of 0-1 sequences (every term is "0" or "1") of length n+1 whose every term is adjacent to a term "1". - Yuda Chen, Apr 06 2022
From Wajdi Maaloul, Jun 23 2022: (Start)
For n > 1, a(n) is the number of ways to tile the figure below using squares and dominoes: a horizontal strip of length n-1 that contains a central vertical strip of length 3. Below are figures for a(2) through a(5):
                                     
      ||      |_|     ||_        |_|_
      ||     ||_|    |||_|     |||_|_|
      ||       ||      ||           ||
(End)
a(n) is the number of compositions of n+2 with 1's, 3's and 4's, with the restriction that you cannot begin with two 1's. - Greg Dresden and Yuan Shen, Aug 10 2024

			All solutions for n=4:
  0   0   1   1   0
  0   0   0   0   1
  0   0   0   0   1
  1   0   1   0   0

R. H. Hardin, Table of n, a(n) for n = 0..200 (corrected by _R. H. Hardin_, Jan 19 2019)
Eric Weisstein's World of Mathematics, Path Graph
Eric Weisstein's World of Mathematics, Total Dominating Set
Index entries for linear recurrences with constant coefficients, signature (1,0,1,1).

Cf. A195978.

a:=[1,3,4,5];; for n in [5..40] do a[n]:=a[n-1]+a[n-3]+a[n-4]; od; Concatenation([0], a); # G. C. Greubel, Apr 03 2019

R:=PowerSeriesRing(Integers(), 40); [0] cat Coefficients(R!( x*(1+x)^2/(1-x-x^3-x^4) )); // G. C. Greubel, Apr 03 2019

Table[(LucasL[n + 3] - 2 Sin[n Pi/2] - 4 Cos[n Pi/2])/5, {n, 0, 40}] (* Eric W. Weisstein, Apr 10 2018 *)
LinearRecurrence[{1, 0, 1, 1}, {0, 1, 3, 4, 5}, 40] (* Eric W. Weisstein, Apr 10 2018; amended for a(0) by Georg Fischer, Apr 03 2019 *)
CoefficientList[Series[x*(1+x)^2/(1-x-x^3-x^4), {x, 0, 40}], x] (* Eric W. Weisstein, Apr 10 2018 *)

my(x='x+O('x^40)); concat([0], Vec(x*(1+x)^2/(1-x-x^3-x^4))) \\ G. C. Greubel, Apr 03 2019

(x*(1+x)^2/(1-x-x^3-x^4)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Apr 03 2019

a(n) = a(n-1) + a(n-3) + a(n-4).
G.f.: x*(1 + x)^2 / ((1 + x^2)*(1 - x - x^2)). - Colin Barker, Feb 17 2018
a(n) = (A000032(n + 3) - 2*sin(n*Pi/2) - 4*cos(n*Pi/2))/5. - Eric W. Weisstein, Apr 10 2018
a(n) = (Lucas(n+3) - (-1)^(floor(n/2))*(3+(-1)^n))/5. - G. C. Greubel, Apr 03 2019
From Wajdi Maaloul, Jun 23 2022: (Start)
a(2n) = A226205(n+1) = - A121646(n+1) = Fibonacci(n+1)^2 - Fibonacci(n)^2 = Fibonacci(n+2)*Fibonacci(n-1);
a(2n+1) = Fibonacci(n+2)^2 = A007598(n+2).
(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

A264080 a(n) = 6F(n)F(n+1) + (-1)^n, where F = A000045.

Original entry on oeis.org

1, 5, 13, 35, 91, 239, 625, 1637, 4285, 11219, 29371, 76895, 201313, 527045, 1379821, 3612419, 9457435, 24759887, 64822225, 169706789, 444298141, 1163187635, 3045264763, 7972606655, 20872555201, 54645058949, 143062621645, 374542805987, 980565796315

Offset: 0

Bruno Berselli, Nov 03 2015

a(n) is prime for n = 1, 2, 5, 7, 14, 15, 29, 40, 49, 57, 70, 87, 105, 127, 175, 279, 362, 647, 727, ...

Bruno Berselli, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000032, A000045; A001654, A002878.

Cf. similar sequences of the type k*F(n)*F(n+1)+(-1)^n: A226205 (k=1); A236428 (k=2); A014742 (k=3); A061647 (k=4); A002878 (k=5).

[6*Fibonacci(n)*Fibonacci(n+1)+(-1)^n: n in [0..30]];

a:= n-> (<<0|1|0>, <0|0|1>, <-1|2|2>>^n. <<1,5,13>>)[1, 1]:
seq(a(n), n=0..30);  # Alois P. Heinz, Sep 28 2016

Table[6 Fibonacci[n] Fibonacci[n + 1] + (-1)^n, {n, 0, 30}]
LinearRecurrence[{2,2,-1},{1,5,13},30] (* Harvey P. Dale, Jul 12 2019 *)

makelist(6*fib(n)*fib(n+1)+(-1)^n, n, 0, 30);

for(n=0, 30, print1(6*fibonacci(n)*fibonacci(n+1)+(-1)^n", "));

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

Vec((1+3*x+x^2)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Sep 28 2016

[6*fibonacci(n)*fibonacci(n+1)+(-1)^n for n in (0..30)]

G.f.: (1+3*x+x^2) / ((1+x)*(1-3*x+x^2)). - Corrected by Colin Barker, Sep 28 2016
a(n) = -a(-n-1) = 2*a(n-1) + 2*a(n-2) - a(n-3) for all n in Z.
a(n) = L(2*n+1) + F(n)*F(n+1) = A002878(n) + A001654(n). See similar identity for A061647.
a(n) = A001654(n+1) + 3*A001654(n) + A001654(n-1).
a(n) - a(n-1) = 2*A099016(n) with a(-1)=-1.
a(n) + a(n-1) = 2*A097134(n) for n>0.
Sum_{i>=0} 1/a(i) = 1.3232560865206157372628688449331...
a(n) = (2^(-n)*(-(-2)^n-3*(3-sqrt(5))^n*(-1+sqrt(5))+3*(1+sqrt(5))*(3+sqrt(5))^n))/5. - Colin Barker, Sep 28 2016
E.g.f.: (1/5)*exp(-x)*(-1 + 6*exp(5*x/2)*(cosh((sqrt(5)*x)/2) + sqrt(5)*sinh((sqrt(5)*x)/2))). - Stefano Spezia, Dec 09 2019

A260259 a(n) = F(n)*F(n+1) - (-1)^n, where F = A000045.

Original entry on oeis.org

-1, 2, 1, 7, 14, 41, 103, 274, 713, 1871, 4894, 12817, 33551, 87842, 229969, 602071, 1576238, 4126649, 10803703, 28284466, 74049689, 193864607, 507544126, 1328767777, 3478759199, 9107509826, 23843770273, 62423800999, 163427632718, 427859097161, 1120149658759

Offset: 0

Bruno Berselli, Oct 31 2015

Primes in sequence for n = 1, 3, 5, 6, 9, 24, 42, 48, 53, 71, 86, 102, 138, 182, 302, 438, 506, 926, ...

Bruno Berselli, Table of n, a(n) for n = 0..500
A. Bremner, R. Høibakk, D. Lukkassen, Crossed ladders and Euler’s quartic, Annales Mathematicae et Informaticae, 36 (2009) pp. 29-41. See p. 33.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

First bisection of A111569.
Cf. A226205: numbers of the form F(n)*F(n+1)+(-1)^n.
Cf. A000045, A001654, A003482, A059929, A089508 (first bisection, without -1), A206351.

[Fibonacci(n)*Fibonacci(n+1)-(-1)^n: n in [0..30]];

with(combinat): A260259:=n->fibonacci(n)*fibonacci(n+1)-(-1)^n: seq(A260259(n), n=0..50); # Wesley Ivan Hurt, Feb 04 2017

Table[Fibonacci[n] Fibonacci[n + 1] - (-1)^n, {n, 0, 30}]

makelist(fib(n)*fib(n+1)-(-1)^n,n,0,30);

for(n=0, 30, print1(fibonacci(n)*fibonacci(n+1)-(-1)^n", "));

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

Vec(-(1-4*x+x^2)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Sep 29 2016

[fibonacci(n)*fibonacci(n+1)-(-1)^n for n in (0..30)]

G.f.: (-1 + 4*x - x^2)/((1 + x)*(1 - 3*x + x^2)).
a(n) = -a(-n-1) = 2*a(n-1) + 2*a(n-2) - a(n-3) for all n in Z.
a(n) = F(n+2)^2 - 2*F(n+1)^2.
a(n) = A059929(n) - A059929(n-1) with A059929(-1)=1.
a(n) = -A001654(n+1) + 4*A001654(n) - A001654(n-1).
a(n) = A206351((n+2)/2)-2 for even n; a(n) = A003482((n-1)/2)+2 for odd n.
Sum_{i>=0} 1/a(i) = .754301907697893871765121109686...
a(n) = (2^(-1-n)*(-3*(-1)^n*2^(2+n)-(3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n))/5. - Colin Barker, Sep 29 2016

A338225 a(n) = F(n+3) * F(n+1) + (-1)^n where F(n) = A000045(n) are the Fibonacci numbers.

Original entry on oeis.org

2, 11, 23, 66, 167, 443, 1154, 3027, 7919, 20738, 54287, 142131, 372098, 974171, 2550407, 6677058, 17480759, 45765227, 119814914, 313679523, 821223647, 2149991426, 5628750623, 14736260451, 38580030722, 101003831723, 264431464439, 692290561602, 1812440220359, 4745030099483

Offset: 1

Burak Muslu, Jan 30 2021

Twice the difference between the areas of consecutive deltoids with cross lengths F(n+3) and F(n).
Also it is the difference between the areas of consecutive rectangles with side lengths F(n+3) and F(n).
Also first differences of A226205 for n > 2.

			For n = 2, a(2) = F(2+3) * F(2+1) + (-1)^2 = 5 * 2 + 1 = 11.

Burak Muslu, Sayılar ve Bağlantılar, Luna, 2021, p. 50.

Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000045, A007598, A226205, A341208.

a[n_] := Fibonacci[n + 3] * Fibonacci[n + 1] + (-1)^n; Array[a, 30] (* Amiram Eldar, Jan 30 2021 *)

a(n) = fibonacci(n+3)*fibonacci(n+1) + (-1)^n; \\ Michel Marcus, Mar 25 2021

a(n) = F(n+3) * F(n+1) + (-1)^n, for n > 0.
G.f.: x*(2 + 7*x - 3*x^2)/(1 - 2*x - 2*x^2 + x^3). - Stefano Spezia, Jan 30 2021
a(n) = F(n+2)^2 + 2*(-1)^n = A007598(n+2) + 2*(-1)^n. - Amiram Eldar, Jan 11 2022

A236165 a(n) = a(n-1) + a(n-2) + a(n-3), with a(0) = a(1) = 1, a(2) = 0.

Original entry on oeis.org

1, 1, 0, 0, 2, 3, 3, 5, 10, 16, 24, 39, 65, 105, 168, 272, 442, 715, 1155, 1869, 3026, 4896, 7920, 12815, 20737, 33553, 54288, 87840, 142130, 229971, 372099, 602069, 974170, 1576240, 2550408, 4126647, 6677057, 10803705, 17480760, 28284464, 45765226, 74049691

Offset: 0

Michael Somos, Jan 19 2014

			G.f. = 1 + x + 2*x^4 + 3*x^5 + 3*x^6 + 5*x^7 + 10*x^8 + 16*x^9 + ...

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,1,1).

Cf. A000045, A059929, A226205.

I:=[1,1,0,0]; [n le 4 select I[n] else Self(n-1)+Self(n-3)+Self(n-4): n in [1..50]]; // Vincenzo Librandi, Jan 20 2015

a[ n_] := Fibonacci[ Quotient[ n, 2] - 1] Fibonacci[ Quotient[ n, 2] + 1 + Mod[n, 2]];
LinearRecurrence[{1,0,1,1},{1,1,0,0},50] (* Harvey P. Dale, Jan 19 2015 *)
CoefficientList[Series[(1 - x^2 - x^3) / (1 - x - x^3 - x^4), {x, 0, 70}], x] (* Vincenzo Librandi, Jan 20 2015 *)

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

G.f.: (1 - x^2 - x^3) / (1 - x - x^3 - x^4).
a(n+1)*a(n+3) = a(n)*a(n+2) + a(n+1)*a(n+2) for all n in Z.
a(n+1) + a(n-1) = A000045(n) for all n in Z.
a(2n) = A059929(n-1), a(2n-1) = A226205(n).

A341928 a(n) = F(n+4) * F(n+2) + 7 * (-1)^n where F(n) = A000045(n) are the Fibonacci numbers.

Original entry on oeis.org

3, 31, 58, 175, 435, 1162, 3019, 7927, 20730, 54295, 142123, 372106, 974163, 2550415, 6677050, 17480767, 45765219, 119814922, 313679515, 821223655, 2149991418, 5628750631, 14736260443, 38580030730, 101003831715, 264431464447, 692290561594, 1812440220367

Offset: 1

Burak Muslu, Feb 23 2021

First differences of A341208.
Second differences of A338225.
Third differences of A226205 n > 2.
Third differences between the areas of consecutive rectangles with side lengths F(n+3) and F(n).
Twice the third differences between the areas of consecutive deltoids with cross lengths F(n+3) and F(n).
Twice the third differences between the areas of consecutive triangles with the height and base length are F(n+3) and F(n).

			For n = 2, a(2) = F(2+4) * F(2+2) + 7 * (-1)^2 = 8 * 3 + 7 = 31.

Burak Muslu, Sayılar ve Bağlantılar, Luna, 2021, p. 51 (in Turkish).

Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000045, A341208, A338225, A226205.

Table[Fibonacci[n + 4] * Fibonacci[n + 2] + 7 * (-1)^n, {n, 1, 28}] (* Amiram Eldar, Feb 23 2021 *)

a(n) = fibonacci(n+4)*fibonacci(n+2) + 7*(-1)^n; \\ Michel Marcus, Feb 23 2021

a(n) = F(n+4) * F(n+2) + 7 * (-1)^n.

G.f.: x*(3 + 25*x - 10*x^2)/(1 - 2*x - 2*x^2 + x^3).

A343008 a(n) = F(n+5) * F(n+2) - 12 * (-1)^n where F(n) = A000045(n) are the Fibonacci numbers.

Original entry on oeis.org

28, 27, 117, 260, 727, 1857, 4908, 12803, 33565, 87828, 229983, 602057, 1576252, 4126635, 10803717, 28284452, 74049703, 193864593, 507544140, 1328767763, 3478759213, 9107509812, 23843770287, 62423800985, 163427632732, 427859097147, 1120149658773

Offset: 1

Burak Muslu, Apr 02 2021

nonn

First differences of A341928.
Second differences of A341208.
Third differences of A338225.
Fourth differences of A226205.
Fourth differences between the areas of consecutive rectangles with side lengths F(n+3) and F(n).
Twice the fourth differences between the areas of consecutive deltoids with cross lengths F(n+3) and F(n).
Twice the fourth differences between the areas of consecutive triangles with the height and base length are F(n+3) and F(n).

			For n = 2, a(2) = F(2+5) * F(2+2) - 12 * (-1)^2 = 13 * 3 - 12 = 27.

B. Muslu, Sayılar ve Bağlantılar, Luna, 2021, p. 52.

Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000045, A226205, A338225, A341208, A341928.

a[n_]:=Fibonacci[n+5]*Fibonacci[n+2]-12(-1)^n
Array[a,30] (* Giorgos Kalogeropoulos, Apr 02 2021 *)

a(n) = F(n+5) * F(n+2) - 12 * (-1)^n.

G.f.: x*(28 - 29*x + 7*x^2)/(1 - 2*x - 2*x^2 + x^3).

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A230448 T(n, k) = T(n-1, k-1) + T(n-1, k) with T(n, 0) = 1 and T(n, n) = A226205(n+1), n >= 0 and 0 <= k <= n.

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A001654 Golden rectangle numbers: F(n) * F(n+1), where F(n) = A000045(n) (Fibonacci numbers).

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A195971 Number of n X 1 0..4 arrays with each element x equal to the number its horizontal and vertical neighbors equal to 2,0,1,3,4 for x=0,1,2,3,4.

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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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A264080 a(n) = 6*F(n)*F(n+1) + (-1)^n, where F = A000045.

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A260259 a(n) = F(n)*F(n+1) - (-1)^n, where F = A000045.

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A264080 a(n) = 6F(n)F(n+1) + (-1)^n, where F = A000045.