A059840 -id:A059840 - 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

A202453 Fibonacci self-fusion matrix, by antidiagonals.

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 5, 5, 6, 5, 5, 8, 8, 9, 9, 8, 8, 13, 13, 15, 15, 15, 13, 13, 21, 21, 24, 24, 24, 24, 21, 21, 34, 34, 39, 39, 40, 39, 39, 34, 34, 55, 55, 63, 63, 64, 64, 63, 63, 55, 55, 89, 89, 102, 102, 104, 104, 104, 102, 102, 89, 89, 144, 144, 165, 165

Offset: 1

Clark Kimberling, Dec 19 2011

The Fibonacci self-fusion matrix, F, is the fusion P**Q, where P and Q are the lower and upper triangular Fibonacci matrices.  See A193722 for the definition of fusion of triangular arrays.
Every term F(n,k) of F is a product of two Fibonacci numbers; indeed,
F(n,k)=F(n)*F(k+1) if k is even;
F(n,k)=F(n+1)*F(k) if k is odd.
antidiagonal sums: (1,2,6,12,...), A054454
diagonal (1,2,6,15,...), A001654
diagonal (1,3,9,24,...), A064831
diagonal (2,5,15,39,..), A059840
diagonal (3,8,24,63,..), A080097
diagonal (5,13,39,102,...), A080143
diagonal (8,21,63,165,...), A080144
principal submatrix sums, A202462
All the principal submatrices are invertible, and the terms in the inverses are in {-3,-2,-1,0,1,2,3}.

			Northwest corner:
1...1....2....3....5....8....13
1...2....3....5....8...13....21
2...3....6....9...15...24....39
3...5....9...15...24...39....63
5...8...15...24...40...64...104

Clark Kimberling, Fusion, Fission, and Factors, Fib. Q., 52 (2014), 195-202.

Cf. A000045, A202451, A202452, A202503 (Fibonacci fission array).

n = 12;
Q = NestList[Most[Prepend[#, 0]] &, #, Length[#] - 1] &[Table[Fibonacci[k], {k, 1, n}]];
P = Transpose[Q]; F = P.Q;
Flatten[Table[P[[i]][[k + 1 - i]], {k, 1, n}, {i, 1, k}]] (* A202452 as a sequence *)
Flatten[Table[Q[[i]][[k + 1 - i]], {k, 1, n}, {i, 1, k}]] (* A202451 as a sequence *)
Flatten[Table[F[[i]][[k + 1 - i]], {k, 1, n}, {i, 1, k}]] (* A202453 as a sequence *)
TableForm[Q]  (* A202451, upper tri. Fibonacci array *)
TableForm[P]  (* A202452, lower tri. Fibonacci array *)
TableForm[F]  (* A202453, Fibonacci fusion array *)
TableForm[FactorInteger[F]]

Matrix product P*Q, where P, Q are the lower and upper triangular Fibonacci matrices, A202451 and A202452.

A059929 a(n) = Fibonacci(n)*Fibonacci(n+2).

Original entry on oeis.org

0, 2, 3, 10, 24, 65, 168, 442, 1155, 3026, 7920, 20737, 54288, 142130, 372099, 974170, 2550408, 6677057, 17480760, 45765226, 119814915, 313679522, 821223648, 2149991425, 5628750624, 14736260450, 38580030723, 101003831722, 264431464440, 692290561601, 1812440220360

Offset: 0

Henry Bottomley, Feb 09 2001

Expansion of golden ratio (1+sqrt(5))/2 as an infinite product: phi = Product_{i>=0} (1+1/(Fibonacci(2*i+1) * Fibonacci(2*i+3)-1)) * (1-1/(Fibonacci(2*i+2) * Fibonacci(2i+4)+1)). - Thomas Baruchel, Nov 11 2003
Each of these is one short of or one over the square of a Fibonacci number (A007598). This means that a rectangle sized F(n) by F(n + 2) units can't be converted into a square with sides of length F(n + 1) units unless one square unit of material is added or removed. - Alonso del Arte, May 03 2011
These are the integer parts of the numerators of the numbers with continued fraction representations [1, 2, 2, 2, ...], [1, 1, 2, 2, 2, ...], [1, 1, 1, 2, 2, 2, ...], etc., that is, sqrt(2), (2+sqrt(2))/2, 3-sqrt(2), (10+sqrt(2))/7, (24-sqrt(2))/14, etc. - Geoffrey Caveney, May 03 2014
a(n) appears also as the third component of the square of [F(n), F(n+1), F(n+2), F(n+3)], for n >= 0, where F(n) = A000045(n), in the Clifford algebra Cl_2 over Euclidean 2-space. The whole quartet of sequences for this square is [-A248161(n), A079472(n+1), a(n), A121801(n+1)]. See the Oct 15 2014 comment in A147973 where also a reference is given. - Wolfdieter Lang, Nov 01 2014
Numbers with a continued fraction expansion with the repeating sequence of length n [1, 1, ..., 1, 2], n-1 ones followed by a single two, for n > = 1, appear to be equal to (F(n) + sqrt(a(n)))/F(n+1), where F(n) = A000045(n). - R. James Evans, Nov 21 2018
The preceding conjecture is true. Proof: For n >= 1 let c(n) := confrac(repeat(1^{n-1}, 2)) where 1^{k} denotes 1 taken k times. This can be computed, e.g. from [Perron, third and fourth eq. on p. 62], as c(n) = (F(n) + sqrt(F(n+1)^2 - (-1)^n)) / F(n+1), which is the conjectured formula because F(n+1)^2 - (-1)^n = a(n). - Wolfdieter Lang, Jan 05 2019

			G.f. = 2*x + 3*x^2 + 10*x^3 + 24*x^4 + 65*x^5 + 168*x^6 + ... - _Michael Somos_, Mar 18 2022

Oskar Perron, Die Lehre von den Kettenbrüchen, Band I, 3. Auflage, B. G. Teubner, Stuttgart, 1954, pp. 61-61.

Muniru A Asiru, Table of n, a(n) for n = 0..2374 (first 501 terms from Harry J. Smith)
Tim Jones (producer), Engineering Bits & Bytes: The Fibonacci Puzzle, Wayne State University College of Engineering (2011).
Eric H. Kuo, Applications of graphical condensation for enumerating matchings and tilings, arXiv:math/0304090 [math.CO], 2003.
Marc Renault, Properties of the Fibonacci Sequence Under Various Moduli, Master's Thesis, Wake Forest University, 1996.
Michel Waldschmidt, Open Diophantine problems, arXiv:math/0312440 [math.NT], 2003-2004.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Bisection of A070550.
First differences of A059840.
Cf. A000045, A001519, A001654, A007598, A033999, A035513, A079472, A121801, A126358, A147973, A248161.

a:=List([0..30],n->Fibonacci(n)*Fibonacci(n+2));; Print(a); # Muniru A Asiru, Jan 05 2019

[Fibonacci(n)*Fibonacci(n+2): n in [0..30]]; // Vincenzo Librandi, Jul 02 2014

with(combinat): a:=n->fibonacci(n)*fibonacci(n+2): seq(a(n), n=0..26); # Zerinvary Lajos, Oct 07 2007

Table[Fibonacci[n]*Fibonacci[n+2],{n,0,60}] (* Vladimir Joseph Stephan Orlovsky, Nov 17 2009 *)

a(n) = fibonacci(n)*fibonacci(n + 2) \\ Harry J. Smith, Jun 30 2009

from sympy import fibonacci
[fibonacci(n)*fibonacci(n+2) for n in range(30)] # Stefano Spezia, Jan 05 2019

[fibonacci(n)*fibonacci(n+2) for n in range(30)] # G. C. Greubel, Nov 21 2018

a(n) = Fibonacci(n+1)^2 - (-1)^n = A007598(n+1) + A033999(n+1) = A000045(n+1)^2 - A033999(n).
G.f.: (2*x-x^2) / ((1+x)*(1-3*x+x^2)).
Sum_{n>=1} 1/a(n) = 1.
Sum_{n>=1} (-1)^n/a(n) = 2 - sqrt(5).
Sum_{n>=1} 1/a(2n-1) = 1/phi = (sqrt(5) - 1)/2. - Franz Vrabec, Sep 15 2005
Sum_{n>=1} 1/a(2n) = (3 - sqrt(5))/2. - Franz Vrabec, Nov 30 2009
a(n) = ((7+3*sqrt(5))/10)*((3+sqrt(5))/2)^(n-1) + ((7-3*sqrt(5))/10)*((3-sqrt(5))/2)^(n-1) + (3/5)*(-1)^(n-1). - Tim Monahan, Aug 09 2011
a(n) = (Lucas(n+1)^2 - Fibonacci(n+1)^2)/4. - Vincenzo Librandi, Aug 02 2014
a(n) = F(n-2)*F(n) + F(n-1)*F(n) + F(n-2)*F(n+1) + F(n-1)*F(n+1), where F=A000045, F(-2)=-1, F(-1)=1. - Bruno Berselli, Nov 03 2015
a(n) = A035513(1,n-1)*A035513(3,n-1)/2 = A035513(1,n-1)*A035513(4,n-1)/3. - R. J. Mathar, Sep 04 2016
a(n)+a(n+1) = A001519(n+2). - R. J. Mathar, Oct 19 2021
a(n) = 2*A001654(n) - A001654(n-1). - R. J. Mathar, Oct 19 2021
a(n)+a(n+3) = 2*F(2n+5) = A126358(n+2). - Andrés Ventas, Oct 25 2021
Sum_{n>=1} Fibonacci(n+1)/a(n) = 2. - Amiram Eldar, Jan 11 2022
a(n) = a(-2-n) and a(n) + a(n+3) = 2*(a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Mar 18 2022

A080097 a(n) = Fibonacci(n+2)^2 - 1.

Original entry on oeis.org

0, 3, 8, 24, 63, 168, 440, 1155, 3024, 7920, 20735, 54288, 142128, 372099, 974168, 2550408, 6677055, 17480760, 45765224, 119814915, 313679520, 821223648, 2149991423, 5628750624, 14736260448, 38580030723, 101003831720

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Jan 29 2003

a(n), a(n)+1 and a(n)+2 are consecutive members of A049997.

G. C. Greubel, Table of n, a(n) for n = 0..1000
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.
Index entries for linear recurrences with constant coefficients, signature (3,0,-3,1).

Equals A007598(n+2) - 1.

Cf. A000032, A000045, A001622, A049997, A059840, A064831, A080144.

List([0..30], n-> Fibonacci(n+2)^2 -1); # G. C. Greubel, Jul 23 2019

[Fibonacci(n+2)^2 -1: n in [0..30]]; // G. C. Greubel, Jul 23 2019

CoefficientList[Series[(3x+2x^2-x^3)/((1-x^2)(1-2x-2x^2+x^3)), {x, 0, 30}], x]
Table[Fibonacci[n+2]^2-1,{n,0,30}] (* Vladimir Joseph Stephan Orlovsky, Apr 03 2011 *)
LinearRecurrence[{3,0,-3,1},{0,3,8,24},40] (* Harvey P. Dale, Nov 23 2024 *)

A080097(n):=fib(n+2)^2-1$ makelist(A080097(n),n,0,30); /* Martin Ettl, Nov 13 2012 */

a(n)=fibonacci(n+2)^2-1 \\ Charles R Greathouse IV, Feb 06 2013

[fibonacci(n+2)^2 -1 for n in (0..30)] # G. C. Greubel, Jul 23 2019

If n is odd, then a(n) = F(n+1)*F(n+3) = F(n)*F(n+4) - 2, else a(n) = F(n)*F(n+4) = F(n+1)*F(n+3) - 2, where F(n) = Fibonacci numbers (A000045).
a(n) = (Lucas(2*n+4) - 2*(-1)^n - 5)/5.
O.g.f.: x*(3-x)/((1-x^2)*(1-3*x+x^2)) (see a comment on A080144). - Wolfdieter Lang, Jul 30 2012
a(n) = Sum_{k=1..n} F(k+3)*F(k) = A027941(n) + 2*A001654(n), n>=0. - Wolfdieter Lang, Jul 27 2012
Sum_{n>=1} 1/a(n) = (43 - 15*sqrt(5))/18 = 29/9 - 5*phi/3, where phi is the golden ratio (A001622). - Amiram Eldar, Oct 20 2020
a(n) = 3*a(n-1) - 3*a(n-3) + a(n-4). - Joerg Arndt, Nov 13 2023

Edited by Ralf Stephan, May 15 2005

A080144 a(n) = F(4)F(n)F(n+1) + F(5)F(n+1)^2 if n odd, a(n) = F(4)F(n)F(n+1) + F(5)F(n+1)^2 - F(5) if n even, where F(n) is the n-th Fibonacci number (A000045).

Original entry on oeis.org

0, 8, 21, 63, 165, 440, 1152, 3024, 7917, 20735, 54285, 142128, 372096, 974168, 2550405, 6677055, 17480757, 45765224, 119814912, 313679520, 821223645, 2149991423, 5628750621, 14736260448, 38580030720, 101003831720

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Jan 31 2003

The present sequence is a member of the m-family of sums b(m,n) := Sum_{k=1..n} F(k+m)*F(k) for m >= 0, n >= 0 (see the B. Cloitre formula given below (m=5)). These sums are (F(m)*A027941(n) + L(m)*A001654(n))/2, with F = A000045 and the L = A000032. Proof by induction on m using the recurrence. - Wolfdieter Lang, Jul 27 2012
The o.g.f. of b(m,n) is A(m,x) = x*(F(m+1) - F(m-1)*x)/((1-x^2)*(1-3*x+x^2)), m >= 0, with F(-1)=1. - Wolfdieter Lang, Jul 30 2012
b(m,n) = ((-1)^(n+1)*L(m) - 5*F(m) + 2*L(m + 2*n + 1))/10. - Ehren Metcalfe, Aug 21 2017

G. C. Greubel, Table of n, a(n) for n = 0..1000
S. 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.
Index entries for linear recurrences with constant coefficients, signature (3, 0, -3, 1).

Cf. A000032, A000045, A059840, A064831, A080143.

F:=Fibonacci;; List([0..30], n-> (2*F(n+3)^2 -5-3*(-1)^n)/2); # G. C. Greubel, Jul 23 2019

F:=Fibonacci; [(2*F(n+3)^2 -5-3*(-1)^n)/2: n in [0..30]]; // G. C. Greubel, Jul 23 2019

CoefficientList[Series[x*(8+5*x-3*x^2)/((1-x^2)*(1-2x-2x^2+x^3)), {x, 0, 30}], x]
With[{F=Fibonacci}, Table[(2*F[n + 3]^2 -5-3*(-1)^n)/2, {n,0,30}]] (* G. C. Greubel, Jul 23 2019 *)

my(x='x+O('x^30)); concat([0],Vec(x*(8-3*x)/((1-x^2)*(1-3*x+x^2)) )) \\ G. C. Greubel, Mar 04 2017

vector(30, n, n--; f=fibonacci; (2*f(n+3)^2 -5-3*(-1)^n)/2) \\ G. C. Greubel, Jul 23 2019

f=fibonacci; [(2*f(n+3)^2 -5-3*(-1)^n)/2 for n in (0..30)] # G. C. Greubel, Jul 23 2019

G.f.: x*(8-3*x)/((1-x^2)*(1-3*x+x^2)) (see the comment section). - Wolfdieter Lang, Jul 30 2012
a(n) = Sum_{i=0..n} A000045(i+5)*A000045(i). - Benoit Cloitre, Jun 14 2004
a(n) = (5*A027941(n) + 11*A001654(n))/2, n >= 0. See A080143 and A080097. See the comment section for the general formula for such sums. - Wolfdieter Lang, Jul 27 2012
a(n) = (2*Lucas(2*n + 6) + 11*(-1)^(n+1) - 25)/10. - Ehren Metcalfe, Aug 21 2017
a(n) = (2*Fibonacci(n+3)^2 - 5 - 3*(-1)^n)/2. - G. C. Greubel, Jul 23 2019

A081018 a(n) = (Lucas(4n+1)-1)/5, or Fibonacci(2n)*Fibonacci(2n+1), or A081017(n)/5.

Original entry on oeis.org

0, 2, 15, 104, 714, 4895, 33552, 229970, 1576239, 10803704, 74049690, 507544127, 3478759200, 23843770274, 163427632719, 1120149658760, 7677619978602, 52623190191455, 360684711361584, 2472169789339634, 16944503814015855, 116139356908771352, 796030994547383610

Offset: 0

R. K. Guy, Mar 01 2003

Another interpretation of this sequence is: nonnegative integers k such that (k + 1)^2 + (2k)^2 is a perfect square. So apart from a(0) = 0, a(n) + 1 and 2a(n) form the legs of a Pythagorean triple. - Nick Hobson, Jan 13 2007
Also solution y of Diophantine equation x^2 + 4*y^2 = k^2 for which x=y+1. - Carmine Suriano, Jun 23 2010
Also the index of the first of two consecutive heptagonal numbers whose sum is equal to the sum of two consecutive triangular numbers. - Colin Barker, Dec 20 2014
Nonnegative integers k such that G(x) = k for some rational number x where G(x) = x/(1-x-x^2) is the generating function of the Fibonacci numbers. - Tom Edgar, Aug 24 2015
The integer solutions of the equation a(b+1) = (a-b)(a-b-1) or, equivalently, binomial(a, b) = binomial(a-1, b+1) are given by (a, b) = (a(n+1), A003482(n)=Fibonacci(2*n) * Fibonacci(2*n+3)) (Lind and Singmaster). - Tomohiro Yamada, May 30 2018

			G.f. = 2*x + 15*x^2 + 104*x^3 + 714*x^4 + 4895*x^5 + 33552*x^6 + ...

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 28.
Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.

Colin Barker, Table of n, a(n) for n = 0..1000
Pridon Davlianidze, Problem B-1279, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 4 (2020), p. 368; An Unusual Generalization, Solution to Problem B-1279 by J. N. Senadheera, ibid., Vol. 59, No. 4 (2021), pp. 370-371.
Dae S. Hong, When is the Generating Function of the Fibonacci Numbers an Integer?, The College Mathematics Journal, Vol. 46, No. 2 (March 2015), pp. 110-112
D. A. Lind, The quadratic field Q(sqrt(5)) and a certain diophantine equation, Fibonacci Quart., Vol. 6, No. 3 (1968), pp. 86-93.
David Singmaster, Repeated binomial coefficients and Fibonacci numbers, Fibonacci Quart., Vol. 13, No. 4 (1973), pp. 295-298.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers), A081017.
Partial sums of A033891. Bisection of A001654 and A059840.
Equals A089508 + 1.
Cf. A001622, A081007, A081016, A178898.

List([0..30], n-> (Lucas(1,-1, 4*n+1)[2] -1)/5); # G. C. Greubel, Jul 14 2019

[(Lucas(4*n+1)-1)/5: n in [0..30]]; // Vincenzo Librandi, Aug 24 2015

luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 25 do printf(`%d,`,(luc(4*n+1)-1)/5) od: # James Sellers, Mar 03 2003

(LucasL[4*Range[0,30]+1]-1)/5 (* or *) LinearRecurrence[{8,-8,1}, {0,2,15}, 30] (* G. C. Greubel, Aug 24 2015, modified Jul 14 2019 *)

concat(0, Vec(x*(2-x)/((1-x)*(1-7*x+x^2)) + O(x^30))) \\ Colin Barker, Dec 20 2014

[(lucas_number2(4*n+1,1,-1) -1)/5 for n in (0..30)] # G. C. Greubel, Jul 14 2019

a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
a(n) = Fibonacci(3) + Fibonacci(7) + Fibonacci(11) + ... + Fibonacci(4n+3).
G.f.: x*(2-x)/((1-x)*(1-7*x+x^2)). - Colin Barker, Mar 30 2012
E.g.f.: (1/5)^(3/2)*((1+phi^2)*exp(phi^4*x) - (1 + (1/phi^2))*exp(x/phi^4) - sqrt(5)*exp(x)), where 2*phi = 1 + sqrt(5). - G. C. Greubel, Aug 24 2015
From - Michael Somos, Aug 27 2015: (Start)
a(n) = -A081016(-1-n) for all n in Z.
0 = a(n) - 7*a(n+1) + a(n+2) - 1 for all n in Z.
0 = a(n)*a(n+2) - a(n+1)^2 + a(n+1) + 2 for all n in Z.
0 = a(n)*(a(n) -7*a(n+1) -1) + a(n+1)*(a(n+1) - 1) - 2 for all n in Z. (End)
a(n) = (k(n) + sqrt(k(n)*(4 + 5*k(n))))/2, where k(n) = A049684(n). - Stefano Spezia, Mar 11 2021
Product_{n>=1} (1 + 1/a(n)) = phi (A001622) (Davlianidze, 2020). - Amiram Eldar, Nov 30 2021

More terms from James Sellers, Mar 03 2003

A097083 Positive values of k such that there is exactly one permutation p of (1,2,3,...,k) such that i+p(i) is a Fibonacci number for 1<=i<=k.

Original entry on oeis.org

1, 2, 3, 5, 9, 15, 24, 39, 64, 104, 168, 272, 441, 714, 1155, 1869, 3025, 4895, 7920, 12815, 20736, 33552, 54288, 87840, 142129, 229970, 372099, 602069, 974169, 1576239, 2550408, 4126647, 6677056, 10803704, 17480760, 28284464, 45765225

Offset: 1

John W. Layman, Jul 23 2004

nonn

Numbers k such that A097082(k) = 1. If f is a Fibonacci number and k < f <= 2k, then a permutation for f-k-1 may be extended to a permutation for k, with p(i) = f-i for f-k < i <= k. This explains the sparseness of this sequence. - David Wasserman, Dec 19 2007

If the formula is correct, the bisections give A059840 and A064831. - David Wasserman, Dec 19 2007

G. C. Greubel, Table of n, a(n) for n = 1..1000
Andreas M. Hinz and Paul K. Stockmeyer, Precious Metal Sequences and Sierpinski-Type Graphs, J. Integer Seq., Vol 25 (2022), Article 22.4.8.
Index entries for linear recurrences with constant coefficients, signature (2,-1,1,0,-1).

Cf. A073364, A000045.

a=b=c=d=0;Table[e=a+b+d+1;a=b;b=c;c=d;d=e,{n,100}] (* Vladimir Joseph Stephan Orlovsky, Feb 26 2011 *)
CoefficientList[Series[x/((x - 1)*(x^2 + 1)*(x^2 + x - 1)), {x,0,50}], x] (* G. C. Greubel, Mar 05 2017 *)
LinearRecurrence[{2,-1,1,0,-1},{1,2,3,5,9},50] (* Harvey P. Dale, Nov 09 2024 *)

x='x+O('x^50); Vec(x/((x - 1)*(x^2 + 1)*(x^2 + x - 1))) \\ G. C. Greubel, Mar 05 2017

It appears that {a(n)} satisfies a(1)=1, a(2)=2 and, for n>2, a(n) = F(n+2) - a(n-2) - 1, where {F(k)} is the sequence of Fibonacci numbers, i.e, that the sequence is the partial sums of A006498.
If the partial sum assumption is correct: a(n) = floor(phi^(n+3)/5), where phi=(1+sqrt(5))/2 = A001622, and a(n) = a(n-1) + a(n-2) + ( (n*(n+1)/2) mod 2). - Gary Detlefs, Mar 12 2011
From R. J. Mathar, Mar 13 2011: (Start)
If the partial sum assumption is correct: a(n)= +2*a(n-1) -a(n-2) +a(n-3) -a(n-5).
G.f.: x/( (x-1)*(x^2+1)*(x^2+x-1) ).
a(n) = A000032(n+3)/5 -(-1)^n*A112030(n)/10 - 1/2. (End)
Conjecture: a(n) = floor(F(n+3)/sqrt(5)), where F(n) = A000045(n) are Fibonacci numbers. - Vladimir Reshetnikov, Nov 05 2015

a(9) from Ray Chandler, Jul 29 2004
More terms from David Wasserman, Dec 19 2007
Terms > 90000 assuming the partial sums formula by Vladimir Joseph Stephan Orlovsky, Feb 26 2011

A080143 a(n) = F(3)F(n)F(n+1) + F(4)F(n+1)^2 - F(4) if n even, F(3)F(n)F(n+1) + F(4)F(n+1)^2 if n odd, where F(n) is the n-th Fibonacci number (A000045).

Original entry on oeis.org

0, 5, 13, 39, 102, 272, 712, 1869, 4893, 12815, 33550, 87840, 229968, 602069, 1576237, 4126647, 10803702, 28284464, 74049688, 193864605, 507544125, 1328767775, 3478759198, 9107509824, 23843770272, 62423800997, 163427632717

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Jan 30 2003

G. C. Greubel, Table of n, a(n) for n = 0..1000
S. 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.
Index entries for linear recurrences with constant coefficients, signature (3,0,-3,1).

Cf. A064831, A059840, A080097, A080144.

F:=Fibonacci;; List([0..30], n-> (2*F(n+2)*F(n+3) -3 -(-1)^n)/2); # G. C. Greubel, Jul 23 2019

F:=Fibonacci; [(2*F(n+2)*F(n+3) -3 -(-1)^n)/2: n in [0..30]]; // G. C. Greubel, Jul 23 2019

CoefficientList[Series[x*(5+3*x-2*x^2)/((1-x^2)*(1-2*x-2*x^2+x^3)), {x, 0, 30}], x]
With[{F=Fibonacci}, Table[(2*F[n+2]*F[n+3] -3 -(-1)^n)/2, {n,0,30}]] (* G. C. Greubel, Jul 23 2019 *)

my(x='x+O('x^30)); concat([0], Vec(x*(5+3*x-2*x^2)/((1-x^2)*(1- 2*x-2*x^2+x^3)))) \\ G. C. Greubel, Mar 05 2017

vector(30, n, n--; f=fibonacci; (2*f(n+2)*f(n+3) -3 -(-1)^n)/2) \\ G. C. Greubel, Jul 23 2019

f=fibonacci; [(2*f(n+2)*f(n+3) -3 -(-1)^n)/2 for n in (0..30)] # G. C. Greubel, Jul 23 2019

G.f.: x*(5-2*x)/((1-x^2)*(1-3*x+x^2)), see a comment on A080144 for A(4,x). - Wolfdieter Lang, Jul 30 2012
a(n) = Sum_{i=0..n} ( A000045(i+4)*A000045(i) ). - Benoit Cloitre, Jun 14 2004
a(n) = (3*A027941(n) + 7*A001654(n))/2, n >= 0. Proof: from the preceding sum given by B. Cloitre via recurrence on the first factor under the sum. See also A080097 and A059840(n+2). - Wolfdieter Lang, Jul 27 2012
a(n) = (2*Lucas(2*n + 5) + 7*(-1)^(n+1) - 15)/10. - Ehren Metcalfe, Aug 21 2017
a(n) = (2*Fibonacci(n+2)*Fibonacci(n+3) - 3 - (-1)^n)/2. - G. C. Greubel, Jul 23 2019

A202503 Fibonacci self-fission matrix, by antidiagonals.

Original entry on oeis.org

1, 1, 2, 2, 3, 3, 3, 5, 5, 5, 5, 8, 9, 8, 8, 8, 13, 14, 15, 13, 13, 13, 21, 23, 24, 24, 21, 21, 21, 34, 37, 39, 39, 39, 34, 34, 34, 55, 60, 63, 64, 63, 63, 55, 55, 55, 89, 97, 102, 103, 104, 102, 102, 89, 89, 89, 144, 157, 165, 167, 168, 168, 165, 165, 144, 144, 144

Offset: 1

Clark Kimberling, Dec 20 2011

The Fibonacci self-fission matrix, F, is the fission P^^Q, where P and Q are the matrices given at A202502 and A202451.  See A193842 for the definition of fission.
antidiagonal sums:  (1, 3, 8, 18, 38, ...), A064831
diagonal (1,  5, 14,  39, ...), A119996
diagonal (2,  8, 23,  63, ...), A180664
diagonal (2,  5, 15,  39, ...), A059840
diagonal (3,  8, 24,  63, ...), A080097
diagonal (5, 13, 39, 102, ...), A080143
diagonal (8, 21, 63, 165, ...), A080144
All the principal submatrices are invertible, and the terms in the inverses are in {-3,-2,-1,0,1,2,3}.

			Northwest corner:
1....1....2....3....5.....8....13...21
2....3....5....8...13....21....34...55
3....5....9...14...23....37....60...97
5....8...15...24...39....63...102...165
8...13...24...39...64...103...167...270

Clark Kimberling, Fusion, Fission, and Factors, Fib. Q., 52 (2014), 195-202.

Cf. A000045, A202451, A202453, A202502.

n = 14;
Q = NestList[Most[Prepend[#, 0]] &, #, Length[#] - 1] &[Table[Fibonacci[k], {k, 1, n}]];
Qt = Transpose[Q]; P1 = Qt - IdentityMatrix[n];
P = P1[[Range[2, n], Range[1, n]]];
F = P.Q;
Flatten[Table[P[[i]][[k + 1 - i]], {k, 1, n - 1}, {i, 1, k}]] (* A202502 as a sequence *)
Flatten[Table[Q[[i]][[k + 1 - i]], {k, 1, n - 1}, {i, 1, k}]] (* A202451 as a sequence *)
Flatten[Table[F[[i]][[k + 1 - i]], {k, 1, n - 1}, {i, 1, k}]] (* A202503 as a sequence *)
TableForm[P]  (* A202502, modified lower triangular Fibonacci array *)
TableForm[Q]  (* A202451, upper tri. Fibonacci array *)
TableForm[F]  (* A202503, Fibonacci fission array *)

A193917 Triangular array: the self-fusion of (p(n,x)), where p(n,x)=sum{F(k+1)*x^(n-k) : 0<=k<=n}, where F=A000045 (Fibonacci numbers).

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 2, 3, 6, 9, 3, 5, 9, 15, 24, 5, 8, 15, 24, 40, 64, 8, 13, 24, 39, 64, 104, 168, 13, 21, 39, 63, 104, 168, 273, 441, 21, 34, 63, 102, 168, 272, 441, 714, 1155, 34, 55, 102, 165, 272, 440, 714, 1155, 1870, 3025, 55, 89, 165, 267, 440, 712, 1155

Offset: 0

Clark Kimberling, Aug 09 2011

See A193722 for the definition of fusion of two sequences of polynomials or triangular arrays.  (Fusion is defined at A193822; fission, at A193742; see A202503 and A202453 for infinite-matrix representations of fusion and fission.)
First five rows of P (triangle of coefficients of polynomials p(n,x)):
1
1...1
1...1...2
1...1...2...3
1...1...2...3...5
First eight rows of A193917:
1
1...1
1...2...3
2...3...6...9
3...5...9...15...24
5...8...15..24...40...64
8...13..24..39...64...104..168
13..21..39..63...104..168..273..441
...
col 1:  A000045
col 2:  A000045
col 3:  A022086
col 4:  A022086
col 5:  A022091
col 6:  A022091
col 7:  A022355
col 8:  A022355
right edge, w(n,n):  A064831
w(n,n-1):  A001654
w(n,n-2):  A064831
w(n,n-3):  A059840
w(n,n-4):  A080097
w(n,n-5):  A080143
w(n,n-6):  A080144
Suppose n is an even positive integer and w(n+1,x) is the polynomial matched to row n+1 of A193917 as in the Mathematica program (and definition of fusion at A193722), where the first row is counted as row 0.

			First six rows:
1
1...1
1...2...3
2...3...6....9
3...5...9....15...24
5...8...15...24...40...64

Cf. A193722, A064831, A193918, A194000, A194001.

z = 12;
p[n_, x_] := Sum[Fibonacci[k + 1]*x^(n - k), {k, 0, n}];
q[n_, x_] := p[n, x];
t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, x] /. x -> 0;
w[n_, x_] := Sum[t[n, k]*q[n + 1 - k, x], {k, 0, n}]; w[-1, x_] := 1
g[n_] := CoefficientList[w[n, x], {x}]
TableForm[Table[Reverse[g[n]], {n, -1, z}]]
Flatten[Table[Reverse[g[n]], {n, -1, z}]]  (* A193917 *)
TableForm[Table[g[n], {n, -1, z}]]
Flatten[Table[g[n], {n, -1, z}]]  (* A193918 *)

Showing 1-10 of 14 results. Next

cp's OEIS Frontend

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

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A202453 Fibonacci self-fusion matrix, by antidiagonals.

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A059929 a(n) = Fibonacci(n)*Fibonacci(n+2).

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A080097 a(n) = Fibonacci(n+2)^2 - 1.

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A080144 a(n) = F(4)*F(n)*F(n+1) + F(5)*F(n+1)^2 if n odd, a(n) = F(4)*F(n)*F(n+1) + F(5)*F(n+1)^2 - F(5) if n even, where F(n) is the n-th Fibonacci number (A000045).

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A080144 a(n) = F(4)F(n)F(n+1) + F(5)F(n+1)^2 if n odd, a(n) = F(4)F(n)F(n+1) + F(5)F(n+1)^2 - F(5) if n even, where F(n) is the n-th Fibonacci number (A000045).

A080143 a(n) = F(3)F(n)F(n+1) + F(4)F(n+1)^2 - F(4) if n even, F(3)F(n)F(n+1) + F(4)F(n+1)^2 if n odd, where F(n) is the n-th Fibonacci number (A000045).