A064170 -id:A064170 - cp's OEIS frontend

A001622 Decimal expansion of golden ratio phi (or tau) = (1 + sqrt(5))/2.

1, 6, 1, 8, 0, 3, 3, 9, 8, 8, 7, 4, 9, 8, 9, 4, 8, 4, 8, 2, 0, 4, 5, 8, 6, 8, 3, 4, 3, 6, 5, 6, 3, 8, 1, 1, 7, 7, 2, 0, 3, 0, 9, 1, 7, 9, 8, 0, 5, 7, 6, 2, 8, 6, 2, 1, 3, 5, 4, 4, 8, 6, 2, 2, 7, 0, 5, 2, 6, 0, 4, 6, 2, 8, 1, 8, 9, 0, 2, 4, 4, 9, 7, 0, 7, 2, 0, 7, 2, 0, 4, 1, 8, 9, 3, 9, 1, 1, 3, 7, 4, 8, 4, 7, 5

Offset: 1

N. J. A. Sloane

Also decimal expansion of the positive root of (x+1)^n - x^(2n). (x+1)^n - x^(2n) = 0 has only two real roots x1 = -(sqrt(5)-1)/2 and x2 = (sqrt(5)+1)/2 for all n > 0. - Cino Hilliard, May 27 2004
The golden ratio phi is the most irrational among irrational numbers; its successive continued fraction convergents F(n+1)/F(n) are the slowest to approximate to its actual value (I. Stewart, in "Nature's Numbers", Basic Books, 1997). - Lekraj Beedassy, Jan 21 2005
Let t=golden ratio. The lesser sqrt(5)-contraction rectangle has shape t-1, and the greater sqrt(5)-contraction rectangle has shape t. For definitions of shape and contraction rectangles, see A188739. - Clark Kimberling, Apr 16 2011
The golden ratio (often denoted by phi or tau) is the shape (i.e., length/width) of the golden rectangle, which has the special property that removal of a square from one end leaves a rectangle of the same shape as the original rectangle. Analogously, removals of certain isosceles triangles characterize side-golden and angle-golden triangles. Repeated removals in these configurations result in infinite partitions of golden rectangles and triangles into squares or isosceles triangles so as to match the continued fraction, [1,1,1,1,1,...] of tau. For the special shape of rectangle which partitions into golden rectangles so as to match the continued fraction [tau, tau, tau, ...], see A188635. For other rectangular shapes which depend on tau, see A189970, A190177, A190179, A180182. For triangular shapes which depend on tau, see A152149 and A188594; for tetrahedral, see A178988. - Clark Kimberling, May 06 2011
Given a pentagon ABCDE, 1/(phi)^2 <= (A*C^2 + C*E^2 + E*B^2 + B*D^2 + D*A^2) / (A*B^2 + B*C^2 + C*D^2 + D*E^2 + E*A^2) <= (phi)^2. - Seiichi Kirikami, Aug 18 2011
If a triangle has sides whose lengths form a geometric progression in the ratio of 1:r:r^2 then the triangle inequality condition requires that r be in the range 1/phi < r < phi. - Frank M Jackson, Oct 12 2011
The graphs of x-y=1 and x*y=1 meet at (tau,1/tau). - Clark Kimberling, Oct 19 2011
Also decimal expansion of the first root of x^sqrt(x+1) = sqrt(x+1)^x. - Michel Lagneau, Dec 02 2011
Also decimal expansion of the root of (1/x)^(1/sqrt(x+1)) = (1/sqrt(x+1))^(1/x). - Michel Lagneau, Apr 17 2012
This is the case n=5 of (Gamma(1/n)/Gamma(3/n))*(Gamma((n-1)/n)/Gamma((n-3)/n)): (1+sqrt(5))/2 = (Gamma(1/5)/Gamma(3/5))*(Gamma(4/5)/Gamma(2/5)). - Bruno Berselli, Dec 14 2012
Also decimal expansion of the only number x>1 such that (x^x)^(x^x) = (x^(x^x))^x = x^((x^x)^x). - Jaroslav Krizek, Feb 01 2014
For n >= 1, round(phi^prime(n)) == 1 (mod prime(n)) and, for n >= 3, round(phi^prime(n)) == 1 (mod 2*prime(n)). - Vladimir Shevelev, Mar 21 2014
The continuous radical sqrt(1+sqrt(1+sqrt(1+...))) tends to phi. - Giovanni Zedda, Jun 22 2019
Equals sqrt(2+sqrt(2-sqrt(2+sqrt(2-...)))). - Diego Rattaggi, Apr 17 2021
Given any complex p such that real(p) > -1, phi is the only real solution of the equation z^p+z^(p+1)=z^(p+2), and the only attractor of the complex mapping z->M(z,p), where M(z,p)=(z^p+z^(p+1))^(1/(p+2)), convergent from any complex plane point. - Stanislav Sykora, Oct 14 2021
The only positive number such that its decimal part, its integral part and the number itself (x-[x], [x] and x) form a geometric progression is phi, with respectively (phi -1, 1, phi) and a ratio = phi. This is the answer to the 4th problem of the 7th Canadian Mathematical Olympiad in 1975 (see IMO link and Doob reference). - Bernard Schott, Dec 08 2021
The golden ratio is the unique number x such that f(n*x)*c(n/x) - f(n/x)*c(n*x) = n for all n >= 1, where f = floor and c = ceiling. - Clark Kimberling, Jan 04 2022
In The Second Scientific American Book Of Mathematical Puzzles and Diversions, Martin Gardner wrote that, by 1910, Mark Barr (1871-1950) gave phi as a symbol for the golden ratio. - Bernard Schott, May 01 2022
Phi is the length of the equal legs of an isosceles triangle with side c = phi^2, and internal angles (A,B) = 36 degrees, C = 108 degrees. - Gary W. Adamson, Jun 20 2022
The positive solution to x^2 - x - 1 = 0. - Michal Paulovic, Jan 16 2023
The minimal polynomial of phi^n, for nonvanishing integer n, is P(n, x) = x^2 - L(n)*x + (-1)^n, with the Lucas numbers L = A000032, extended to negative arguments with L(n) = (-1)^n*L(n). P(0, x) = (x - 1)^2 is not minimal. - Wolfdieter Lang, Feb 20 2025
This is the largest real zero x of (x^4 + x^2 + 1)^2 = 2*(x^8 + x^4 + 1). - Thomas Ordowski, May 14 2025

			1.6180339887498948482045868343656381177203091798057628621...

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 24, 112, 123, 184, 190, 203.
Michael Doob, The Canadian Mathematical Olympiad & L'Olympiade Mathématique du Canada 1969-1993 - Canadian Mathematical Society & Société Mathématique du Canada, Problem 4, 1975, pages 76-77, 1993.
Richard A. Dunlap, The Golden Ratio and Fibonacci Numbers, World Scientific, River Edge, NJ, 1997.
Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, Vol. 94, Cambridge University Press, 2003, Section 1.2.
Martin Gardner, The Second Scientific American Book Of Mathematical Puzzles and Diversions, "Phi: The Golden Ratio", Chapter 8, Simon & Schuster, NY, 1961.
Martin Gardner, Weird Water and Fuzzy Logic: More Notes of a Fringe Watcher, "The Cult of the Golden Ratio", Chapter 9, Prometheus Books, 1996, pages 90-97.
Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.5 The Fibonacci and Related Sequences, p. 287.
H. E. Huntley, The Divine Proportion, Dover, NY, 1970.
Mario Livio, The Golden Ratio, Broadway Books, NY, 2002. [see the review by G. Markowsky in the links field]
Gary B. Meisner, The Golden Ratio: The Divine Beauty of Mathematics, Race Point Publishing (The Quarto Group), 2018. German translation: Der Goldene Schnitt, Librero, 2023.
Scott Olsen, The Golden Section, Walker & Co., NY, 2006.
Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 137-139.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Hans Walser, The Golden Section, Math. Assoc. of Amer. Washington DC 2001.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See pp. 36-40.
Claude-Jacques Willard, Le nombre d'or, Magnard, Paris, 1987.

Robert G. Wilson v, Table of n, a(n) for n = 1..100000
Mohammad K. Azarian, Problem 123, Missouri Journal of Mathematical Sciences, Vol. 10, No. 3 (Fall 1998), p. 176; Solution, ibid., Vol. 12, No. 1 (Winter 2000), pp. 61-62.
John Baez, This week's finds in mathematical physics, Week 203.
John Baez, The Rankin Lectures 2008, My Favorite Numbers: 5. [video]
Murray Berg, Phi, the golden ratio (to 4599 decimal places) and Fibonacci numbers, Fib. Quart., Vol. 4, No. 2 (1961), pp. 157-162.
Ömür Deveci, Zafer Adıgüzel and Taha Doğan, On the Generalized Fibonacci-circulant-Hurwitz numbers, Notes on Number Theory and Discrete Mathematics (2020) Vol. 26, No. 1, 179-190.
T. Eveilleau, Le nombre d'or (in French).
Abdul Gaffar, Anand B. Joshi, Sonali Singh, and Keerti Srivastava, A high capacity multi-image steganography technique based on golden ratio and non-subsampled contourlet transform, Multimedia Tools and Applications (2022).
Gutenberg Project, The golden ratio to 20000 places.
ICON Project, The golden ratio to 50000 places.
The IMO Compendium, Problem 4, 7th Canadian Mathematical Olympiad 1975.
L. B. W. Jolley, Summation of Series, Dover, 1961
Franklin H. J. Kenter, It's good to be phi: a solution to a problem of Gosper and Knuth, arXiv:1712.04856 [math.HO], 2017.
Clark Kimberling, Two kinds of golden triangles, generalized to match continued fractions, Journal for Geometry and Graphics, Vol. 11 (2007), pp. 165-171.
Clark Kimberling, Lucas Representations of Positive Integers, J. Int. Seq., Vol. 23 (2020), Article 20.9.5.
Ron Knott, Fibonacci numbers and the golden section.
Wolfdieter Lang, A list of representative simple difference sets of the Singer type for small orders m, Karlsruher Institut für Technologie (Karlsruhe, Germany 2020).
Wolfdieter Lang, Cantor's List of Real Algebraic Numbers of Heights 1 to 7, arXiv:2307.10645 [math.NT], 2023.
Simon Litsyn and Vladimir Shevelev, Irrational Factors Satisfying the Little Fermat Theorem, International Journal of Number Theory, Vol. 1, No. 4 (2005), pp. 499-512.
Gary B. Meisner, Phi, The Golden Number.
George Markowsky, Misconceptions About the Golden Ratio, College Mathematics Journal, 23:1 (January 1992), 2-19.
George Markowsky, Book review: The Golden Ratio, Notices of the AMS, 52:3 (March 2005), 344-347.
R. S. Melham and A. G. Shannon, Inverse Trigonometric Hyperbolic Summation Formulas Involving Generalized Fibonacci Numbers, The Fibonacci Quarterly, Vol. 33, No. 1 (1995), pp. 32-40.
Jean-Christophe Michel, Le nombre d'or.
J. J. O'Connor and E. F. Robertson, The Golden ratio.
Hugo Pfoertner, 1 million digits of phi, Computed using A. J. Yee's y-cruncher.
Simon Plouffe, Plouffe's Inverter, The golden ratio to 10 million digits. [Only announcement, file truncated]
Simon Plouffe, The golden ratio:(1+sqrt(5))/2 to 20000 places.
Fred Richman, Fibonacci sequence with multiprecision Java, Successive approximations to phi from ratios of consecutive Fibonacci numbers.
Herman P. Robinson, The CSR Function, Popular Computing (Calabasas, CA), Vol. 4, No. 35 (Feb 1976), pages PC35-3 to PC35-4. Annotated and scanned copy.
E. F. Schubert, The Fibonacci series.
Vladimir Shevelev, A property of n-bonacci constant, Seqfan (Mar 23 2014).
Jonathan Sondow, Evaluation of Tachiya's algebraic infinite products involving Fibonacci and Lucas numbers, Diophantine Analysis and Related Fields 2011 - AIP Conference Proceedings, Vol. 1385, pp. 97-100; arXiv:1106.4246 [math.NT], 2011.
Matthew R. Watkins, The "Golden Mean" in number theory.
Eric Weisstein's World of Mathematics, Golden Ratio.
Eric Weisstein's World of Mathematics, Fibonacci n-Step Number.
Eric Weisstein's World of Mathematics, Pisot Number.
Eric Weisstein's World of Mathematics, Silver Ratio.
Wikipedia, Mark Barr.
Wikipedia, Golden ratio.
Wikipedia, Kronecker-Weber theorem.
Wikipedia, Metallic mean.
Alexander J. Yee, y-cruncher - A Multi-Threaded Pi-Program.
Index entries for algebraic numbers, degree 2.
Index to sequences related to Olympiads.

Cf. A000012 (continued fraction coefficients), A000032, A000045, A006497, A080039, A104457, A188635, A192222, A192223, A145996, A139339, A197762, A002163, A094874, A134973.
Cf. A102208, A102769, A131595.
Cf. A302973, A303069, A304022.

Digits:=1000; evalf((1+sqrt(5))/2); # Wesley Ivan Hurt, Nov 01 2013

RealDigits[(1 + Sqrt[5])/2, 10, 130] (* Stefan Steinerberger, Apr 02 2006 *)
RealDigits[ Exp[ ArcSinh[1/2]], 10, 111][[1]] (* Robert G. Wilson v, Mar 01 2008 *)
RealDigits[GoldenRatio,10,120][[1]] (* Harvey P. Dale, Oct 28 2015 *)

default(realprecision, 20080); x=(1+sqrt(5))/2; for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b001622.txt", n, " ", d));  \\ Harry J. Smith, Apr 19 2009

/* Digit-by-digit method: write it as 0.5+sqrt(1.25) and start at hundredths digit */
r=11; x=400; print(1); print(6);
for(dig=1, 110, {d=0; while((20*r+d)*d <= x, d++);
d--; /* while loop overshoots correct digit */
print(d); x=100*(x-(20*r+d)*d); r=10*r+d})
\\ Michael B. Porter, Oct 24 2009

a(n) = floor(10^(n-1)*(quadgen(5))%10);
alist(len) = digits(floor(quadgen(5)*10^(len-1))); \\ Chittaranjan Pardeshi, Jun 22 2022

from sympy import S
def alst(n): # truncate extra last digit to avoid rounding
  return list(map(int, str(S.GoldenRatio.n(n+1)).replace(".", "")))[:-1]
print(alst(105)) # Michael S. Branicky, Jan 06 2021

Equals Sum_{n>=2} 1/A064170(n) = 1/1 + 1/2 + 1/(2*5) + 1/(5*13) + 1/(13*34) + ... - Gary W. Adamson, Dec 15 2007
Equals Hypergeometric2F1([1/5, 4/5], [1/2], 3/4) = 2*cos((3/5)*arcsin(sqrt(3/4))). - Artur Jasinski, Oct 26 2008
From Hieronymus Fischer, Jan 02 2009: (Start)
The fractional part of phi^n equals phi^(-n), if n is odd. For even n, the fractional part of phi^n is equal to 1-phi^(-n).
General formula: Provided x>1 satisfies x-x^(-1)=floor(x), where x=phi for this sequence, then:
for odd n: x^n - x^(-n) = floor(x^n), hence fract(x^n) = x^(-n),
for even n: x^n + x^(-n) = ceiling(x^n), hence fract(x^n) = 1 - x^(-n),
for all n>0: x^n + (-x)^(-n) = round(x^n).
x=phi is the minimal solution to x - x^(-1) = floor(x) (where floor(x)=1 in this case).
Other examples of constants x satisfying the relation x - x^(-1) = floor(x) include A014176 (the silver ratio: where floor(x)=2) and A098316 (the "bronze" ratio: where floor(x)=3). (End)
Equals 2*cos(Pi/5) = e^(i*Pi/5) + e^(-i*Pi/5). - Eric Desbiaux, Mar 19 2010
The solutions to x-x^(-1)=floor(x) are determined by x=(1/2)*(m+sqrt(m^2+4)), m>=1; x=phi for m=1. In terms of continued fractions the solutions can be described by x=[m;m,m,m,...], where m=1 for x=phi, and m=2 for the silver ratio A014176, and m=3 for the bronze ratio A098316. - Hieronymus Fischer, Oct 20 2010
Sum_{n>=1} x^n/n^2 = Pi^2/10 - (log(2)*sin(Pi/10))^2 where x = 2*sin(Pi/10) = this constant here. [Jolley, eq 360d]
phi = 1 + Sum_{k>=1} (-1)^(k-1)/(F(k)*F(k+1)), where F(n) is the n-th Fibonacci number (A000045). Proof. By Catalan's identity, F^2(n) - F(n-1)*F(n+1) = (-1)^(n-1). Therefore,(-1)^(n-1)/(F(n)*F(n+1)) = F(n)/F(n+1) - F(n-1)/F(n). Thus Sum_{k=1..n} (-1)^(k-1)/(F(k)*F(k+1)) = F(n)/F(n+1). If n goes to infinity, this tends to 1/phi = phi - 1. - Vladimir Shevelev, Feb 22 2013
phi^n = (A000032(n) + A000045(n)*sqrt(5)) / 2. - Thomas Ordowski, Jun 09 2013
Let P(q) = Product_{k>=1} (1 + q^(2*k-1)) (the g.f. of A000700), then A001622 = exp(Pi/6) * P(exp(-5*Pi)) / P(exp(-Pi)). - Stephen Beathard, Oct 06 2013
phi = i^(2/5) + i^(-2/5) = ((i^(4/5))+1) / (i^(2/5)) = 2*(i^(2/5) - (sin(Pi/5))i) = 2*(i^(-2/5) + (sin(Pi/5))i). - Jaroslav Krizek, Feb 03 2014
phi = sqrt(2/(3 - sqrt(5))) = sqrt(2)/A094883. This follows from the fact that ((1 + sqrt(5))^2)*(3 - sqrt(5)) = 8, so that ((1 + sqrt(5))/2)^2 = 2/(3 - sqrt(5)). - Geoffrey Caveney, Apr 19 2014
exp(arcsinh(cos(Pi/2-log(phi)*i))) = exp(arcsinh(sin(log(phi)*i))) = (sqrt(3) + i) / 2. - Geoffrey Caveney, Apr 23 2014
exp(arcsinh(cos(Pi/3))) = phi. - Geoffrey Caveney, Apr 23 2014
cos(Pi/3) + sqrt(1 + cos(Pi/3)^2). - Geoffrey Caveney, Apr 23 2014
2*phi = z^0 + z^1 - z^2 - z^3 + z^4, where z = exp(2*Pi*i/5). See the Wikipedia Kronecker-Weber theorem link. - Jonathan Sondow, Apr 24 2014
phi = 1/2 + sqrt(1 + (1/2)^2). - Geoffrey Caveney, Apr 25 2014
Phi is the limiting value of the iteration of x -> sqrt(1+x) on initial value a >= -1. - Chayim Lowen, Aug 30 2015
From Isaac Saffold, Feb 28 2018: (Start)
1 = Sum_{k=0..n} binomial(n, k) / phi^(n+k) for all nonnegative integers n.
1 = Sum_{n>=1} 1 / phi^(2n-1).
1 = Sum_{n>=2} 1 / phi^n.
phi = Sum_{n>=1} 1/phi^n. (End)
From Christian Katzmann, Mar 19 2018: (Start)
phi = Sum_{n>=0} (15*(2*n)! + 8*n!^2)/(2*n!^2*3^(2*n+2)).
phi = 1/2 + Sum_{n>=0} 5*(2*n)!/(2*n!^2*3^(2*n+1)). (End)
phi = Product_{k>=1} (1 + 2/(-1 + 2^k*(sqrt(4+(1-2/2^k)^2) + sqrt(4+(1-1/2^k)^2)))). - Gleb Koloskov, Jul 14 2021
Equals Product_{k>=1} (Fibonacci(3*k)^2 + (-1)^(k+1))/(Fibonacci(3*k)^2 + (-1)^k) (Melham and Shannon, 1995). - Amiram Eldar, Jan 15 2022
From Michal Paulovic, Jan 16 2023: (Start)
Equals the real part of 2 * e^(i * Pi / 5).
Equals 2 * sin(3 * Pi / 10) = 2*A019863.
Equals -2 * sin(37 * Pi / 10).
Equals 1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / ...)))).
Equals (2 + 3 * (2 + 3 * (2 + 3 * ...)^(1/4))^(1/4))^(1/4).
Equals (1 + 2 * (1 + 2 * (1 + 2 * ...)^(1/3))^(1/3))^(1/3).
Equals (1 + phi + (1 + phi + (1 + phi + ...)^(1/3))^(1/3))^(1/3).
Equals 13/8 + Sum_{k=0..oo} (-1)^(k+1)*(2*k+1)!/((k+2)!*k!*4^(2*k+3)).
(End)
phi^n = phi * A000045(n) + A000045(n-1). - Gary W. Adamson, Sep 09 2023
The previous formula holds for integer n, with F(-n) = (-1)^(n+1)*F(n), for n >= 0, with F(n) = A000045(n), for n >= 0. phi^n are integers in the quadratic number field Q(sqrt(5)). - Wolfdieter Lang, Sep 16 2023
Equals Product_{k>=0} ((5*k + 2)*(5*k + 3))/((5*k + 1)*(5*k + 4)). - Antonio Graciá Llorente, Feb 24 2024
From Antonio Graciá Llorente, Apr 21 2024: (Start)
Equals Product_{k>=1} phi^(-2^k) + 1, with phi = A001622.
Equals Product_{k>=0} ((5^(k+1) + 1)*(5^(k-1/2) + 1))/((5^k + 1)*(5^(k+1/2) + 1)).
Equals Product_{k>=1} 1 - (4*(-1)^k)/(10*k - 5 + (-1)^k) = Product_{k>=1} A047221(k)/A047209(k).
Equals Product_{k>=0} ((5*k + 7)*(5*k + 1 + (-1)^k))/((5*k + 1)*(5*k + 7 + (-1)^k)).
Equals Product_{k>=0} ((10*k + 3)*(10*k + 5)*(10*k + 8)^2)/((10*k + 2)*(10*k + 4)*(10*k + 9)^2).
Equals Product_{k>=5} 1 + 1/(Fibonacci(k) - (-1)^k).
Equals Product_{k>=2} 1 + 1/Fibonacci(2*k).
Equals Product_{k>=2} (Lucas(k)^2 + (-1)^k)/(Lucas(k)^2 - 4*(-1)^k). (End)

Additional links contributed by Lekraj Beedassy, Dec 23 2003
More terms from Gabriel Cunningham (gcasey(AT)mit.edu), Oct 24 2004
More terms from Stefan Steinerberger, Apr 02 2006
Broken URL to Project Gutenberg replaced by Georg Fischer, Jan 03 2009
Edited by M. F. Hasler, Feb 24 2014

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

Original entry on oeis.org

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

A049684 a(n) = Fibonacci(2n)^2.

Original entry on oeis.org

0, 1, 9, 64, 441, 3025, 20736, 142129, 974169, 6677056, 45765225, 313679521, 2149991424, 14736260449, 101003831721, 692290561600, 4745030099481, 32522920134769, 222915410843904, 1527884955772561, 10472279279564025, 71778070001175616, 491974210728665289

Offset: 0

Clark Kimberling

This is the r=9 member of the r-family of sequences S_r(n) defined in A092184 where more information can be found.

Apparently, this sequence consists of those nonnegative integers k for which x*(x^2-1)*y*(y^2-1) = k*(k^2-1) has a solution in nonnegative integers x, y. If k = a(n), x = A000045(2*n-1) and y = A000045(2*n+1) are a solution. See A374375 for numbers k*(k^2-1) that can be written as a product of two or more factors of the form x*(x^2-1). - Pontus von Brömssen, Jul 14 2024

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 27.
H. J. H. Tuenter, Fibonacci summation identities arising from Catalan's identity, Fib. Q., 60:4 (2022), 312-319.

Vincenzo Librandi, Table of n, a(n) for n = 0..500
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876.
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums II, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 42, 2012, pp. 2053-2059.
S. Barbero, U. Cerruti, and N. Murru, On polynomial solutions of the Diophantine equation (x + y - 1)^2 = wxy, Rendiconti Sem. Mat. Univ. Pol. Torino (2020) Vol. 78, No. 1, 5-12.
Pridon Davlianidze, Problem B-1264, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 1 (2020), p. 82; It's All About Catalan, Solution to Problem B-1264, ibid., Vol. 59, No. 1 (2021), pp. 87-88.
E. Kilic, Y. T. Ulutas, and N. Omur, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq. 14 (2011) #11.5.6, table 1, k=2.
R. Stephan, Boring proof of a nonlinearity
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

First differences give A033890.
First differences of A103434.
Bisection of A007598 and A064841.
a(n) = A064170(n+2) - 1 = (1/5) A081070.
Cf. A000032, A000045, A001622, A001906, A056854, A065563, A092184, A374375.

Join[{a=0, b=1}, Table[c=7*b-1*a+2; a=b; b=c, {n, 60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 18 2011 *)
Fibonacci[Range[0, 40, 2]]^2 (* Harvey P. Dale, Mar 22 2012 *)
Table[Fibonacci[n - 1] Fibonacci[n + 1] - 1, {n, 0, 40, 2}] (* Bruno Berselli, Feb 12 2015 *)
LinearRecurrence[{8, -8, 1},{0, 1, 9},21] (* Ray Chandler, Sep 23 2015 *)

numlib::fibonacci(2*n)^2 $ n = 0..35; // Zerinvary Lajos, May 13 2008

PARI
```
a(n)=fibonacci(2*n)^2
```

[fibonacci(2*n)^2 for n in range(0, 21)] # Zerinvary Lajos, May 15 2009

G.f.: (x+x^2) / ((1-x)*(1-7*x+x^2)).
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3) with n>2, a(0)=0, a(1)=1, a(2)=9.
a(n) = 7*a(n-1) - a(n-2) + 2 = A001906(n)^2.
a(n) = (A000032(4*n)-2)/5. [This is in Koshy's book (reference under A065563) on p. 88, attributed to Lucas 1876.] - Wolfdieter Lang, Aug 27 2012
a(n) = 1/5*(-2 + ( (7+sqrt(45))/2 )^n + ( (7-sqrt(45))/2 )^n). - Ralf Stephan, Apr 14 2004
a(n) = 2*(T(n, 7/2)-1)/5 with twice the Chebyshev polynomials of the first kind evaluated at x=7/2: 2*T(n, 7/2)= A056854(n). - Wolfdieter Lang, Oct 18 2004
a(n) = F(2*n-1)*F(2*n+1)-1, see A064170 - Bruno Berselli, Feb 12 2015
a(n) = Sum_{i=1..n} F(4*i-2) for n>0. - Bruno Berselli, Aug 25 2015
From Peter Bala, Nov 20 2019: (Start)
Sum_{n >= 1} 1/(a(n) + 1) = (sqrt(5) - 1)/2.
Sum_{n >= 1} 1/(a(n) + 4) = (3*sqrt(5) - 2)/16. More generally, it appears that
Sum_{n >= 1} 1/(a(n) + F(2*k+1)^2) = ((2*k+1)*F(2*k+1)*sqrt(5) - Lucas(2*k+1))/ (2*F(2*k+1)*F(4*k+2)) for k = 0,1,2,....
Sum_{n >= 2} 1/(a(n) - 1) = (8 - 3*sqrt(5))/9. (End)
E.g.f.: (1/5)*(-2*exp(x) + exp((16*x)/(1 + sqrt(5))^4) + exp((1/2)*(7 + 3*sqrt(5))*x)). - Stefano Spezia, Nov 23 2019
Product_{n>=2} (1 - 1/a(n)) = phi^2/3, where phi is the golden ratio (A001622) (Davlianidze, 2020). - Amiram Eldar, Dec 01 2021
a(n) = A092521(n-1)+A092521(n). - R. J. Mathar, Nov 22 2024

Better description and more terms from Michael Somos

A350922 a(0) = 2, a(1) = 5, and a(n) = 7*a(n-1) - a(n-2) - 4 for n >= 2.

Original entry on oeis.org

2, 5, 29, 194, 1325, 9077, 62210, 426389, 2922509, 20031170, 137295677, 941038565, 6449974274, 44208781349, 303011495165, 2076871684802, 14235090298445, 97568760404309, 668746232531714, 4583654867317685, 31416837838692077, 215334210003526850, 1475922632185995869

Offset: 0

Max Alekseyev, Jan 22 2022

One of 10 linear second-order recurrence sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4 and together forming A350916.
From William P. Orrick, Dec 20 2023: (Start)
Every term is a Markov number (see A002559) and, for n > 1, corresponds to a node of the Markov tree A368546 whose sibling and ancestors are all odd-indexed Fibonacci numbers. For n > 1, a(n) is the label of the node obtained from the root by going left n - 2 times and then right. Its Farey index, described in the comments to A368546, is 2 / (2*n - 1).
For instance, a(3) = 194 comes from going left once from the root node of the Markov tree and then right, which corresponds to the sequence of Markov numbers 5, 13, 194.  The corresponding sequence of Farey indices is 1/2, 1/3, 2/5. The sibling of the final node corresponds to Markov number 34 and Farey index 1/4. (End)

Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A049684, A064170, A350916, A002559, A000045.

Other sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4: A103974, A350917, A350919, A350920, A350921, A350923, A350924, A350925, A350926.

CoefficientList[Series[(2 - x)*(1 - 5*x)/((1 - x)*(1 - 7*x + x^2)), {x, 0, 22}],x] (* James C. McMahon, Dec 22 2023 *)

G.f.: (2 - x)*(1 - 5*x)/((1 - x)*(1 - 7*x + x^2)). - Stefano Spezia, Jan 22 2022
a(n) = 3*A049684(n) + 2 = 3*A064170(n+2) - 1. - Hugo Pfoertner, Jan 22 2022
a(n) = 3*A000045(2*n - 1) * A000045(2*n + 1) - 1 = A000045(2*n - 1)^2 + A000045(2*n + 1)^2. - William P. Orrick, Jan 08 2023

A337928 Numbers w such that (F(2n+1)^2, -F(2n)^2, -w) are primitive solutions of the Diophantine equation 2x^3 + 2y^3 + z^3 = 1, where F(n) is the n-th Fibonacci number (A000045).

Original entry on oeis.org

1, 5, 31, 209, 1429, 9791, 67105, 459941, 3152479, 21607409, 148099381, 1015088255, 6957518401, 47687540549, 326855265439, 2240299317521, 15355239957205, 105246380382911, 721369422723169, 4944339578679269, 33889007628031711, 232278713817542705

Offset: 0

XU Pingya, Sep 30 2020

			2*(F(5)^2)^3 + 2*(-F(4)^2)^3 + (-31)^3 = 2*(25)^3 + 2*(-9)^3 + (-31)^3 = 1, a(2) = 31.

Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045, A000578, A056573, A081018, A337929.

Table[(2*Fibonacci[2n+1]^6 - 2*Fibonacci[2n]^6 - 1)^(1/3), {n, 0, 21}]
Table[(Fibonacci[2n+1]*Fibonacci[2n+2]- Fibonacci[2n]^2), {n, 0, 21}] (* Wolfgang Berndt, May 26 2023 *)
LinearRecurrence[{8,-8,1},{1,5,31},30] (* Harvey P. Dale, Dec 17 2023 *)

Vec((1 - 3*x - x^2) / ((1 - x)*(1 - 7*x + x^2)) + O(x^20)) \\ Colin Barker, Oct 01 2020

a(n) = (2*F(2*n+1)^6 - 2*F(2*n)^6 - 1)^(1/3).
From Colin Barker, Oct 01 2020: (Start)
G.f.: (1 - 3*x - x^2) / ((1 - x)*(1 - 7*x + x^2)).
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3) for n>2.
(End)
a(n) = 2*A081018(n) + 1. - Hugo Pfoertner, Oct 01 2020
a(n) = A064170(n+2) + A033888(n). - Flávio V. Fernandes, Jan 10 2021
a(n) = F(2*n+1)*F(2*n+2) - F(2*n)^2. - Wolfgang Berndt, May 26 2023
a(2*n-1) = 5 + 6*Sum_{k=1..n-1} F(8*k+1), a(2*n) = 1 + 6*Sum_{k=1..n} F(8*k-3). - XU Pingya, Jun 09 2024

A337929 Numbers w such that (F(2n-1)^2, -F(2n)^2, w) are primitive solutions of the Diophantine equation 2x^3 + 2y^3 + z^3 = 1, where F(n) is the n-th Fibonacci number (A000045).

Original entry on oeis.org

1, 11, 79, 545, 3739, 25631, 175681, 1204139, 8253295, 56568929, 387729211, 2657535551, 18215019649, 124847601995, 855718194319, 5865179758241, 40200540113371, 275538601035359, 1888569667134145, 12944449068903659, 88722573815191471, 608113567637436641

Offset: 1

XU Pingya, Sep 30 2020

			2*(F(3)^2)^3 + 2*(-F(4)^2)^3 + 11^3 = 2*4^3 + 2*(-9)^3 + 11^3 = 1, 11 is a term.

Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045, A000578, A003482, A056573, A337928.

Table[(2*Fibonacci[2n]^6 - 2*Fibonacci[2n-1]^6 + 1)^(1/3), {n, 22}]
LinearRecurrence[{8,-8,1},{1,11,79},30] (* Harvey P. Dale, Aug 23 2021 *)

a(n) = (2*F(2*n)^6 - 2*F(2*n-1)^6 + 1)^(1/3).
From Colin Barker, Oct 01 2020: (Start)
G.f.: x*(1 + 3*x - x^2) / ((1 - x)*(1 - 7*x + x^2)).
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3) for n>3.
(End)
a(n) = 2*A003482(n) + 1. - Hugo Pfoertner, Oct 01 2020
a(n) = A033888(n) - A064170(n+2). - Flávio V. Fernandes, Jan 10 2021

A094568 Triangle of binary products of Fibonacci numbers.

Original entry on oeis.org

2, 3, 5, 8, 10, 13, 21, 24, 26, 34, 55, 63, 65, 68, 89, 144, 165, 168, 170, 178, 233, 377, 432, 440, 442, 445, 466, 610, 987, 1131, 1152, 1155, 1157, 1165, 1220, 1597, 2584, 2961, 3016, 3024, 3026, 3029, 3050, 3194, 4181, 6765, 7752, 7896, 7917, 7920, 7922

Offset: 1

Clark Kimberling, May 12 2004

Start with the triangle in A094566: starting with row 2, expel from each row the term that is a square of a Fibonacci number (A007598). The remaining triangle is this sequence.
In each row, the difference between neighboring terms is a Fibonacci number. For n>1, row n consists of n numbers, first F(2n) and last F(2n+1).
Central numbers: (2,10,65,442,...), essentially A064170.
Alternating row sums: 2,2,11,11,78,78,...; the sequence b=(2,11,78,...) is A094569.

			First four rows:
2
3 5
8 10 13
21 24 26 34

Clark Kimberling, Orderings of products of Fibonacci numbers, Fibonacci Quarterly 42:1 (2004), pp. 28-35.

Cf. A000045, A007598, A094565, A094566, A094569.

pef(k, n) = fibonacci(2*k)*fibonacci(2*n-2*k);
pof(k, n) = fibonacci(2*n-2*k+1)*fibonacci(2*k-1);
isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8)); \\ from A010056
isfib2(x) = issquare(x) && isfib(sqrtint(x));
tabl(nn) = {for (n=2, nn, if (n % 2 == 0, for (k=1, n/2, if (! isfib2(x = pef(k,n)), print1(x, ", "));); forstep (k=n/2, 1, -1, if (! isfib2(x = pof(k,n)), print1(x, ", "));), for (k=1, n\2, if (! isfib2(x = pef(k,n)), print1(x, ", "));); forstep (k=n\2+1, 1, -1, if (! isfib2(x = pof(k,n)), print1(x, ", ")););); print(););} \\ Michel Marcus, May 04 2016

A081076 a(n) = Lucas(4n) + 3, or 5Fibonacci(2n-1)Fibonacci(2n+1).

Original entry on oeis.org

5, 10, 50, 325, 2210, 15130, 103685, 710650, 4870850, 33385285, 228826130, 1568397610, 10749957125, 73681302250, 505019158610, 3461452808005, 23725150497410, 162614600673850, 1114577054219525, 7639424778862810

Offset: 0

R. K. Guy, Mar 04 2003

Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000032 (Lucas numbers), A000045 (Fibonacci numbers), A056854.

List([0..30], n-> Lucas(1, -1, 4*n)[2] +3 ); # G. C. Greubel, May 26 2020

[Lucas(4*n) +3: n in [0..30]]; // G. C. Greubel, May 26 2020

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 30 do printf(`%d,`,luc(4*n)+3) od: # James Sellers, Mar 05 2003

Table[LucasL[4 n] + 3, {n,0,30}] (* Wesley Ivan Hurt, Nov 20 2014 *)

Vec(-5*(2*x^2-6*x+1)/((x-1)*(x^2-7*x+1)) + O(x^30)) \\ Michel Marcus, Dec 23 2014

[lucas_number2(4*n,1,-1) + 3 for n in (0..30)] # G. C. Greubel, May 26 2020

a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
G.f.: 5*(1 -6*x +2*x^2)/((1-x)*(1-7*x+x^2)). - Colin Barker, Jun 22 2012
Sum_{n>=0} 1/a(n) = phi/5, where phi = A001622 = (1 + sqrt(5))/2. - Diego Rattaggi, Apr 22 2020
From G. C. Greubel, May 26 2020: (Start)
a(n) = 5*A064170(n+1).
a(n) = Lucas(n)^4 - 4*(-1)^n*Lucas(n)^2 + 5.
E.g.f.: 3*exp(x) + 2*exp(7*x/2)*cosh(3*sqrt(5)*x/2). (End)

More terms from James Sellers, Mar 05 2003

A262342 Area of Lewis Carroll's paradoxical F(2n+1) X F(2n+3) rectangle.

Original entry on oeis.org

10, 65, 442, 3026, 20737, 142130, 974170, 6677057, 45765226, 313679522, 2149991425, 14736260450, 101003831722, 692290561601, 4745030099482, 32522920134770, 222915410843905, 1527884955772562, 10472279279564026, 71778070001175617, 491974210728665290, 3372041405099481410

Offset: 1

Jonathan Sondow, Oct 16 2015

Warren Weaver (1938): "In a familiar geometrical paradox a square of area 8 X 8 = 64 square units is cut into four parts which may be refitted to form a rectangle of apparent area 5 X 13 = 65 square units.... Lewis Carroll generalized this paradox...."
Carroll cuts a F(2n+2) X F(2n+2) square into four parts, where F(n) is the n-th Fibonacci number. Two parts are right triangles with legs F(2n) and F(2n+2); two are right trapezoids three of whose sides are F(2n), F(2n+1), and F(2n+1). (Thus n > 0.) The paradox (or dissection fallacy) depends on Cassini's identity F(2n+1) * F(2n+3) = F(2n+2)^2 + 1.
For an extension of the paradox to a F(2n+1) X F(2n+1) square using Cassini's identity F(2n) * F(2n+2) = F(2n+1)^2 - 1, see Dudeney (1970), Gardner (1956), Horadam (1962), Knott (2014), Kumar (1964), and Sillke (2004). Sillke also has many additional references and links.

			F(3) * F(5) = 2 * 5 = 10 = 3^2 + 1 = F(4)^2 + 1, so a(1) = 10.
G.f. = 10*x + 65*x^2 + 442*x^3 + 3026*x^4 + 20737*x^5 + 142130*x^6 + 974170*x^7 + ...

W. W. Rouse Ball and H. S. M. Coxeter, Mathematical Recreations and Essays, 13th edition, Dover, 1987, p. 85.
Henry E. Dudeney, 536 Puzzles and Curious Problems, Scribner, reprinted 1970, Problems 352-353 and their answers.
Martin Gardner, Mathematics, Magic and Mystery, Dover, 1956, Chap. 8.
Edward Wakeling, Rediscovered Lewis Carroll Puzzles, Dover, 1995, p. 12.
David Wells, The Penguin Book of Curious and Interesting Puzzles, Penguin, 1997, Puzzle 143.

Colin Barker, Table of n, a(n) for n = 1..1000
Margherita Barile, Dissection Fallacy, MathWorld.
A. F. Horadam, Fibonacci Sequences and a Geometrical Paradox, Math. Mag., Vol. 35, No. 1 (1962), pp. 1-11.
Ron Knott, Fibonacci jigsaw puzzles, 2014.
Santosh Kumar, On Fibonacci Sequences and a Geometrical Paradox, Math. Mag., Vol. 37, No. 4 (1964), pp. 221-223.
Oskar Schlömilch, Ein geometrisches Paradoxon, Zeitschrift für Mathematik und Physik, Vol. 13 (1868), p. 162.
Torsten Sillke, Jigsaw paradox, 2004.
David Singmaster, Vanishing area puzzles, Recreational Math. Mag., Vol. 1 (2014), pp. 10-21.
Warren Weaver, Lewis Carroll and a Geometrical Paradox, American Math. Monthly, Vol. 45, No. 4 (1938), pp. 234-236.
Wikipedia, Cassini and Catalan identities.
Wikipedia, Fibonacci number.
Wikipedia, Missing square puzzle; also see External Links.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045, A001519, A059929, A064170, A070550, A166516, A176055, A190018, A236165, A245306.

[Fibonacci(2*n+1)*Fibonacci(2*n+3) : n in [1..30]]; // Wesley Ivan Hurt, Oct 16 2015

with(combinat): A262342:=n->fibonacci(2*n+1)*fibonacci(2*n+3): seq(A262342(n), n=1..30); # Wesley Ivan Hurt, Oct 16 2015

Table[Fibonacci[2 n + 1] Fibonacci[2 n + 3], {n, 22}]
LinearRecurrence[{8,-8,1},{10,65,442},30] (* Harvey P. Dale, Aug 06 2024 *)

Vec(-x*(2*x^2-15*x+10)/((x-1)*(x^2-7*x+1)) + O(x^30)) \\ Colin Barker, Oct 17 2015

a(n) = fibonacci(2*n+1) * fibonacci(2*n+3) \\ Altug Alkan, Oct 17 2015

a(n) = Fibonacci(2n+1) * Fibonacci(2n+3) = Fibonacci(2n+2)^2 + 1 for n > 0.
From Colin Barker, Oct 17 2015: (Start)
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
G.f.: -x*(2*x^2-15*x+10) / ((x-1)*(x^2-7*x+1)).
(End)
a(3*k-2) mod 2 = 0; a(3*k-1) mod 2 = 1; a(3*k) mod 2 = 0, k > 0. - Altug Alkan, Oct 17 2015
a(n) = A059929(2*n+1) = A070550(4*n+1) = A166516(2*n+2) = A190018(8*n) = A236165(4*n+4) = A245306(2*n+2). - Bruno Berselli, Oct 17 2015
a(n) = A064170(n+3). - Alois P. Heinz, Oct 17 2015
E.g.f.: (1/5)*((1/phi*r)*exp(b*x) + (phi^4/r)*exp(a*x) + 3*exp(x) - 10), where r = 2*phi+1, 2*a=7+3*sqrt(5), 2*b=7-3*sqrt(5). - G. C. Greubel, Oct 17 2015
a(n) = (A337928(n+1) - A337929(n+1)) / 2. - Flávio V. Fernandes, Feb 06 2021
Sum_{n>=1} 1/a(n) = sqrt(5)/2 - 1 = A176055 - 2. - Amiram Eldar, Mar 04 2021

cp's OEIS Frontend

A001622 Decimal expansion of golden ratio phi (or tau) = (1 + sqrt(5))/2.

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

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A049684 a(n) = Fibonacci(2n)^2.

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A350922 a(0) = 2, a(1) = 5, and a(n) = 7*a(n-1) - a(n-2) - 4 for n >= 2.

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A337928 Numbers w such that (F(2n+1)^2, -F(2n)^2, -w) are primitive solutions of the Diophantine equation 2*x^3 + 2*y^3 + z^3 = 1, where F(n) is the n-th Fibonacci number (A000045).

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A337929 Numbers w such that (F(2*n-1)^2, -F(2*n)^2, w) are primitive solutions of the Diophantine equation 2*x^3 + 2*y^3 + z^3 = 1, where F(n) is the n-th Fibonacci number (A000045).

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A337928 Numbers w such that (F(2n+1)^2, -F(2n)^2, -w) are primitive solutions of the Diophantine equation 2x^3 + 2y^3 + z^3 = 1, where F(n) is the n-th Fibonacci number (A000045).

A337929 Numbers w such that (F(2n-1)^2, -F(2n)^2, w) are primitive solutions of the Diophantine equation 2x^3 + 2y^3 + z^3 = 1, where F(n) is the n-th Fibonacci number (A000045).

A081076 a(n) = Lucas(4n) + 3, or 5Fibonacci(2n-1)Fibonacci(2n+1).

A360467 a(n) = Fibonacci(4n+2) + 3Fibonacci(2*n+1)^2.