A104457 -id:A104457 - cp's OEIS frontend

A139341 Decimal expansion of e^((1+sqrt(5))/2).

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

Offset: 1

Mohammad K. Azarian, Apr 14 2008

By the Lindemann-Weierstrass theorem, this constant is transcendental. - Charles R Greathouse IV, May 13 2019

			5.04316564336002865131188218928542471032359017541384...

Index entries for transcendental numbers.

Cf. A001622, A094214, A104457, A098317, A002390, A139339, A139340, A139342.

phi := (1+sqrt(5))/2 ; evalf(exp(phi)) ; # R. J. Mathar, Oct 16 2015

RealDigits[Exp[GoldenRatio], 10, 100][[1]] (* Amiram Eldar, Feb 08 2022 *)

exp((sqrt(5)+1)/2) \\ Charles R Greathouse IV, May 13 2019

From Amiram Eldar, Feb 08 2022: (Start)
Equals exp(A001622).
Equals 1/A139342. (End)

A132338 Decimal expansion of 1 - 1/phi.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane, Nov 07 2007

Density of 1's in Fibonacci word A003849.
Also decimal expansion of Sum_{n>=1} ((-1)^(n+1))*1/phi^n. - Michel Lagneau, Dec 04 2011
The Lambert series evaluated at this point is 0.8828541617125076... [see André-Jeannin]. - R. J. Mathar, Oct 28 2012
Because this equals 2 - phi, this is an integer in the quadratic number field Q(sqrt(5)). (Note that this is also sqrt(5 - 3*phi).) - Wolfdieter Lang, Jan 08 2018
When m >= 1, the equation m*x^m + (m-1)*x^(m-1) + ... + 2*x^2 + x - 1 = 0 has only one positive root, u(m) (say); then lim_{m->oo} u(m) = (3-sqrt(5))/2 (see Aubonnet). - Bernard Schott, May 12 2019
Cosine of the zenith angle at which a string should be cut so that a ball tied to one of its ends, set moving without friction around a vertical circle with the minimum speed in a uniform gravitational field, will then travel through the fixed center of the circle. - Stefano Spezia, Oct 25 2020
Algebraic number of degree 2 with minimal polynomial x^2 - 3*x + 1. The other root is 1 + phi = A104457. - Wolfdieter Lang, Aug 29 2022

			0.38196601125010515179541316563436188...

F. Aubonnet, D. Guinin and A. Ravelli, Oral, Concours d'entrée des Grandes Ecoles Scientifiques, Exercices résolus, "Crus" 1982-83, Bréal, 1983, Exercice 210, 40-42.

Ivan Panchenko, Table of n, a(n) for n = 0..1000
R. André-Jeannin, Lambert series and the summation of reciprocals in certain Fibonacci-Lucas-Type sequences, Fib. Quart. 28 (1990) 223-226.
Yiyan Ni, Myron Hlynka, and Percy H. Brill, Urn Models and Fibonacci Series, arXiv:1806.09150 [math.CO], 2018. See (9) p. 7.
Index entries for algebraic numbers, degree 2

Cf. A001622, A003849, A094874, A079585, A094214, A104457.

RealDigits[N[1/GoldenRatio^2,200]] (* Vladimir Joseph Stephan Orlovsky, May 27 2010 *)
RealDigits[1-1/GoldenRatio,10,120][[1]] (* Harvey P. Dale, Mar 30 2024 *)

(3-sqrt(5))/2 \\ Michel Marcus, Oct 26 2020

Equals 1 - 1/phi = 2 - phi, with phi from A001622.
Equals A094874 - 1, or A079585 - 2, or the square of A094214.
Equals (5-sqrt(5))^2/20 = 1/phi^2 = 1/A104457. - Joost Gielen, Sep 28 2013 [corrected by Joerg Arndt, Sep 29 2013]
Equals (3-sqrt(5))/2. - Bernard Schott, May 12 2019
Equals Sum_{k >= 2} (-1)^k/(Fibonacci(k)*Fibonacci(k+1)). See Ni et al. - Michel Marcus, Jun 26 2018

A296184 Decimal expansion of 2 + phi, with the golden section phi from A001622.

Original entry on oeis.org

3, 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

Offset: 1

Wolfdieter Lang, Jan 08 2018

In a regular pentagon, inscribed in a unit circle this equals twice the largest distance between a vertex and a midpoint of a side.
This is an integer in the quadratic number field Q(sqrt(5)).
Only the first digit differs from A001622.

			3.618033988749894848204586834365638117720309179805762862135448622705260462...

Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 5.25, p. 417.

Sumit Kumar Jha, Two complementary relations for the Rogers-Ramanujan continued fraction, arXiv:2112.12081 [math.NT], 2021.
Index entries for algebraic numbers, degree 2.

Cf. A001622, A094214, A104457, A176055, A020837.

2 + 2*cos(2*Pi/n): A104457 (n = 5), A116425 (n = 7), A332438 (n = 9), A019973 (n = 12).

First@ RealDigits[2 + GoldenRatio, 10, 77] (* Michael De Vlieger, Jan 13 2018 *)

(5 + sqrt(5))/2 \\ Altug Alkan, Mar 19 2018

Equals 2 + A001622 = 1 + A104457 = 3 + A094214.
From Christian Katzmann, Mar 19 2018: (Start)
Equals Sum_{n>=0} (15*(2*n)!+40*n!^2)/(2*n!^2*3^(2*n+2)).
Equals 5/2 + Sum_{n>=0} 5*(2*n)!/(2*n!^2*3^(2*n+1)). (End)
Constant c = 2 + 2*cos(2*Pi/10). The linear fractional transformation z -> c - c/z has order 10, that is, z = c - c/(c - c/(c - c/(c - c/(c - c/(c - c/(c - c/(c - c/(c - c/(c - c/(z)))))))))). - Peter Bala, May 09 2024

A139340 Decimal expansion of the cube root of the golden ratio. That is, the decimal expansion of ((1+sqrt(5))/2)^(1/3).

Original entry on oeis.org

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

Offset: 1

Mohammad K. Azarian, Apr 14 2008

Larger of the real roots of x^6 - x^3 - 1. - Charles R Greathouse IV, Apr 14 2014

			1.1739849967053285...

Chai Wah Wu, Table of n, a(n) for n = 1..10000
Index entries for algebraic numbers, degree 6

Cf. A001622, A094214, A104457, A098317, A002390, A139339.

phi := (sqrt(5)+1)/2 ; evalf(root[3](phi)) ; # R. J. Mathar, Oct 16 2015

RealDigits[N[GoldenRatio^(1/3),200]][[1]] (* Vladimir Joseph Stephan Orlovsky, Feb 06 2012 *)

polrootsreal(x^6-x^3-1)[2] \\ Charles R Greathouse IV, Apr 14 2014

A116425 Decimal expansion of 2 + 2cos(2Pi/7).

Original entry on oeis.org

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

Offset: 1

Eric W. Weisstein, Feb 15 2006

A root of the equation x^3 - 5*x^2 + 6*x - 1 = 0. - Arkadiusz Wesolowski, Jan 13 2016

The other two roots of this minimal polynomial of the present algebraic number (rho(7))^2, with rho(7) = 2*cos(Pi/7) = A160389 are (2*cos(3*Pi/7))^2 = (A255241)^2 and (2*cos(5*Pi/7))^2 = (-A255249)^2. - Wolfdieter Lang, Mar 30 2020

			3.246979603717467061...

Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 5.25 Tutte-Beraha Constants, p. 417.

Jesús Salas and Alan D. Sokal, Transfer matrices and partition functions zeros for antiferromagnetic Potts models, arXiv:cond-mat/0004330 [cond-mat.stat-mech], 2000-2001, p. 64.
Eric Weisstein's World of Mathematics, Logistic Map
Eric Weisstein's World of Mathematics, Silver Constant
Index entries for algebraic numbers, degree 3

Cf. A231187, A160389.
Cf. A003558, A054142, A255241, A255249.
2 + 2*cos(2*Pi/n): A104457 (n = 5), A332438 (n = 9), A296184 (n = 10), A019973 (n = 12).

First@ RealDigits[N[2 + 2 Cos[2 Pi/7], 120]] (* Michael De Vlieger, Jan 13 2016 *)

2 + 2*cos(2*Pi/7) \\ Michel Marcus, Jan 13 2016

Equals (2*cos(Pi/7))^2 = (A160389)^2.
Equals 2 + i^(4/7) - i^(10/7). - Peter Luschny, Apr 04 2020
Let  c = 2 + 2*cos(2*Pi/7). The linear fractional transformation z -> c - c/z has order 7, that is, z = c - c/(c - c/(c - c/(c - c/(c - c/(c - c/(c - c/z)))))). - Peter Bala, May 09 2024

A332438 Decimal expansion of (2*cos(Pi/9))^2 = A332437^2.

Original entry on oeis.org

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

Offset: 1

Wolfdieter Lang, Mar 31 2020

This algebraic number rho(9)^2 of degree 3 is a root of its minimal polynomial x^3 - 6*x^2 + 9*x - 1.
The other two roots are x2 = (2*cos(5*Pi/9))^2 = (2*cos(4*Pi/9))^2 = (R(4,rho(9)))^2 = 2 - rho(9) = 0.120614758..., and x3 = (2*cos(7*Pi/9))^2 = (2*cos(7*Pi/9))^2 = (R(7,rho(9)))^2 = 4 + rho(9) - rho(9)^2 = 2.347296355... = A130880 + 2, with rho(9) = 2*cos(Pi/9) = A332437, the monic Chebyshev polynomials R (see A127672), and the computation is done modulo the minimal polynomial of rho(9) which is x^3 - 3*x - 1 (see A187360).
This gives the representation of these roots in the power basis of the simple field extension Q(rho(9)). See the linked W. Lang paper in A187360, sect. 4.
This number rho(9)^2 appears as limit of the quotient of consecutive numbers af various sequences, e.g., A094256 and A094829.
The algebraic number rho(9)^2 - 2 = 1.532088898... of Q(rho(9)) has minimal polynomial x^3 - 3*x + 1 over Q. The other roots are -rho(9) = -A332437 and 2 + rho(9) - rho(9)^2 = A130880. - Wolfdieter Lang, Sep 20 2022

			3.5320888862379560704047853011108333478716649...

Index entries for algebraic numbers, degree 3.

Cf. A019879, A063289, A094256, A094829, A127672, A130880, A187360, A332437.

2 + 2*cos(2*Pi/n): A104457 (n = 5), A116425 (n = 7), A296184 (n = 10), A019973 (n = 12).

RealDigits[(2*Cos[Pi/9])^2, 10, 100][[1]] (* Amiram Eldar, Mar 31 2020 *)

(2*cos(Pi/9))^2 \\ Michel Marcus, Sep 23 2022

Equals (2*cos(Pi/9))^2 = rho(9)^2 = A332437^2.
Equals 2 + i^(4/9) - i^(14/9). - Peter Luschny, Apr 04 2020
Equals 2 + w1^(1/3) + w2^(1/3), where w1 = (-1 + sqrt(3)*i)/2 = exp(2*Pi*i/3) and w2 = (-1 - sqrt(3)*i)/2 are the complex roots of x^3 - 1. - Wolfdieter Lang, Sep 20 2022
Constant c = 2 + 2*cos(2*Pi/9). The linear fractional transformation z -> c - c/z has order 9, that is, z = c - c/(c - c/(c - c/(c - c/(c - c/(c - c/(c - c/(c - c/(c - c/(z))))))))). - Peter Bala, May 09 2024
From Amiram Eldar, Nov 22 2024: (Start)
Equals 3 + sec(Pi/9)/2 = 3 + 1/(2*A019879).
Equals 3 + Product_{k>=3} (1 + (-1)^k/A063289(k)). (End)

A032908 One of four 3rd-order recurring sequences for which the first derived sequence and the Galois transformed sequence coincide.

Original entry on oeis.org

2, 2, 3, 6, 14, 35, 90, 234, 611, 1598, 4182, 10947, 28658, 75026, 196419, 514230, 1346270, 3524579, 9227466, 24157818, 63245987, 165580142, 433494438, 1134903171, 2971215074, 7778742050, 20365011075, 53316291174, 139583862446, 365435296163, 956722026042

Offset: 0

Michele Elia (elia(AT)polito.it)

a(n) is also a sequence with the property that the difference between the sum and product of two consecutive terms is equal to the square of the difference between those terms, i.e., a(n)*a(n+1) - (a(n)+ a(n+1)) = (a(n) - a(n + 1))^2. The difference between those two terms, a(n + 1) - a(n) = F(2n -2), the (2n - 2)th Fibonacci number. - John Baker, May 18 2010
Conjecture: consecutive terms of this sequence and consecutive terms of A101265 provide all the positive integer solutions of (a+b)*(a+b+1) == 0 (mod (a*b)). - Robert Israel, Aug 26 2015
Conjecture is true: see Mathematics Stack Exchange link. - Robert Israel, Sep 06 2015
Consecutive terms of this sequence and consecutive terms of A101879 provide all the positive integer pairs for which K = (a+1)/b + (b+1)/a is an integer. For this sequence, K = 3. - Andrey Vyshnevyy, Sep 18 2015

L. E. Dickson, History of the Theory of Numbers, Dover, New York, 1971.

Robert Israel, Table of n, a(n) for n = 0..2154
Michele Elia, A Note on derived linear recurring sequences, pp. 83-92 of Proceedings Seventh Int. Conference on Fibonacci Numbers and their Applications (Austria, 1996), Applications of Fibonacci Numbers, Volume 7.
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 919.
Robert Israel, Will Jagy et al., Diophantine equation (x+y)(x+y+1)-kxy=0, Mathematics Stack Exchange, Sep 1 2015.
Hieu D. Nguyen and Douglas Taggart, Mining the OEIS: Ten Experimental Conjectures, 2013. Mentions this sequence. - From _N. J. A. Sloane_, Mar 16 2014
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Cf. A001519, A001622, A049310, A101265, A104457.

[2] cat [n le 1 select 2 else Floor((3*Self(n-1) + Sqrt(5*Self(n-1)^2 - 10*Self(n-1) + 1) - 1)/2): n in [1..30]]; // Vincenzo Librandi, Aug 27 2015

f:= proc(n) option remember; local x;
  x:= procname(n-1); (3*x + sqrt(5*x^2 - 10*x + 1) - 1)/2 end proc:
f(0):= 2: f(1):= 2:
map(f, [$0..30]); # Robert Israel, Aug 26 2015

Table[Fibonacci[2 n - 1] + 1, {n, 0, 27}] (* Michael De Vlieger, Aug 26 2015 *)
LinearRecurrence[{4,-4,1},{2,2,3},40] (* Harvey P. Dale, Apr 11 2018 *)

Vec((2-6*x+3*x^2)/(1-4*x+4*x^2-x^3)+O(x^66)) \\ Joerg Arndt, Jul 02 2013

G.f.: (2 - 6*x + 3*x^2)/((1 - x)*(1 - 3*x + x^2)).
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3).
a(n) = Fibonacci(2*n-1) + 1 = A001519(n) + 1. - Vladeta Jovovic, Mar 19 2003
a(n) = 3*a(n - 1) - a(n - 2) - 1. - N. Sato, Jan 21 2010
From  Wolfdieter Lang, Aug 27 2014: (Start)
a(n) = 1 + S(n-1, 3) - S(n-2, 3) =  1 + A001519(n), with Chebyshev S-polynomials (see A049310). For n < 0, we have S(-1, x) = 0 and S(-2, x) = -1.
This follows from the partial fraction decomposition of the g.f., 1/(1 - x) + (1 - 2*x)/ (1 - 3*x + x^2), using the recurrence for S, or from A001519. (End)
From Robert Israel, Aug 26 2015: (Start)
(a(n) + a(n+1))*(a(n) + a(n+1) + 1) = 5*a(n)*a(n+1).
a(n+1) = (3*a(n) + sqrt(5*a(n)^2 - 10*a(n) + 1) - 1)/2 for n >= 1. (End)
a(n) = 1 + (2^(-1-n) * ((3 - sqrt(5))^n * (1 + sqrt(5)) + (-1 + sqrt(5)) * (3 + sqrt(5))^n)) / sqrt(5). - Colin Barker, Nov 02 2016
Sum_{n>=0} 1/a(n) = phi (A001622). - Amiram Eldar, Oct 05 2020
Product_{n>=1} (1 + 1/a(n)) = (3+sqrt(5))/2 = phi^2 (A104457). - Amiram Eldar, Nov 28 2024

More terms from Ralf Stephan, Mar 10 2003

Index for Chebyshev polynomials and cross reference added by Wolfdieter Lang, Aug 27 2014

A106729 Sum of two consecutive squares of Lucas numbers (A001254).

Original entry on oeis.org

5, 10, 25, 65, 170, 445, 1165, 3050, 7985, 20905, 54730, 143285, 375125, 982090, 2571145, 6731345, 17622890, 46137325, 120789085, 316229930, 827900705, 2167472185, 5674515850, 14856075365, 38893710245, 101825055370, 266581455865

Offset: 0

Lekraj Beedassy, May 14 2005

Positive values of x (or y) satisfying x^2 - 3xy + y^2 + 25 = 0. - Colin Barker, Feb 08 2014
Positive values of x (or y) satisfying x^2 - 7xy + y^2 + 225 = 0. - Colin Barker, Feb 09 2014
Positive values of x (or y) satisfying x^2 - 18xy + y^2 + 1600 = 0. - Colin Barker, Feb 26 2014

Bruno Berselli, Table of n, a(n) for n = 0..300
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (3,-1).

Cf. A000032, A000045, A000204, A001254, A081071, A104457.

[Fibonacci(n-2)^2+Fibonacci(n+3)^2: n in [0..30]]; // Vincenzo Librandi, Jul 09 2011

seq(combinat:-fibonacci(n-2)^2 + combinat:-fibonacci(n+3)^2, n=0..100); # Robert Israel, Nov 23 2014

Table[LucasL[n]^2 + LucasL[n+1]^2, {n, 0, 30}] (* Wesley Ivan Hurt, Nov 23 2014 *)
Total/@Partition[LucasL[Range[0,30]]^2,2,1] (* Harvey P. Dale, Jun 26 2022 *)

a(n) = fibonacci(n-2)^2 + fibonacci(n+3)^2;
vector(30, n, a(n-1)) \\ G. C. Greubel, Dec 17 2017

[fibonacci(n-2)^2 + fibonacci(n+3)^2 for n in (0..30)] # G. C. Greubel, Sep 10 2021

a(n) = Lucas(n)^2 + Lucas(n+1)^2 = 5*(Fibonacci(n)^2 + Fibonacci(n+1)^2) = 5*A001519(n+1).
a(n) = 3*a(n-1) - a(n-2). - T. D. Noe, Dec 11 2006
G.f.: 5*(1-x)/(1-3*x+x^2). - Philippe Deléham, Nov 16 2008
a(n) = Fibonacci(n-2)^2 + Fibonacci(n+3)^2. - Gary Detlefs, Dec 28 2010
a(n) = [1,1; 1,2]^(n-2).{3,4}.{3,4}, for n>=3. - John M. Campbell, Jul 09 2011
a(n) = Lucas(2n) + Lucas(2n+2). - Richard R. Forberg, Nov 23 2014
From Robert Israel, Nov 23 2014: (Start)
a(n) = 5*A000045(2*n+1).
E.g.f.: (5+sqrt(5))/2 * exp((3+sqrt(5))*x/2) + (5-sqrt(5))/2 * exp((3-sqrt(5))*x/2). (End)
From Enrique Navarrete, Mar 24 2025: (Start)
a(n)^2 = 20 + 5*A081071(n).
Limit_{n->oo} a(n+1)/a(n) = (3 + sqrt(5))/2 (see A104457). (End)

Corrected by T. D. Noe, Dec 11 2006

More terms from Bruno Berselli, Jul 17 2011

A109134 Decimal expansion of Phi, the real root of the equation 1/x = (x-1)^2.

Original entry on oeis.org

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

Offset: 1

Lekraj Beedassy, Aug 17 2005

The silver number (A060006) is equal to Phi*(Phi-1).
Also Phi*(Phi-1) = 1/(Phi-1). - Richard R. Forberg, Oct 08 2014
Equations to which this is a root can also be written as: x = sqrt(x + sqrt(x)); x^2 - x - sqrt(x) = 0; or this form where n = 1: x = n + 1/sqrt(x). When n = 2 then the root is 2.618033988... = A104457 = 1 + A001622 or 1 + "Golden Ratio" called phi. - Richard R. Forberg, Oct 08 2014
Also equals the largest root (negated) of the Mandelbrot polynomial P_2(z) = 1+z*(1+z)^2. - Jean-François Alcover, Apr 16 2015
Suppose that r is a real number in the interval [3/2, 5/3). Let C(r) = (c(k)) be the sequence of coefficients in the Maclaurin series for 1/(Sum_{k>=0} floor((k+1)*r))(-x)^k). Conjectures: the limit L(r) of c(k+1)/c(k) as k -> oo exists, L(r) is discontinuous at 5/3 (cf. A279676), and the left limit of L(r) as r->5/3 is Phi. - Clark Kimberling, Jul 11 2017
From Wolfdieter Lang, Nov 07 2022: (Start)
This equals r + 2/3 where r is the real root of y^3 - (1/3)*y - 25/27.
The other roots of x^3 - 2*x^2 + x - 1 are (2 + w1*((25 + 3*sqrt(69))/2)^(1/3) + w2*((25 - 3*sqrt(69))/2)^(1/3))/3 = 0.1225611668... + 0.7448617668...*i, and its complex conjugate, where w1 = (-1 + sqrt(3)*i)/2 = exp(2*Pi*i/3) and w2 = (-1 - sqrt(3)*i)/2 are the complex conjugate roots of x^3 - 1.
Using hyperbolic functions these roots are (2 - cosh((1/3)*arccosh(25/2)) + sqrt(3)*sinh((1/3)*arccosh(25/2))*i)/3, and its complex conjugate. (End)

			1.75487766624669276004950889635852869189460661777279314398928397064...

Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 5.11, p. 340.
Martin Gardner, A Gardner's Workout, pp. 124-126, A. K. Peters MA 2001.

Vincenzo Librandi, Table of n, a(n) for n = 1..5000
Simon Baker, On small bases which admit countably many expansions, Journal of Number Theory, Volume 147, February 2015, Pages 515-532.
Simon Plouffe, Plouffe's Inverter .
Nikita Sidorov, Expansions in non-integer bases: Lower, middle and top orders, Journal of Number Theory, Volume 129, Issue 4, April 2009, Pages 741-754. See Prop. 2.3 p. 744.
Yuru Zou and Derong Kong, On a problem of countable expansions, Journal of Number Theory, Volume 158, January 2016, Pages 134-150. See Theorem 1.1 p. 135.
Index entries for algebraic numbers, degree 3.

Cf. A001622, A001685, A060006, A075778, A104457, A358181.

FindRoot[x^3 - 2x^2 + x - 1 == 0, {x, 1.75}, WorkingPrecision -> 128][[1, 2]] (* Robert G. Wilson v, Aug 19 2005 *)
Root[x^3-2x^2+x-1, x, 1] // RealDigits[#, 10, 105]& // First (* Jean-François Alcover, Mar 05 2013 *)

d=104;default(realprecision,d);print(k=solve(x=1,2,(x-1)^2-1/x)); for(c=0,d,z=floor(k);print1(z,",",);k=10*(k-z))

polrootsreal(x^3-2*x^2+x-1)[1] \\ Charles R Greathouse IV, Aug 15 2014

Equals 1+A075778. - R. J. Mathar, Aug 20 2008
Equals (1/6*(108+12*sqrt(69))^(1/3) + 2/(108+12*sqrt(69))^(1/3))^2. - Vaclav Kotesovec, Oct 08 2014
Equals Rho^2 where Rho is the plastic number 1.3247179572...(see A060006). - Philippe Deléham, Sep 29 2020
From Wolfdieter Lang, Nov 07 2022: (Start)
Equals (2 + ((25 + 3*sqrt(69))/2)^(1/3) + ((25 + 3*sqrt(69))/2)^(-1/3))/3.
Equals (2 + ((25 + 3*sqrt(69))/2)^(1/3) + ((25 - 3*sqrt(69))/2)^(1/3))/3.
Equals 2*(1 + cosh((1/3)*arccosh(25/2)))/3. (End)
Equals - Sum_{k>=1} Gamma(k - k/5 - 1)*Gamma(k/5 + 1)*sin(3*k*Pi/5)/(k*Pi*Gamma(k)). - Antonio Graciá Llorente, Dec 14 2024

Extended by Klaus Brockhaus and Robert G. Wilson v, Aug 19 2005

A112373 a(n+2) = (a(n+1)^3 + a(n+1)^2)/a(n) with a(0)=1, a(1)=1.

Original entry on oeis.org

1, 1, 2, 12, 936, 68408496, 342022190843338960032, 584861200495456320274313200204390612579749188443599552

Offset: 0

Andrew Hone, Dec 02 2005

nonn

As n tends to infinity, log(log(a(n)))/n tends to log((3+sqrt(5))/2) = A104457.
The Laurent property is satisfied by any second-order recurrence of the form a(n+2)=f(a(n+1))/a(n) where f is a polynomial of the form f(x)=x^m*p(x) with m a positive integer >= 2 and p arbitrary. In that case a(0)=a(1)=1 generates a sequence of integers and the ratios a(n+1)/a(n) and a(n+1)*a(n-1)/a(n)^2 are integers for all n. - Andrew Hone, Dec 12 2005
Also denominators of Sum_{k=0..n} 1/a(k) with numerators = A259644. - Reinhard Zumkeller, Jul 02 2015
The next term (a(8)) has 141 digits. - Harvey P. Dale, Apr 05 2019

Reinhard Zumkeller, Table of n, a(n) for n = 0..10
Joshua Alman, Cesar Cuenca, and Jiaoyang Huang, Laurent phenomenon sequences, Journal of Algebraic Combinatorics, Vol. 43, No. 3 (2015), pp. 589-633.
Sergey Fomin and Andrei Zelevinsky, The Laurent Phenomenon, Advances in Applied Mathematics, Vol. 28, No. 2 (2002) pp. 119-144.
Andrew N. W. Hone, Curious continued fractions, nonlinear recurrences and transcendental numbersJournal of Integer Sequences, Vol. 18 (2015), Article #15.8.4; arXiv preprint, arXiv:1507.00063 [math.NT], 2015.
Andrew N. W. Hone, Continued fractions for some transcendental numbers, Monatshefte für Mathematik, Vol. 182 (2017), pp. 33-38; arXiv preprint, arXiv:1509.05019 [math.NT], 2015-2016.

Cf. A114550, A114551, A114552.

Cf. A259644.

a112373 n = a112373_list !! n
a112373_list = 1 : 1 : zipWith (\u v -> (u^3 + u^2) `div` v)
                               (tail a112373_list) a112373_list
-- Reinhard Zumkeller, Jul 02 2015

I:=[1,1]; [n le 2 select I[n] else (Self(n-1)^3+Self(n-1)^2)/Self(n-2): n in [1..10]]; // Vincenzo Librandi, Jul 02 2015

a[0]:=1; a[1]:=1; f(x):=x^3+x^2; for n from 0 to 8 do a[n+2]:=simplify(subs(x=a[n+1],f(x))/a[n]) od; s[3]:=ln(2^2*3); s[4]:=ln(2^3*3^2*13); for n from 3 to 10000 do s[n+2]:=evalf(3*s[n+1]+ln(1+exp(-s[n+1]))-s[n]): od: print(evalf(ln(s[10002])/(10002))): evalf(ln((3+sqrt(5))/2));
# s[n]=ln(a[n]); ln(s[n])/n converges slowly to 0.962...

RecurrenceTable[{a[n] == (a[n - 1]^3 + a[n - 1]^2)/a[n - 2], a[0] == a[1] == 1}, a, {n, 0, 7}] (* Michael De Vlieger, Jul 02 2015 *)
nxt[{a_,b_}]:={b,(b^3+b^2)/a}; NestList[nxt,{1,1},8][[All,1]] (* Harvey P. Dale, Apr 05 2019 *)

{a(n) = my(a=self()); if(n<0, a(1-n), n<2, 1, a(n-1)^2 * (1 + a(n-1)) / a(n-2))}; /* Michael Somos, Apr 19 2017 */

a(n+1) / a(n) = A114552(n), a(n) = a(1-n) for all n in Z. - Michael Somos, Apr 19 2017

Sum_{n>=0} 1/a(n) = A114550. - Amiram Eldar, Nov 13 2020

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