A000058 -id:A000058 - cp's OEIS frontend

A144784 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 11.

11, 111, 12211, 149096311, 22229709804712411, 494159998001727075769152612720511, 244194103625066907517263589918036880566782292998362610615987380611

Offset: 1

Artur Jasinski, Sep 21 2008

nonn

For the "exact" formula, compare the Aho-Sloane reference in A000058. - N. J. A. Sloane, Apr 07 2014

A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437.
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437 (original plus references that F.Q. forgot to include - see last page!)
Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Problem 958, College Mathematics Journal, Vol. 42, No. 4, September 2011, p. 330.
Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Solution College Mathematics Journal, Vol. 43, No. 4, September 2012, pp. 340-342.

Cf. A000058, A082732, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A144786, A144787, A144788, A144808.

See A239900 for another version.

a = {}; r = 11; Do[AppendTo[a, r]; r = r^2 - r + 1, {n, 1, 10}]; a

a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 11.

a(n) ~ c^(2^n) where c = 3.242214... (see A144808).

A001999 a(n) = a(n-1)*(a(n-1)^2 - 3).

Original entry on oeis.org

3, 18, 5778, 192900153618, 7177905237579946589743592924684178

Offset: 0

N. J. A. Sloane

The next terms in the sequence contain 102 and 305 digits. - Harvey P. Dale, Jun 09 2011
From Peter Bala, Nov 13 2012: (Start)
The present sequence is the case x = 3 of the following general remarks. For other cases see A219160 (x = 4), A219161 (x = 5) and A112845 (x = 6).
Let x > 2 and let alpha := {x + sqrt(x^2 - 4)}/2. Define a sequence a(n) (which depends on x) by setting a(n) = alpha^(3^n) + (1/alpha)^(3^n). Then it is easy to verify that the sequence a(n) satisfies the recurrence equation a(n+1) = a(n)^3 - 3*a(n) with the initial condition a(0) = x.
We have the following identity, valid for x > 2: sqrt((x + 2)/(x - 2)) = (1 + 2/(x-1))*sqrt((y + 2)/(y - 2)), where y = x^3 - 3*x. Iterating the identity produces the product expansion sqrt((x+2)/(x-2)) = Product_{n = 0..oo} (1 + 2/(a(n) - 1)), with a(0) = x and a(n+1) = a(n)^3 - 3*a(n). The rate of convergence is cubic (Fine).
For similar results to the above see A001566 and A219162. (End)
Let b(n) = a(n) - 3. The sequence {b(n)} appears to be a strong divisibility sequence, that is, gcd(b(n),b(m)) = b(gcd(n,m)) for n, m >= 1. - Peter Bala, Dec 08 2022

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..7
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437.
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437 (original plus references that F.Q. forgot to include - see last page!)
E. B. Escott, Rapid method for extracting a square root, Amer. Math. Monthly, Vol. 44, No. 10 (1937), pp. 644-646.
N. J. Fine, Infinite products for k-th roots, Amer. Math. Monthly Vol. 84, No. 8 (Oct. 1977), pp. 629-630.
Walther Janous, Problem B-916, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 39, No. 2 (2001), p. 181; Subscript Is Power, Solution to Problem B-916 by H.-J. Seiffert, ibid., Vol. 40, No. 1 (2002), p. 86.
Hideyuki Ohtsu, Problem B-1316, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 60, No. 4 (2022), p. 365.
Eric Weisstein's World of Mathematics, Pierce Expansion.

Cf. A000032, A006276, A001566, A002814, A045529, A112845, A219160, A219161, A219162.

NestList[#(#^2-3)&,3,6] (* Harvey P. Dale, Jun 09 2011 *)
RecurrenceTable[{a[n] == a[n - 1]^3 - 3*a[n - 1], a[0] == 3}, a, {n,
  0, 5}] (* G. C. Greubel, Dec 30 2016 *)

a(n)=2*fibonacci(2*3^n+1)-fibonacci(2*3^n)

a(n) = 2*F(2*3^n+1) - F(2*3^n) = ceiling(tau^(2*3^n)) where F(k) = A000045(k) is the k-th Fibonacci number and tau is the golden ratio. - Benoit Cloitre, Nov 29 2002
From Peter Bala, Nov 13 2012: (Start)
a(n) = ((3 + sqrt(5))/2)^(3^n) + ((3 - sqrt(5))/2)^(3^n).
Product_{n >= 0} (1 + 2/(a(n) - 1)) = sqrt(5).
a(n) = A002814(n+1) + 1. (End)
a(n) = 2*T(3^n,3/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. Cf. A219161. - Peter Bala, Feb 01 2017
From Amiram Eldar, Jan 12 2022: (Start)
a(n) = A000032(2*3^n).
a(n) = A006267(n)^2 + 2.
Product_{k=0..n} (a(k)-1) = Fibonacci(3^(n+1)) = A045529(n+1) (Janous, 2001). (End)
Sum_{n>=0} arctanh(1/a(n)) = log(5)/4 (Ohtsu, 2022). - Amiram Eldar, Dec 15 2022

A007660 a(n) = a(n-1)*a(n-2) + 1 with a(0) = a(1) = 0.

Original entry on oeis.org

0, 0, 1, 1, 2, 3, 7, 22, 155, 3411, 528706, 1803416167, 953476947989903, 1719515742866809222961802, 1639518622529236077952144318816050685207, 2819178082162327154499022366029959843954512194276761760087463015

Offset: 0

N. J. A. Sloane, Robert G. Wilson v

If we omit the first three terms of the sequence, a(n)/a(n-1) can be expressed as the continued fraction [a(n-2); a(n-1)]. - Eric Angelini, Feb 10 2005
This may be regarded as a multiplicative dual of the Fibonacci sequence A000045. Write Fibonacci's formula as F(0)=0, F(1)=1; F(n)=[F(n-1)+F(n-2)]*1 with n>1. Swap '+' and '*' and we have the present sequence! - B. Joshipura (bhushit(AT)yahoo.com), Aug 29 2007
a(n+1) divides a(2n+1), a(3n+1), a(4n+1), etc., this is because modulo a(n+1): a(1)=a(n+1)=0 and a(2)=a(n+2)=1 so the sequence repeats modulo a(n+1) with period n. - Isaac Kaufmann, Sep 04 2020

			b(10) / b(5) = 1803416167 / 7 = 257630881. - _Michael Somos_, Dec 29 2012

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..19 (shortened by _N. J. A. Sloane_, Jan 13 2019)
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437.
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437 (original plus references that F.Q. forgot to include - see last page!)
Bhushit Joshipura, 1, 1, 2, 3, 7, ... Multiplication dual of Fibonacci?, posting in newsgroup sci.math, Jul 28 2007.
S. Kak, The Golden Mean and the Physics of Aesthetics, arXiv:physics/0411195 [physics.hist-ph], 2004.

Cf. A250309, A253853, A258113.

a007660 n = a007660_list !! n
a007660_list = 0 : 0 : map (+ 1)
                       (zipWith (*) a007660_list $ tail a007660_list)
-- Reinhard Zumkeller, Jan 17 2015

I:=[0,0]; [n le 2 select I[n] else Self(n-1)*Self(n-2)+1: n in [1..20]]; // Vincenzo Librandi, Nov 14 2011

a[0] = a[1] = 0; a[n_] := a[n - 1]*a[n - 2] + 1; Table[ a[n], {n, 0, 15} ]
RecurrenceTable[{a[0]==a[1]==0,a[n]==a[n-1]a[n-2]+1},a,{n,20}] (* Harvey P. Dale, Nov 12 2011 *)

a(n) := if (n=0 or n=1) then 0 else a(n-1)*a(n-2)+1 $
makelist(a(n),n,0,18); /* Emanuele Munarini, Mar 24 2017 */

a(n) is asymptotic to c^(phi^n) where phi = (1 + sqrt(5))/2 and c = A258113 = 1.1130579759029319... - Benoit Cloitre, Sep 26 2003

b(n) = a(n+1) is a divisibility sequence. - Michael Somos, Dec 29 2012

A144779 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 5.

Original entry on oeis.org

5, 21, 421, 176821, 31265489221, 977530816197201697621, 955566496615167328821993756200407115362021, 913107329453384594090655605142589591944556891901674138343716072975722193082773842421

Offset: 1

Artur Jasinski, Sep 21 2008

nonn

			a(0) = 4, a(1) = 4+1 = 5, a(2) = 4*5+1 = 21, a(3) = 4*5*21+1 = 421, a(4) = 4*5*21*421+1 = 176821, ... - _Philippe Deléham_, Apr 19 2013

Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Problem 958, College Mathematics Journal, Vol. 42, No. 4, September 2011, p. 330.
Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Solution, College Mathematics Journal, Vol. 43, No. 4, September 2012, pp. 340-342.

Cf. A000058, A082732, A144780, A144781, A144782, A144783, A144784, A144785, A144786, A144787, A144788.

a = {}; k = 5; Do[AppendTo[a, k]; k = k^2 - k + 1, {n,1,10}]; a (* Artur Jasinski, Sep 21 2008 *)
NestList[#^2-#+1&,5,8] (* Harvey P. Dale, Jan 17 2012 *)

a(n) = round(2.127995907464107054577351...)^(2^n) = round(A144803^(2^n)). [corrected by Joerg Arndt, Jan 15 2021]

a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 5.

A144781 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 8.

Original entry on oeis.org

8, 57, 3193, 10192057, 103878015699193, 10790642145601683494645152057, 116437957914435303575899742229333045108448631998006179193, 13557798043283806950297045269968250387897834581711367551819275131055206893868524458302302046950954641412419952057

Offset: 1

Artur Jasinski, Sep 21 2008

nonn

Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Problem 958, College Mathematics Journal, Vol. 42, No. 4, September 2011, p. 330.
Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Solution College Mathematics Journal, Vol. 43, No. 4, September 2012, pp. 340-342.

Cf. A000058, A082732, A144805.

a = {}; k = 8; Do[AppendTo[a, k]; k = k^2 - k + 1, {n, 1, 10}]; a
NestList[#^2-#+1&,8,10] (* Harvey P. Dale, Jan 29 2017 *)

a(n) ~ c^(2^n) where is c is 2.74167747444233776776... (A144805).

A144782 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 9.

Original entry on oeis.org

9, 73, 5257, 27630793, 763460694178057, 582872231554839914154126117193, 339740038317718918529575265905277902175236102890836244082057

Offset: 1

Artur Jasinski, Sep 21 2008

nonn

Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Problem 958, College Mathematics Journal, Vol. 42, No. 4, September 2011, p. 330.
Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Solution College Mathematics Journal, Vol. 43, No. 4, September 2012, pp. 340-342.

Cf. A000058, A082732, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A144786, A144787, A144788.

a = {}; r = 9; Do[AppendTo[a, r]; r = r^2 - r + 1, {n, 1, 10}]; a
NestList[#^2-#+1&,9,10] (* Harvey P. Dale, Aug 31 2014 *)

a(n) ~ c^(2^n) with c = 2.918012...

a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 9.

A144783 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 10.

Original entry on oeis.org

10, 91, 8191, 67084291, 4500302031888391, 20252718378218776104731448680491, 410172601707440572557971589875869064610540321970215293555320591, 168241563191450680898537024308131628447885486994777537422995633998657738457104605412468520116391629012196009150161991233268691

Offset: 1

Artur Jasinski, Sep 21 2008

nonn

Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Problem 958, College Mathematics Journal, Vol. 42, No. 4, September 2011, p. 330. Solution published in Vol. 43, No. 4, September 2012, pp. 340-342

Cf. A000058, A082732, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A144786, A144787, A144788.

a = {}; r = 10; Do[AppendTo[a, r]; r = r^2 - r + 1, {n, 1, 10}]; a (* or *)
Table[Round[3.08435104906918990233569320020272148875011089837398848476442237096569188195734783139337492942278549518507672786196650938869338548385641623^(2^n)], {n, 1, 8}] (* Artur Jasinski *)
NestList[#^2-#+1&,10,8] (* Harvey P. Dale, May 07 2017 *)

a(n) = round((3.08435104906918990233569320020272148875011089837398848476442237096569...)^(2^n)) = round(A144807^(2^n)). [corrected by Joerg Arndt, Jan 15 2021]

a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 10.

A144785 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 12.

Original entry on oeis.org

12, 133, 17557, 308230693, 95006159799029557, 9026170399758739819525199160586693, 81471752085480849000657595909467634426991447160798281416700808089557

Offset: 1

Artur Jasinski, Sep 21 2008

nonn

Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Problem 958, College Mathematics Journal, Vol. 42, No. 4, September 2011, p. 330. Solution published in Vol. 43, No. 4, September 2012, pp. 340-342

A000058, A082732, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A144786, A144787, A144788

a = {}; r = 12; Do[AppendTo[a, r]; r = r^2 - r + 1, {n, 1, 10}]; a or Table[Round[3.39277252592669675143137065018187376847206615308598784654603692312172475924599026837940758013759324881455503678006543568111163817496672898^(2^n)], {n, 1, 8}] (*Artur Jasinski*)
NestList[#^2-#+1&,12,6] (* Harvey P. Dale, Jan 01 2016 *)

a(n) =3.39277252592669675143137065018187376847206615308598784654603692312172475924599026837940758013759324881455503678006543568111163817496672898^(2^n) a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 11

A006585 Egyptian fractions: number of solutions to 1 = 1/x_1 + ... + 1/x_n in positive integers x_1 < ... < x_n.

Original entry on oeis.org

1, 0, 1, 6, 72, 2320, 245765, 151182379

Offset: 1

N. J. A. Sloane

All denominators in the expansion 1 = 1/x_1 + ... + 1/x_n are bounded by A000058(n-1), i.e., 0 < x_1 < ... < x_n < A000058(n-1). Furthermore, for a fixed n, x_i <= (n+1-i)*(A000058(i-1)-1). - Max Alekseyev, Oct 11 2012

If on the other hand, x_k need not be unique, see A002966. - Robert G. Wilson v, Jul 17 2013

			The 6 solutions for n=4 are 2,3,7,42; 2,3,8,24; 2,3,9,18; 2,3,10,15; 2,4,5,20; 2,4,6,12.

Marc LeBrun, personal communication.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

M. Le Brun, Email to N. J. A. Sloane, Jul 1991
S. V. Konyagin, Double exponential lower bound for the number of representations of unity by Egyptian fractions. Mathematical Notes, 95:1-2 (2014), 277-281.
T. D. Browning and C. Elsholtz, The number of representations of rationals as a sum of unit fractions, Illinois J. Math. 55:2 (2011), 685-696.
Joel Louwsma, On solutions of Sum_{i=1..n} 1/x_i = 1 in integers of the form 2^a*k^b, where k is a fixed odd positive integer, arXiv:2402.09515 [math.NT], 2024.
Index entries for sequences related to Egyptian fractions

Cf. A000058, A002966, A002967, A280518.

a(n) = A280520(n,1).

a(1)-a(7) are confirmed by Jud McCranie, Dec 11 1999

a(8) from John Dethridge (jcd(AT)ms.unimelb.edu.au), Jan 08 2004

A076393 Decimal expansion of Vardi constant arising in the Sylvester sequence.

Original entry on oeis.org

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

Offset: 1

Benoit Cloitre, Nov 06 2002

Vardi showed A000058(n) = floor(c^(2^(n+1))+1/2) where c=1.26408473...
Named after the Canadian mathematician Ilan Vardi (b. 1957). - Amiram Eldar, Jun 22 2021
This constant is transcendental. - Quanyu Tang, Mar 20 2025

			1.26408473530530111307959958416466949111456...

Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 443-448.
Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., 1994, exercise 4.37, p. 518.
Ilan Vardi, Computational Recreations in Mathematica, Addison-Wesley, 1991, pp. 82-89.

A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437.
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437 (original plus references that F.Q. forgot to include - see last page!)
Thomas Bloom, Let F(k) be the number of solutions to 1 = 1/n_1 + ... + 1/n_k, where 1 <= n_1 < ... < n_k are distinct integers. Find good estimates for F(k), Erdős Problems.
Matthew Brendan Crawford, On the Number of Representations of One as the Sum of Unit Fractions, Master's Thesis, Virginia Polytechnic Institute and State University (2019).
Artūras Dubickas, Transcendency of some constants related to integer sequences of polynomial iterations, Ramanujan J, Vol. 57, 2022, pp. 569-581.
Steven Finch, Exercises in Iterational Asymptotics, arXiv:2411.16062 [math.NT], 2024. See p. 10.
Zheng Li and Quanyu Tang, On a conjecture of Erdős and Graham about the Sylvester's sequence, arXiv:2503.12277 [math.NT], 2025. See p. 3.
Benjamin Nill, Volume and lattice points of reflexive simplices, Discrete & Computational Geometry, Vol. 37, No. 2 (2007), pp. 301-320; arXiv preprint, arXiv:math/0412480 [math.AG], 2004-2007.
Terence Tao, Erdős problem database, see no. 148.
Eric Weisstein's World of Mathematics, Sylvester's Sequence.

Cf. A000058.

digits = 105; For[c = 2; olds = -1; s = 0; j = 1, RealDigits[olds, 10, digits+5] != RealDigits[s, 10, digits+5], j++; c = c^2-c+1, olds = s; s = s + 2^(-j-1)*Log[1+(2*c-1)^-2] // N[#, digits+5]&]; chi = Sqrt[6]/2*Exp[s]; RealDigits[chi, 10, digits] // First (* Jean-François Alcover, Jun 05 2014 *)

Equals lim_{n->oo} A000058(n)^(1/2^(n+1)). - Robert FERREOL, Feb 15 2019

Equals sqrt((3/2) * Product_{k>=0} (1 + 1/(2*A000058(k)-1)^2)^(1/2^(k+1))). - Amiram Eldar, Jun 22 2021

cp's OEIS Frontend

A144784 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 11.

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A001999 a(n) = a(n-1)*(a(n-1)^2 - 3).

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A007660 a(n) = a(n-1)*a(n-2) + 1 with a(0) = a(1) = 0.

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A144779 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 5.

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A144781 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 8.

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A144782 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 9.

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A144783 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 10.

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A144785 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 12.

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