A219161 -id:A219161 - cp's OEIS frontend

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

3, 18, 5778, 192900153618, 7177905237579946589743592924684178

Offset: 0

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

A112845 Recurrence a(n) = a(n-1)^3 - 3*a(n-1) with a(0) = 6.

Original entry on oeis.org

6, 198, 7761798, 467613464999866416198, 102249460387306384473056172738577521087843948916391508591105798

Offset: 0

Eric W. Weisstein, Sep 21 2005

Identical to A006243 apart from the initial term. For some general remarks on this recurrence see A001999. - Peter Bala, Nov 13 2012

G. C. Greubel, Table of n, a(n) for n = 0..6
E. B. Escott, Rapid method for extracting a square root, Amer. Math. Monthly, 44 (1937), 644-646.
N. J. Fine, Infinite products for k-th roots, Amer. Math. Monthly Vol. 84, No. 8, Oct. 1977, 629-630.
Eric Weisstein's World of Mathematics, Pierce Expansion

Cf. A006275, A006276.
Cf. A006243. - R. J. Mathar, Aug 15 2008
Cf. A001999, A219160, A219161.

RecurrenceTable[{a[n] == a[n - 1]^3 - 3*a[n - 1], a[0] == 6}, a, {n,
  0, 5}] (* G. C. Greubel, Dec 30 2016 *)
NestList[#^3-3#&,6,5] (* Harvey P. Dale, Jul 23 2025 *)

a(n) = -2*cos(3^n*arccos(-3)).
From Peter Bala, Nov 13 2012: (Start)
a(n) = (3 + 2*sqrt(2))^(3^n) + (3 - 2*sqrt(2))^(3^n).
Product {n = 0..inf} (1 + 2/(a(n) - 1)) = sqrt(2).
(End)

A219160 Recurrence equation a(n+1) = a(n)^3 - 3*a(n) with a(0) = 4.

Original entry on oeis.org

4, 52, 140452, 2770663499604052, 21269209556953516583554114034636483645584976452

Offset: 0

Peter Bala, Nov 13 2012

For some general remarks on this recurrence see A001999.

G. C. Greubel, Table of n, a(n) for n = 0..6
E. B. Escott, Rapid method for extracting a square root, Amer. Math. Monthly, 44 (1937), 644-646.
N. J. Fine, Infinite products for k-th roots, Amer. Math. Monthly Vol. 84, No. 8, Oct. 1977.

Cf. A001999, A112845, A219161.

RecurrenceTable[{a[n] == a[n - 1]^3 - 3*a[n - 1], a[0] == 4}, a, {n,
  0, 5}] (* G. C. Greubel, Dec 30 2016 *)

a(n) = (2 + sqrt(3))^(3^n) + (2 - sqrt(3))^(3^n).

Product {n = 0..inf} (1 + 2/(a(n) - 1)) = sqrt(3). The rate of convergence is cubic. Fine remarks that taking the first twelve factors of the product would give well over 300,000 correct decimals for sqrt(3).

A002000 a(n+1) = a(n)*(a(n)^2 - 3) with a(0) = 7.

Original entry on oeis.org

7, 322, 33385282, 37210469265847998489922, 51522323599677629496737990329528638956583548304378053615581043535682

Offset: 0

N. J. A. Sloane

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).

Vincenzo Librandi, Table of n, a(n) for n = 0..6
E. B. Escott, Rapid method for extracting a square root, Amer. Math. Monthly, 44 (1937), 644-646.

Cf. A000032, A001999, A006267, A219161, A271223.

[n eq 1 select 7 else Self(n-1)^3 - 3*Self(n-1): n in [1..6]]; // Vincenzo Librandi, Feb 09 2017

a := proc(n) option remember; if n = 0 then 7 else a(n-1)^3 - 3*a(n-1) end if; end;
seq(a(n), n = 0..4); # Peter Bala, Nov 15 2022

RecurrenceTable[{a[0] == 7, a[n] == a[n - 1]^3 - 3 a[n - 1]}, a, {n, 0, 8}]
(* Vincenzo Librandi, Feb 09 2017 *)
NestList[#(#^2-3)&,7,4] (* Harvey P. Dale, Aug 11 2021 *)

From Peter Bala, Feb 01 2017: (Start)
a(n) = ((7 + sqrt(45))/2)^(3^n) + ((7 - sqrt(45))/2)^(3^n).
a(n) = 2*T(3^n,7/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
Product_{n >= 0} (1 + 2/(a(n) - 1)) = 3*sqrt(5)/5.
Cf. A001999 and A219161. (End)
From Peter Bala, Nov 15 2022: (Start)
a(n) = Lucas(4*(3^n)).
a(n+1) == a(n) (mod 3^(n+1)) (a particular case of the Gauss congruences for the Lucas numbers).
Conjecture: a(n+1) == a(n) (mod 3^(n+r+2)) for n >= r.
The least positive residue of a(n) mod(3^n) = 3^n - 2 = A058481(n). In the ring of 3-adic integers the limit_{n -> oo} a(n) exists and is equal to -2.
Product_{k = 0..n} (a(k) - 1) = (1/3)*Lucas(6*(3^n)). (End)

A282180 a(n+1) = a(n)*(a(n)^2 - 3) with a(0) = 8.

Original entry on oeis.org

8, 488, 116212808, 1569502402942700328379688, 3866214585126515728777536857817155683642224883875510905654220958052649608

Offset: 0

Vincenzo Librandi, Feb 10 2017

nonn

E. B. Escott, Rapid method for extracting a square root, Amer. Math. Monthly, 44 (1937), 644-646.

Cf. similar sequences with initial value k: A001999 (k=3), A219160 (k=4), A219161 (k=5), A112845 (k=6), A002000 (k=7), this sequence (k=8), A282181 (k=9), A006242 (k=10), A006243 (k=198).

[n eq 1 select 8 else Self(n-1)^3 - 3*Self(n-1): n in [1..6]];

RecurrenceTable[{a[0] == 8, a[n] == a[n-1]^3 - 3 a[n-1]}, a, {n, 8}]

a(n) = (4 + sqrt(15))^(3^n) + (4 - sqrt(15))^(3^n). - Bruno Berselli, Feb 10 2017

a(n) = -2*cos(3^n * arccos(-4)). - Daniel Suteu, Feb 10 2017

cp's OEIS Frontend

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

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A112845 Recurrence a(n) = a(n-1)^3 - 3*a(n-1) with a(0) = 6.

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A219160 Recurrence equation a(n+1) = a(n)^3 - 3*a(n) with a(0) = 4.

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A002000 a(n+1) = a(n)*(a(n)^2 - 3) with a(0) = 7.

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A282180 a(n+1) = a(n)*(a(n)^2 - 3) with a(0) = 8.

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