A112845 -id:A112845 - 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

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).

A219161 Recurrence equation a(n+1) = a(n)^3 - 3*a(n) with a(0) = 5.

Original entry on oeis.org

5, 110, 1330670, 2356194280407770990, 13080769480548649962914459850235688797656360638877986030

Offset: 0

Peter Bala, Nov 13 2012

For some general remarks on this recurrence see A001999.

The next term (a(5)) has 166 digits. - Harvey P. Dale, Apr 23 2019

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.

Cf. A001999, A112845, A219160.

RecurrenceTable[{a[n] == a[n - 1]^3 - 3*a[n - 1], a[0] == 5}, a, {n,
  0, 5}] (* G. C. Greubel, Dec 30 2016 *)
NestList[#^3-3#&,5,5] (* Harvey P. Dale, Apr 23 2019 *)

a(n) = (1/2*(5 + sqrt(21)))^(3^n) + (1/2*(5 - sqrt(21)))^(3^n).
Product_{n = 0..inf} (1 + 2/(a(n) - 1)) = sqrt(7/3).
a(n) = 2*T(3^n,5/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. Cf. A001999. - Peter Bala, Feb 01 2017

A006243 Extracting a square root.

Original entry on oeis.org

198, 7761798, 467613464999866416198, 102249460387306384473056172738577521087843948916391508591105798

Offset: 1

N. J. A. Sloane

nonn

Jeffrey Shallit, personal communication.
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 = 1..6
E. B. Escott, Rapid method for extracting a square root, Amer. Math. Monthly, 44 (1937), 644-646.

Cf. A006242, A112845 (differs only for the initial 6).

[n eq 1 select 198 else Self(n-1)^3-3*Self(n-1): n in [1..5]]; // Vincenzo Librandi, Feb 09 2017

RecurrenceTable[{a[1]==198, a[n]==a[n-1]^3 - 3 a[n-1]}, a, {n, 8}] (* Vincenzo Librandi, Feb 09 2017 *)

a(1) = 198, a(n) = a(n-1)^3 - 3*a(n-1). - Sean A. Irvine, Feb 08 2017

a(n) = (99 + 70*sqrt(2))^(3^(n-1)) + (99 - 70*sqrt(2))^(3^(n-1)). - Bruno Berselli, Feb 10 2017

Offset changed by Sean A. Irvine, Feb 08 2017

A219506 Pierce expansion of 2 - sqrt(3).

Original entry on oeis.org

3, 5, 51, 53, 140451, 140453, 2770663499604051, 2770663499604053, 21269209556953516583554114034636483645584976451, 21269209556953516583554114034636483645584976453

Offset: 0

Peter Bala, Nov 22 2012

For x in the open interval (0,1) define the map f(x) = 1 - x*floor(1/x). The n-th term (n >= 0) in the Pierce expansion of x is given by floor(1/f^(n)(x)), where f^(n)(x) denotes the n-th iterate of the map f, with the convention that f^(0)(x) = x.
The present sequence is the case x = 2 - sqrt(3).
Shallit has shown that the Pierce expansion of the quadratic irrational (c - sqrt(c^2 - 4))/2 has the form [c(0) - 1, c(0) + 1, c(1) - 1, c(1) + 1, c(2) - 1, c(2) + 1, ...], where c(0) = c and c(n+1) = c(n)^3 - 3*c(n). This is the case c = 4. For other cases see A006276 (c = 3), A219507 (c = 5) and A006275 (essentially c = 6 apart from the initial term).
The Pierce expansion of ((c - sqrt(c^2 - 4))/2)^(3^n) is [c(n) - 1, c(n) + 1, c(n+1) - 1, c(n+1) + 1, c(n+2) - 1, c(n+2) + 1, ...].

			We have the alternating series expansions
2 - sqrt(3) = 1/3 - 1/(3*5) + 1/(3*5*51) - 1/(3*5*51*53) + ...
(2 - sqrt(3))^3 = 1/51 - 1/(51*53) + 1/(51*53*140451) - ...
(2 - sqrt(3))^9 = 1/140451 - 1/(140451*140453) + ....

G. C. Greubel, Table of n, a(n) for n = 0..13
F. L. Bauer, Letters to the editor: An Infinite Product for Square-Rooting with Cubic Convergence, The Mathematical Intelligencer, Vol. 20, Issue 1, (1998), 12-14.
T. A. Pierce, On an algorithm and its use in approximating roots of algebraic equations, Amer. Math. Monthly, Vol. 36 No. 10, (1929) p.523-525.
Jeffrey Shallit, Some predictable Pierce expansions, Fib. Quart., 22 (1984), 332-335.
Eric Weisstein's World of Mathematics, Pierce Expansion

Cf. A006275, A006276, A112845, A219160, A219507, A219508.

PierceExp[A_, n_] := Join[Array[1 &, Floor[A]], First@Transpose@ NestList[{Floor[1/Expand[1 - #[[1]] #[[2]]]], Expand[1 - #[[1]] #[[2]]]} &, {Floor[1/(A - Floor[A])], A - Floor[A]}, n - 1]]; PierceExp[N[2 - Sqrt[3] , 7!], 10] (* G. C. Greubel, Nov 14 2016 *)

a(2*n) = (2 + sqrt(3))^(3^n) + (2 - sqrt(3))^(3^n) - 1.
a(2*n + 1) = (2 + sqrt(3))^(3^n) + (2 - sqrt(3))^(3^n) + 1.
From Peter Bala, Jan 18 2022: (Start)
a(2*n+2) = a(2*n)^3 + 3*a(2*n)^2 - 3; a(2*n+1) = a(2*n-1)^3 - 3*a(2*n-1)^2 + 3.
a(2*n) = 6*(Product_{k = 1..n-1} a(2*k))^2 - 3, with a(0) = 1;
a(2*n+1) = 2*(Product_{k = 0..n-1} a(2*k+1))^2 + 3, with a(1) = 5.
sqrt(3) = (1 + 2/3)*(1 + 2/51)*(1 + 2/140451)*(1 + 2/2770663499604051)* ....  See Bauer.
1/sqrt(3) = (1 - 2/5)*(1 - 2/53)*(1 - 2/140453)*(1 - 2/2770663499604053)* .... (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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A219160 Recurrence equation a(n+1) = a(n)^3 - 3*a(n) with a(0) = 4.

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A219161 Recurrence equation a(n+1) = a(n)^3 - 3*a(n) with a(0) = 5.

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A006243 Extracting a square root.

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A219506 Pierce expansion of 2 - sqrt(3).

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

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