A271223 -id:A271223 - cp's OEIS frontend

A006267 Continued cotangent for the golden ratio.

1, 4, 76, 439204, 84722519070079276, 608130213374088941214747405817720942127490792974404

Offset: 0

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

Harry J. Smith, Table of n, a(n) for n = 0..7
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.
Jeffrey Shallit, Predictable regular continued cotangent expansions, J. Res. Nat. Bur. Standards Sect. B, Vol. 80B, No. 2 (1976), pp. 285-290.
Zalman Usiskin, Problem B-266, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 11, No. 3 (1973), p. 334; Lucas Numbers for Powers of 3, Solution to Problem B-266 by David Zeitlin, ibid., Vol. 12, No. 3 (1974), p. 315-316.
Eric Weisstein's World of Mathematics, Lehmer Cotangent Expansion.

Cf. A000032, A000204, A001622, A001999, A002666, A002667, A002668, A006266, A002813, A045529, A271223, A268924.

a := proc(n) option remember; if n = 1 then 4 else a(n-1)^3 + 3*a(n-1) end if; end: seq(a(n), n = 1..5); # Peter Bala, Nov 15 2022

c = N[GoldenRatio, 1000]; Table[Round[c^(3^n)], {n, 1, 8}] (* Artur Jasinski, Sep 22 2008 *)
a = {}; x = 4; Do[AppendTo[a, x]; x = x^3 + 3 x, {n, 1, 10}]; a (* Artur Jasinski, Sep 24 2008 *)

a(n)=fibonacci(3^n+1) + fibonacci(3^n-1) \\ Andrew Howroyd, Dec 30 2024

a(n)={my(t=1); for(i=1, n, t = t^3 + 3*t); t} \\ Andrew Howroyd, Dec 30 2024

(1+sqrt(5))/2 = cot(Sum_{n>=0} (-1)^n*acot(a(n))); let b(0) = (1+sqrt(5))/2, b(n) = (b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1))) then a(n) = floor(b(n)). - Benoit Cloitre, Apr 10 2003
a(n) = A000204(3^n). - Benoit Cloitre, Sep 18 2005
a(n) = round(c^(3^n)) where c = GoldenRatio = 1.6180339887498948482... = (sqrt(5)+1)/2 (A001622). - Artur Jasinski, Sep 22 2008
a(n) = a(n-1)^3 + 3*a(n-1), a(0) = 1. - Artur Jasinski, Sep 24 2008
a(n+1) = Product_{k = 0..n} A002813(k). Thus a(n) divides a(n+1). - Peter Bala, Nov 22 2012
Sum_{n>=0} a(n)^2/A045529(n+1) = 1. - Amiram Eldar, Jan 12 2022
a(n) = Product_{k=0..n-1} (Lucas(2*3^k) + 1) (Usiskin, 1973). - Amiram Eldar, Jan 29 2022
From Peter Bala, Nov 15 2022: (Start)
a(n) = Lucas(3^n) for n >= 1.
a(n) == 1 (mod 3) for n >= 1.
a(n+1) == a(n) (mod 3^(n+1)) for n >= 1 (a particular case of the Gauss congruences for the Lucas numbers).
The smallest positive residue of a(n) mod 3^n = A268924(n).
In the ring of 3-adic integers the limit_{n -> oo} a(n) exists and is equal to A271223. Cf. A006266. (End)

The next term is too large to include.

A268924 One of the two successive approximations up to 3^n for the 3-adic integer sqrt(-2). These are the numbers congruent to 1 mod 3 (except for n = 0).

Original entry on oeis.org

0, 1, 4, 22, 22, 22, 508, 508, 2695, 2695, 2695, 2695, 356989, 888430, 4077076, 4077076, 18425983, 18425983, 147566146, 534986635, 534986635, 7508555437, 28429261843, 28429261843, 122572440670, 405001977151

Offset: 0

Wolfdieter Lang, Apr 05 2016

The other approximation for the 3-adic integer sqrt(-2) with numbers 2 (mod 3) is given in A271222.
For the digits of this 3-adic integer sqrt(-2), that is the scaled first differences, see A271223. This 3-adic number has the digits read from the right to the left ...2202101200022211102201101021200010200211 = u.
The companion 3-adic number has digits ...20020121022200011120021121201022212022012 = -u. See A271224.
For details see the W. Lang link ``Note on a Recurrence or Approximation Sequences of p-adic Square Roots'' given under A268922, also for the Nagell reference and Hensel lifting. Here p = 3, b = 2, x_1 = 1 and  z_1 = 1.

			n=2: 4^2 + 2 = 18 == 0 (mod 3^2), and 4 is the only solution from {0, 1, ..., 8} which is congruent to 1 modulo 3.
n=3: the only solution of  X^2 + 2 == 0 (mod 3^3) with X from {0, ..., 26} and X == 1 (mod 3) is 22. The number 5 = A271222(3)  also satisfies the first congruence but not the second one: 5  == 2 (mod 3).
n=4: the only solution of X^2 + 2 == 0 (mod 3^4) with X from {0, ..., 80} and X == 1 (mod 3) is also 22. The number 59 = A271222(4) also satisfies the first congruence but not the second one: 59  == 2 (mod 3).

Trygve Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964, p. 87.

Seiichi Manyama, Table of n, a(n) for n = 0..2096
Peter Bala, A note on A268924 and A271222, Nov 28 2022.
Wikipedia, Hensel's Lemma.

Cf. A000032, A006267, A268922, A271222, A271223, A271224.

with(padic):D1:=op(3,op([evalp(RootOf(x^2+2),3,20)][1])): 0,seq(sum('D1[k]*3^(k-1)','k'=1..n), n=1..20);
# alternative program based on the Lucas numbers L(3^n) = A006267(n)
a := proc(n) option remember; if n = 1 then 1 else irem(a(n-1)^3 + 3*a(n-1), 3^n) end if; end: seq(a(n), n = 1..22); # Peter Bala, Nov 15 2022

a(n) = truncate(sqrt(-2+O(3^(n)))); \\ Michel Marcus, Apr 09 2016

def a268924(n):
    ary=[0]
    a, mod = 1, 3
    for i in range(n):
          b=a%mod
          ary.append(b)
          a=b**2 + b + 2
          mod*=3
    return ary
print(a268924(100)) # Indranil Ghosh, Aug 04 2017, after Ruby

def A268924(n)
  ary = [0]
  a, mod = 1, 3
  n.times{
    b = a % mod
    ary << b
    a = b * b + b + 2
    mod *= 3
  }
  ary
end
p A268924(100) # Seiichi Manyama, Aug 03 2017

a(n)^2 + 2 == 0 (mod 3^n), and a(n) == 1 (mod 3). Representatives of the complete residue system {0, 1, ..., 3^n-1} are taken.
Recurrence for n >= 1: a(n) = modp(a(n-1) +  a(n-1)^2 + 2, 3^n), n >= 2, with a(1) = 1. Here modp(a, m) is used to pick the representative of the residue class a modulo m from the smallest nonnegative complete residue system {0, 1, ..., m-1}.
a(n) = 3^n - A271222(n), n >= 1.
a(n) == Lucas(3^n) (mod 3^n). - Peter Bala, Nov 10 2022

A271224 Digits of one of the two 3-adic integers sqrt(-2). Here the sequence with first digit 2.

Original entry on oeis.org

2, 1, 0, 2, 2, 0, 2, 1, 2, 2, 2, 0, 1, 0, 2, 1, 2, 1, 1, 2, 0, 0, 2, 1, 1, 1, 0, 0, 0, 2, 2, 2, 0, 1, 2, 1, 0, 2, 0, 0, 2, 0, 2, 1, 0, 2, 1, 0, 0, 0, 1, 2, 0, 2, 1, 0, 2, 0, 2, 2, 1

Offset: 0

Wolfdieter Lang, Apr 05 2016

This is the scaled first difference sequence of A271222. See the formula.
The digits of the other 3-adic integer sqrt(-2), are given in A271223. See also a comment on A268924 for the two 3-adic numbers sqrt(-2), called there u and -u.
a(n) is the unique solution of the linear congruence 2*A271222(n)*a(n) + A271226(n) == 0 (mod 3), n>=1. Therefore only the values 0, 1, and 2 appear. See the Nagell reference given in A268922, eq. (6) on p. 86, adapted to this case.
  a(0) = 2 follows from the formula given below.
For details see the Wolfdieter Lang link under A268992.
The first k digits in the base 3 representation of A002203(3^k) = A006266(k) give the first k terms of the sequence. - Peter Bala, Nov 26 2022

			a(4) = 2 because 2*59*2 + 43 = 279 == 0 (mod 3).
a(4) = - 43*(2*59) (mod 3) = -1*(2*(-1)) (mod 3) = 2.
A271222(5) = 221  = 2*3^0 + 1*3^1 + 0*3^2 + 2*3^3 + 2*3^4.

Trygve Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964, pp. 86 and 77-78.

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Peter Bala, A note on A268924 and A271222, Nov 28 2022.
BCMATH Congruence Programs, Finding a p-adic square root of a quadratic residue (mod p), p an odd prime.

Cf. A268924, A268992, A271222, A271226, A271223 (companion).

a(n) = truncate(-sqrt(-2+O(3^(n+1))))\3^n; \\ Michel Marcus, Apr 09 2016

a(n) = (b(n+1) - b(n))/3^n, n >= 0, with b(n) = A271222(n), n >= 0.
a(n) = - A271226(n)*2*A271222(n) (mod 3), n >= 1. Solution of the linear congruence given above in a comment. See, e.g., Nagell, Theorem 38 pp. 77-78.
A271222(n+1) = sum(a(k)*3^k, k=0..n), n >= 0.

A271222 One of the two successive approximations up to 3^n for the 3-adic integer sqrt(-2). These are the numbers congruent to 2 mod 3 (except for the initial 0).

Original entry on oeis.org

0, 2, 5, 5, 59, 221, 221, 1679, 3866, 16988, 56354, 174452, 174452, 705893, 705893, 10271831, 24620738, 110714180, 239854343, 627274832, 2951797766, 2951797766, 2951797766, 65713916984, 159857095811, 442286632292

Offset: 0

Wolfdieter Lang, Apr 05 2016

The other approximation for the 3-adic integer sqrt(-2) with numbers 1 (mod 3) is given in A268924.
For the digits of this 3-adic integer sqrt(-2), that is the scaled first differences, see A271224. This 3-adic number has the digits read from the right to the left ... 20020121022200011120021121201022212022012 = -u. For the digits of u see A271223.
For details see the W. Lang link ``Note on a Recurrence or Approximation Sequences of p-adic Square Roots'' given under A268922, also for the Nagell reference and Hensel lifting. Here p = 3,  b = 2, x_2 = 2 and  z_2 = 2.

			n=2: 5^2 + 2 = 27 == 0 (mod 3^2), and 5 is the only solution from {0, 1, ..., 8} which is congruent to 2 modulo 3.
n=3: the only solution of X^2 + 2 == 0 (mod 3^3) with X from {0, ..., 26} and X == 2(mod 3) is 5. The number 22 = A268924(3) also satisfies the first congruence but not the second one: 22  == 1 (mod 3).
n=4: the only solution of X^2 + 2 == 0 (mod 3^4) with X from {0, ..., 80} and X == 2 (mod 3) is 59. The number 22 = A268924(4) also satisfies the first congruence but not the second one: 59  == 1 (mod 3).

Trygve Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964, p. 87.

Seiichi Manyama, Table of n, a(n) for n = 0..2095
Peter Bala, A note on A268924 and A271222, Nov 28 2022.
Wikipedia, Hensel's Lemma.

Cf. A002293, A268922, A268924, A271223, A271224.

with(padic):D2:=op(3,op([evalp(RootOf(x^2+2),3,20)][2])): 0,seq(sum('D2[k]*3^(k-1)','k'=1..n), n=1..20);

a(n) = if (n, 3^n - truncate(sqrt(-2+O(3^(n)))), 0); \\ Michel Marcus, Apr 09 2016

def a271222(n):
      ary=[0]
      a, mod = 2, 3
      for i in range(n):
          b=a%mod
          ary.append(b)
          a=2*b**2 + b + 4
          mod*=3
      return ary
print(a271222(100)) # Indranil Ghosh, Aug 04 2017, after Ruby

def A271222(n)
  ary = [0]
  a, mod = 2, 3
  n.times{
    b = a % mod
    ary << b
    a = 2 * b * b + b + 4
    mod *= 3
  }
  ary
end
p A271222(100) # Seiichi Manyama, Aug 03 2017

a(n)^2 + 2 == 0 (mod 3^n), and a(n) == 2 (mod 3). Representatives of the complete residue system {0, 1, ..., 3^n-1} are taken.
Recurrence for n >= 1: a(n) = modp(a(n-1) +  2*(a(n-1)^2 + 2), 3^n), n >= 2, with a(1) = 2. Here modp(a, m) is used to pick the representative of the residue class a modulo m from the smallest nonnegative complete residue system {0, 1, ... , m-1}.
a(n) = 3^n - A268924(n), n >= 1.
a(n) == A002203(3^n) (mod 3^n). - Peter Bala, Nov 10 2022

A318962 Digits of one of the two 2-adic integers sqrt(-7) that ends in 01.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1

Offset: 0

Jianing Song, Sep 06 2018

Over the 2-adic integers there are 2 solutions to x^2 = -7, one ends in 01 and the other ends in 11. This sequence gives the former one. See A318960 for detailed information.

			...10110001110011100100110001100000010110101.

Jianing Song, Table of n, a(n) for n = 0..1000

Cf. A318960.
Digits of p-adic integers:
this sequence, A318963 (2-adic, sqrt(-7));
A271223, A271224 (3-adic, sqrt(-2));
A269591, A269592 (5-adic, sqrt(-4));
A210850, A210851 (5-adic, sqrt(-1));
A290566 (5-adic, 2^(1/3));
A290563 (5-adic, 3^(1/3));
A290794, A290795 (7-adic, sqrt(-6));
A290798, A290799 (7-adic, sqrt(-5));
A290796, A290797 (7-adic, sqrt(-3));
A212152, A212155 (7-adic, (1+sqrt(-3))/2);
A051277, A290558 (7-adic, sqrt(2));
A286838, A286839 (13-adic, sqrt(-1));
A309989, A309990 (17-adic, sqrt(-1)).
Also there are numerous sequences related to digits of 10-adic integers.

a(n) = truncate(-sqrt(-7+O(2^(n+2))))\2^n

a(0) = 1, a(1) = 0; for n >= 2, a(n) = 0 if A318960(n)^2 + 7 is divisible by 2^(n+2), otherwise 1.
a(n) = 1 - A318963(n) for n >= 1.
For n >= 2, a(n) = (A318960(n+1) - A318960(n))/2^n.

Corrected by Jianing Song, Aug 28 2019

A318963 Digits of one of the two 2-adic integers sqrt(-7) that ends in 11.

Original entry on oeis.org

1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0

Offset: 0

Jianing Song, Sep 06 2018

Over the 2-adic integers there are 2 solutions to x^2 = -7, one ends in 01 and the other ends in 11. This sequence gives the latter one. See A318961 for detailed information.

			...01001110001100011011001110011111101001011.

Jianing Song, Table of n, a(n) for n = 0..1000

Cf. A318961.
Digits of p-adic integers:
A318962, this sequence (2-adic, sqrt(-7));
A271223, A271224 (3-adic, sqrt(-2));
A269591, A269592 (5-adic, sqrt(-4));
A210850, A210851 (5-adic, sqrt(-1));
A290566 (5-adic, 2^(1/3));
A290563 (5-adic, 3^(1/3));
A290794, A290795 (7-adic, sqrt(-6));
A290798, A290799 (7-adic, sqrt(-5));
A290796, A290797 (7-adic, sqrt(-3));
A212152, A212155 (7-adic, (1+sqrt(-3))/2);
A051277, A290558 (7-adic, sqrt(2));
A286838, A286839 (13-adic, sqrt(-1));
A309989, A309990 (17-adic, sqrt(-1)).
Also there are numerous sequences related to digits of 10-adic integers.

a(n) = if(n==1, 1, truncate(sqrt(-7+O(2^(n+2))))\2^n)

a(0) = a(1) = 1; for n >= 2, a(n) = 0 if A318961(n)^2 + 7 is divisible by 2^(n+2), otherwise 1.
a(n) = 1 - A318962(n) for n >= 1.
For n >= 2, a(n) = (A318961(n+1) - A318961(n))/2^n.

Corrected by Jianing Song, Aug 28 2019

A309989 Digits of one of the two 17-adic integers sqrt(-1).

Original entry on oeis.org

4, 2, 10, 5, 12, 16, 12, 8, 13, 3, 14, 0, 6, 1, 0, 15, 1, 8, 14, 5, 7, 16, 14, 1, 5, 13, 9, 6, 5, 12, 16, 15, 9, 16, 14, 12, 16, 1, 3, 6, 4, 10, 15, 5, 16, 12, 2, 1, 5, 4, 0, 15, 2, 11, 14, 9, 5, 1, 11, 16, 15, 7, 5, 6, 14, 3, 12, 0, 0, 11, 12, 13, 9, 5, 4, 16, 13

Offset: 0

Jianing Song, Aug 26 2019

This square root of -1 in the 17-adic field ends with digit 4. The other, A309990, ends with digit 13 (D when written as a 17-adic number).

			The solution to x^2 == -1 (mod 17^4) such that x == 4 (mod 17) is x == 27493 (mod 17^4), and 27493 is written as 5A24 in heptadecimal, so the first four terms are 4, 2, 10 and 5.

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Wikipedia, p-adic number

Cf. A286877, A286878.
Digits of p-adic square roots:
A318962, A318963 (2-adic, sqrt(-7));
A271223, A271224 (3-adic, sqrt(-2));
A269591, A269592 (5-adic, sqrt(-4));
A210850, A210851 (5-adic, sqrt(-1));
A290794, A290795 (7-adic, sqrt(-6));
A290798, A290799 (7-adic, sqrt(-5));
A290796, A290797 (7-adic, sqrt(-3));
A051277, A290558 (7-adic, sqrt(2));
A321074, A321075 (11-adic, sqrt(3));
A321078, A321079 (11-adic, sqrt(5));
A322091, A322092 (13-adic, sqrt(-3));
A286838, A286839 (13-adic, sqrt(-1));
A322087, A322088 (13-adic, sqrt(3));
this sequence, A309990 (17-adic, sqrt(-1)).

a(n) = truncate(sqrt(-1+O(17^(n+1))))\17^n

a(n) = (A286877(n+1) - A286877(n))/17^n.

For n > 0, a(n) = 16 - A309990(n).

A309990 Digits of one of the two 17-adic integers sqrt(-1).

Original entry on oeis.org

13, 14, 6, 11, 4, 0, 4, 8, 3, 13, 2, 16, 10, 15, 16, 1, 15, 8, 2, 11, 9, 0, 2, 15, 11, 3, 7, 10, 11, 4, 0, 1, 7, 0, 2, 4, 0, 15, 13, 10, 12, 6, 1, 11, 0, 4, 14, 15, 11, 12, 16, 1, 14, 5, 2, 7, 11, 15, 5, 0, 1, 9, 11, 10, 2, 13, 4, 16, 16, 5, 4, 3, 7, 11, 12, 0

Offset: 0

Jianing Song, Aug 26 2019

This square root of -1 in the 17-adic field ends with digit 13 (D when written as a 17-adic number). The other, A309989, ends with digit 4.

			The solution to x^2 == -1 (mod 17^4) such that x == 13 (mod 17) is x == 56028 (mod 17^4), and 56028 is written as B6ED in heptadecimal, so the first four terms are 13, 14, 6 and 11.

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Wikipedia, p-adic number

Cf. A286877, A286878.
Digits of p-adic square roots:
A318962, A318963 (2-adic, sqrt(-7));
A271223, A271224 (3-adic, sqrt(-2));
A269591, A269592 (5-adic, sqrt(-4));
A210850, A210851 (5-adic, sqrt(-1));
A290794, A290795 (7-adic, sqrt(-6));
A290798, A290799 (7-adic, sqrt(-5));
A290796, A290797 (7-adic, sqrt(-3));
A051277, A290558 (7-adic, sqrt(2));
A321074, A321075 (11-adic, sqrt(3));
A321078, A321079 (11-adic, sqrt(5));
A322091, A322092 (13-adic, sqrt(-3));
A286838, A286839 (13-adic, sqrt(-1));
A322087, A322088 (13-adic, sqrt(3));
A309989, this sequence (17-adic, sqrt(-1)).

a(n) = truncate(-sqrt(-1+O(17^(n+1))))\17^n

a(n) = (A286878(n+1) - A286878(n))/17^n.

For n > 0, a(n) = 16 - A309989(n).

A309474 Digits of one of the two 3-adic integers sqrt(-1/2).

Original entry on oeis.org

1, 2, 1, 2, 2, 1, 2, 0, 1, 1, 1, 0, 2, 1, 2, 0, 1, 2, 0, 1, 0, 0, 1, 2, 0, 2, 1, 1, 1, 2, 2, 2, 1, 0, 1, 2, 1, 2, 1, 1, 2, 1, 2, 0, 0, 1, 2, 1, 1, 1, 0, 1, 0, 1, 2, 1, 2, 1, 2, 2, 0, 2, 1, 0, 1, 2, 1, 0, 2, 1, 2, 1, 0, 2, 1, 2, 2, 0, 2, 1, 1, 2, 0, 2, 0, 2, 0, 2, 2, 0, 2, 2, 0, 0, 0, 1, 2

Offset: 0

Seiichi Manyama, Aug 04 2019

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A271223, A309475, A309476.

T:= select(t -> padic:-ratvaluep(t,1)=1, [padic:-rootp(x^2+1/2,3,100)]):
op([1,1,3],T); # Robert Israel, Aug 05 2019

Vecrev(digits(truncate(sqrt(-1/2+O(3^100))), 3))

p           = ...122121, p^2   = ...111111.
q = A271223 = ...200211, p * q = ...000001.
a(n) = (b(n+1) - b(n))/3^n, with b(n) = A309476(n).

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)

cp's OEIS Frontend

A006267 Continued cotangent for the golden ratio.

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A268924 One of the two successive approximations up to 3^n for the 3-adic integer sqrt(-2). These are the numbers congruent to 1 mod 3 (except for n = 0).

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A271224 Digits of one of the two 3-adic integers sqrt(-2). Here the sequence with first digit 2.

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A271222 One of the two successive approximations up to 3^n for the 3-adic integer sqrt(-2). These are the numbers congruent to 2 mod 3 (except for the initial 0).

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A318962 Digits of one of the two 2-adic integers sqrt(-7) that ends in 01.

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A318963 Digits of one of the two 2-adic integers sqrt(-7) that ends in 11.

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