A006267 -id:A006267 - cp's OEIS frontend

A045529 a(n+1) = 5a(n)^3 - 3a(n), a(0) = 1.

1, 2, 34, 196418, 37889062373143906, 271964099255182923543922814194423915162591622175362

Offset: 0

Jose Eduardo Blazek, Dec 11 1999

The next term, a(6), has 153 digits. - Harvey P. Dale, Oct 24 2011

Seiichi Manyama, Table of n, a(n) for n = 0..7
Daniel Duverney and Takeshi Kurosawa, Transcendence of infinite products involving Fibonacci and Lucas numbers, Research in Number Theory, Vol. 8 (2002), Article 68.
Zalman Usiskin, Problem B-265, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 11, No. 3 (1973), p. 333; Fibonacci Numbers for Powers of 3, Solution to Problem B-265 by Ralph Garfield and David Zeitlin, ibid., Vol. 12, No. 3 (1974), p. 315.

Cf. (k^n)-th Fibonacci number: A058635 (k=2), this sequence (k=3), A145231 (k=4), A145232 (k=5), A145233 (k=6), A145234 (k=7), A250487 (k=8), A250488 (k=9), A250489 (k=10).

Cf. A000032, A000045, A001622, A001999, A002814, A006267, A094214.

a := proc(n) option remember; if n = 0 then 1 else 5*a(n-1)^3 - 3*a(n-1) end if; end:
seq(a(n), n = 0..5); # Peter Bala, Nov 24 2022

G = (1 + Sqrt[5])/2; Table[Expand[(G^(3^n) - (1 - G)^(3^n))/Sqrt[5]], {n, 1, 7}] (* Artur Jasinski, Oct 05 2008 *)
Table[Round[(4/5)^(1/2)*Cosh[3^n*ArcCosh[((5/4)^(1/2))]]], {n, 1, 4}] (* Artur Jasinski, Oct 05 2008 *)
RecurrenceTable[{a[0]==1,a[n]==5a[n-1]^3-3a[n-1]},a[n],{n,6}] (* Harvey P. Dale, Oct 24 2011 *)
NestList[5#^3-3#&,1,5] (* Harvey P. Dale, Dec 21 2014 *)

A045529(n):=fib(3^n)$
makelist(A045529(n),n,0,10); /* Martin Ettl, Nov 12 2012 */

The first example I know in which a(n) can be expressed as (4/5)^(1/2)*cosh(3^n*arccosh((5/4)^(1/2))).
a(n) = Fibonacci(3^n). - Leroy Quet, Mar 17 2002
a(n+1) = a(n)*A002814(n+1). - Lekraj Beedassy, Jun 16 2003
a(n) = (phi^(3^n) - (1 - phi)^(3^n))/sqrt(5), where phi is the golden ratio (A001622). - Artur Jasinski, Oct 05 2008
a(n) = Product_{k=0..n-1} (Lucas(2*3^k) - 1) (Usiskin, 1973). - Amiram Eldar, Jan 29 2022
From Peter Bala, Nov 24 2022: (Start)
a(2*n+2) == a(2*n) (mod 3^(2*n+1)); a(2*n+3) == a(2*n+1) (mod 3^(2*n+2));
a(2*n+1) + a(2*n) == 0 (mod 3^(2*n+1)).
a(2*n) == 1 (mod 3) and a(2*n+1) == 2 (mod 3).
5*a(n)^2 == 2 (mod 3^(n+1)).
In the ring of 3-adic integers, the sequences {a(2*n)} and {a(2*n+1)} are both Cauchy sequences and converge to the pair of 3-adic roots of the quadratic equation 5*x^2 - 2 = 0. (End)
From Amiram Eldar, Jan 07 2023: (Start)
Product_{n>=1} (1 + 2/(sqrt(5)*a(n)-1)) = phi (A001622).
Product_{n>=1} (1 - 2/(sqrt(5)*a(n)+1)) = 1/phi (A094214).
Both formulas are from Duverney and Kurosawa (2022). (End)

A006266 A continued cotangent.

Original entry on oeis.org

2, 14, 2786, 21624372014, 10111847525912679844192131854786, 1033930953043290626825587838528711318150300040875029341893199068078185510802565166824630504014

Offset: 0

N. J. A. Sloane

The next (6th) term is 280 digits long. - M. F. Hasler, Oct 06 2014

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
Jeffrey Shallit, Predictable regular continued cotangent expansions, J. Res. Nat. Bur. Standards Sect. B 80B (1976), no. 2, 285-290.

Cf. A002203, A006267, A114525, A145451, A145452, A271222, A271224.

[Evaluate(DicksonFirst(3^n, -1), 2): n in [0..7]]; // G. C. Greubel, Mar 25 2022

a := proc(n) option remember; if n = 1 then 14 else a(n-1)^3 + 3*a(n-1) end if; end: seq(a(n), n = 1..5); # Peter Bala Nov 15 2022

Table[Round[(1+Sqrt[2])^(3^n)],{n,0,10}] (* Artur Jasinski, Sep 24 2008 *)
LucasL[3^Range[0, 7], 2] (* G. C. Greubel, Mar 25 2022 *)

a(n,s=2)=for(i=2,n,s*=(s^2+3));s \\ M. F. Hasler, Oct 06 2014

[lucas_number2(3^n,2,-1) for n in (0..7)] # G. C. Greubel, Mar 25 2022

From Artur Jasinski, Sep 24 2008: (Start)
a(n+1) = a(n)^3 + 3*a(n) with a(0) = 2.
a(n) = round((1+sqrt(2))^(3^n)). [Corrected by M. F. Hasler, Oct 06 2014] (End)
From Peter Bala, Nov 15 2022: (Start)
a(n) = A002203(3^n).
a(n) = L(3^n,2), where L(n,x) denotes the n-th Lucas polynomial of A114525.
a(n) == 2 (mod 3).
a(n+1) == a(n) (mod 3^(n+1)) for n >= 1 (a particular case of the Gauss congruences for the companion Pell numbers).
The smallest positive residue of a(n) mod(3^n) = A271222(n).
In the ring of 3-adic integers the limit_{n -> oo} a(n) exists and is equal to A271224. Cf. A006267. (End)

Edited by M. F. Hasler, Oct 06 2014

Offset corrected by G. C. Greubel, Mar 25 2022

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

A006268 A continued cotangent.

Original entry on oeis.org

3, 36, 46764, 102266868132036, 1069559300034650646049671039050649693658764

Offset: 0

N. J. A. Sloane

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

Jeffrey Shallit, Predictable regular continued cotangent expansions, J. Res. Nat. Bur. Standards Sect. B 80B (1976), no. 2, 285-290.

Continued cotangents: A006267, A006266, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189 (k = 1 to 15 with k=4 being A006267(n+1)).

Cf. A002813, A006273.

a = {}; k = 3; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a (* Artur Jasinski, Oct 03 2008 *)
Table[Round[N[(3/2 + Sqrt[13]/2)^(3^(n - 1)), 1000]], {n, 1, 8}] (* Artur Jasinski, Oct 03 2008 *)

a(n) = if (n==0, 3, a(n-1)^3 + 3*a(n-1)); \\ Michel Marcus, Aug 28 2020

From Artur Jasinski, Oct 03 2008: (Start)
a(n+1) = a(n)^3 + 3*a(n) and a(0)=3.
a(n) = round((3/2 + sqrt(13)/2)^(3^(n - 1))). (End)
From  Peter Bala, Jan 19 2022: (Start)
a(n) = (3/2 + sqrt(13)/2)^(3^(n-1)) + (3/2 - sqrt(13)/2)^(3^(n-1))
a(n) divides a(n+1) and b(n) = a(n+1)/a(n) satisfies the recurrence b(n+1) = b(n)^3 - 3*b(n-1)^2 + 3. For remarks about this recurrence see A002813.
1 + a(n)^2 = A006273(n+1). (End)

A145180 Continued cotangent recurrence a(n+1) = a(n)^3 + 3*a(n) and a(1) = 6.

Original entry on oeis.org

6, 234, 12813606, 2103846732371087589834, 9311985549495522884757461748592522243432897275494229148348315206

Offset: 1

Artur Jasinski, Oct 03 2008

General formula for continued cotangent recurrences type:
a(n+1) = a(n)3 + 3*a(n) and a(1)=k is following:
a(n) = Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))].
The next term (a(6)) has 192 digits. - Harvey P. Dale, Mar 09 2013

J. Shallit, Predictable regular continued cotangent expansions, J. Res. Nat. Bur. Standards Sect. B 80B (1976), no. 2, 285-290.
Eric W. Weisstein, MathWorld: Lehmer Cotangent Expansion

Cf. A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189 (k = 1 to 15 with k=4 being A006267(n+1)).

Cf. A002813.

a = {}; k = 6; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((6 + Sqrt[40])/2)^(3^(n - 1))], {n, 1, 5}] (* Artur Jasinski *)
NestList[#^3+3#&,6,5] (* Harvey P. Dale, Mar 09 2013 *)

a(n+1)=a(n)^3 + 3*a(n) and a(1)=6
a(n)=Floor[((6+Sqrt[6^2+4])/2)^(3^(n-1))]
a(n) divides a(n+1) and b(n) = a(n+1)/a(n) satisfies the recurrence b(n+1) = b(n)^3 - 3*b(n-1)^2 + 3. See A002813. - Peter Bala, Nov 23 2012

A145189 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=15.

Original entry on oeis.org

15, 3420, 40001698260, 64008151994095341241755497070780, 262244184463346778261182615794616508638576477409715732397097802610370956164308073990185129764340

Offset: 1

Artur Jasinski, Oct 03 2008

nonn

General formula for continued cotangent recurrences type:
a(n+1)=a(n)3+3*a(n) and a(1)=k is following:
a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
k=1 see A006267
k=2 see A006266
k=3 see A006268
k=4 see A006267(n+1)
k=5 see A006269
k=6 see A145180
k=7 see A145181
k=8 see A145182
k=9 see A145183
k=10 see A145184
k=11 see A145185
k=12 see A145186
k=13 see A145187
k=14 see A145188
k=15 see A145189

A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189

a = {}; k = 15; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((15 + Sqrt[229])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
NestList[#^3+3#&,15,5] (* Harvey P. Dale, Aug 20 2017 *)

a(n+1)=a(n)3+3*a(n) and a(1)=14

a(n)=Floor[((14+Sqrt[14^2+4])/2)^(3^(n-1))]

A271223 Digits of one of the two 3-adic integers sqrt(-2).

Original entry on oeis.org

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

Offset: 0

Wolfdieter Lang, Apr 05 2016

This is the scaled first difference sequence of A268924. See the formula.
The digits of the other 3-adic integer sqrt(-2), are given in A271224. See also 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*A268924(n)*a(n) + A271225(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) = 1 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 Lucas(3^n) give the first k terms of the sequence. For example, the base 3 representation of Lucas(3^5) = 84722519070079276 begins 1 + 1*3 + 2*(3^2) + 0*(3^3) + 0*(3^4) + ... so the sequence begins [1, 1, 2, 0, 0, ...]. - Peter Bala, Nov 15 2022

			a(4) = 0 because 2*22*3 + 6 = 138 == 0 (mod 3).
a(4) = - 6*(2*22) (mod 3) = -0*(2*1) (mod 3) = 0.
A268924(5) =  22 = 1*3^0 + 1*3^1 + 2*3^2 + 0*3^3 + 0*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
BCMATH Congruence Programs, Finding a p-adic square root of a quadratic residue (mod p), p an odd prime.

Cf. A006267, A268924, A268992, A271224, A271225.

# uses properties of the numbers Lucas(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 proc:
convert(a(70), base, 3); # Peter Bala, Nov 15 2022

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) = A268924(n), n >= 0.
a(n) = - A271225(n)*2*A268924(n) (mod 3), n >= 1. Solution of the linear congruence given above in a comment. See, e.g., Nagell, Theorem 38 pp. 77-78.
A268924(n+1) = sum(a(k)*3^k, k=0..n), n >= 0.

A145181 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=7.

Original entry on oeis.org

7, 364, 48229636, 112186849649044142700364, 1411971263214164889494039458947084336929208169473485667118006013929636

Offset: 1

Artur Jasinski, Oct 03 2008

nonn

General formula for continued cotangent recurrences type:
a(n+1)=a(n)3+3*a(n) and a(1)=k is following:
a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
k=1 see A006267
k=2 see A006266
k=3 see A006268
k=4 see A006267(n+1)
k=5 see A006269
k=6 see A145180
k=7 see A145181
k=8 see A145182
k=9 see A145183
k=10 see A145184
k=11 see A145185
k=12 see A145186
k=13 see A145187
k=14 see A145188
k=15 see A145189

A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189

a = {}; k = 7; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((7 + Sqrt[53])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)

a(n+1)=a(n)^3 + 3*a(n) and a(1)=7

a(n)=Floor[((7+Sqrt[7^2+4])/2)^(3^(n-1))]

A145182 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=8.

Original entry on oeis.org

8, 536, 153992264, 3651713626720249047672536, 48695646535829720063008633136610768101443687873746944465180200686293744264

Offset: 1

Artur Jasinski, Oct 03 2008

nonn

General formula for continued cotangent recurrences type:
a(n+1)=a(n)3+3*a(n) and a(1)=k is following:
a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
k=1 see A006267
k=2 see A006266
k=3 see A006268
k=4 see A006267(n+1)
k=5 see A006269
k=6 see A145180
k=7 see A145181
k=8 see A145182
k=9 see A145183
k=10 see A145184
k=11 see A145185
k=12 see A145186
k=13 see A145187
k=14 see A145188
k=15 see A145189
The next term has 222 digits. - Harvey P. Dale, Mar 02 2018

A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189

a = {}; k = 7; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((8 + Sqrt[68])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
RecurrenceTable[{a[1]==8,a[n]==a[n-1]^3+3a[n-1]},a,{n,5}] (* Harvey P. Dale, Mar 02 2018 *)

a(n+1)=a(n)3+3*a(n) and a(1)=8

a(n)=Floor[((8+Sqrt[8^2+4])/2)^(3^(n-1))]

A145183 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=9.

Original entry on oeis.org

9, 756, 432083484, 80668317387203269343374356, 524939187888223206865848253384923777974930416670334526494423558665677185033084

Offset: 1

Artur Jasinski, Oct 03 2008

nonn

General formula for continued cotangent recurrences type:
a(n+1)=a(n)3+3*a(n) and a(1)=k is following:
a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
For k=1 see A006267
k=2 see A006266
k=3 see A006268
k=4 see A006267(n+1)
k=5 see A006269
k=6 see A145180
k=7 see A145181
k=8 see A145182
k=9 see A145183
k=10 see A145184
k=11 see A145185
k=12 see A145186
k=13 see A145187
k=14 see A145188
k=15 see A145189

A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189

a = {}; k = 9; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((9 + Sqrt[85])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
NestList[#^3+3#&,9,5] (* Harvey P. Dale, Nov 02 2011 *)

a(n+1)=a(n)3+3*a(n) and a(1)=9

a(n)=Floor[((9+Sqrt[9^2+4])/2)^(3^(n-1))]

cp's OEIS Frontend

A045529 a(n+1) = 5*a(n)^3 - 3*a(n), a(0) = 1.

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A006266 A continued cotangent.

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

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A006268 A continued cotangent.

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A145180 Continued cotangent recurrence a(n+1) = a(n)^3 + 3*a(n) and a(1) = 6.

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A145189 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=15.

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Mathematica

Formula

A271223 Digits of one of the two 3-adic integers sqrt(-2).

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A045529 a(n+1) = 5a(n)^3 - 3a(n), a(0) = 1.