A006267 -id:A006267 - cp's OEIS frontend

A145184 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=10.

10, 1030, 1092730090, 1304784252725333839617919270, 2221345538213703371536935622204403026741331806706388823688859272519059871168740810

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 = 10; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((10 + Sqrt[104])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)

a(n+1)=a(n)3+3*a(n) and a(1)=10

a(n)=Floor[((10+Sqrt[10^2+4])/2)^(3^(n-1))]

A145185 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=11.

Original entry on oeis.org

11, 1364, 2537720636, 16342986943522226847837781364, 4365101043708483494615466932242949707161871659736799144058331102381689400753867700636

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 = 11; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((11 + Sqrt[125])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
RecurrenceTable[{a[1]==11,a[n]==a[n-1]^3+3*a[n-1]},a,{n,5}] (* Harvey P. Dale, Jul 23 2012 *)

a(n+1)=a(n)^3+3*a(n) and a(1)=11

a(n)=Floor[((11+Sqrt[11^2+4])/2)^(3^(n-1))]

A145186 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=12.

Original entry on oeis.org

12, 1764, 5489037036, 165382092777963331246695013764, 4523404750894779548516344022127873154658656755028228436816797201835023951822441803129036

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 (a(6)) has 263 digits. - Harvey P. Dale, May 21 2018

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

a = {}; k = 12; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((12 + Sqrt[148])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
NestList[#^3+3#&,12,5] (* Harvey P. Dale, May 21 2018 *)

a(n+1)=a(n)3+3*a(n) and a(1)=12.

a(n)=Floor[((12+Sqrt[12^2+4])/2)^(3^(n-1))].

A145187 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) with a(1)=13.

Original entry on oeis.org

13, 2236, 11179326964, 1397162674037779847605429310236, 2727350312258670490076364505418491429134385511825631286349491134548728023756939667650354964

Offset: 1

Artur Jasinski, Oct 03 2008

nonn

General formula for continued cotangent recurrences of the form a(n+1)=a(n)^3+3*a(n) with a(1)=k is 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

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

a = {}; k = 13; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((13 + Sqrt[173])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)

a(n+1)=a(n)^3+3*a(n) and a(1)=13

a(n)=Floor[((13+Sqrt[13^2+4])/2)^(3^(n-1))]

A145188 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=14.

Original entry on oeis.org

14, 2786, 21624372014, 10111847525912679844192131854786, 1033930953043290626825587838528711318150300040875029341893199068078185510802565166824630504014

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
Essentially the same as A006266. [From R. J. Mathar, Mar 18 2009]

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

a = {}; k = 14; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((14 + Sqrt[200])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)

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))]

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

A144837 a(n) = Lucas(5^n).

Original entry on oeis.org

11, 167761, 132878596168524201724674011

Offset: 1

Artur Jasinski, Sep 22 2008

Previous name was: a(n) = round(phi^(5^n)) where phi = 1.6180339887... = (sqrt(5) + 1)/2 = A001622.

a(4), a 131-digit number, is too large to show here.

			The base 5 representation of a(3) = 132878596168524201724674011 begins 1 + 2*5 + 0*(5^2) + 2*(5^3) + 3*(5^4) + 0*(5^5) + 4*(5^6) + O(5^7) so A269591 begins [1, 2, 0, 2, 3, 0, 4, ...]. - _Peter Bala_, Nov 14 2022

Amiram Eldar, Table of n, a(n) for n = 1..5

Cf. A000032, A000351, A001622, A001566, A006267, A144836, A144838, A144839, A268922, A269591.

a := proc(n) option remember; if n = 1 then 11 else a(n-1)^5 + 5*a(n-1)^3 + 5*a(n-1) end if; end;
seq(a(n), n = 1..5); # Peter Bala, Nov 14 2022

Table[Round[GoldenRatio^(5^n)], {n, 1, 5}]
c = (1 + Sqrt[5])/2; Table[Expand[c^(5^n) + (1 - c)^(5^n)], {n, 1, 5}] (* Artur Jasinski, Oct 05 2008 *)
LucasL[5^Range[5]] (* Harvey P. Dale, Apr 01 2023 *)

a(n) = phi^(5^n) + (1-phi)^(5^n) = phi^(5^n) + (-phi)^(-5^n). - Artur Jasinski, Oct 05 2008
From Peter Bala, Nov 14 2022: (Start)
a(n) = A000032(5^n).
a(n) = a(n-1)^5 + 5*a(n-1)^3 + 5*a(n-1) with a(1) = 11.
a(n) == 1 (mod 5).
a(n+1) == a(n) (mod 5^(n+1)) for n >= 1 (a particular case of the Gauss congruences for the Lucas numbers).
Conjecture: a(n+1) == a(n) (mod 5^(n+r+1)) for n >= r.
The smallest positive residue of a(n) mod(5^n) = A268922(n).
In the ring of 5-adic integers the limit_{n -> oo} a(n) exists and is equal to A269591. An example is given below. (End)

New name from Peter Bala, Nov 10 2022

A244847 Decimal expansion of rho_c = (5-sqrt(5))/10, the asymptotic critical density for the hard hexagon model.

Original entry on oeis.org

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

Offset: 0

Jean-François Alcover, Nov 12 2014

The vertical distance between the accumulation point and the outermost point of a golden spiral inscribed inside a golden rectangle with dimensions phi and 1 along the x and y axes, respectively (the horizontal distance is A176015). - Amiram Eldar, May 18 2021

The limiting frequency of the digit 1 in the base phi representation of real numbers in the range [0,1], where phi is the golden ratio (A001622) (Rényi, 1957). - Amiram Eldar, Mar 18 2025

			0.2763932022500210303590826331268723764559381640388474275729102754589479...

Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 1.2 The Golden Mean, phi, p. 7.
Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 5.12.1 Phase transitions in Lattice Gas Models, p. 347.

Rodney J. Baxter Hard hexagons: exact solution, Journal of Physics A: Mathematical and General, Vol. 13, No. 3 (1980), pp. L61-L70, alternative link.
P. S. Bruckman and I. J. Good, A Generalization of a Series of De Morgan with Applications of Fibonacci Type, The Fibonacci Quarterly, Vol. 14, No. 3 (1976), pp. 193-196.
Alfréd Rényi, Representations for real numbers and their ergodic properties, Acta Math. Acad. Sci. Hungar., Vol.8, No. 3-4 (1957), pp. 477-493.
Eric Weisstein's MathWorld, Hard Hexagon Entropy Constant.
Wikipedia, Hard Hexagon Model.
Index entries for algebraic numbers, degree 2.

Cf. A242671, A244593.
Cf. A000032, A000045.
Essentially the same sequence of digits as A229760 and A187799.

RealDigits[(5 - Sqrt[5])/10, 10, 102] // First

Equals 1/(sqrt(5)*phi), where phi = (1+sqrt(5))/2 = A001622. - Vaclav Kotesovec, Nov 13 2014
Equals lim_{n -> infinity} A000045(n)/A000032(n+1). - Bruno Berselli, Jan 22 2018
Equals Sum_{n>=1} A000045(3^(n-1))/A000032(3^n) = Sum_{n>=1} A045529(n-1)/A006267(n). - Amiram Eldar, Dec 20 2018
Equals 1 - A242671. - Amiram Eldar, Mar 18 2025

A002813 a(0) = 4; for n > 0, a(n) = a(n-1)^3 - 3*a(n-1)^2 + 3.

Original entry on oeis.org

4, 19, 5779, 192900153619, 7177905237579946589743592924684179, 369822356418414944143680173221426891716916679027557977938929258031490127514207143830378340325399155219

Offset: 0

N. J. A. Sloane

An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004
The next term, a(7), has 305 digits. - Harvey P. Dale, Jul 19 2011
From Peter Bala, Nov 22 2012: (Start)
The present sequence is the case x = 1 of the following general remarks about the recurrence a(n+1) = a(n)^3 - 3*a(n-1)^2 + 3. Cf. A002814.
Define a sequence of polynomials P(n,x) inductively by setting P(0,x) = x^2 + 3 and P(n+1,x) = P(n,x^3 + 3*x) for n >= 0. Then P(n,x) satisfies the cubic recurrence P(n+1,x) = P(n,x)^3 - 3*P(n-1,x)^2 + 3 with the initial condition P(0,x) = x^2 + 3.
An explicit formula is P(n,x) = Q(3^(n+1),x)/Q(3^n,x), where Q(n,x) = ((x + sqrt(x^2 + 4))/2)^n + ((x - sqrt(x^2 + 4))/2)^n.
Alternatively, P(n,x) = ((x^2 + 2 + sqrt(x^4 + 4*x^2))/2)^(3^n) + ((x^2 + 2 - sqrt(x^4 + 4*x^2))/2)^(3^n) + 1.
Iterating the algebraic identity x/sqrt(x^2 + 4) = (1 - 2/(x^2 + 3))*y/sqrt(y^2 + 4), where y = x^3 + 3*x, leads to the product expansion x/sqrt(x^2 + 4) = Product_{n = 0..oo} (1 - 2/P(n,x)). See Escott and also Fine.
The sequence A(n,x) := x*Product_{k = 0..n} P(k,x) satisfies the recurrence A(n+1,x) = A(n,x)^3 + 3*A(n,x). These sequences occur in the continued cotangent expansions of Lehmer. Cases currently in the database are A006267 (x = 1), A006266 (x = 2), A006268 (x = 3), A006269 (x = 5) and A145180 through A145189 (x = 6 through x = 15).
(End)

L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 1, p. 397.
E. Lucas, Nouveaux théorèmes d'arithmétique supérieure, Comptes Rend., 83 (1876), 1286-1288.
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..8
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.
E. Lucas, Nouveaux théorèmes d'arithmétique supérieure (annotated scanned copy)
J. Shallit Predictable regular continued cotangent expansions, J. Res. Nat. Bur. Standards Sect. B 80B (1976), no. 2, 285-290.
Eric Weisstein's World of Mathematics, Lehmer Cotangent Expansion

Cf. A002814, A006267.

[Lucas(2*3^n)+1: n in [0..5]]; // Vincenzo Librandi, Jul 20 2011

NestList[#^3-3#^2+3&,4,6] (* Harvey P. Dale, Jul 19 2011 *)

a(n)=if(n<1,4*(n==0),a(n-1)^3-3*a(n-1)^2+3)

a(n)=if(n<0,0,n=2*3^n;fibonacci(n+1)+fibonacci(n-1)+1)

a(n) = L(2*3^n)+1 where L=Lucas numbers.
a(n) = L(3^(n+1))/L(3^n). - Benoit Cloitre, Sep 18 2005
a(n) = A001999(n)+1. - R. J. Mathar, Apr 22 2007
From Peter Bala, Nov 22 2012: (Start)
a(n) = ((3 + sqrt(5))/2)^(3^n) + ((3 - sqrt(5))/2)^(3^n) + 1.
(1/5)*sqrt(5) = Product_{n = 0..oo} (1 - 2/a(n)).
A006267(n+1) = Product_{k = 0..n} a(k).
A002814(n+1) = a(n) - 2. (End)

A144836 a(n) = round(phi^(4^n)) where phi is the golden ratio (A001622).

Original entry on oeis.org

2, 7, 2207, 23725150497407, 316837008400094222150776738483768236006420971486980607

Offset: 0

Artur Jasinski, Sep 22 2008

Amiram Eldar, Table of n, a(n) for n = 0..6

Cf. A000032, A001566, A001622, A006267, A144837, A144838, A144839.

Cf. A374883 (phi^4).

a := proc(n) option remember; if n = 0 then 2 elif n = 1 then 7 else a(n-1)^4 - 4*a(n-1)^2 + 2 end if; end proc: seq(a(n), n = 0..4); # Peter Bala, Nov 28 2022

Table[Round[GoldenRatio^(4^n)], {n, 0, 5}]
c = (1 + Sqrt[5])/2; Join[{2}, Table[Expand[c^(4^n) + (1 - c)^(4^n)], {n, 1, 5}]] (* Artur Jasinski, Oct 05 2008 *)
Table[Round[2*Cosh[4^n*ArcCosh[Sqrt[5]/2]]], {n, 0, 5}] (* Artur Jasinski, Oct 09 2008 *)
a[n_] := LucasL[4^n]; a[0] = 2; Array[a, 5, 0] (* Amiram Eldar, Jul 12 2025 *)

a(n)=round(((1+sqrt(5))/2)^4^n) \\ Charles R Greathouse IV, Jul 29 2011

a(n) = Lucas(4^n) = A000032(4^n), n>0.
a(n) = phi^(4^n) + (1 - phi)^(4^n) = phi^(4^n) + (-phi)^(-4^n), where phi is golden ratio = (1 + sqrt(5))/2 = 1.6180339887..., n>0. - Artur Jasinski, Oct 05 2008
a(n) = 2*cosh(4^n*arccosh(sqrt(5)/2)), n>0. - Artur Jasinski, Oct 09 2008
a(n+1) = a(n)^4 - 4*a(n-1)^2 + 2 with a(1) = 7. - Peter Bala, Nov 28 2022

Offset corrected by Charles R Greathouse IV, May 15 2013
Offset changed to 0 by Georg Fischer, Sep 02 2022
New name from Peter Bala, Nov 18 2022
Revised by editors, Jul 12 2025

cp's OEIS Frontend

A145184 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=10.

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A145185 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=11.

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A145186 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=12.

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A145187 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) with a(1)=13.

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A145188 Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=14.

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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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A144837 a(n) = Lucas(5^n).

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Maple

Mathematica

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A244847 Decimal expansion of rho_c = (5-sqrt(5))/10, the asymptotic critical density for the hard hexagon model.

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