A153727 Period 3: repeat [1, 4, 2] ; Trajectory of 3x+1 sequence starting at 1.

1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2

Offset: 0

Author

Philippe Deléham, Dec 30 2008

Keywords

• nonn
• easy

Comments

From Klaus Brockhaus, May 23 2010: (Start)

Continued fraction expansion of (7+sqrt(229))/18.

Decimal expansion of 142/999. (End)

a(A008585(n)) = 1; a(A016777(n)) = 4; a(A016789(n)) = 2. - Reinhard Zumkeller, Oct 08 2011

References

• C. A. Pickover, The Math Book, 2009; Collatz Conjecture, pp 374-375.

Links

• Eric Weisstein's World of Mathematics, Collatz Problem
• Wikipedia, Collatz conjecture
• Index entries for sequences related to 3x+1 (or Collatz) problem
• Index entries for linear recurrences with constant coefficients, signature (0,0,1).

Crossrefs

Row 1 of A347270.

Cf. A178236 (decimal expansion of (7+sqrt(229))/18). Appears in A179133 (n>=1).

Cf. A006370, A080425, A130196, A131534.

Programs

Haskell

a153727 n = a153727_list !! n
a153727_list = iterate a006370 1 -- Reinhard Zumkeller, Oct 08 2011

Magma

[2^(-n mod 3) : n in [0..50]]; // Wesley Ivan Hurt, Jun 20 2014

Maple

A153727:=n->2^(-n mod 3); seq(A153727(n), n=0..50); # Wesley Ivan Hurt, Jun 20 2014

Mathematica

a[n_] := {1, 4, 2}[[Mod[n, 3] + 1]]; Table[a[n], {n, 0, 104}] (* Jean-François Alcover, Jul 19 2013 *)
LinearRecurrence[{0, 0, 1},{1, 4, 2},105] (* Ray Chandler, Aug 25 2015 *)
PadRight[{},120,{1,4,2}] (* Harvey P. Dale, Jul 20 2025 *)

PARI

A153727(n)=[1,4,2][n%3+1] \\ M. F. Hasler, Feb 10 2011

Sage

[power_mod(2,-n,7)for n in range(0, 105)] # Zerinvary Lajos, Jun 07 2009

Sage

[power_mod(4, n, 7)for n in range(0, 105)] # Zerinvary Lajos, Nov 25 2009

Formula

a(3n)=1, a(3n+1)=4, a(3n+2)=2.

G.f.: (1+4*x+2*x^2)/(1-x^3).

a(n) = 4^n mod 7. - Zerinvary Lajos, Nov 25 2009

a(n) = A130196(n) * A131534(n). - Reinhard Zumkeller, Nov 12 2009

a(n) = 2^(-n mod 3) = 2^A080425(n). - Wesley Ivan Hurt, Jun 20 2014

a(n) = sqrt(4^(5*n) mod 21). - Gary Detlefs, Jul 07 2014

From Wesley Ivan Hurt, Jun 30 2016: (Start)

a(n) = a(n-3) for n>2.

a(n) = (7 - 4*cos(2*n*Pi/3) + 2*sqrt(3)*sin(2*n*Pi/3))/3. (End)