A185590 Iterate the map in A006369 starting at 44.

44, 59, 79, 105, 70, 93, 62, 83, 111, 74, 99, 66, 44, 59, 79, 105, 70, 93, 62, 83, 111, 74, 99, 66, 44, 59, 79, 105, 70, 93, 62, 83, 111, 74, 99, 66, 44, 59, 79, 105, 70, 93, 62, 83, 111, 74, 99, 66, 44, 59, 79, 105, 70, 93, 62, 83, 111, 74, 99, 66, 44, 59, 79, 105, 70, 93, 62, 83, 111, 74, 99, 66, 44

Offset: 1

Author

N. J. A. Sloane, Feb 04 2011

Keywords

• nonn
• easy

Comments

Periodic with period length 12.

References

• A. O. L. Atkin, Comment on Problem 63-13, SIAM Rev., 8 (1966), 234-236.
• J. C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, Amer. Math. Soc., 2010; see page 270. (beware of typo: Lagarias says the orbit of 144 (not 44) has period 12.)

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A006369, A028394, A028396, A094328, A094329, A185589.

Trajectories under A006368 and A006369: A180853, A217218, A185590, A180864, A028393, A028394, A094328, A094329, A028396, A028395, A217729, A182205, A223083-A223088, A185589, A185590.

Programs

Haskell

a185590 n = a185590_list !! (n-1)
a185590_list = iterate a006369 44  -- Reinhard Zumkeller, Dec 31 2011

Mathematica

f[n_] := If[Mod[n, 3] == 0, 2*n/3, Round[4*n/3]]; a[1] = 44; a[n_] := a[n] = f[a[n - 1]]; Table[a[n], {n, 1, 73}] (* Jean-François Alcover, Jun 10 2013 *)