A365478 In the Collatz problem, largest value in the trajectory of n in the 3x+1 function (denoted by T(x) in the literature, and defined as T(x) = (3x+1)/2 if x is odd, T(x) = x/2 if x is even), or -1 if the trajectory is divergent.

1, 2, 8, 4, 8, 8, 26, 8, 26, 10, 26, 12, 20, 26, 80, 16, 26, 26, 44, 20, 32, 26, 80, 24, 44, 26, 4616, 28, 44, 80, 4616, 32, 50, 34, 80, 36, 56, 44, 152, 40, 4616, 42, 98, 44, 68, 80, 4616, 48, 74, 50, 116, 52, 80, 4616, 4616, 56, 98, 58, 152, 80, 92, 4616, 4616

Offset: 1

Author

Paolo Xausa, Sep 05 2023

Keywords

• nonn

Comments

This sequence differs from A025586, where the division by 2 does not immediately follow the 3x+1 step when x is odd.

Here by definition the trajectory ends when 1 is reached, so a(1) = 1.

Kontorovich and Lagarias (2009, 2010) call these values the maximum excursion values.

Examples

			a(11) = 26 because 26 is the largest value in the trajectory 11 -> 17 -> 26 -> 13 -> 20 -> 10 -> 5 -> 8 -> 4 -> 2 -> 1.
		

Links

• Paolo Xausa, Table of n, a(n) for n = 1..10000
• Alex V. Kontorovich and Jeffrey C. Lagarias, Stochastic Models for the 3x+1 and 5x+1 Problems, arXiv:0910.1944 [math.NT], 2009, pp. 11-14, and in Jeffrey C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, American Mathematical Society, 2010, pp. 140-142.
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A014682, A025586 (equivalent for the Collatz function), A166245.

Programs

Mathematica

A365478[n_]:=Max[NestWhileList[If[OddQ[#],(3#+1)/2,#/2]&,n,#>1&]];Array[A365478,100]

Formula

a(n) <= A025586(n).