A060445 "Dropping time" in 3x+1 problem starting at 2n+1 (number of steps to reach a lower number than starting value). Also called glide(2n+1).

0, 6, 3, 11, 3, 8, 3, 11, 3, 6, 3, 8, 3, 96, 3, 91, 3, 6, 3, 13, 3, 8, 3, 88, 3, 6, 3, 8, 3, 11, 3, 88, 3, 6, 3, 83, 3, 8, 3, 13, 3, 6, 3, 8, 3, 73, 3, 13, 3, 6, 3, 68, 3, 8, 3, 50, 3, 6, 3, 8, 3, 13, 3, 24, 3, 6, 3, 11, 3, 8, 3, 11, 3, 6, 3, 8, 3, 65, 3, 34, 3, 6, 3, 47, 3, 8, 3, 13, 3, 6, 3, 8, 3

Offset: 0

Author

N. J. A. Sloane, Apr 07 2001

Keywords

• nonn
• easy
• nice

Comments

If the starting value is even then of course the next step in the trajectory is smaller (cf. A102419).

The dropping time can be made arbitrarily large: If the starting value is of form n(2^m)-1 and m > 1, the next value is 3n(2^m)-3+1. That divided by 2 is 3n(2^(m-1))-1. It is bigger than the starting value and of the same form - substitute 3n -> n and m-1 -> m, so recursively get an increasing subsequence of m odd values. The dropping time is obviously longer than that. This holds even if Collatz conjecture were refuted. For example, m=5, n=3 -> 95, 286, 143, 430, 215, 646, 323, 970, 485, 1456, 728, 364, 182, 91. So the subsequence in reduced Collatz variant is 95, 143, 215, 323, 485. - Juhani Heino, Jul 21 2017

Examples

			3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2, taking 6 steps, so a(1) = 6.
		

Links

• T. D. Noe, Table of n, a(n) for n = 0..10000
• Jason Holt, Plot of first 1 billion terms, log scale on x axis
• Jason Holt, Plot of first 10 billion terms, log scale on x axis
• Eric Roosendaal, On the 3x + 1 problem
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

A060565 gives the first lower number that is reached. Cf. A060412-A060415, A217934.

See A074473, A102419 for other versions of this sequence.

Cf. A122437 (allowable dropping times), A122442 (least k having dropping time A122437(n)).

Cf. A070165.

Programs

Haskell

a060445 0 = 0
a060445 n = length $ takeWhile (>= n') $ a070165_row n'
            where n' = 2 * n + 1
-- Reinhard Zumkeller, Mar 11 2013

Mathematica

nxt[n_]:=If[OddQ[n],3n+1,n/2]; Join[{0},Table[Length[NestWhileList[nxt, n,#>=n&]]-1, {n,3,191,2}]]  (* Harvey P. Dale, Apr 23 2011 *)

Python

def a(n):
    if n<1: return 0
    n=2*n + 1
    N=n
    x=0
    while True:
        if n%2==0: n//=2
        else: n = 3*n + 1
        x+=1
        if nIndranil Ghosh, Apr 22 2017

Extensions

More terms from Jason Earls, Apr 08 2001 and from Michel ten Voorde Apr 09 2001

Still more terms from Larry Reeves (larryr(AT)acm.org), Apr 12 2001