A102419 "Dropping time" in 3x+1 problem starting at n (number of steps to reach a lower number than starting value); a(1) = 0 by convention. Also called glide(n).

0, 1, 6, 1, 3, 1, 11, 1, 3, 1, 8, 1, 3, 1, 11, 1, 3, 1, 6, 1, 3, 1, 8, 1, 3, 1, 96, 1, 3, 1, 91, 1, 3, 1, 6, 1, 3, 1, 13, 1, 3, 1, 8, 1, 3, 1, 88, 1, 3, 1, 6, 1, 3, 1, 8, 1, 3, 1, 11, 1, 3, 1, 88, 1, 3, 1, 6, 1, 3, 1, 83, 1, 3, 1, 8, 1, 3, 1, 13, 1, 3, 1, 6, 1, 3, 1, 8, 1, 3, 1, 73, 1, 3, 1, 13, 1, 3, 1, 6

Offset: 1

Author

N. J. A. Sloane, Sep 15 2006

Keywords

• nonn

Examples

			1: 0 steps
2 1: 1 step
3 10 5 16 8 4 2 1: 6 steps (before it drops below n)
4 2 1: 1 step
5 16 8 4 2 1: 3 steps
6 3 ...: 1 step
7 22 11 34 17 52 26 13 40 20 10 5 ...: 11 steps
...
Records: 0.1.6.11.96.132...171... (A217934)
at.......1.2.3..7.27.703.10087... (A060412)
		

Links

• N. J. A. Sloane, Table of n, a(n) for n = 1..10000
• Eric Roosendaal, On the 3x + 1 problem
• N. J. A. Sloane, First 36 terms of A217934 and A060412 [From Roosendaal web site]

Crossrefs

Cf. A060445, A074473, A126241.

For records see A060412, A217934. - N. J. A. Sloane, Oct 20 2012

Programs

Mathematica

Prepend[Table[Length[NestWhileList[If[EvenQ[#],#/2,3#+1]&,n,#>=n&]],{n,2,99}],1]-1 (* Jayanta Basu, May 28 2013 *)

Python

def a(n):
    if n<3: return 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

Formula

a(2n) = 1; a(2n+1) = A060445(n).

a(n) = A074473(n)-1 for n>1.

a(n) = floor(A126241(n)*(1+log(2)/log(3))). - K. Spage, Oct 22 2009