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

1,3

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]

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

Example

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)

Mathematica

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

Prog

(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 n<N: return x
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Apr 22 2017

Crossrefs

Cf. A060445, A074473, A126241.

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

Sequence in context: A152935 A253686 A290101 * A352891 A339387 A074193

Adjacent sequences: A102416 A102417 A102418 * A102420 A102421 A102422

Keyword

nonn

Author

N. J. A. Sloane, Sep 15 2006

Status

approved