A217934 -id:A217934 - cp's OEIS frontend

A074473 Dropping time for the 3x+1 problem: for n >= 2, number of iteration that first becomes smaller than the initial value if Collatz-function (A006370) is iterated starting at n; a(1)=1 by convention.

1, 2, 7, 2, 4, 2, 12, 2, 4, 2, 9, 2, 4, 2, 12, 2, 4, 2, 7, 2, 4, 2, 9, 2, 4, 2, 97, 2, 4, 2, 92, 2, 4, 2, 7, 2, 4, 2, 14, 2, 4, 2, 9, 2, 4, 2, 89, 2, 4, 2, 7, 2, 4, 2, 9, 2, 4, 2, 12, 2, 4, 2, 89, 2, 4, 2, 7, 2, 4, 2, 84, 2, 4, 2, 9, 2, 4, 2, 14, 2, 4, 2, 7, 2, 4, 2, 9, 2, 4, 2, 74, 2, 4, 2, 14, 2, 4, 2, 7

Offset: 1

Labos Elemer, Sep 19 2002

nonn

Here we call the starting value iteration number 1, although usually the count is started at 0, which would subtract 1 from the values for n >= 2 - see A060445, A102419.

			n=2k: then a(2k)=2 because the second iterate is k<n=2k, the first iterate below 2k; n=4k+1, k>1: the list = {4k+1, 12k+4, 6k+2, 3k+1, ...} i.e. the 4th term is always the first below initial value, so a(4k+1)=4;
n=15: the list={15, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1} and 12th term is first sinks below iv=15, so a(15)=12; relatively larger values occur at n=4k+3.
n=3: the list is {3, 10, 5, 16, 8, 4, 2, 1, ..}, the 7th term is 2, which is the first smaller than 3, so a(3)=7.

N. J. A. Sloane, Table of n, a(n) for n = 1..10000

Cf. A006370, A075476, A075477, A075478, A075479, A075480, A075481, A075482, A075483, A060445, A060412, A217934.

Equals A102419(n)+1.

nextx[x_Integer] := If[OddQ@x, 3x + 1, x/2]; f[1] = 1; f[n_] := Length@ NestWhileList[nextx, n, # >= n &]; Array[f, 83] (* Bobby R. Treat (drbob(at)bigfoot.com), Sep 16 2006 *)

A074473(n) = if (n<3, n,  my(N=n, x=1); while (1, if (n%2==0, n/=2, n = 3*n + 1); x++; if (nMichel Marcus, Aug 15 2025

def a(n):
    if n<3: return n
    N=n
    x=1
    while True:
        if n%2==0: n/=2
        else: n = 3*n + 1
        x+=1
        if nIndranil Ghosh, Apr 15 2017

Edited by N. J. A. Sloane, Sep 15 2006

A060412 In the '3x+1' problem, these values for the starting value set new records for the "dropping time", number of steps to reach a lower value than the start.

Original entry on oeis.org

2, 3, 7, 27, 703, 10087, 35655, 270271, 362343, 381727, 626331, 1027431, 1126015, 8088063, 13421671, 20638335, 26716671, 56924955, 63728127, 217740015, 1200991791, 1827397567, 2788008987, 12235060455

Offset: 1

N. J. A. Sloane, Apr 06 2001; b-file added Nov 27 2007

nonn

The (3x+1)/2 steps and the halving steps are counted. - Don Reble, May 13 2006

Where records occur in A102419 (could be prefixed by an initial 1). - N. J. A. Sloane, Oct 20 2012

			See A102419.

N. J. A. Sloane, Table of n, a(n) for n = 1..35 (from the web page of Tomás Oliveira e Silva)
Tomás Oliveira e Silva, Tables
Eric Roosendaal, On the 3x + 1 problem
N. J. A. Sloane, First 36 terms of A217934 and A060412 [From Roosendaal web site]
Index entries for sequences related to 3x+1 (or Collatz) problem

A060413 gives associated "dropping times", A060414 the maximal values and A060415 the steps at which the maxima occur. See also A217934.

Cf. A060445, A008884, A161021, A161022, A161023, A014682, A126241.

dcoll[n_]:=Length[NestWhileList[If[EvenQ[#],#/2,3#+1]&,n,#>=n&]]; t={max=2}; Do[If[(y=dcoll[n])>max,max=y; AppendTo[t,n]],{n,3,1130000,4}]; t (* Jayanta Basu, May 28 2013 *)

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).

Original entry on oeis.org

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

N. J. A. Sloane, Apr 07 2001

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

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

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

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.

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

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 *)

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

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

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).

Original entry on oeis.org

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

N. J. A. Sloane, Sep 15 2006

nonn

			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)

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]

Cf. A060445, A074473, A126241.

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

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

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

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

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A074473 Dropping time for the 3x+1 problem: for n >= 2, number of iteration that first becomes smaller than the initial value if Collatz-function (A006370) is iterated starting at n; a(1)=1 by convention.

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A060412 In the '3x+1' problem, these values for the starting value set new records for the "dropping time", number of steps to reach a lower value than the start.

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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).

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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).

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