A075480 -id:A075480 - 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

A361733 Length of the Collatz (3x + 1) trajectory from k = 10^n - 1 to a term less than k, or -1 if the trajectory never goes below k.

Original entry on oeis.org

4, 7, 17, 12, 113, 17, 79, 22, 51, 33, 64, 35, 128, 56, 110, 53, 84, 128, 107, 115, 175, 82, 477, 172, 141, 182, 188, 110, 159, 167, 301, 206, 151, 146, 128, 195, 190, 299, 208, 276, 180, 185, 500, 203, 229, 190, 265, 270, 288, 252, 299, 208, 350, 348, 459, 330, 314, 268, 490, 361, 578

Offset: 1

Paul M. Bradley, Mar 22 2023

nonn

k = 10^n - 1 = A002283(n) is the repdigit consisting of n digits, all 9s.
The sequence seems to be chaotic but broadly increasing.
By contrast, repdigits of 1, 3, 5, or 7, have constant dropping times after a few initial values each.

			a(1) = 4 as for k = 9, the Collatz trajectory begins 9, 28, 14, 7, ...;
a(2) = 7 as for k = 99, the Collatz trajectory begins 99, 298, 149, 448, 224, 112, 56, ...;
a(3) = 17 as for k = 999, the Collatz trajectory begins 999, 2998, 1499, 4498, 2249, 6748, 3374, 1687, 5062, 2531, 7594, 3797, 11392, 5696, 2848, 1424, 712, ... .

Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A074473, A002283, A006370, A074474, A075483, A075480.

collatzLen[a_Integer] := Module[{len = 1, x = a},
  While[x >= a,    If[Mod[x, 2] > 0,
      x = 3 x + 1,
      x = Quotient[x, 2]
    ];
    len++
  ];
  Return[len]
]

f(n) = if (n%2, 3*n+1, n/2); \\ A006370
b(n) = if (n<3, return(n)); my(m=n, nb=0); while (1, m = f(m); nb++; if (m < n, return(nb+1));); \\ A074473
a(n) = b(10^n-1); \\ Michel Marcus, Mar 28 2023

def collatz_len(a):
    length = 1
    x = a
    while x >= a:
        if x % 2 > 0:
            x = 3 * x + 1
        else:
            x = x // 2
        length += 1
    return length

a(n) = A074473(10^n-1).

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.

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