A075477 -id:A075477 - 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

A075482 Number of iteration that first becomes smaller than the initial value if Collatz-function (A006370) is iterated, starting with numbers of the form 64n + 59.

Original entry on oeis.org

12, 14, 12, 45, 12, 14, 12, 17, 12, 14, 12, 33, 12, 14, 12, 20, 12, 14, 12, 25, 12, 14, 12, 17, 12, 14, 12, 20, 12, 14, 12, 30, 12, 14, 12, 25, 12, 14, 12, 17, 12, 14, 12, 30, 12, 14, 12, 22, 12, 14, 12, 69, 12, 14, 12, 17, 12, 14, 12, 22, 12, 14, 12, 22, 12, 14, 12, 82, 12

Offset: 0

Labos Elemer, Sep 24 2002

nonn

1stSubmergeLengths[=A074473] with initial values belonging to other residue classes modulo 64 are either listed in A075476-A075483 or can be easily determined. For 64k+2s the first sink below initial value is at 2nd iterate; for 64k+4s+1 the first submerge below initial value comes at 4th term of iteration list; finally if initial value is of 64k+4s+3 form or moreover initial value = 64k+r, r = 3, 11, 19, 23, 35, 43, 51, 55, then for all k first sink emerges at the 7th, 9th, 7th, 9th, 7th, 9th, 7th, 9th iterates, respectively.

			n=0: 64n + 59 = 59, the list = {59,178,89,268,134,67,202,101,304,152,76,38,...} the 12th term = 38 < 59 = the initial value, so a(0)=12.

Antti Karttunen, Table of n, a(n) for n = 0..16384
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A006370, A074472, A074473, A075476, A075477-A075483.

Table[Function[m, Length@ NestWhileList[If[EvenQ@ #, #/2, 3 # + 1] &, m, # >= m - 1 &]][64 n + 59], {n, 0, 84}] (* Michael De Vlieger, Mar 25 2017 *)

A006370(n) = if(n%2, 3*n+1, n/2);
A074473(n) = if(1==n,n,my(org_n=n); for(i=1,oo,if(nA006370(n)));
A075482(n) = A074473((64*n)+59); \\ Antti Karttunen, Oct 09 2018

a(n) = A074473(64n + 59).

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Python

Extensions