A125731 -id:A125731 - cp's OEIS frontend

A006577 Number of halving and tripling steps to reach 1 in '3x+1' problem, or -1 if 1 is never reached.

0, 1, 7, 2, 5, 8, 16, 3, 19, 6, 14, 9, 9, 17, 17, 4, 12, 20, 20, 7, 7, 15, 15, 10, 23, 10, 111, 18, 18, 18, 106, 5, 26, 13, 13, 21, 21, 21, 34, 8, 109, 8, 29, 16, 16, 16, 104, 11, 24, 24, 24, 11, 11, 112, 112, 19, 32, 19, 32, 19, 19, 107, 107, 6, 27, 27, 27, 14, 14, 14, 102, 22

Offset: 1

N. J. A. Sloane, Bill Gosper

The 3x+1 or Collatz problem is as follows: start with any number n. If n is even, divide it by 2, otherwise multiply it by 3 and add 1. Do we always reach 1? This is a famous unsolved problem. It is conjectured that the answer is yes.
It seems that about half of the terms satisfy a(i) = a(i+1). For example, up to 10000000, 4964705 terms satisfy this condition.
n is an element of row a(n) in triangle A127824. - Reinhard Zumkeller, Oct 03 2012
The number of terms that satisfy a(i) = a(i+1) for i being a power of ten from 10^1 through 10^10 are: 0, 31, 365, 4161, 45022, 477245, 4964705, 51242281, 526051204, 5378743993. - John Mason, Mar 02 2018
5 seems to be the only number whose value matches its total number of steps (checked to n <= 10^9). - Peter Woodward, Feb 15 2021

			a(5)=5 because the trajectory of 5 is (5,16,8,4,2,1).

R. K. Guy, Unsolved Problems in Number Theory, E16.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

N. J. A. Sloane, Table of n, a(n) for n = 1..10000
David Eisenbud and Brady Haran, UNCRACKABLE? The Collatz Conjecture, Numberphile video, 2016.
Geometry.net, Links on Collatz Problem
Christian Hercher, There are no Collatz m-Cycles with m <= 91, J. Int. Seq. (2023) Vol. 26, Article 23.3.5.
Jason Holt, Log-log plot of first billion terms
Jason Holt, Plot of 1 billion values of the number of steps to drop below n (A060445), log scale on x axis
Jason Holt, Plot of 10 billion values of the number of steps to drop below n (A060445), log scale on x axis
A. Krowne, Collatz problem, PlanetMath.org.
J. C. Lagarias, The 3x+1 problem and its generalizations, Amer. Math. Monthly, 92 (1985), 3-23.
J. C. Lagarias, How random are 3x+1 function iterates?, in The Mathemagician and the Pied Puzzler - A Collection in Tribute to Martin Gardner, Ed. E. R. Berlekamp and T. Rogers, A. K. Peters, 1999, pp. 253-266.
J. C. Lagarias, The 3x+1 Problem: an annotated bibliography, II (2000-2009), arXiv:0608208 [math.NT], 2006-2012.
J. C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, Amer. Math. Soc., 2010.
Jeffrey C. Lagarias, The 3x+1 Problem: An Overview, arXiv:2111.02635 [math.NT], 2021.
M. Le Brun, Email to N. J. A. Sloane, Jul 1991
Mathematical BBS, Biblography on Collatz Sequence
P. Picart, Algorithme de Collatz et conjecture de Syracuse
E. Roosendaal, On the 3x+1 problem
J. L. Simons, On the nonexistence of 2-cycles for the 3x+1 problem, Math. Comp. 75 (2005), 1565-1572.
N. J. A. Sloane, "A Handbook of Integer Sequences" Fifty Years Later, arXiv:2301.03149 [math.NT], 2023, p. 8.
G. Villemin's Almanach of Numbers, Cycle of Syracuse
Eric Weisstein's World of Mathematics, Collatz Problem
Wikipedia, Collatz Conjecture
Index entries for sequences related to 3x+1 (or Collatz) problem

See A070165 for triangle giving trajectories of n = 1, 2, 3, ....
Cf. A006370, A125731, A127885, A127886, A008908, A112695, A135282, A208981, A025586.
See also A008884, A161021, A161022, A161023.

import Data.List (findIndex)
import Data.Maybe (fromJust)
a006577 n = fromJust $ findIndex (n `elem`) a127824_tabf
-- Reinhard Zumkeller, Oct 04 2012, Aug 30 2012

A006577 := proc(n)
        local a,traj ;
        a := 0 ;
        traj := n ;
        while traj > 1 do
                if type(traj,'even') then
                        traj := traj/2 ;
                else
                        traj := 3*traj+1 ;
                end if;
                a := a+1 ;
        end do:
        return a;
end proc: # R. J. Mathar, Jul 08 2012

f[n_] := Module[{a=n,k=0}, While[a!=1, k++; If[EvenQ[a], a=a/2, a=a*3+1]]; k]; Table[f[n],{n,4!}] (* Vladimir Joseph Stephan Orlovsky, Jan 08 2011 *)
Table[Length[NestWhileList[If[EvenQ[#],#/2,3#+1]&,n,#!=1&]]-1,{n,80}] (* Harvey P. Dale, May 21 2012 *)

a(n)=if(n<0,0,s=n; c=0; while(s>1,s=if(s%2,3*s+1,s/2); c++); c)

step(n)=if(n%2,3*n+1,n/2);
A006577(n)=if(n==1,0,A006577(step(n))+1); \\ Michael B. Porter, Jun 05 2010

def a(n):
    if n==1: return 0
    x=0
    while True:
        if n%2==0: n//=2
        else: n = 3*n + 1
        x+=1
        if n<2: break
    return x
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jun 05 2017

def A006577(n):
    ct = 0
    while n != 1: n = A006370(n); ct += 1
    return ct # Ya-Ping Lu, Feb 22 2024

collatz<-function(n) ifelse(n==1,0,1+ifelse(n%%2==0,collatz(n/2),collatz(3*n+1))); sapply(1:72, collatz) # Christian N. K. Anderson, Oct 09 2024

a(n) = A006666(n) + A006667(n).
a(n) = A112695(n) + 2 for n > 2. - Reinhard Zumkeller, Apr 18 2008
a(n) = A008908(n) - 1. - L. Edson Jeffery, Jul 21 2014
a(n) = A135282(n) + A208981(n) (after Alonso del Arte's comment in A208981), if 1 is reached, otherwise a(n) = -1. - Omar E. Pol, Apr 10 2022
a(n) = 2*A007814(n + 1) + a(A085062(n)) + 1 for n > 1. - Wing-Yin Tang, Jan 06 2025

More terms from Larry Reeves (larryr(AT)acm.org), Apr 27 2001

"Escape clause" added to definition by N. J. A. Sloane, Jun 06 2017

A127885 a(n) = minimal number of steps to get from n to 1, where a step is x -> 3x+1 if x is odd, or x -> either x/2 or 3x+1 if x is even; or -1 if 1 is never reached.

Original entry on oeis.org

0, 1, 7, 2, 5, 8, 16, 3, 11, 6, 14, 9, 9, 17, 17, 4, 12, 12, 20, 7, 7, 15, 15, 10, 23, 10, 23, 10, 18, 18, 31, 5, 18, 13, 13, 13, 13, 21, 26, 8, 21, 8, 21, 16, 16, 16, 29, 11, 16, 16, 24, 11, 11, 24, 24, 11, 24, 19, 24, 19, 19, 32, 32, 6, 19, 19, 27, 14, 14, 14, 27, 14, 27, 14, 14, 22, 22, 27, 27, 9, 22, 22, 22, 9, 9, 22, 22, 17, 22, 17, 30, 17, 17, 30, 30, 12, 30, 17, 17, 17

Offset: 1

David Applegate and N. J. A. Sloane, Feb 04 2007

nonn

In contrast to the "3x+1" problem (see A006577), here you are free to choose either step if x is even.

See A125731 for the number of steps in the reverse direction, from 1 to n.

			Several early values use the path:
6 -> 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1.
The first path where choosing 3x+1 for even x helps is:
9 -> 28 -> 85 -> 256 -> 128 -> 64 -> 32 -> 16 -> 8 -> 4 -> 2 -> 1.

M. J. Halm, Sequences (Re)discovered, Mpossibilities 81 (Aug. 2002), p. 1.

David Applegate, Table of n, a(n) for n = 1..1000
Index entries for sequences related to 3x+1 (or Collatz) problem

A127886 gives the difference between A006577 and this sequence.
Cf. A006577, A125731, A127887, A125195, A125686, A125719, A261870.
Cf. A290100 (size of the final set when using Alekseyev's algorithm).
Cf. also A257265.

# Code from David Applegate: Be careful - the function takes an iteration limit and returns the limit if it wasn't able to determine the answer (that is, if A127885(n,lim) == lim, all you know is that the value is >= lim). To use it, do manual iteration on the limit.
A127885 := proc(n,lim) local d,d2; options remember;
if (n = 1) then return 0; end if;
if (lim <= 0) then return 0; end if;
if (n > 2 ^ lim) then return lim; end if;
if (n mod 2 = 0) then
d := A127885(n/2,lim-1);
d2 := A127885(3*n+1,d);
if (d2 < d) then d := d2; end if;
else
d := A127885(3*n+1,lim-1);
end if;
return 1+d;
end proc;

Table[-1 + Length@ NestWhileList[Flatten[# /. {k_ /; OddQ@ k :> 3 k + 1, k_ /; EvenQ@ k :> {k/2, 3 k + 1}}] &, {n}, FreeQ[#, 1] &], {n, 126}] (* Michael De Vlieger, Aug 20 2017 *)

{ A127885(n) = my(S,k); S=[n]; k=0; while( S[1]!=1, k++; S=vecsort( concat(apply(x->3*x+1,S), apply(x->x\2, select(x->x%2==0,S) )),,8);  ); k } /* Max Alekseyev, Sep 03 2015 */

a(1) = 0; and for n > 1, if n is odd, a(n) = 1 + a(3n+1), and if n is even, a(n) = 1 + min(a(3n+1),a(n/2)). [But with a similar caveat as in A257265] - Antti Karttunen, Aug 20 2017

Escape clause added to definition by N. J. A. Sloane, Aug 20 2017

A375494 a(n) = minimum number of operations chosen from f(x) = 3x+1 and g(x) = floor(x/2) needed to reach n when starting from 1.

Original entry on oeis.org

1, 0, 2, 4, 1, 6, 3, 3, 8, 5, 5, 5, 10, 2, 7, 7, 7, 7, 12, 4, 4, 9, 4, 9, 9, 9, 9, 14, 6, 6, 6, 6, 11, 6, 6, 11, 11, 11, 11, 11, 3, 16, 8, 8, 8, 8, 8, 8, 13, 8, 8, 8, 8, 13, 13, 13, 13, 13, 5, 13, 5, 5, 18, 10, 10, 10, 10, 5, 10, 10, 10, 10, 15, 10, 10, 10, 10, 10, 10, 10

Offset: 0

Russell Y. Webb, Aug 18 2024

nonn

The Collatz problems (related problems known by various names, see references) involve iterating the Collatz Mapping (A006370) which applies either 3n+1 or n/2 iteratively when n is odd or even respectively. Considering f(x)=3x+1 and g(x)=x/2 as integer operations of interest, we ask what is the shortest sequence of these operation that produces the nonnegative integers starting with x_0=1. One is chosen as the starting value since producing the number one is the stopping condition for the Hailstone sequences (A006577). The number of distinct shortest sequences of operations (A375495) and the numbers with unique, shortest constructions (A375496) are also of interest.

Seems likely there is a sequence of f and g starting from 1 to reach each nonnegative integer, but a proof has not been proposed.

			For example, to start with 1 and produce the number 7. The shortest sequence is unique and length 3: (3*x+1, floor(x/2), 3*x+1) = f(g(f(x_0))), which follows the trajectory x_0=1, x_1=4, x_2=2, x_3=7.

Eric Weisstein's World of Mathematics, Collatz Problem

Cf. A375495 (number of ways), A375496 (with unique way).

Cf. A125731.

from itertools import product
seq = [None for _ in range(200)]
num = [   0 for _ in range(len(seq))]
for L in range(0, 23):
   for P in product((True, False), repeat=L):
      x = 1
      for upward in P:
         x = 3*x+1 if upward else x//2
      if x < len(seq):
         if num[x] == 0 or L < seq[x]:
            seq[x], num[x] = L, 1
         elif L == seq[x]:
            num[x] += 1
print(', '.join([str(x) for x in seq]))

cp's OEIS Frontend

A006577 Number of halving and tripling steps to reach 1 in '3x+1' problem, or -1 if 1 is never reached.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

PARI

Python

Python

R

Formula

Extensions

A127885 a(n) = minimal number of steps to get from n to 1, where a step is x -> 3x+1 if x is odd, or x -> either x/2 or 3x+1 if x is even; or -1 if 1 is never reached.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

A375494 a(n) = minimum number of operations chosen from f(x) = 3x+1 and g(x) = floor(x/2) needed to reach n when starting from 1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Python