A060412 -id:A060412 - cp's OEIS frontend

A014682 The Collatz or 3x+1 function: a(n) = n/2 if n is even, otherwise (3n+1)/2.

0, 2, 1, 5, 2, 8, 3, 11, 4, 14, 5, 17, 6, 20, 7, 23, 8, 26, 9, 29, 10, 32, 11, 35, 12, 38, 13, 41, 14, 44, 15, 47, 16, 50, 17, 53, 18, 56, 19, 59, 20, 62, 21, 65, 22, 68, 23, 71, 24, 74, 25, 77, 26, 80, 27, 83, 28, 86, 29, 89, 30, 92, 31, 95, 32, 98, 33, 101, 34, 104

Offset: 0

Mohammad K. Azarian

This is the function usually denoted by T(n) in the literature on the 3x+1 problem. See A006370 for further references and links.
Intertwining of sequence A016789 '2,5,8,11,... ("add 3")' and the nonnegative integers.
a(n) = log_2(A076936(n)). - Amarnath Murthy, Oct 19 2002
The average value of a(0), ..., a(n-1) is A004526(n). - Amarnath Murthy, Oct 19 2002
Partial sums are A093353. - Paul Barry, Mar 31 2008
Absolute first differences are essentially in A014681 and A103889. - R. J. Mathar, Apr 05 2008
Only terms of A016789 occur twice, at positions given by sequences A005408 (odd numbers) and A016957 (6n+4): (1,4), (3,10), (5,16), (7,22), ... - Antti Karttunen, Jul 28 2017
a(n) represents the unique congruence class modulo 2n+1 that is represented an odd number of times in any 2n+1 consecutive oblong numbers (A002378). This property relates to Jim Singh's 2018 formula, as n^2 + n is a relevant oblong number. - Peter Munn, Jan 29 2022

			a(3) = -3*(-1) - 2*1 - 1*(-1) - 0*1 + 1*(-1) + 2*1 + 3*(-1) + 4*1 + 5*(-1) + 6*1 = 5. - _Bruno Berselli_, Dec 14 2015

J. C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, Amer. Math. Soc., 2010.

N. J. A. Sloane (terms 0..1000) & Antti Karttunen, Table of n, a(n) for n = 0..100000
C. J. Everett, Iteration of the number-theoretic function f(2n) = n, f(2n + 1) = 3n + 2, Advances in Mathematics, Volume 25, Issue 1, July 1977, Pages 42-45.
Jeffrey C. Lagarias, The 3x+1 Problem: An Overview, arXiv:2111.02635 [math.NT], 2021.
Keenan Monks, Kenneth G. Monks, Kenneth M. Monks, and Maria Monks, Strongly sufficient sets and the distribution of arithmetic sequences in the 3x+1 graph, arXiv:1204.3904v1 [math.DS], Apr 17 2012.
R. Terras, A stopping time problem on the positive integers, Acta Arith. 30 (1976) 241-252.
Eric Weisstein's World of Mathematics, Collatz Problem
Wikipedia, Collatz conjecture
Index entries for sequences related to 3x+1 (or Collatz) problem
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A002378, A004526, A076936, A139391, A016116, A126241, A060412, A060413, A006370, A070168 (iterations), A005408, A016957, A064455, A153285, A008619, A193356.
Bisections: A001477, A016789.
Cf. A093353.

a014682 n = if r > 0 then div (3 * n + 1) 2 else n'
            where (n', r) = divMod n 2
-- Reinhard Zumkeller, Oct 03 2014

[IsOdd(n) select (3*n+1)/2 else n/2: n in [0..52]]; // Vincenzo Librandi, Sep 28 2018

T:=proc(n) if n mod 2 = 0 then n/2 else (3*n+1)/2; fi; end; # N. J. A. Sloane, Jan 31 2011
A076936 := proc(n) option remember ; local apr,ifr,me,i,a ; if n <=2 then n^2 ; else apr := mul(A076936(i),i=1..n-1) ; ifr := ifactors(apr)[2] ; me := -1 ; for i from 1 to nops(ifr) do me := max(me, op(2,op(i,ifr))) ; od ; me := me+ n-(me mod n) ; a := 1 ; for i from 1 to nops(ifr) do a := a*op(1,op(i,ifr))^(me-op(2,op(i,ifr))) ; od ; if a = A076936(n-1) then me := me+n ; a := 1 ; for i from 1 to nops(ifr) do a := a*op(1,op(i,ifr))^(me-op(2,op(i,ifr))) ; od ; fi ; RETURN(a) ; fi ; end: A014682 := proc(n) log[2](A076936(n)) ; end: for n from 1 to 85 do printf("%d, ",A014682(n)) ; od ; # R. J. Mathar, Mar 20 2007

Collatz[n_?OddQ] := (3n + 1)/2; Collatz[n_?EvenQ] := n/2; Table[Collatz[n], {n, 0, 79}] (* Alonso del Arte, Apr 21 2011 *)
LinearRecurrence[{0, 2, 0, -1}, {0, 2, 1, 5}, 70] (* Jean-François Alcover, Sep 23 2017 *)
Table[If[OddQ[n], (3 n + 1) / 2, n / 2], {n, 0, 60}] (* Vincenzo Librandi, Sep 28 2018 *)

a(n)=if(n%2,3*n+1,n)/2 \\ Charles R Greathouse IV, Sep 02 2015

a(n)=if(n<2,2*n,(n^2-n-1)%(2*n+1)) \\ Jim Singh, Sep 28 2018

def a(n): return n//2 if n%2==0 else (3*n + 1)//2
print([a(n) for n in range(101)]) # Indranil Ghosh, Jul 29 2017

From Paul Barry, Mar 31 2008: (Start)
G.f.: x*(2 + x + x^2)/(1-x^2)^2.
a(n) = (4*n+1)/4 - (2*n+1)*(-1)^n/4. (End)
a(n) = -a(n-1) + a(n-2) + a(n-3) + 4. - John W. Layman
For n > 1 this is the image of n under the modified "3x+1" map (cf. A006370): n -> n/2 if n is even, n -> (3*n+1)/2 if n is odd. - Benoit Cloitre, May 12 2002
O.g.f.: x*(2+x+x^2)/((-1+x)^2*(1+x)^2). - R. J. Mathar, Apr 05 2008
a(n) = 5/4 + (1/2)*((-1)^n)*n + (3/4)*(-1)^n + n. - Alexander R. Povolotsky, Apr 05 2008
a(n) = Sum_{i=-n..2*n} i*(-1)^i. - Bruno Berselli, Dec 14 2015
a(n) = Sum_{k=0..n-1} Sum_{i=0..k} C(i,k) + (-1)^k. - Wesley Ivan Hurt, Sep 20 2017
a(n) = (n^2-n-1) mod (2*n+1) for n > 1. - Jim Singh, Sep 26 2018
The above formula can be rewritten to show a pattern: a(n) = (n*(n+1)) mod (n+(n+1)). - Peter Munn, Jan 29 2022
Binary: a(n) = (n shift left (n AND 1)) - (n shift right 1) = A109043(n) - A004526(n). - Rudi B. Stranden, Jun 15 2021
From Rudi B. Stranden, Mar 21 2022: (Start)
a(n) = A064455(n+1) - 1, relating the number ON cells in row n of cellular automaton rule 54.
a(n) = 2*n - A071045(n).
(End)
E.g.f.: (1 + x)*sinh(x)/2 + 3*x*cosh(x)/2 = ((4*x+1)*e^x + (2*x-1)*e^(-x))/4. - Rénald Simonetto, Oct 20 2022
a(n) = n*(n mod 2) + ceiling(n/2) = A193356(n) + A008619(n+1). - Jonathan Shadrach Gilbert, Mar 12 2023
a(n) = 2*a(n-2) - a(n-4) for n > 3. - Chai Wah Wu, Apr 17 2024

Edited by N. J. A. Sloane, Apr 26 2008, at the suggestion of Artur Jasinski

Edited by N. J. A. Sloane, Jan 31 2011

A006884 In the '3x+1' problem, these values for the starting value set new records for highest point of trajectory before reaching 1.

Original entry on oeis.org

1, 2, 3, 7, 15, 27, 255, 447, 639, 703, 1819, 4255, 4591, 9663, 20895, 26623, 31911, 60975, 77671, 113383, 138367, 159487, 270271, 665215, 704511, 1042431, 1212415, 1441407, 1875711, 1988859, 2643183, 2684647, 3041127, 3873535, 4637979, 5656191

Offset: 1

N. J. A. Sloane, Robert Munafo

Both the 3x+1 steps and the halving steps are counted.

Where records occur in A025586: A006885(n) = A025586(a(n)) and A025586(m) < A006885(n) for m < a(n). - Reinhard Zumkeller, May 11 2013

R. B. Banks, Slicing Pizzas, Racing Turtles and Further Adventures in Applied Mathematics, Princeton Univ. Press, 1999. See p. 96.
D. R. Hofstadter, Goedel, Escher, Bach: an Eternal Golden Braid, Random House, 1980, p. 400.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

David Barina, Table of n, a(n) for n = 1..98 (terms 1..84 from T. D. Noe, terms 85..89 from N. J. A. Sloane).
David Barina, Path records.
David Barina, Improved verification limit for the convergence of the Collatz conjecture, J. Supercomputing (2025) Vol. 81, Art. No. 810. See p. 12.
Brian Hayes, Computer Recreations: On the ups and downs of hailstone numbers, Scientific American, 250 (No. 1, 1984), pp. 10-16.
J. C. Lagarias, The 3x+1 problem and its generalizations, Amer. Math. Monthly, 92 (1985), 3-23.
G. T. Leavens and M. Vermeulen, 3x+1 search problems, Computers and Mathematics with Applications, 24 (1992), 79-99.
G. T. Leavens and M. Vermeulen, 3x+1 search programs, Computers and Mathematics with Applications, 24 (1992), 79-99. (Annotated scanned copy)
Tomás Oliveira e Silva, Tables (gives many more terms).
Eric Roosendaal, 3x+1 Path Records.
Olivier Rozier and Claude Terracol, Paradoxical behavior in Collatz sequences, arXiv:2502.00948 [math.GM], 2025. See p. 15.
Robert G. Wilson v, Letter to N. J. A. Sloane with attachments, Jan. 1989.
Robert G. Wilson v, Tables of A6877, A6884, A6885, Jan. 1989.
Index entries for sequences related to 3x+1 (or Collatz) problem
Index entries for sequences from "Goedel, Escher, Bach"

A060409 gives associated "dropping times", A060410 the maximal values and A060411 the steps at which the maxima occur.

Cf. A006885, A006877, A006878, A033492, A060412-A060415, A132348.

a006884 n = a006884_list !! (n-1)
a006884_list = f 1 0 a025586_list where
   f i r (x:xs) = if x > r then i : f (i + 1) x xs else f (i + 1) r xs
-- Reinhard Zumkeller, May 11 2013

mcoll[n_]:=Max@@NestWhileList[If[EvenQ[#],#/2,3#+1]&,n,#>1&]; t={1,max=2}; Do[If[(y=mcoll[n])>max,max=y; AppendTo[t,n]],{n,3,705000,4}]; t (* Jayanta Basu, May 28 2013 *)
DeleteDuplicates[Parallelize[Table[{n,Max[NestWhileList[If[EvenQ[#],#/2,3#+1]&,n,#>1&]]},{n,57*10^5}]],GreaterEqual[#1[[2]],#2[[2]]]&][[;;,1]] (* Harvey P. Dale, Apr 23 2023 *)

A025586(n)=my(r=n); while(n>2, if(n%2, n=3*n+1; if(n>r, r=n)); n>>=1); r
r=0; for(n=1,1e6, t=A025586(n); if(t>r, r=t; print1(n", "))) \\ Charles R Greathouse IV, May 25 2016

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.

Original entry on oeis.org

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

A126241 Dropping times in the 3n+1 problem (or the Collatz problem). Let T(n):=n/2 if n is even, (3n+1)/2 otherwise (A014682). Let a(n) be the smallest integer k such that T^(k)(n)

Original entry on oeis.org

0, 1, 4, 1, 2, 1, 7, 1, 2, 1, 5, 1, 2, 1, 7, 1, 2, 1, 4, 1, 2, 1, 5, 1, 2, 1, 59, 1, 2, 1, 56, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 5, 1, 2, 1, 54, 1, 2, 1, 4, 1, 2, 1, 5, 1, 2, 1, 7, 1, 2, 1, 54, 1, 2, 1, 4, 1, 2, 1, 51, 1, 2, 1, 5, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 5, 1, 2, 1, 45, 1, 2, 1, 8, 1, 2, 1, 4

Offset: 1

Christof Menzel (christof.menzel(AT)hs-niederrhein.de), Mar 08 2007

nonn

Also called "stopping times", although that term is usually reserved for A006666.
From K. Spage, Oct 22 2009, corrected Aug 21 2014: (Start)
Congruency relationship: For n>1 and m>1, all m congruent to n mod 2^(a(n)) have a dropping time equal to a(n).
By refining the definition of the dropping time to "starting with x=n, iterate x until (abs(x) <= abs(n))" the above congruency relationship holds for all nonnegative values of n and all positive or negative values of m including zero.
By this refined definition, a(1)=2 rather than the usual zero set by convention. All other values of positive a(n) remain unchanged. (End)
Terras defines a coefficient stopping time (definition 1.5) tau(n) = d which is the smallest d for which 3^u/2^d < 1 where u is the number of tripling steps among the first d steps starting from n. Clearly tau(n) <= a(n), and Terras conjectures (conjecture 2.9) that tau(n) = a(n) for n>=2. - Olivier Rozier, May 13 2024

			s(15) = 7, since the trajectory {T^(k)(15)} (k=1,2,3,...) equals 23,35,53,80,40,20,10.

J. C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, Amer. Math. Soc., 2010. See p. 33.

T. D. Noe, Table of n, a(n) for n = 1..10000
J. C. Lagarias, The 3x+1 Problem: An Annotated Bibliography (1963-1999), arXiv:math/0309224 [math.NT], 2003-2011.
Olivier Rozier and Claude Terracol, Paradoxical behavior in Collatz sequences, arXiv:2502.00948 [math.GM], 2025. See p. 2.
Riho Terras, A stopping time problem on the positive integers, Acta Arith. 30 (1976) 241-252, with definition 0.1 chi(n) = a(n).
Index entries for sequences related to 3x+1 (or Collatz) problem

See A074473, which is the main entry for dropping times.
Cf. A014682, A006666, A006577.
Records: A060412, A060413.
Cf. A020914 (allowable dropping times). - K. Spage, Aug 22 2014

Collatz2[n_] := If[n<2, {}, Rest[NestWhileList[If[EvenQ[#], #/2, (3 # + 1)/2] &, n, # >= n &]]]; Table[Length[Collatz2[n]], {n, 1, 1000}]

a(n) = ceiling(A102419(n)/(1+log(2)/log(3))). - K. Spage, Aug 22 2014

Broken link fixed by K. Spage, Oct 22 2009

A008884 3x+1 sequence starting at 27.

Original entry on oeis.org

27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079

Offset: 0

N. J. A. Sloane

27=A060412(4); a(A006577(27))=a(111)=1; a(n)=A161021(n+59) for n with 103<=n<=111. - Reinhard Zumkeller, Jun 03 2009

At step 109 enters the loop 4 2 1 4 2 1 4 2 1 ... - N. J. A. Sloane, Jul 27 2019

R. K. Guy, Unsolved Problems in Number Theory, E16.
H.-O. Peitgen et al., Chaos and Fractals, Springer, p. 33.

T. D. Noe, Table of n, a(n) for n = 0..111
F. Oort, Prime numbers, 2013, ICCM Notices, Talk at Academia Sinica and National Taiwan University, 17-XII-2012.
Index entries for sequences related to 3x+1 (or Collatz) problem
Index entries for linear recurrences with constant coefficients, signature (0,0,1).

Cf. A161022, A161023.

Row 27 of A347270.

[ n eq 1 select 27 else IsOdd(Self(n-1)) select 3*Self(n-1)+1 else Self(n-1) div 2: n in [1..70] ]; // Klaus Brockhaus, Dec 25 2010

f := proc(n) option remember; if n = 0 then 27; elif f(n-1) mod 2 = 0 then f(n-1)/2 else 3*f(n-1)+1; fi; end;

NestList[If[EvenQ[#],#/2,3#+1]&,27,70] (* Harvey P. Dale, Jun 30 2011 *)

Collatz(n,lim=0)={
my(c=n,e=0,L=List(n)); if(lim==0, e=1; lim=n*10^6);
for(i=1,lim, if(c%2==0, c=c/2, c=3*c+1); listput(L,c); if(e&&c==1, break));
return(Vec(L)); }
print(Collatz(27)) \\ A008884 (from 27 to the first 1)
\\ Anatoly E. Voevudko, Mar 26 2016

a(0) = 27, a(n) = 3*a(n-1)+1 if a(n-1) is odd, a(n) = a(n-1)/2 if a(n-1) is even. - Vincenzo Librandi, Dec 24 2010; corrected by Klaus Brockhaus, Dec 25 2010

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

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

cp's OEIS Frontend

A060413 In the '3x+1' problem, take the sequence of starting values which set new records for the "dropping time" (A060412); sequence gives associated dropping times.

Views

Author

Keywords

Links

Crossrefs

A060415 In the '3x+1' problem, take the sequence of starting values which set new records for the "dropping time" (A060412); sequence gives associated iterate where maximal value is reached in the trajectory with that start.

Views

Author

Keywords

Links

Crossrefs

Extensions

A060414 In the '3x+1' problem, take the sequence of starting values which set new records for the "dropping time" (A060412); sequence gives associated maximal value reached in the trajectory with that start.

Views

Author

Keywords

Links

A014682 The Collatz or 3x+1 function: a(n) = n/2 if n is even, otherwise (3n+1)/2.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

PARI

PARI

Python

Formula

Extensions

A006884 In the '3x+1' problem, these values for the starting value set new records for highest point of trajectory before reaching 1.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Haskell

Mathematica

PARI

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

A126241 Dropping times in the 3n+1 problem (or the Collatz problem). Let T(n):=n/2 if n is even, (3n+1)/2 otherwise (A014682). Let a(n) be the smallest integer k such that T^(k)(n)

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A008884 3x+1 sequence starting at 27.

Views

Author

Keywords