A006666 -id:A006666 - cp's OEIS frontend

A302175 a(n) = [2^A006666(n)/3^A006667(n)], where [x] = floor(x).

1, 2, 3, 4, 5, 7, 8, 8, 11, 10, 12, 14, 14, 16, 16, 16, 18, 22, 22, 21, 21, 25, 25, 28, 29, 28, 32, 33, 33, 33, 36, 32, 39, 37, 37, 44, 44, 44, 47, 42, 48, 42, 53, 50, 50, 50, 54, 56, 59, 59, 59, 56, 56, 64, 64, 67, 71, 67, 71, 67, 67, 72, 72, 64, 79, 79, 79, 75

Offset: 1

Michel Lagneau, Apr 03 2018

The sequence contains A211981 and the powers of 2 (A000079).
There exists a subset E = { 1, 2, 3, 4, 5, 8, 10, 16, 21, 32, 42, 64, 85, 128, 170, 227, 256, 341, 512, 682, 1024, 2048, ...} in {a(n)} such that each element m of E generates the Collatz sequence of iterates m -> T_1(m) -> T_2(m) -> T_3(m) -> ... -> 1 where any T_i(m) is an element of E of the form [2^i /3^j] where i = A006666(m), or A006666(m)-1, or ... and j = A006667(m), or A006667(m)-1, or ..., but with A006667(m) <= 3. If m is even then m/2 is in E.
For example, the statement that "3 is an element of E" implies that each element of the trajectory 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1 belongs to E. Thus the trajectory of the number 3 can be represented by [2^5/3^2] -> [2^5/3^1] -> [2^4/3^1] -> [2^4/3^0] -> [2^3/3^0] -> [2^2/3^0] -> [2^1/3^0] -> [2^0/3^0].

			a(39) = [2^A006666(39)/3^A006667(39)] = [2^23/3^11] = [47.353937...] = 47.

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

Cf. A000079, A006666, A006667, A211981, A225089, A265099.

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; nn = 70; t = {}; n = 0; While[Length[t] < nn, n++; c = Collatz[n]; ev = Length[Select[c, EvenQ]]; od = Length[c] - ev - 1; AppendTo[t, Floor[2^ev/3^od]]]; t

a(n) = my(t, h); while(n>1, if(n%2, n=3*n+1; t++, n>>=1; h++)); 2^h\3^t; \\ Michel Marcus, May 05 2018

A303811 Least k such that A006666(k)/A006667(k) = prime(n).

Original entry on oeis.org

159, 6, 10, 40, 640, 2560, 40960, 163840, 2621440, 167772160, 671088640, 42949672960, 687194767360, 2748779069440, 43980465111040, 2814749767106560, 180143985094819840, 720575940379279360, 46116860184273879040, 737869762948382064640, 2951479051793528258560

Offset: 1

Michel Lagneau, Sep 10 2018

nonn

A006666 and A006667 are respectively the number of halving and tripling steps in the '3x+1' problem.

For n > 2, it seems that a(n) is of the form a(n) = 5*2^q with q = 1, 3, 7, 9, 13, 15, 19, 25, 27, 33, 37, 39, 43, 49, 55, 57, 63, 67, 69, ... (Numbers q such that q+4 is prime: A172367)

			a(4) = 40 because A006666(40)/A006667(40) = 7/1 = prime(4).

Cf. A006577, A006666, A006667, A172367, A287798.

nn:=10^20:
for n from 1 to 10 do:
ii:=0:
   for k from 1 to nn while(ii=0) do:
    it0:=0:it1:=0:m:=k:
      for i from 1 to nn while(m<>1) do:
        if irem(m, 2)=0
         then
         m:=m/2:it0:=it0+1:
         else
         m:=3*m+1:it1:=it1+1:
       fi:
     od:
      if it1<>0 and it0/it1 = ithprime(n)
       then
       ii:=1:printf(`%d %d \n`,n,k):
       else
     fi:
od:
od:

A304123 Denominators of records low values of the ratio n*3^A006667(n)/2^A006666(n).

Original entry on oeis.org

1, 32, 2048, 8192, 2199023255552, 144115188075855872, 576460752303423488, 2305843009213693952

Offset: 1

Michel Lagneau, May 06 2018

			For n=1 to 10 the ratios are 1, 1, 27/32, 1, 15/16, 27/32, 1701/2048, 1, 6561/8192, 15/16, so the low records are 1, 27/32, 1701/2048, 6561/8192, ...

Cf. A127789 (for the indices where these records occur).
Cf. A304119 (for the corresponding numerators).
Cf. A006666, A006667.

q=1;Collatz[n_]:=NestWhileList[If[EvenQ[#],#/2,3 #+1]&,n,#>1&];nn=5000;t={};n=0;While[Length[t]

ht(n) = my(h, t); while(n>1, if(n%2, n=3*n+1; t++, n>>=1; h++)); return([h, t]);
lista(nn) = {m = 2; for (n=1, nn, v = ht(n); newm = n*3^v[2]/2^v[1]; if (newm < m, print1(denominator(newm), ", "); m = newm)); } \\ Michel Marcus, May 06 2018

A304524 Consider the ratio res(p) = 2^A006666(p) / (p*3^A006667(p)) where p is prime. The prime numbers in this sequence are those for which res(p) sets a new record.

Original entry on oeis.org

2, 3, 7, 37, 43, 229, 271, 379, 673, 839, 1987, 5297, 25111, 44641, 50221, 94057, 334423, 1189057, 1759579, 2505337, 28153249, 46869157, 87780541, 584543567, 768901097

Offset: 1

Michel Lagneau, May 14 2018

Is the sequence finite?
In the general case, the residue of a number n in the 3x+1 problem is defined as the ratio res(n) = 2^A006666(n) / (n*3^A006667(n)) (see A127789).
Conjecture: for all prime p, res(p) < res(993) = 2^61/(3^32*993) = 1.253142... (see A304174).

			From _Jon E. Schoenfield_, May 23 2018: (Start)
Let D = A006666(p) and U = A006667(p); then res(p) = 2^D/(p*3^U). It seems clear that res(993) - res(p) is converging toward a positive value:
.
          p |   D |  U |     res(p)      | res(993)-res(p)
  ----------+-----+----+-----------------+----------------
          2 |   1 |  0 | 1               | 0.2531421443...
          3 |   5 |  2 | 1.1851851851... | 0.0679569592...
          7 |  11 |  5 | 1.2039976484... | 0.0491444959...
         37 |  15 |  6 | 1.2148444741... | 0.0382976702...
         43 |  20 |  9 | 1.2389111604... | 0.0142309838...
        229 |  24 | 10 | 1.2407145246... | 0.0124276197...
        271 |  29 | 13 | 1.2425797507... | 0.0105623936...
        379 |  39 | 19 | 1.2480350469... | 0.0051070974...
        673 |  43 | 21 | 1.2494773856... | 0.0036647587...
        839 |  56 | 29 | 1.2514151532... | 0.0017269911...
       1987 |  62 | 32 | 1.2525114739... | 0.0006306704...
       5297 |  65 | 33 | 1.2529055685... | 0.0002365758...
      25111 |  72 | 36 | 1.2529406796... | 0.0002014647...
      44641 |  76 | 38 | 1.2529625095... | 0.0001796348...
      50221 |  73 | 36 | 1.2529656281... | 0.0001765162...
      94057 |  85 | 43 | 1.2529812032... | 0.0001609411...
     334423 |  90 | 45 | 1.2529882803... | 0.0001538640...
    1189057 |  95 | 47 | 1.2529909733... | 0.0001511710...
    1759579 | 113 | 58 | 1.2529910420... | 0.0001511023...
    2505337 | 104 | 52 | 1.2529915763... | 0.0001505680...
   28153249 | 117 | 58 | 1.2529917096... | 0.0001504347...
   46869157 | 132 | 67 | 1.2529917720... | 0.0001503722...
   87780541 | 144 | 74 | 1.2529919281... | 0.0001502162...
  584543567 | 161 | 83 | 1.2529919325... | 0.0001502118...
  768901097 | 182 | 96 | 1.2529919396... | 0.0001502047...
(End)

Eric Roosendaal, On the 3x+1 Problem

Cf. A006666, A006667, A127789, A304174.

lst={2};Print["a(n)"," ","A006667(a(n))"," ","A006666(a(n))","       ","res(a(n))"];q=1;Collatz[n_]:=NestWhileList[If[EvenQ[#],#/2,3 #+1]&,Prime[n],#>1&];nn=10000;t={};n=0;While[Length[t]5000,Break[]];q=Prime[n]*3^od/2^ev]];lst

a(23)-a(24) from Jon E. Schoenfield, May 19 2018

A318907 Numbers m such that A006666(m)/A006667(m) is an integer.

Original entry on oeis.org

5, 6, 10, 17, 20, 21, 24, 26, 40, 42, 44, 45, 46, 80, 84, 85, 96, 104, 106, 112, 113, 116, 117, 120, 122, 136, 138, 140, 141, 150, 151, 159, 160, 168, 170, 283, 288, 296, 298, 304, 308, 309, 320, 321, 324, 325, 326, 331, 336, 340, 341, 377, 384, 416, 424, 426

Offset: 1

Michel Lagneau, Sep 06 2018

nonn

A006666 and A006667 are respectively the number of halving and tripling steps in the '3x+1' problem.
The corresponding integers are 4, 3, 5, 3, 6, 6, 4, 4, 7, 7, 3, 3, 3, 8, 8, 8, 5, 5, 5, 3, 5, 3, 3, 3, ...
The numbers of the form (4^k - 1)/3 for k > 1 (A002450) are in the sequence.
We observe subsets of consecutive numbers: (5, 6), (20, 21), (44, 45, 46), (84, 85), (112, 113), ...

			17 is in the sequence because A006666(17)/A006667(17) = 9/3 = 3 is an integer.

Cf. A002450, A006666, A006667, A277367, A281938.

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; nn = 70; t = {}; n = 0; While[Length[t] < nn, n++; c = Collatz[n]; ev = Length[Select[c, EvenQ]]; od = Length[c] - ev - 1; If[od>0 && IntegerQ[ev/od],AppendTo[t, n]]]; t

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

Original entry on oeis.org

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

A006667 Number of tripling steps to reach 1 from n in '3x+1' problem, or -1 if 1 is never reached.

Original entry on oeis.org

0, 0, 2, 0, 1, 2, 5, 0, 6, 1, 4, 2, 2, 5, 5, 0, 3, 6, 6, 1, 1, 4, 4, 2, 7, 2, 41, 5, 5, 5, 39, 0, 8, 3, 3, 6, 6, 6, 11, 1, 40, 1, 9, 4, 4, 4, 38, 2, 7, 7, 7, 2, 2, 41, 41, 5, 10, 5, 10, 5, 5, 39, 39, 0, 8, 8, 8, 3, 3, 3, 37, 6, 42, 6, 3, 6, 6, 11, 11, 1, 6, 40, 40, 1, 1, 9, 9, 4, 9, 4, 33, 4, 4, 38

Offset: 1

N. J. A. Sloane and Bill Gosper

A075680, which gives the values for odd n, isolates the essential behavior of this sequence. - T. D. Noe, Jun 01 2006

A033959 and A033958 give record values and where they occur. - Reinhard Zumkeller, Jan 08 2014

J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 204, Problem 22.
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).

T. D. Noe, Table of n, a(n) for n = 1..10000
J. C. Lagarias, The 3x+1 problem and its generalizations, Amer. Math. Monthly, 92 (1985), 3-23.
Eric Weisstein's World of Mathematics, Collatz Problem.
Wikipedia, Collatz conjecture
Index entries for sequences related to 3x+1 (or Collatz) problem

Equals A078719(n)-1.

Cf. A000079, A000265, A006370, A006577, A006666 (halving steps), A092893, A127789, A139391, A209229.

a006667 = length . filter odd . takeWhile (> 2) . (iterate a006370)
a006667_list = map a006667 [1..]
-- Reinhard Zumkeller, Oct 08 2011

a:= proc(n) option remember; `if`(n<2, 0,
      `if`(n::even, a(n/2), 1+a(3*n+1)))
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Aug 08 2023

Table[Count[Differences[NestWhileList[If[EvenQ[#],#/2,3#+1]&,n,#>1&]], ?Positive], {n,100}] (* _Harvey P. Dale, Nov 14 2011 *)

for(n=2,100,s=n; t=0; while(s!=1,if(s%2==0,s=s/2,s=(3*s+1)/2; t++); if(s==1,print1(t,","); ); ))

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, Apr 14 2017

a(1) = 0, a(n) = a(n/2) if n is even, a(n) = a(3n+1)+1 if n>1 is odd. The Collatz conjecture is that this defines a(n) for all n >= 1.
a(n) = A078719(n) - 1; a(A000079(n))=0; a(A062052(n))=1; a(A062053(n))=2; a(A062054(n))=3; a(A062055(n))=4; a(A062056(n))=5; a(A062057(n))=6; a(A062058(n))=7; a(A062059(n))=8; a(A062060(n))=9. - Reinhard Zumkeller, Oct 08 2011
a(n*2^k) = a(n), for all k >= 0. - L. Edson Jeffery, Aug 11 2014
a(n) = floor(log(2^A006666(n)/n)/log(3)). - Joe Slater, Aug 30 2017
a(n) = a(A085062(n)) + A007814(n+1) for n >= 2. - Alan Michael Gómez Calderón, Feb 07 2025
From  Alan Michael Gómez Calderón, Mar 31 2025: (Start)
a(n) = a(A139391(n)) + (n mod 2) for n >= 2;
a(n) = a(A139391(A000265(n))) - A209229(n) + 1 for n >= 2;
a(n) = a(A000265(A139391(n))) + (n mod 2) for n >= 2. (End)

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

A008908 a(n) = (1 + number of halving and tripling steps to reach 1 in the Collatz (3x+1) problem), or -1 if 1 is never reached.

Original entry on oeis.org

1, 2, 8, 3, 6, 9, 17, 4, 20, 7, 15, 10, 10, 18, 18, 5, 13, 21, 21, 8, 8, 16, 16, 11, 24, 11, 112, 19, 19, 19, 107, 6, 27, 14, 14, 22, 22, 22, 35, 9, 110, 9, 30, 17, 17, 17, 105, 12, 25, 25, 25, 12, 12, 113, 113, 20, 33, 20, 33, 20, 20, 108, 108, 7, 28, 28, 28, 15, 15, 15, 103

Offset: 1

N. J. A. Sloane, Bill Gosper

The number of steps (iterations of the map A006370) to reach 1 is given by A006577, this sequence counts 1 more. - M. F. Hasler, Nov 05 2017

When Collatz 3N+1 function is seen as an isometry over the dyadics, the halving step necessarily following each tripling is not counted, hence N -> N/2, if even, but N -> (3N+1)/2, if odd. Counting iterations of this map until reaching 1 leads to sequence A064433. - Michael Vielhaber (vielhaber(AT)gmail.com), Nov 18 2009

R. K. Guy, Unsolved Problems in Number Theory, E16.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
J. C. Lagarias, The 3x+1 problem and its generalizations, Amer. Math. Monthly, 92 (1985), 3-23.
Nitrxgen, Collatz Calculator
Wikipedia, Collatz conjecture
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A006577, A006370, A006667, A075677.

a008908 = length . a070165_row
-- Reinhard Zumkeller, May 11 2013, Aug 30 2012, Jul 19 2011

a:= proc(n) option remember; 1+`if`(n=1, 0,
      a(`if`(n::even, n/2, 3*n+1)))
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Jan 29 2021

Table[Length[NestWhileList[If[EvenQ[ # ], #/2, 3 # + 1] &, i, # != 1 &]], {i, 75}]

a(n)=my(c=1); while(n>1, n=if(n%2, 3*n+1, n/2); c++); c \\ Charles R Greathouse IV, May 18 2015

def a(n):
    if n==1: return 1
    x=1
    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, Apr 15 2017

a(n) = A006577(n) + 1.
a(n) = f(n,1) with f(n,x) = if n=1 then x else f(A006370(n),x+1).
a(A033496(n)) = A159999(A033496(n)). - Reinhard Zumkeller, May 04 2009
a(n) = A006666(n) + A078719(n).
a(n) = length of n-th row in A070165. - Reinhard Zumkeller, May 11 2013

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
Edited by M. F. Hasler, Nov 05 2017

A070168 Irregular triangle of Terras-modified Collatz problem.

Original entry on oeis.org

1, 2, 1, 3, 5, 8, 4, 2, 1, 4, 2, 1, 5, 8, 4, 2, 1, 6, 3, 5, 8, 4, 2, 1, 7, 11, 17, 26, 13, 20, 10, 5, 8, 4, 2, 1, 8, 4, 2, 1, 9, 14, 7, 11, 17, 26, 13, 20, 10, 5, 8, 4, 2, 1, 10, 5, 8, 4, 2, 1, 11, 17, 26, 13, 20, 10, 5, 8, 4, 2, 1, 12, 6, 3, 5, 8, 4, 2, 1, 13, 20, 10, 5, 8, 4, 2, 1, 14, 7, 11

Offset: 1

Eric W. Weisstein, Apr 23 2002

The row length of this irregular triangle is A006666(n) + 1 = A064433(n+1), n >= 1. - Wolfdieter Lang, Mar 20 2014

			The irregular triangle begins:
n\k   0   1   2   3   4   5   6   8  9 10  11  12  13  14 ...
1:    1
2:    2   1
3:    3   5   8   4   2   1
4:    4   2   1
5:    5   8   4   2   1
6:    6   3   5   8   4   2   1
7:    7  11  17  26  13  20  10   5  8  4   2   1
8:    8   4   2   1
9:    9  14   7  11  17  26  13  20 10  5   8   4   2   1
10:  10   5   8   4   2   1
11:  11  17  26  13  20  10   5   8  4  2   1
12:  12   6   3   5   8   4   2   1
13:  13  20  10   5   8   4   2   1
14:  14   7  11  17  26  13  20  10  5  8   4   2   1
15:  15  23  35  53  80  40  20  10  5  8   4   2   1
...  formatted by _Wolfdieter Lang_, Mar 20 2014
-------------------------------------------------------------

Reinhard Zumkeller, Rows n = 1..250 of triangle, flattened
J. C. Lagarias, The 3x+1 Problem and its Generalizations, Amer. Math. Monthly 92 (1985) 3-23.
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

Cf. A070165 (ordinary Collatz case).

Cf. A014682, A248573, A285098 (row sums).

a070168 n k = a070168_tabf !! (n-1) !! (k-1)
a070168_tabf = map a070168_row [1..]
a070168_row n = (takeWhile (/= 1) $ iterate a014682 n) ++ [1]
a070168_list = concat a070168_tabf
-- Reinhard Zumkeller, Oct 03 2014

f[n_] := If[EvenQ[n], n/2, (3 n + 1)/2];
Table[NestWhileList[f, n, # != 1 &], {n, 1, 30}] // Grid (* Geoffrey Critzer, Oct 18 2014 *)

def a(n):
    if n==1: return [1]
    l=[n, ]
    while True:
        if n%2==0: n//=2
        else: n = (3*n + 1)//2
        l.append(n)
        if n<2: break
    return l
for n in range(1, 16): print(a(n)) # Indranil Ghosh, Apr 15 2017

From Wolfdieter Lang, Mar 20 2014: (Start)
See Lagarias, pp. 4-7, eqs. (2.1), (2.4) with (2.5) and (2.6).
T(n,k) = T^{(k)}(n), with the iterations of the Terras-modified Collatz map: T(n) = n/2 if n is even and otherwise (3*n+1)/2, n >= 1. T^{(0)}(n) = n.
T(n,k) = lambda(n,k)*n + rho(n,k), with lambda(n,k) = (3^X(n,k,-1))/2^k and rho(n,k) = sum(x(n,j)*(3^X(n,k,j))/ 2^(k-j), j=0..(k-1)) with X(n,k,j) = sum(x(n,j+p), p=1.. (k-1-j)) where x(n,j) = T^{(j)}(n) (mod 2). The parity sequence suffices to determine T(n,k).
(End)

Name shortened, tabl changed into tabf, Cf. added by Wolfdieter Lang, Mar 20 2014

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

cp's OEIS Frontend

A302175 a(n) = [2^A006666(n)/3^A006667(n)], where [x] = floor(x).

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A303811 Least k such that A006666(k)/A006667(k) = prime(n).

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A304123 Denominators of records low values of the ratio n*3^A006667(n)/2^A006666(n).

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A304524 Consider the ratio res(p) = 2^A006666(p) / (p*3^A006667(p)) where p is prime. The prime numbers in this sequence are those for which res(p) sets a new record.

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A318907 Numbers m such that A006666(m)/A006667(m) is an integer.

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

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A006667 Number of tripling steps to reach 1 from n in '3x+1' problem, or -1 if 1 is never reached.

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