A006577 -id:A006577 - cp's OEIS frontend

A239126 Rectangular array showing the starting values M(n, k), k >= 1, for the Collatz operation (ud)^n, n >= 1, ending in an odd number, read by antidiagonals.

3, 7, 7, 11, 15, 15, 15, 23, 31, 31, 19, 31, 47, 63, 63, 23, 39, 63, 95, 127, 127, 27, 47, 79, 127, 191, 255, 255, 31, 55, 95, 159, 255, 383, 511, 511, 35, 63, 111, 191, 319, 511, 767, 1023, 1023, 39, 71, 127, 223, 383, 639, 1023, 1535, 2047, 2047

Offset: 1

Wolfdieter Lang, Mar 13 2014

The companion array and triangle for the odd end numbers N(n, k) is given in A239127.
The two operations on natural numbers m used in the Collatz 3x+1 conjecture are here (following the M. Trümper paper given in the link) denoted by u for 'up' and d for 'down': u m = 3*m+1, if m is odd, and d m = m/2 if m is even. The present array gives all start numbers M(n, k) for the Collatz word (ud)^n = s^n (s = ud is useful because, except for the one letter word u, at least one d follows a letter u), with n >= 1, and k >= 1. Such Collatz sequences have the maximal number of u's (grow fastest).
This rectangular array is M of Example 2.2. with x=y = n, n >= 1, of the M. Trümper reference, pp. 7-8, written as a triangle by taking NE-SW diagonals. The Collatz sequence starting with M(n, k) has length 2*n+1 for each k and it ends in the odd number N(n, k) given in A239127.
The first row sequences of the array M (columns of triangle TM) are A004767, A004771, A125169, A239128, ...

			The rectangular array M(n, k) begins:
n\k     1    2    3    4     5     6     7     8     9    10 ...
1:      3    7   11   15    19    23    27    31    35    39
2:      7   15   23   31    39    47    55    63    71    79
3:     15   31   47   63    79    95   111   127   143   159
4:     31   63   95  127   159   191   223   255   287   319
5:     63  127  191  255   319   383   447   511   575   639
6:    127  255  383  511   639   767   895  1023  1151  1279
7:    255  511  767 1023  1279  1535  1791  2047  2303  2559
8:    511 1023 1535 2047  2559  3071  3583  4095  4607  5119
9:   1023 2047 3071 4095  5119  6143  7167  8191  9215 10239
10:  2047 4095 6143 8191 10239 12287 14335 16383 18431 20479
...
The triangle TM(m, n) begins (zeros are not shown):
m\n   1    2     3     4     5     6      7      8      9    10 ...
1:    3
2:    7    7
3:   11   15    15
4:   15   23    31    31
5:   19   31    47    63    63
6:   23   39    63    95   127   127
7:   27   47    79   127   191   255    255
8:   31   55    95   159   255   383    511    511
9:   35   63   111   191   319   511    767   1023   1023
10:  39   71   127   223   383   639   1023   1535   2047  2047
...
---------------------------------------------------------------------
n=1, ud, k=1: M(1, 1) = 3 = TM(1, 1), N(1,1) = 5 with the Collatz sequence  [3, 10, 5] of length 3.
n=1, ud, k=2: M(1, 2) = 7 = TM(2, 1), N(1,2) = 11 with the Collatz sequence  [7, 22, 11] of length 3.
n=4, (ud)^4, k=2: M(4, 2) = 63 = TM(5, 4), N(4,2) = 323 with the Collatz sequence  [63, 190, 95, 286, 143, 430, 215, 646, 323] of length 9.
n=5, (ud)^5, k=1: M(5, 1) = 63 =  TM(5, 5), N(5,1) = 485 with the Collatz sequence  [63, 190, 95, 286, 143, 430, 215, 646, 323, 970, 485] of length 11.

Wolfdieter Lang, On Collatz' Words, Sequences, and Trees, J. of Integer Sequences, Vol. 17 (2014), Article 14.11.7.
Manfred Trümper, The Collatz Problem in the Light of an Infinite Free Semigroup, Chinese Journal of Mathematics, Vol. 2014, Article ID 756917, 21 pages.
Eric Weisstein's World of Mathematics, Collatz Problem.
Wikipedia, Collatz Conjecture.

Cf. A006577, A139399, A112695, A238475, A238476, A004767, A004771, A125169, A239127, A239128.

The array: M(n, k) = 2^(n+1)*k - 1 for n >= 1 and k >= 1.
The triangle: TM(m, n) = M(n, m-n+1) = 2^(n+1)*(m-n+1) - 1 for m >= n >= 1 and 0 for m < n.
a(n) = 4*A087808(A130328(n-1)) - 1 (conjectured). - Christian Krause, Jun 15 2021

A061313 Minimal number of steps to get from 1 to n by (a) subtracting 1 or (b) multiplying by 2.

Original entry on oeis.org

0, 1, 3, 2, 5, 4, 4, 3, 7, 6, 6, 5, 6, 5, 5, 4, 9, 8, 8, 7, 8, 7, 7, 6, 8, 7, 7, 6, 7, 6, 6, 5, 11, 10, 10, 9, 10, 9, 9, 8, 10, 9, 9, 8, 9, 8, 8, 7, 10, 9, 9, 8, 9, 8, 8, 7, 9, 8, 8, 7, 8, 7, 7, 6, 13, 12, 12, 11, 12, 11, 11, 10, 12, 11, 11, 10, 11, 10, 10, 9, 12, 11, 11, 10, 11, 10, 10, 9, 11, 10

Offset: 1

Henry Bottomley, Jun 06 2001

Also number of steps to get from n to 1 by process of adding 1 if odd, or dividing by 2 if even.
It is straightforward to prove that the number n appears F(n) times in this sequence, where F(n) is the n-th Fibonacci number (A000045). - Gary Gordon, May 31 2019
Conjecture: a(n)+2 is the sum of the terms of the Hirzebruch (negative) continued fraction for the Stern-Brocot tree fraction A007305(n)/A007306(n). - Andrey Zabolotskiy, Apr 17 2020

			a(2) = 1 since 2 = 1*2, a(3) = 3 since 3 = 1*2*2-1, a(11) = 6 since 11 = (1*2*2-1)*2*2-1.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Ralf Stephan, Divide-and-conquer generating functions. I. Elementary sequences, arXiv:math/0307027 [math.CO], 2003.
Index to sequences related to the complexity of n

Cf. A056792, A006577, A029837, A000045.

Cf. A007305, A007306, A008687.

a061313 n = fst $ until ((== 1) . snd) (\(u, v) -> (u + 1, f v)) (0, n)
   where f n = if r == 0 then n' else n + 1  where (n', r) = divMod n 2
-- Reinhard Zumkeller, Sep 05 2015

f[n_] := Block[{c = 0, m = n}, While[m != 1, If[ EvenQ[m], While[ EvenQ[m], m = m/2; c++ ], m++; c++ ]]; Return[c]]; Table[f[n], {n, 1, 100}]

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

xpcount(n) = { p = 1; for(x=1,n, p1 = x; ct=0; while(p1>1, if(p1%2==0,p1/=2; ct++,p1 = p1*p+1; ct++) ); print1(ct, ", ") ) }

a(n) = if(n--,2*(logint(n,2)+1)) - hammingweight(n); \\ Kevin Ryde, Oct 21 2021

a(2n) = a(n)+1; a(2n+1) = a(n+1)+2; a(1) = 0.
Is Sum_{k=1..n} a(k) asymptotic to C*n*log(n) where 3 > C > 2? - Benoit Cloitre, Aug 31 2002
G.f.: x/(1-x) * Sum_{k>=0} (x^2^k + x^2^(k+1)/(1+x^2^k)). - Ralf Stephan, Jun 14 2003
a(n) = A080791(n-1) + A029837(n). - Ralf Stephan, Jun 14 2003
a(n) = 2*A023416(n-1) + A000120(n-1) = A023416(A062880(n)) = A023416(A000695(n)) + 1. - Ralf Stephan, Jul 16 2003
a(n) = A119477(n) - 1. - Philippe Deléham, Nov 03 2008

A112695 Number of steps needed to reach 4,2,1 in Collatz' 3*n+1 conjecture.

Original entry on oeis.org

1, 2, 5, 0, 3, 6, 14, 1, 17, 4, 12, 7, 7, 15, 15, 2, 10, 18, 18, 5, 5, 13, 13, 8, 21, 8, 109, 16, 16, 16, 104, 3, 24, 11, 11, 19, 19, 19, 32, 6, 107, 6, 27, 14, 14, 14, 102, 9, 22, 22, 22, 9, 9, 110, 110, 17, 30, 17, 30, 17, 17, 105, 105, 4, 25, 25, 25, 12, 12, 12

Offset: 1

Wolfdieter Lang, Oct 31 2005

a(n) = number of iterations of the Collatz 3*x+1 map applied to n until the conjectured 4,2,1 sequence is reached.

			a(1)=1 because the sequence for n=1 is 1,4,2,1.
a(4)=0 from 4,2,1.
a(7)=14 from 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.

C. A. Pickover, Dr. Googols wundersame Welt der Zahlen, Deutscher Taschenbuch Verlag, Kap. 14, pp. 87, 193. German translation of: Wonders of numbers - Adventures in Mathematics, Mind and Meaning, Oxford University Press 2003.

Ken Conrow, Collatz 3n+1 Problem.
Eric Weisstein's World of Mathematics, Collatz Problem
Index entries for sequences related to 3x+1 (or Collatz) problem

Essentially the same sequence as A139399.

Cf. A006370, A006577.

f[n_] := If[EvenQ[n], n/2, 3 n + 1];
a[n_] := Length[NestWhileList[f, n, {#1, #2, #3} != {4, 2, 1}&, 3]] - 3;
Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Aug 08 2022 *)

a(n) = A006577(n) - 2, n >= 3, a(1)=1, a(2)=2.

A161021 Collatz (or 3x+1) trajectory starting at 703.

Original entry on oeis.org

703, 2110, 1055, 3166, 1583, 4750, 2375, 7126, 3563, 10690, 5345, 16036, 8018, 4009, 12028, 6014, 3007, 9022, 4511, 13534, 6767, 20302, 10151, 30454, 15227, 45682, 22841, 68524, 34262, 17131, 51394, 25697, 77092, 38546, 19273, 57820, 28910

Offset: 0

Reinhard Zumkeller, Jun 03 2009

703 = A060412(5); a(A006577(703)) = a(170) = 1;
a(n) = A008884(n-59) for n with 162 <= n <= 170;
a(n) = A161022(n+53) for n with 155 <= n <= 170;
a(n) = A161023(n+153) for n with 144 <= n <= 170.

R. Zumkeller, Table of n, a(n) for n = 0..170
Index entries for sequences related to 3x+1 (or Collatz) problem
Index entries for linear recurrences with constant coefficients, signature (0,0,1).

NestList[If[EvenQ[#],#/2,3#+1]&,703,40] (* Harvey P. Dale, Nov 27 2011 *)

Edited by N. J. A. Sloane, Jul 27 2019

A220237 Triangle read by rows: sorted terms of Collatz trajectories.

Original entry on oeis.org

1, 1, 2, 1, 2, 3, 4, 5, 8, 10, 16, 1, 2, 4, 1, 2, 4, 5, 8, 16, 1, 2, 3, 4, 5, 6, 8, 10, 16, 1, 2, 4, 5, 7, 8, 10, 11, 13, 16, 17, 20, 22, 26, 34, 40, 52, 1, 2, 4, 8, 1, 2, 4, 5, 7, 8, 9, 10, 11, 13, 14, 16, 17, 20, 22, 26, 28, 34, 40, 52, 1, 2, 4, 5, 8, 10, 16

Offset: 1

Reinhard Zumkeller, Jan 03 2013

n-th row = sorted list of {A070165(n,k): k = 1..A006577(n)};

T(n,1) = 1 if Collatz conjecture is true.

			The table begins:
.   1:  [1]
.   2:  [1,2]
.   3:  [1,2,3,4,5,8,10,16]
.   4:  [1,2,4]
.   5:  [1,2,4,5,8,16]
.   6:  [1,2,3,4,5,6,8,10,16]
.   7:  [1,2,4,5,7,8,10,11,13,16,17,20,22,26,34,40,52]
.   8:  [1,2,4,8]
.   9:  [1,2,4,5,7,8,9,10,11,13,14,16,17,20,22,26,28,34,40,52]
.  10:  [1,2,4,5,8,10,16]
.  11:  [1,2,4,5,8,10,11,13,16,17,20,26,34,40,52]
.  12:  [1,2,3,4,5,6,8,10,12,16] .

Reinhard Zumkeller, Rows n = 1..120 of triangle, flattened
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A006577 (row lengths), A025586(right edge), A033493 (row sums).

import Data.List (sort)
a220237 n k = a220237_tabf !! (n-1) !! (k-1)
a220237_row n = a220237_tabf !! (n-1)
a220237_tabf = map sort a070165_tabf

T:= proc(n) option remember; `if`(n=1, 1,
      sort([n, T(`if`(n::even, n/2, 3*n+1))])[])
    end:
seq(T(n), n=1..10);  # Alois P. Heinz, Oct 16 2021

Flatten[Table[Sort[NestWhileList[If[EvenQ[#],#/2,3#+1]&,n,#>1&]],{n,12}]] (* Harvey P. Dale, Jan 28 2013 *)

A277109 Starting from 2^n+1, the length of the longest sequence of consecutive numbers which all take the same number of steps to reach 1 in the Collatz (or '3x+1') problem.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 3, 1, 3, 1, 3, 1, 1, 1, 3, 1, 3, 7, 15, 30, 1, 1, 3, 7, 15, 26, 26, 1, 1, 1, 3, 1, 3, 1, 3, 1, 1, 1, 3, 1, 3, 7, 15, 1, 1, 1, 3, 1, 3, 1, 3, 1, 1, 1, 3, 1, 3, 7, 15, 1, 1, 3, 7, 15, 26, 1, 3, 7, 15, 31, 63, 26, 30, 26, 26, 26, 26, 30, 46, 26, 26, 26, 26, 1, 1, 1, 3, 7, 15, 1, 3, 1

Offset: 0

Dmitry Kamenetsky, Sep 30 2016

a(500) was found by Guo-Gang Gao (see links).
Interestingly, this sequence has many sets of consecutive terms that are increasing powers of 2 minus 1. For example: a(291) to a(307), a(447) to a(467), and a(603) to a(625). It is not clear why this is the case.
The largest value known in this sequence is a(3951) = 2^47-1 = 140737488355327.
Conjecture: If a(n) = 2^k - 1 for some k > 1, then a(n-1) = 2^(k-1) - 1. Conjecture holds for n <= 1812.
From Hartmut F. W. Hoft, Aug 16 2018: (Start)
The conjecture is true. Let the lengths of the Collatz runs equal q for all numbers 2^n + 1, 2^n + 2, 2^n + 3, 2^n + 4, ..., 2^n + 2^k - 2, 2^n + 2^k - 1. Then dividing the 2^(k-1) - 1 even numbers by two gives rise to the sequence 2^(n-1) + 1, 2^(n-1) + 2, ..., 2^(n-1) + 2^(k-1) - 1 of numbers for which the lengths of the Collatz runs equals q-1. Furthermore, let the length of the Collatz run of 2^n + 2^k be r != q then the length of the Collatz run of 2^(n-1) + 2^(k-1) is r-1 != q-1, i.e., a(n-1) = 2^(k-1) - 1.
Conjecture: Let a(k), ..., a(k+m), m >= 0, be a subsequence of this sequence such that a(k)=a(k+m+1)=1 and a(k+i) > 1, 1 <= i <= m. Then the lengths of the Collatz runs of a(k+i), 0 <= i <= m, increase by 1. In addition, there is an initial segment of increasing numbers a(k), ..., a(k+j), for some 0 <= j <= m, in each such subsequence having the form 2^i - 1, 0 < i <= j. (End)

			a(6) = 3, because 2^6+1, 2^6+2 and 2^6+3 all take 27 steps to reach 1.
From _Hartmut F. W. Hoft_, Aug 16 2018: (Start)
Two examples for the conjecture (L(n) denotes the length of the Collatz run):
n      a(n)    L(n)          n      a(n)    L(n)
64     26      483           20      1       72
------------------           ------------------
65      1      559           21      1      166
66      3      560           22      3      167
67      7      561           23      7      168
68     15      562           24     15      169
69     31      563           ------------------
70     63      564           25     26      170
------------------           26     26      171
71     26      565           ------------------
72     30      566           27      1      247
73     26      567
74     26      568
75     26      569
76     26      570
77     30      571
78     46      572
79     26      573
80     26      574
81     26      575
82     26      576
------------------
83      1      626
The "power of 2 minus 1" initial section of any such subsequence of a(n) is always increasing. However, there is no apparent ordering in the second section when that is present. (End)

Karl Frinkle, Table of n, a(n) for n = 0..3951 (terms 0..1812 from Dmitry Kamenetsky).
Guo-Gang Gao, On consecutive numbers of the same height in the Collatz problem, Discrete Mathematics, Volume 112, pages 261-267, 1993.
Dmitry Kamenetsky, C program to compute the sequence.
Pureferret, Longest known sequence of identical consecutive Collatz sequence lengths, Mathematics StackExchange, 2013.
Carlos Rivera, Puzzle 847: Consecutive primes with the same Collatz length, The Prime Puzzles and Problems Connection.
Carlos Rivera, Puzzle 851: Puzzle 847 revisited, The Prime Puzzles and Problems Connection.

Cf. A006577, A078441, A179118, A277684.

f[n_] := Length[NestWhileList[If[EvenQ@ #, #/2, 3 # + 1] &, n, # != 1 &]] - 1; Table[k = 1; While[f[2^n + k] == f[2^n + k + 1], k++]; k, {n, 120}] (* Michael De Vlieger, Oct 03 2016 *)

nbsteps(n)= s=n; c=0; while(s>1, s=if(s%2, 3*s+1, s/2); c++); c;
a(n) = {my(ns = 2^n+1); my(nbs = nbsteps(ns)); while(nbsteps(ns+1) == nbs, ns++); ns - 2^n;} \\ Michel Marcus, Oct 30 2016

A125711 In the "3x+1" problem, let 1 denote a halving step and 0 denote an x->3x+1 step. Then a(n) is obtained by writing the sequence of steps needed to reach 1 from 2n and reading it as a decimal number.

Original entry on oeis.org

1, 3, 175, 7, 47, 431, 87791, 15, 743151, 111, 22255, 943, 751, 218863, 175087, 31, 5871, 1791727, 1431279, 239, 191, 55023, 44015, 1967, 11917039, 1775, 3515647479163389605506303638875119, 481007, 382703, 437231, 108202665749908974283165422824431, 63, 95803119

Offset: 1

N. J. A. Sloane, Feb 01 2007

			6 -> 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1, so a(3) is the decimal equivalent of 10101111, which is 175.

Alois P. Heinz, Table of n, a(n) for n = 1..10000
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A006577, A125710, A125626.

b:= proc(n) option remember; `if`(n=1, [0$2], `if`(n::even,
     (p-> p+[1, 2^p[1]])(b(n/2)), (p-> p+[1, 0])(b(3*n+1))))
    end:
a:= n-> b(2*n)[2]:
seq(a(n), n=1..33);  # Alois P. Heinz, Apr 02 2025

f[x_] := If[EvenQ[x], x/2, 3x + 1];g[n_] := FromDigits[Mod[Most[NestWhileList[f, 2n, # > 1 &]], 2, 1] - 1, 2];Table[g[n], {n, 40}] (* Ray Chandler, Feb 02 2007 *)

Extended by Ray Chandler, Feb 02 2007

A127789 Record indices of 2^h(n)/(3^t(n)*n), where h and t are the number of halving resp. tripling steps in the '3x + 1' problem.

Original entry on oeis.org

1, 3, 7, 9, 505, 559, 745, 993

Offset: 1

Paul Boddington, Apr 04 2007

Original name: In the '3x + 1' problem we would expect a positive integer n to be approximately equal to 2^j / 3^k where j is the number of halving steps and k is the number of '3x + 1' steps required to reach 1. The numbers in this sequence are those for which 2^j / (n*3^k) sets a new record.
Eric Roosendaal calls 2^j / (n*3^k) the residue of n and conjectures that 993 yields the highest residue. - T. D. Noe, Apr 08 2007
It has been verified that the next value, if it exists, is larger than 2^32 ~ 4.3e9. We do not need an "escape clause" (as for A006577, A006666, A006667, ...) in this sequence since the unlikely case of a possibly undefined ratio is irrelevant for the list of records. - M. F. Hasler, May 07 2018

Eric Roosendaal, On the 3x+1 Problem

Cf. A006370 (Collatz map), A014682 (condensed version using (3n+1)/2).

Cf. A006666 (halving steps), A006667 (tripling steps), A006577 (total).

(c(n,c=[0,0])=while(n>1,bittest(n,0)&&c[1]++&&(n=n*3+1)&&next;n\=2;c[2]++);c); m=1;for(n=1,oo,m<<(t=c(n))[2]>n*3^t[1]||next;m=n*3^t[1]/2^t[2];print1(n", ")) \\ M. F. Hasler, May 07 2018

New name from M. F. Hasler, May 07 2018

A210468 Collatz (3x+1) problem with rational numbers: number of steps to reach the end of the cycle starting with 1/(2n+1).

Original entry on oeis.org

0, 2, 3, 6, 5, 7, 4, 17, 8, 31, 10, 76, 24, 8, 5, 35, 28, 17, 16, 11, 27, 13, 22, 44, 72, 54, 90, 15, 16, 38, 6, 35, 39, 113, 86, 99, 92, 33, 53, 63, 13, 54, 170, 79, 56, 71, 41, 107, 23, 11, 96, 67, 30, 158, 87, 9, 40, 49, 22, 116, 62, 60, 7, 54, 71, 44, 67

Offset: 0

Michel Lagneau, Jan 22 2013

This variation of the "3x+1" problem with a class of rational numbers is as follows: start with any number 1/(2n+1). If the numerator is even, divide it by 2, otherwise multiply it by 3 and add 1. Do we always reach the end of a cycle with a rational number? It is conjectured that the answer is yes.
If x is of the form 1/2n, each trajectory is divergent because the numerator is always odd and tends into infinity.
If x is of the form x = m/(2n+1) where m is an integer, it is conjectured that the number of steps tends into the end of a finite cycle.
In this sequence, it appears that the last number of the cycle is 1 when 2n+1 is a power of 3. For example, starting with 1/9, the trajectory is 1/9 -> 4/3 -> 2/3 -> 1/3 -> 2 -> 1 with 5 iterations.
Observations: the last number of each trajectory has three possible forms: 1, 2/(2n+1) or a form m/(2n+1) where m> 2.

			For n = 3, a(3) = 6 because the corresponding trajectory of 1/7 requires 6 iterations to reach the last term of the cycle:
1/7 -> 10/7 -> 5/7 -> 22/7 -> 11/7 -> 40/7 -> 20/7 and 20/7 is the last term because 20/7 -> 10/7 is already in the trajectory. The rational 10/7 has two antecedents: 1/7 and 20/7 are in the same trajectory (this property is conjecturally impossible in the classical 3x + 1 problem with x integer). The periodic nontrivial loop contains 6 distinct rational numbers (20/7 ->10/7->5/7 ->22/7 -> 11/7 -> 40/7).

Michel Lagneau, Table of n, a(n) for n = 0..10000
J. C. Lagarias, The set of rational cycles for the 3x+1 problem, Acta Arith. 56 (1990), 33-53.

Cf. A006577.

with(numtheory): z:={1}:for m from 0 to 80 do: n:=2*m+1:lst:={1/n}:x0:=1: x:=x0*3+n: lst:=lst union {x/n}:for i from 1 to 10000 do: x:=x/2: lst:=lst union {x/n}:  if irem(x,2)=1 then x0:=x:x:=x0*3+n: lst:=lst union {x/n}:else fi:od: n0:=nops(lst):if lst intersect z = {1} then n1:=n0-2: printf(`%d, `,n1): else n1:=n0-1: printf(`%d, `,n1):fi: od:

Collatz[n_] := NestWhileList[If[EvenQ[Numerator[#]], #/2, 3 # + 1] &, n, UnsameQ, All]; Join[{0}, Table[s = Collatz[1/(2 n + 1)]; len = Length[s] - 2; If[s[[-1]] == 2, len = len - 1]; len, {n, 100}]] (* T. D. Noe, Jan 22 2013 *)

A238476 Rectangular array with all start numbers Mo(n, k), k >= 1, for the Collatz operation ud^(2*n-1), n >= 1, ending in an odd number, read by antidiagonals.

Original entry on oeis.org

3, 7, 13, 11, 29, 53, 15, 45, 117, 213, 19, 61, 181, 469, 853, 23, 77, 245, 725, 1877, 3413, 27, 93, 309, 981, 2901, 7509, 13653, 31, 109, 373, 1237, 3925, 11605, 30037, 54613, 35, 125, 437, 1493, 4949, 15701, 46421, 120149, 218453

Offset: 1

Wolfdieter Lang, Mar 10 2014

The two operations on natural numbers m used in the Collatz 3x+1 conjecture are here denoted (with M. Trümper, see the link) by u for 'up' and d for 'down': u m = 3*m+1, if m is odd, and d m = m/2 if m is even. The present array gives all start numbers Mo(n, k), k >= 1, for Collatz sequences following the pattern (word) ud^(2*n-1), with n >= 1, ending in an odd number. This end number does not depend on n and it is given by No(k) = 6*k - 1. This Collatz sequence has length 1 + (1 + 2*n - 1) = 2*n + 1.
This rectangular array is Example 2.1. with x = 2*n-1, n >= 1, of the M. Trümper reference, pp. 4-5, written as a triangle by taking NE-SW diagonals. The case x = 2*n, n >= 1, for the word ud^(2*n) appears as array and triangle A238475.
The first rows of array Mo (columns of triangle To) are A004767, A082285, A239124, ...

			The rectangular array Mo(n, k) begins:
n\k      1        2        3        4        5        6        7        8        9        10 ...
1:       3        7       11       15       19       23       27       31       35        39
2:      13       29       45       61       77       93      109      125      141       157
3:      53      117      181      245      309      373      437      501      565       629
4:     213      469      725      981     1237     1493     1749     2005     2261      2517
5:     853     1877     2901     3925     4949     5973     6997     8021     9045     10069
6:    3413     7509    11605    15701    19797    23893    27989    32085    36181     40277
7:   13653    30037    46421    62805    79189    95573   111957   128341   144725    161109
8:   54613   120149   185685   251221   316757   382293   447829   513365   578901    644437
9:  218453   480597   742741  1004885  1267029  1529173  1791317  2053461  2315605   2577749
10: 873813  1922389  2970965  4019541  5068117  6116693  7165269  8213845  9262421  10310997
...
---------------------------------------------------------------------------------------------
The triangle To(m, n) begins (zeros are not shown):
m\n    1    2    3     4     5      6      7       8       9      10 ...
1:     3
2:     7   13
3:    11   29   53
4:    15   45  117   213
5:    19   61  181   469   853
6:    23   77  245   725  1877   3413
7:    27   93  309   981  2901   7509  13653
8:    31  109  373  1237  3925  11605  30037   54613
9:    35  125  437  1493  4949  15701  46421  120149  218453
10:   39  141  501  1749  5973  19797  62805  185685  480597  873813
...
n=1, ud, k=1: Mo(1, 1) = 3 = To(1, 1), No(1) = 5 with the Collatz sequence [3, 10, 5] of length 3.
n=1, ud, k=2: Mo(1, 2) = 7 = Te(2, 1), No(2) = 11 with the Collatz sequence [7, 22, 11] of length 3.
n=5, ud^9, k=2: Mo(5, 2) = 1877 = Te(6,5), No(2) = 11 with the Collatz sequence [1877, 5632, 2816, 1408, 704, 352, 176, 88, 44, 22, 11] of length 11.

W. Lang, On Collatz' Words, Sequences, and Trees, J. of Integer Sequences, Vol. 17 (2014), Article 14.11.7.
Manfred Trümper, The Collatz Problem in the Light of an Infinite Free Semigroup, Chinese Journal of Mathematics, Vol. 2014, Article ID 756917, 21 pages.
Eric Weisstein's World of Mathematics, Collatz Problem.
Wikipedia, Collatz Conjecture.

Cf. A006577, A139399, A112695, A238475, A004767, A082285, A239124.

Mo(n, k) = 2^(2*n)*k - (2^(2*n-1)+1)/3 for n >= 1 and k >= 1.

To(m, n) = Mo(n, m-n+1) = 2^(2*n)*(m-n+1) - (2^(2*n-1)+1)/3 for m >= n >= 1 and 0 for m < n.

cp's OEIS Frontend

A239126 Rectangular array showing the starting values M(n, k), k >= 1, for the Collatz operation (ud)^n, n >= 1, ending in an odd number, read by antidiagonals.

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A061313 Minimal number of steps to get from 1 to n by (a) subtracting 1 or (b) multiplying by 2.

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A112695 Number of steps needed to reach 4,2,1 in Collatz' 3*n+1 conjecture.

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A161021 Collatz (or 3x+1) trajectory starting at 703.

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A220237 Triangle read by rows: sorted terms of Collatz trajectories.

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A277109 Starting from 2^n+1, the length of the longest sequence of consecutive numbers which all take the same number of steps to reach 1 in the Collatz (or '3x+1') problem.

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A125711 In the "3x+1" problem, let 1 denote a halving step and 0 denote an x->3x+1 step. Then a(n) is obtained by writing the sequence of steps needed to reach 1 from 2n and reading it as a decimal number.

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