Collatz Conjecture (OEIS A006577)

Question

This is the Collatz Conjecture (OEIS A006577):

• Start with an integer n > 1.
• Repeat the following steps:
  • If n is even, divide it by 2.
  • If n is odd, multiply it by 3 and add 1.

It is proven that for all positive integers up to 5 * 2⁶⁰, or about 5764000000000000000, n will eventually become 1.

Your task is to find out how many iterations it takes (of halving or tripling-plus-one) to reach 1.

Relevant xkcd :)

Rules:

• Shortest code wins.
• If a number < 2 is input, or a non-integer, or a non-number, output does not matter.

Test cases

2  -> 1
16 -> 4
5  -> 5
7  -> 16

Answer 1

C - 50 47 characters

Poor little C unfortunately requires an awful amount of code for basic I/O, so shorting all that down has made the UI slightly unintuitive.

b;main(a){return~-a?b++,main(a&1?3*a+1:a/2):b;}

Compile it with for example gcc -o 1 collatz.c. The input is in unary with space-separated digits, and you will find the answer in the exit code. An example with the number 17:

$> ./1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
$> echo $?
12
$>

Answer 2

GolfScript, 24 23 21 20 18 chars

~{(}{3*).2%6\?/}/,

Assumes input on stdin. Online test

Answer 3

FRACTRAN, 8 fractions (45 bytes)

$$\frac{5120}{33}, \frac{15}{11}, \frac{22}{105}, \frac{33}{560}, \frac{13}{5}, \frac{3}{2}, \frac{7}{9}, \frac{11}{7}$$

Takes input as \$3^n\$; halts at \$3 \cdot 13^m\$ for a sequence with \$m\$ iterations.

Try it online!

How it works

If the current state is \$3^n \cdot 13^m\$, if \$n = 1\$, the program halts immediately; if \$n = 2k\$ is even, we have

$$\begin{multline*}3^{2k} \cdot 13^m \xrightarrow{\left(\frac{7}{9}\right)^k} 7^k \cdot 13^m \xrightarrow{\frac{11}{7}} 7^{k - 1}\cdot 11 \cdot 13^m \xrightarrow{\frac{15}{11}} 3 \cdot 5 \cdot 7^{k - 1} \cdot 13^m \\ \xrightarrow{\left(\frac{22}{105} \cdot \frac{15}{11}\right)^{k - 1}} 2^{k - 1} \cdot 3 \cdot 5 \cdot 13^m \xrightarrow{\frac{13}{5}} 2^{k - 1} \cdot 3 \cdot 13^{m + 1} \xrightarrow{\left(\frac{3}{2}\right)^{k - 1}} 3^k \cdot 13^{m + 1};\end{multline*}$$

and if \$n = 2k + 1\$ is odd, we have

$$\begin{multline*} 3^{2k + 1} \cdot 13^m \xrightarrow{\left(\frac{7}{9}\right)^k} 3 \cdot 7^k \cdot 13^m \xrightarrow{\frac{11}{7}} 3 \cdot 7^{k - 1} \cdot 11 \cdot 13^m \xrightarrow{\frac{5120}{33}} 2^{10} \cdot 5 \cdot 7^{k - 1} \cdot 13^m \\ \xrightarrow{\left(\frac{33}{560} \cdot \frac{5120}{33}\right)^{k - 1}} 2^{6k + 4} \cdot 5 \cdot 13^m \xrightarrow{\frac{13}{5}} 2^{6k + 4} \cdot 13^{m + 1} \xrightarrow{\left(\frac{3}{2}\right)^{6k + 4}} 3^{6k + 4} \cdot 13^{m + 1},\end{multline*}$$

where \$6k + 4 = 3n + 1\$.

Answer 4

Perl 34 (+1) chars

$\++,$_*=$_&1?3+1/$_:.5while$_>1}{

Abusing $\ for final output, as per usual. Run with the -p command line option, input is taken from stdin.

Saved one byte due to Elias Van Ootegem. Specifically, the observeration that the following two are equivalent:

$_=$_*3+1
$_*=3+1/$_

Although one byte longer, it saves two bytes by shortening $_/2 to just .5.

Sample usage:

$ echo 176 | perl -p collatz.pl
18

---

PHP 54 bytes

<?for(;1<$n=&$argv[1];$c++)$n=$n&1?$n*3+1:$n/2;echo$c;

Javascript's archnemesis for the Wooden Spoon Award seems to have fallen a bit short in this challenge. There's not a whole lot of room for creativity with this problem, though. Input is taken as a command line argument.

Sample usage:

$ php collatz.php 176
18

Answer 5

Mathematica (35)

If[#>1,#0@If[OddQ@#,3#+1,#/2]+1,0]&

Usage:

If[#>1,#0[If[OddQ@#,3#+1,#/2]]+1,0]&@16
>> 4

Answer 6

Python repl, 48

I'm not convinced that there isn't a shorter expression than n=3*n+1;n/=1+n%2*5;. I probably found a dozen different expressions of all the same length...

i=0
n=input()
while~-n:n=3*n+1;n/=1+n%2*5;i+=1
i

edit: I've found another solution that will never contend, but is too fun not to share.

s='s'
i=s
n=i*input()
while 1:
 while n==n[::2]+n[::2]:i+=s;n=n[::2]
 if n==s:i.rindex(s);break
 n=3*n+s
 i+=s

Answer 7

Java, 165, 156, 154,134,131,129,128,126 (verbose languages need some love too)

class a{public static void main(String[]a){for(int x=Short.valueOf(a[0]),y=0;x>1;x=x%2<1?x/2:x*3+1,System.out.println(++y));}}

All is done inside the for

for(int x=Short.valueOf(a[0]),y=0;x>1;x=x%2<1?x/2:x*3+1,System.out.println(++y))

That's freaking beautiful man. Thanks to Pater Taylor!!!, and the idea of using a for loop was stolen from ugoren

I replaced Integer for Short.

Answer 8

As I usually do, I will start the answers off with my own.

JavaScript, 46 44 chars (run on console)

for(n=prompt(),c=1;n>1;n=n%2?n*3+1:n/2,++c)c

Answer 9

Rebmu: 28

u[++jE1 AeEV?a[d2A][a1M3a]]j

On a problem this brief and mathy, GolfScript will likely win by some percent against Rebmu (if it's not required to say, read files from the internet or generate JPG files). Yet I think most would agree the logic of the Golfscript is nowhere near as easy to follow, and the total executable stack running it is bigger.

Although Rebol's own creator Carl Sassenrath told me he found Rebmu "unreadable", he is busy, and hasn't time to really practice the pig-latin-like transformation via unmushing. This really is merely transformed to:

u [
    ++ j
    e1 a: e ev? a [
        d2 a
    ] [
        a1 m3 a
    ]
]
j

Note that the space was required to get an a: instead of an a. This is a "set-word!" and the evaluator notices that symbol type to trigger assignment.

If it were written in unabbreviated (yet awkwardly-written Rebol), you'd get:

until [
    ++ j
    1 == a: either even? a [
        divide a 2
    ] [
        add 1 multiply 3 a
    ]
 ]
 j

Rebol, like Ruby, evaluates blocks to their last value. The UNTIL loop is a curious form of loop that takes no loop condition, it just stops looping when its block evaluates to something not FALSE or NONE. So at the point that 1 == the result of assigning A (the argument to rebmu) to the result of the Collatz conditional (either is an IF-ELSE which evaluates to the branch it chooses)... the loop breaks.

J and K are initialized to integer value zero in Rebmu. And as aforementioned, the whole thing evaluates to the last value. So a J reference at the end of the program means you get the number of iterations.

Usage:

>> rebmu/args [u[++jE1 AeEV?a[d2A][a1M3a]]j] 16
== 4

Answer 10

Brachylog, 16 bytes

1b|{/₂ℕ|×₃+₁}↰+₁

Try it online!

Explanation

         Either:
  1        The input is 1.
  b        In which case we unify the output with 0 by beheading the 1
           (which removes the leading digit of the 1, and an "empty integer"
           is the same as zero).
|        Or:
  {        This inline predicate evaluates a single Collatz step on the input.
           Either:
    /₂       Divide the input by 2.
    ℕ        And ensure that the result is a natural number (which is
             equivalent to asserting that the input was even).
  |        Or:
    ×₃+₁     Multiply the input by 3 and add 1.
  }
  ↰        Recursively call the predicate on this result.
  +₁       And add one to the output of the recursive call.

An alternative solution at the same byte count:

;.{/₂ℕ|×₃+₁}ⁱ⁾1∧

Try it online!

;.          The output of this is a pair [X,I] where X is the input and
            I will be unified with the output.
{/₂ℕ|×₃+₁}  This is the Collatz step predicate we've also used above.
ⁱ⁾          We iterate this predicate I times on X. Since we haven't actually
            specified I, it is still a free variable that Brachylog can backtrack
            over and it will keep adding on iterations until the next
            constraint can be satisfied.
1           Require the result of the iteration to be 1. Once this is
            satisfied, the output variable will have been unified with
            the minimum number of iterations to get here.
∧           This AND is just used to prevent the 1 from being implicitly
            unified with the output variable as well.

Answer 11

J, 30 characters

<:#-:`(1+3&*)`]@.(2&|+1&=)^:a:

Turned out quite a bit longer than desired

usage:

   <:#-:`(1+3&*)`]@.(2&|+1&=)^:a:2
1
   <:#-:`(1+3&*)`]@.(2&|+1&=)^:a:16
4
   <:#-:`(1+3&*)`]@.(2&|+1&=)^:a:5
5
   <:#-:`(1+3&*)`]@.(2&|+1&=)^:a:7
16
   <:#-:`(1+3&*)`]@.(2&|+1&=)^:a:27
111

• -:`(1+3&*)`] is a gerund composed of three verbs, used on three occasions. -: means "halve", (1+3&*) or (1+3*]) encodes the multiplication step and ] (identity) aids termination.

• 2&|+1&= forms an index to the gerund. literally, "the remainder after division by two plus whether it equals one".

• #verb^:a: iterates the function until the result is stable (here, forced explicitly), while collecting the steps, then counts them. Stolen from @JB. <: decrements the step count by one to align with the question requirements.

Answer 12

80386 assembly, 16 bytes

This example uses AT&T syntax and the fastcall calling convention, the argument goes into ecx:

collatz:
        or $-1,%eax              # 3 bytes, eax = -1;
.Loop:  inc %eax                 # 1 byte,  eax += 1;
        lea 1(%ecx,%ecx,2),%edx  # 4 bytes, edx = 3*ecx + 1;
        shr %ecx                 # 2 bytes, CF = ecx & 1;
                                 #          ecx /= 2;
                                 #          ZF = ecx == 0;
        cmovc %edx,%ecx          # 3 bytes, if (CF) ecx = edx;
        jnz .Loop                # 2 bytes, if (!ZF) goto .Loop;
        ret                      # 1 byte,  return (eax);

Here are the resulting 16 bytes of machine code:

83 c8 ff 40 8d 54 49 01 d1 e9 0f 42 ca 75 f4 c3

Answer 13

FRACTRAN, 24 fractions

Uses 180 bytes, for the more conventional counters...

68/13, 133/102, 341/51, 115/17, 17/19, 87/161, 17/23, 23/29, 53/93,
26973/217, 410/259, 43/111, 976/37, 37/41, 329/215, 37/43, 43/47,
118/265, 1/53, 53/59, 67/305, 1/61, 61/67, 117/4

Try it online!

Explanation

Am a bit lazy to add an explanation now; happy to do so if it gets attention/upvotes! However, I do have some notes for when I first wrote up the Collatz Conjecture code which you can run here. It's almost the same, but the TIO command line arguments are set to print every number in the sequence along the way, which makes it less of a blackbox!

State Diagrams for COLLATZGAME

Here are the state diagrams, for which I used Conway's original notation which he presents in this article.

Change Made

The above is simply to calculate the Collatz sequence for n given an input of the form 2^n. The only change I made to also keep count of the steps taken was to make the 1/3 from states Q -> D into an 11/3, 11 being the smallest unused prime. This fraction is only executed once for every number in the sequence; it's the state that figures out whether the number is even or odd to figure out what's next. Therefore, the 11 prime register is incremented once per number in the sequence, except one, yielding the number of steps.

Note

I simply encoded the state diagram as below and wrote an interpreter which did the dirty work. However, the work done to convert a state diagram to FRACTRAN is also detailed in Conway's article above:

• A: 9/4 -> T
• T: 4/1 -> Q
• Q: 7/6*, 11/3 -> D, 5/1 -> R
• R: 3/7*|Q
• D: 1/3 -> E, 729/7 -> M
• M: 10/7*, 1/3 -> N, 16/1 -> O
• N: 7/5*|M
• E: 2/5*|A
• O: 1/5*|A

Answer 14

APL (31)

A←0⋄A⊣{2⊤⍵:1+3×⍵⋄⍵÷2}⍣{⍺=A+←1}⎕

Answer 15

PowerShell: 77 74 71 70 61

Golfed code:

for($i=(read-host);$i-ne1;$x++){$i=(($i/2),(3*$i+1))[$i%2]}$x

---

Notes:

I originally tried taking the user input without forcing it to an integer, but that broke in an interesting way. Any odd inputs would process inaccurately, but even inputs would work fine. It took me a minute to realize what was going on.

When performing multiplication or addition, PowerShell treats un-typed input as a string first. So, '5'*3+1 becomes '5551' instead of 16. The even inputs behaved fine because PowerShell doesn't have a default action for division against strings. Even the even inputs which would progress through odd numbers worked fine because, by the time PowerShell got to an odd number in the loop, the variable was already forced to an integer by the math operations anyway.

Thanks to Danko Durbic for pointing out I could just invert the multiplication operation, and not have to cast read-host to int since PowerShell bases its operations on the first object.

PowerShell Golfer's Tip: For some scenarios, like this one, switch beats if/else. Here, the difference was 2 characters.

Protip courtesy of Danko Durbic: For this particular scenario, an array can be used instead of switch, to save 8 more characters!

There's no error checking for non-integer values, or integers less than two.

If you'd like to audit the script, put ;$i just before the last close brace in the script.

I'm not sure exactly how well PowerShell handles numbers that progress into very large values, but I expect accuracy is lost at some point. Unfortunately, I also expect there's not much that can be done about that without seriously bloating the script.

---

Ungolfed code, with comments:

# Start for loop to run Collatz algorithm.
# Store user input in $i.
# Run until $i reaches 1.
# Increment a counter, $x, with each run.
for($i=(read-host);$i-ne1;$x++)
{
    # New $i is defined based on an array element derived from old $i.
    $i=(
        # Array element 0 is the even numbers operation.
        ($i/2),
        # Array element 1 is the odd numbers operation.
        (3*$i+1)
    # Array element that defines the new $i is selected by $i%2.
    )[$i%2]
}

# Output $x when the loop is done.
$x

# Variable cleanup. Don't include in golfed code.
rv x,i

---

Test cases:

Below are some samples with auditing enabled. I've also edited the output some for clarity, by adding labels to the input and final count and putting in spacing to set apart the Collatz values.

---
Input: 2

1

Steps: 1

---
Input: 16

8
4
2
1

Steps: 4

---
Input: 5

16
8
4
2
1

Steps: 5

---
Input: 7

22
11
34
17
52
26
13
40
20
10
5
16
8
4
2
1

Steps: 16

---
Input: 42

21
64
32
16
8
4
2
1

Steps: 8

---
Input: 14

7
22
11
34
17
52
26
13
40
20
10
5
16
8
4
2
1

Steps: 17

---
Input: 197

592
296
148
74
37
112
56
28
14
7
22
11
34
17
52
26
13
40
20
10
5
16
8
4
2
1

Steps: 26

---
Input: 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
3238
1619
4858
2429
7288
3644
1822
911
2734
1367
4102
2051
6154
3077
9232
4616
2308
1154
577
1732
866
433
1300
650
325
976
488
244
122
61
184
92
46
23
70
35
106
53
160
80
40
20
10
5
16
8
4
2
1

Steps: 106

---
Input: 6174

3087
9262
4631
13894
6947
20842
10421
31264
15632
7816
3908
1954
977
2932
1466
733
2200
1100
550
275
826
413
1240
620
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
3238
1619
4858
2429
7288
3644
1822
911
2734
1367
4102
2051
6154
3077
9232
4616
2308
1154
577
1732
866
433
1300
650
325
976
488
244
122
61
184
92
46
23
70
35
106
53
160
80
40
20
10
5
16
8
4
2
1

Steps: 111

---
Input: 8008135

24024406
12012203
36036610
18018305
54054916
27027458
13513729
40541188
20270594
10135297
30405892
15202946
7601473
22804420
11402210
5701105
17103316
8551658
4275829
12827488
6413744
3206872
1603436
801718
400859
1202578
601289
1803868
901934
450967
1352902
676451
2029354
1014677
3044032
1522016
761008
380504
190252
95126
47563
142690
71345
214036
107018
53509
160528
80264
40132
20066
10033
30100
15050
7525
22576
11288
5644
2822
1411
4234
2117
6352
3176
1588
794
397
1192
596
298
149
448
224
112
56
28
14
7
22
11
34
17
52
26
13
40
20
10
5
16
8
4
2
1

Steps: 93
---

Interesting bits about the input numbers which are not from the question's test cases:

• 14 and 197 are Keith Numbers
• 31 is a Mersenne Prime
• 6174 is Kaprekar's Constant
• And lastly, I just like 8008135. And Numberphile, obviously.

Answer 16

brainfuck, 59 56 bytes

,-[<->[[>]+<[-<]>>]>[-<<[++>+<]>->]<<[+>+++<]<<+>>>]<<<.

Try it online! (Slightly modified for ease of use)

Input and output as character codes. This is more useful with arbitrarily sized cells, but can still work with small values in limited cell sizes.

How It Works

Tape Format:
Counter 0 Copy Number Binary...
^End           ^Start

,-[ Get input, decrement by 1 and start loop
  <->                  Initialises the copy of the value at -1
  [[>]+<[-<]>>]        Converts the input to binary while preserving a negative copy
  <+>>[-<<[++>+<]>->] If the last digit of the binary is 1 (n-1 is odd), divide by 2 and decrement
  <<[+>+++<]            If the last digit of the binary is 0 (n-1 is even), multiply by 3
  <<+>>>               Increment counter and end on n-1
]<<<.                 End loop and print counter

Answer 17

Jelly, 10 bytes

×3‘ƊHḂ?Ƭi2

Try it online!

How it works

×3‘ƊHḂ?Ƭi2    Main link (monad). Input: integer >= 2
      ?       Create a "ternary-if" function:
     Ḃ          If the input is odd,
×3‘Ɗ            compute 3*n+1;
    H           otherwise, halve it.
       Ƭ      Repeat until results are not unique; collect all results
        i2    Find one-based index of 2

Example: The result of ...Ƭ for input 5 is [5, 16, 8, 4, 2, 1]. The one-based index of 1 is 6, which is 1 higher than expected. So we choose the index of 2 (which is guaranteed to come right before 1) instead.

Answer 18

ngn/k, 21 bytes

#{(2|-2!x;1+3*x)2!x}\

I converted the operation to be x = max(2, x/2) so the sequence had a fixed point, and converges (\) handled the rest. # counts the number of steps, including the beginning.

Answer 19

Jelly, 12 bytes

×3‘$HḂ?ß0’?‘

Try it online!

How it works

×3‘$HḂ?ß0’?‘  Main link. Argument: n (integer)

     Ḃ?       Yield the last bit of n is 1:
   $            Evaluate the three links to the left as a monadic chain:
×3                Multiply n by 3.
  ‘               Increment the product by 1.
    H           Else, halve n.
         ’?   If n-1 is non-zero:
       ß        Recursively call the main link.
        0     Else, yield 0.
           ‘  Increment the result by 1.

Answer 20

Gambit scheme, 106 98 characters, 40 parentheses

(let((f(lambda(x)(cond((= x 1) 0)((odd? x)(+ 1(f(+ 1(* 3 x)))))(else(+ 1(f(/ x 2))))))))(f(read)))

91 89 chars with define directly

(define(f x)(cond((= x 1)0)((odd? x)(+ 1(f(+ 1(* 3 x)))))(else(+ 1(f(/ x 2))))))(f(read))

Answer 21

Python 2, 68 58 54 52 bytes

f=lambda n:1+(n-2and f((n/2,3*n+1)[n%2]));f(input())

Thanks to @Bakuriu and @boothby for the tips :)

Answer 22

GolfScript (23 chars)

~{.1&{.3*)}*.2/.(}do;],

Online test

Answer 23

F# - 65 chars

let rec c n=function 1->n|i->c(n+1)(if i%2=0 then i/2 else i*3+1)

Answer 24

dc, 27 characters

Applying boothby's black magic:

?[d3*1+d2%5*1+/d1<x]dsxxkzp

I'm not really sure if I understand how - or that - it works.

Usage:

$ dc collatz.dc <<< 7
16

dc, 36 characters

My own creation; a somewhat more traditional approach, even tho I had to wrangle with the language a fair bit to overcome the lack of an else part to if statements:

?[2/2Q]se[dd2%[0=e3*1+]xd1<x]dsxxkzp

Internally it produces all numbers of the sequence and stores them on the stack, then pops the final 1 and displays the stack height.

Answer 25

Ruby 1.9, 49 characters

Rubyfied Valentin CLEMENT's Python answer, using the stabby lambda syntax. Sqeezed it into one statement for added unreadability.

(f=->n{n>1&&1+f[[n/2,3*n+1][n%2]]||0})[gets.to_i]

Some overhead because Ruby, unlike Python, is not happy about mixing numbers with booleans.

Answer 26

Retina, 43 bytes

11
2
(2+)1
$1$1$0$0$0$0
2.*
$0x
)`2
1
1?x
1

Takes input and prints output in unary.

Each line should go to its own file. 1 byte per extra file added to byte-count.

You can run the code as one file with the -s flag. E.g.:

> echo -n 1111111|retina -s collatz
1111111111111111

The algorithm is a loop of doing a Collatz step with the unary number and adding a new step-marker x at the end of the string if the number isn't 1.

When the loop ends with 1, we convert the markers to a unary number (removing the leading 1) which is the desired output.

Answer 27

Hexagony, 48 44 bytes

?(]$_)"){{?{*')}/&!/={:<$["/>&_(.<@2'%<>./>=

Try it online!

Expanded:

     ? ( ] $ _
    ) " ) { { ?
   { * ' ) } / &
  ! / = . { < $ [
 " / > & _ ( . < @
  2 ' % < > : / >
   = . . . . . .
    . . . . . .
     . . . . .

Note that this fails on 1 for uhh... reasons. Honestly, I'm not really sure how this works anymore. All I know is that the code for odd numbers is run backwards for even numbers? Somehow?

The new version is much cleaner than the previous, but has a few more directionals in comparison and also ends in a divide-by-zero error. The only case it doesn't error is when it actually handles 1 correctly.

Answer 28

><>, 27 26 23 bytes

\ln;
\::2%:@5*1+2,*+:2=?

Like the other ><> answers, this builds the sequence on the stack. Once the sequence reaches 2, the size of the stack is the number of steps taken.

Thanks to @Hohmannfan, saved 3 bytes by a very clever method of computing the next value directly. The formula used to calculate the next value in the sequence is:

$$f(n)=n\cdot\frac{5(n\bmod2)+1}{2}+(n\bmod2)$$

The fraction maps even numbers to 0.5, and odd numbers to 3. Multiplying by n and adding n%2 completes the calculation - no need to choose the next value at all!

Edit 2: Here's the pre-@Hohmannfan version:

\ln;
\:::3*1+@2,@2%?$~:2=?

The trick here is that both 3n+1 and n/2 are computed at each step in the sequence, and the one to be dropped from the sequence is chosen afterwards. This means that the code doesn't need to branch until 1 is reached, and the calculation of the sequence can live on one line of code.

Edit: Golfed off another character after realising that the only positive integer that can lead to 1 is 2. As the output of the program doesn't matter for input < 2, the sequence generation can end when 2 is reached, leaving the stack size being the exact number of steps required.

Previouser version:

\~ln;
\:::3*1+@2,@2%?$~:1=?

Answer 29

Julia 0.6, 29 27 bytes

!n=n>1&&1+!(n%2>0?3n+1:n/2)

I can't seem to compile Julia 0.1 on my machine, so there's a chance this is non-competing.

Try it online!

Answer 30

Haskell, 39 38 bytes

saved 1 byte thanks to DLosc

f 2=1
f n=1+f(cycle[div n 2,n*3+1]!!n)

Attempt This Online!