Output the Juggler sequence

Question

The Juggler sequence is described as follows. Beginning with an input \$a_1\$, the next term is defined by the recurrence relation

$$a_{k+1} = \begin{cases} \left\lfloor a_k ^ \frac 1 2 \right\rfloor,\text{ if } a_k \text{ is even} \\ \left\lfloor a_k ^ \frac 3 2 \right\rfloor,\text{ if } a_k \text{ is odd} \\ \end{cases}$$

The sequence terminates when it reaches 1, as all subsequent terms would then be 1.

Task

Given an input \$n\$ greater than or equal to 2, write a program/function/generator/etc. that outputs/returns the respective juggler sequence. The output can be in any reasonable form. You may not use a built-in that computes the juggler sequence, or any built-in that directly yields the result. You may assume that the sequence terminates in \$1\$.

Test Cases

Input: output
2: 2, 1
3: 3, 5, 11, 36, 6, 2, 1
4: 4, 2, 1
5: 5, 11, 36, 6, 2, 1

This is a code golf. Shortest code in bytes wins.

Answer 1

Julia, 64 50 48 42 32 30 bytes

g(x)=[x;x<3||g(x^(x%2+.5)÷1)]

This is a recursive function that accepts an integer and returns a float array.

We build an array by concatenating the input with the next term of the sequence, computed as x to the power of its parity plus 1/2. This gives us either x^1/2 or x^1+1/2 = x^3/2. Integer division by 1 gets the floor. When the condition x < 3 is true, the final element will be a Boolean rather than a numeric value, but since the array is not of type Any, this is cast to have the same type as the rest of the array.

Saved 14 bytes thanks to Dennis!

Answer 2

Jelly, 12 11 10 bytes

*BṪ×½Ḟµ’п

Thanks to @Sp3000 for golfing off 1 byte!

Try it online!

How it works

*BṪ×½Ḟµ’п    Main link. Input: n

*B            Elevate n to all of its digits in base 2.
  Ṫ           Extract the last power.
              This yields n ** (n % 2).
   ×½         Multiply with sqrt(n). This yields n ** (n % 2 + 0.5).
     Ḟ        Floor.

      µ       Push the previous chain on the stack and begin a new, monadic chain.
        п    Repeat the previous chain while...
       ’        n - 1 is non-zero.
              Collect all intermediate results in an array.

Answer 3

JavaScript (ES7), 45 33 bytes

f=n=>n<2?n:n+","+f(n**(.5+n%2)|0)

Explanation

Recursive approach. Returns a comma-separated string of numbers.

f=n=>
  n<2?n:          // stop when n == 1
  n               // return n at the start of the list
  +","+f(         // add the rest of the sequence to the list
    n**(.5+n%2)|0 // juggler algorithm
  )

Test

** not used in test for browser compatibility.

f=n=>n<2?n:n+","+f(Math.pow(n,.5+n%2)|0)

<input type="number" oninput="result.textContent=f(+this.value)" />
<pre id="result"></pre>

Answer 4

Mathematica, 40 39 bytes

Thanks to Martin Büttner for saving 1 byte.

NestWhileList[⌊#^.5#^#~Mod~2⌋&,#,#>1&]&

Test case

%[5]
(* {5,11,36,6,2,1} *)

Answer 5

Pyth, 14 12 bytes

.us@*B@N2NNQ

Demonstration

We start with a cumulative reduce, .u, which in this case starts at the input and applies a function until the result repeats, at which point it outputs all of the intermediate results.

The function takes the previous value as N. It starts by taking its square root with @N2. Next, it bifurcates that value on multiplication by N with *B ... N. This creates the list [N ** .5, (N ** .5) * N], the unfloored results for the even and odd cases. Next, the appropriate unfloored result is selected by indexing into the list with @ ... N. Since Pyth has modular indexing, no out-of-bounds errors are thrown. Finally, the result is floored with s.

Answer 6

MATL, 13 12 bytes

`tt2\.5+^ktq

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Explanation

`     % do...while loop
tt   % duplicate top of stack twice, takes implicit input on first iteration
2\    % take a_k mod 2
.5+^  % adds 0.5, to give 1.5 if odd, 0.5 if even, and takes a_k^(0.5 or 1.5)
kt    % Rounds down, and duplicates
q     % Decrement by 1 and use for termination condition---if it is 0, loop will finish

Thanks Luis for saving a byte!

Answer 7

Minkolang 0.15, 25 bytes

ndN(d$7;r2%2*1+;YdNd1=,).

Try it here!

Explanation

n                            Take number from input => n
 dN                          Duplicate and output as number
   (                         Open while loop
    d                        Duplicate top of stack => n, n
     $7                      Push 0.5
       ;                     Pop b,a and push a**b => n, sqrt(n)
        r                    Reverse stack => sqrt(n), n
         2%                  Modulo by 2
           2*                Multiply by 2
             1+              Add 1 => sqrt(n), [1 if even, 3 if odd]
               ;             Pop b,a and push a**b => sqrt(n)**{1,3}
                Y            Floor top of stack
                 dN          Duplicate and output as number
                   d1=,      Duplicate and => 0 if 1, 1 otherwise
                       ).    Pop top of stack and end while loop if 0, then stop.

Answer 8

TSQL, 89 bytes

Input goes in @N:

DECLARE @N INT = 5;

Code:

WITH N AS(SELECT @N N UNION ALL SELECT POWER(N,N%2+.5) N FROM N WHERE N>1)SELECT * FROM N

Answer 9

APL, 28 24 16 bytes

{⌊⍵*.5+2|⎕←⍵}⍣=⎕

This is a program that accepts an integer and prints the successive outputs on separate lines.

Explanation:

{           }⍣=⎕   ⍝ Apply the function until the result is the input
 ⌊⍵*.5+2|⎕←⍵       ⍝ Print the input, compute floor(input^(input % 2 + 0.5))

Try it online

Saved 8 bytes thanks to Dennis!

Answer 10

Java 7, 83 71 bytes

void g(int a){System.out.println(a);if(a>1)g((int)Math.pow(a,a%2+.5));}

I originally used a typical for loop, but I had to jump through hoops to get it working right. After stealing borrowing user81655's idea to recurse instead, I got it down twelve bytes.

Answer 11

Haskell, 70 bytes

Haskell doesn't have integer sqrt built-in, but I think there may be something shorter than floor.sqrt.fromInteger.

s=floor.sqrt.fromInteger
f n|odd n=s$n^3|1<2=s n
g 1=[1]
g n=n:g(f n) 

Answer 12

Oracle SQL 11.2, 128 bytes

WITH v(i)AS(SELECT :1 FROM DUAL UNION ALL SELECT FLOOR(DECODE(MOD(i,2),0,SQRT(i),POWER(i,1.5)))FROM v WHERE i>1)SELECT i FROM v;

Un-golfed

WITH v(i) AS
(
  SELECT :1 FROM DUAL
  UNION ALL
--  SELECT FLOOR(POWER(i,0.5+MOD(i,2))) FROM v WHERE i>1
  SELECT FLOOR(DECODE(MOD(i,2),0,SQRT(i),POWER(i,1.5))) FROM v WHERE i>1 
)
SELECT * FROM v;

Adding MOD(i,2) to .5 is shorter but there is a bug with POWER(2,.5) :

SELECT POWER(4,.5), FLOOR(POWER(4,.5)), TO_CHAR(POWER(4,.5)) FROM DUAL

gives

2   1   1,99999999999999999999999999999999999999

Answer 13

R, 54 51 bytes

z=n=scan();while(n>1){n=n^(.5+n%%2)%/%1;z=c(z,n)};z

Saved 3 bytes thanks to plannapus.

Answer 14

CJam, 18 bytes

ri{__2%.5+#i_(}g]p

Test it here

Similar to David's MATL answer.

Answer 15

Python 3, 57, 45, 43, 41 bytes

Better Solution with suggestion from @mathmandan

def a(n):print(n);n<2or a(n**(.5+n%2)//1)

This method will print each number on a new line

Previous Solution: Cut down to 43 bytes after xnor's recommendation

a=lambda n:[n][:n<2]or[n]+a(n**(n%2+.5)//1)

You can call the above by doing a(10) which returns [10, 3.0, 5.0, 11.0, 36.0, 6.0, 2.0, 1.0]

The above will output the values as floats. If you want them as integers, then we can just add an extra 2 bytes for 43 bytes:

def a(n):print(n);n<2or a(int(n**(.5+n%2)))

Answer 16

TI-Basic, 30 Bytes

Prompt A
Repeat A=1
Disp A
int(A^(remainder(A,2)+.5->A
End
1

Answer 17

Vyxal, 10 bytes

λ∷d›½e⌊;İJ

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Answer 18

JavaScript ES6, 109 102 bytes

s=>(f=n=>n==1?n:n%2?Math.pow(n,3/2)|0:Math.sqrt(n)|0,a=[s],eval("while(f(s)-1)a.push(s=f(s))"),a+",1")

I know this can be golfed. Returns a string of comma-separated numbers.

Answer 19

C++, 122 bytes

#include <iostream>
void f(int n){int i;while(n^1){std::cout<<n<<',';for(i=n*n;i*i>(n%2?n*n*n:n);i--);n=i;}std::cout<<1;}

Answer 20

Retina, 144 bytes

Input and output are in unary.

The 2nd-to-last line contains a space, and the two middle lines and the last line are empty.

{`(\b|)11+$
$&¶$&
m-1=`^(?=^(11)*(1?)).*$
$&,$2
(1+),1$
$1;,
1(?=1*;)
$%_
1+;
$%_
;|,

m-1=`^
1:
+`(1+):(11\1)
1 $2:
1+:$|:1+

-1=`(1+\b)
$#1
 

Try it online

Explanation

{`(\b|)11+$                 # Loop, Duplicate last line
$&¶$&
m-1=`^(?=^(11)*(1?)).*$     # Append ,n%2 to that line (number modulo 2)
$&,$2
(1+),1$                     # Cube that number if odd
$1;,
1(?=1*;)
$%_
1+;
$%_
;|,                         # (Last stage of cubing number)

m-1=`^                      # Integer square root of that number, 
1:                          #   borrowed and modified from another user's answer
+`(1+):(11\1)
1 $2:
1+:$|:1+

-1=`(1+\b)
$#1
 

Integer square root in Retina, by Digital Trauma

Answer 21

C, 64 63 61 bytes

t;q(n){for(;!t;n=pow(n,.5+n%2))printf("%d%c ",n,n^1?44:t++);}

Answer 22

Husk, ¹⁴ 12 bytes

U¡o⌊?^1.5√%2

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Explanation

U¡ȯ⌊?^1.5√%2
 ¡             iterate over the input infinitely, creating a list
  ȯ            with the following three functions:
               take the previous result
          %2   modulo 2
    ?^1.5      if truthy, raise to 3/2th power
         √     else take square root
   ⌊           floor the result
U              cut the list at fixed point

Answer 23

05AB1E, 12 9 bytes

Δ=DÉ·>;mï

-3 bytes thanks to @ovs.

Outputs each result on a separate line.

Try it online or verify some more test cases.

Explanation:

Δ          # Loop until the result no longer changes:
 =         #  Print the current value with trailing newline (without popping)
           #  (which will be the implicit input-integer in the first iteration)
  D        #  Duplicate the value
   É       #  Pop the copy, and check if it's odd (1 if odd; 0 if even)
    ·      #  Double that
     >     #  Increase it by 1
      ;    #  Halve it
       m   #  Take the current value to the power this 1/2 or 3/2
        ï  #  And then cast it to an integer to floor it

Answer 24

Jelly, 7 bytes

*Ḃ×½ḞµƬ

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Basically Dennis' method, but updated to use the modern version of Jelly. I initially had 9 bytes, but switching to the ×½ method saved 2 bytes.

How it works

*Ḃ×½ḞµƬ - Main link. Takes n on the left
     µ  - Group the previous links into a monad f(n):
 Ḃ      -   Bit; n % 2
*       -   n ** (n % 2)
   ½    -   n ** (1 / 2)
  ×     -   n ** (n % 2 + 1 / 2)
    Ḟ   -   Floor
      Ƭ - Repeatedly apply f(n), yielding [n, f(n), f(f(n)), ..., 1]

Answer 25

Factor, 52 bytes

[ [ 3 dupn . odd? 3/2 .5 ? ^ 1 /i dup 1 > ] loop . ]

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Answer 26

Lua, 49 bytes

b=...repeat a=b print(a)b=b^(.5+b%2)//1until a==b

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Answer 27

Vyxal, 55 bits^v2, 6.875 bytes

∷.+e⌊)↔

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Explained

∷.+e⌊)↔­⁡​‎‎⁡⁠⁢⁢‏⁠‎⁡⁠⁢⁣‏‏​⁡⁠⁡‌⁢​‎‎⁡⁠⁡‏⁠‎⁡⁠⁢‏⁠‎⁡⁠⁣‏‏​⁡⁠⁡‌⁣​‎‎⁡⁠⁤‏‏​⁡⁠⁡‌⁤​‎‎⁡⁠⁢⁡‏‏​⁡⁠⁡‌­
     )↔  # ‎⁡Generate until fixed point, starting with the input:
∷.+      # ‎⁢  Add the bit parity of the argument to 0.5 (0.5 if even, 1.5 if odd)
   e     # ‎⁣  And raise the argument to that number
    ⌊    # ‎⁤  Floor it.
💎

Created with the help of Luminespire.

Answer 28

Nibbles, 7.5 bytes (15 nibbles)

`.$^^$^3%$~-2

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`.              # iterate while unique
  $             # starting with input:
        %$~     #  result-so-far modulo-2
      ^3        #  3 to the power of that (so: 3 or 1)
    ^$          #  result-so-far to the power of that
   ^       -2   #  floor square-root

Answer 29

TI BASIC, 43 bytes

I'm pulling a Thomas Kwa and answering this one on my mobile.

Input N
Repeat N=1
Disp N
remainder(N,2->B
If not(B:int(sqrt(N->N
If B:int(N^1.5->N
End
1

Replace sqrt with the actual symbol on your calculator. Displays a linefeed separated list of numbers, which is a reasonable format.

Answer 30

JavaScript ES6, 76 bytes

Is a generator named j. To use, set a = j(<your value>);. To see the next value in the sequence, enter a.next().value.

function*j(N){for(yield N;N-1;)yield N=(N%2?Math.pow(N,3/2):Math.sqrt(N))|0}

Ungolfed:

function* juggler(N){
    yield N;
    while(N!=1){
        N = Math.floor(N % 2 ? Math.pow(N,3/2) : Math.sqrt(N));
        yield N;
    }
}