Challenges

# Juggler sequences

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A Juggler sequence is a sequence that begins with a positive integer $a_0$ and each subsequent term is calculated as:

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

Eventually, once $a_k$ equals $1$, the sequence ends, as all subsequent terms will be $1$. It has been conjectured, but not proven, that all Juggler sequences reach 1.

Given a positive integer $n \ge 2$, output the Juggler sequence beginning with $a_0 = n$ and ending in $1$. You may assume it will always terminate. You should only output a single $1$, and the sequence should be in calculated order ($a_0, a_1, a_2,$ etc.)

This is code golf, so the shortest code in bytes wins

## Test cases

2: 2, 1
3: 3, 5, 11, 36, 6, 2, 1
4: 4, 2, 1
5: 5, 11, 36, 6, 2, 1
6: 6, 2, 1
7: 7, 18, 4, 2, 1
8: 8, 2, 1
9: 9, 27, 140, 11, 36, 6, 2, 1
10: 10, 3, 5, 11, 36, 6, 2, 1
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#### 1 comment

Brownie points for beating or matching my 7 byte Jelly answer caird coinheringaahing‭ about 1 month ago

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# BQN, 17 bytesSBCS

{𝕩∾1𝕊⍟≠⌊𝕩⋆2|𝕩+÷2}

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BQN primitives alone can't express unbounded iteration, so this must be a recursive block function.

{𝕩∾1𝕊⍟≠⌊𝕩⋆2|𝕩+÷2}  # Function 𝕊 taking argument 𝕩
÷2   # Reciprocal of 2 (one half)
𝕩+     # Plus 𝕩
2|       # Modulo 2
𝕩⋆         # 𝕩 power
⌊           # Floor
𝕊              # Keep going
1 ⍟≠            # ...if not 1
𝕩∾                # Prepend 𝕩

1𝕊⍟≠… passes 1 as the left argument 𝕨, which is harmless.

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# Jelly, 7 bytes

*Ḃ×½ḞµƬ
*Ḃ×½ḞµƬ
*Ḃ        input ^ (input % 2)
×½      multiply sqrt(input)
Ḟ     floor
µƬ   iterate until converged and collect results

Try it online!

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# Scala, 78 67 64 bytes

Saved 3 bytes thanks to Razetime

Stream.iterate(_)(x=>math.pow(x,x%2+.5).toInt).takeWhile(_>1):+1

Try it in Scastie!

Stream.iterate(_) //Make an infinite list by repeatedly applying
(x=>              //the following function to the input
math.pow(x,     //Raise x to the power of
x%2           //x%2 (0 or 1)
+.5           //plus .5 (.5 or 1.5)
).toInt         //Floor it
).takeWhile(_>1)  //Take all elements until it hits 1
:+1               //Append a 1
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1/2.0 can be .5 Razetime‭ about 1 month ago

@Razetime I'm an idiot, thanks! user‭ about 1 month ago

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# Husk, 11 10 bytes

U¡λ⌊^+.%2¹

Try it online!

This can probably be trivially ported to Jelly.

## Explanation

U¡o⌊Ṡ^o+.%2
¡            iterate over the input infinitely, creating a list
o           with the following two functions:
Ṡ^        take the previous result to the power
%2   result modulo 2
+.     + 1/2
⌊          floor it
U             longest unique prefix
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## JavaScript (Node.js), 73 66 bytes

This returns a Generator, which calculates the sequence on the fly when you use it!

function*f(a){while(a>1){yield a
a=~~(a%2?a**1.5:a**0.5)}
yield a}

Original, more aesthetically pleasing but slightly longer version:

function*f(a){while(1){yield a
if(a==1){return}
a=~~(a%2?a**1.5:a**0.5)}}

To use it, do something like this, which calculates the whole thing and makes it into an array.

[...f(5)] // returns [ 5, 11, 36, 6, 2, 1 ]

Explanation:

function* f(a) { // makes a generator function
while (a > 1) {
yield a // return this bit of the sequence
a = ~~( // truncates the decimals? apparently? and it's shorter than Math.floor()/Math.trunc()
a%2 ? a**1.5 : a**0.5
)
}
yield a // return the final 1
}
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# APL (Dyalog Unicode), 15 bytesSBCS

With many thanks to Marshall Lochbaum and dzaima for their help debugging and golfing this answer

{⍵∪⌊⍵*2|⍵+.5}⍣≡
{⍵∪⌊⍵*2|⍵+.5}⍣≡  ⍝ ⍵ is our input of either n or our intermediate results
{            }⍣≡  ⍝ run until we reach a fixed point (in this case, 1)
⍵+.5     ⍝ ⍵ plus one half
2|         ⍝ mod 2 gives us 0.5 if our ⍵ was even, else 1.5
⍵*           ⍝ ⍵ to the power of the above exponent
⌊              ⍝ floor
⍵∪              ⍝ union of the answer with the previous results
⍝ The set union will either remove an extra one,
⍝ or if we find a cycle that disproves the conjecture,
⍝ will remove the extra members of the cycle

Try it online!

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# Python 3, 64 bytes

def f(n):
z=[n]
while n>1:n=int(n**((n+.5)%2));z+=n,
return z

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