Juggler sequences
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
[Jelly], 7 bytes Ḃ×½ḞµƬ …
3y ago
BQN, 17 bytesSBCS ``` {𝕩∾1 …
3y ago
Scala, 78 67 64 bytes Saved …
3y ago
[Husk], 11 10 bytes U¡λ …
3y ago
JavaScript (Node.js), 73 66 by …
3y ago
[APL (Dyalog Unicode)], 15 byt …
3y ago
Japt, 17 bytes Needs some m …
3y ago
JavaScript, 37 32 bytes Out …
3y ago
[Python 3], 64 bytes …
3y ago
9 answers
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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
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BQN, 17 bytesSBCS
{𝕩∾1𝕊⍟≠⌊𝕩⋆2|𝕩+÷2}
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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Husk, 11 10 bytes
U¡λ⌊^+.%2¹
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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Scala, 78 67 64 bytes
Saved 3 bytes thanks to Razetime
Stream.iterate(_)(x=>math.pow(x,x%2+.5).toInt).takeWhile(_>1):+1
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
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
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JavaScript, 37 32 bytes
Outputs a comma delimited string.
f=n=>n-1?n+[,f(n**(.5+n%2)|0)]:n
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