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
You are accessing this answer with a direct link, so it's being shown above all other answers regardless of its score. You can return to the normal view.
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
0 comment threads
Jelly, 7 bytes
*Ḃ×½ḞµƬ
*Ḃ×½ḞµƬ
*Ḃ input ^ (input % 2)
×½ multiply sqrt(input)
Ḟ floor
µƬ iterate until converged and collect results
0 comment threads
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.
0 comment threads
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
0 comment threads
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
}
0 comment threads
JavaScript, 37 32 bytes
Outputs a comma delimited string.
f=n=>n-1?n+[,f(n**(.5+n%2)|0)]:n
0 comment threads
1 comment thread