Juggler Sequence | Set 2 (Using Recursion)

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Juggler Sequence is a series of integer number in which the first term starts with a positive integer number a and the remaining terms are generated from the immediate previous term using the below recurrence relation :

$a_{k+1}=\begin{Bmatrix} \lfloor a_{k}^{1/2} \rfloor & for \quad even \quad a_k\\ \lfloor a_{k}^{3/2} \rfloor & for \quad odd \quad a_k \end{Bmatrix}$
Juggler Sequence starting with number 3:
5, 11, 36, 6, 2, 1
Juggler Sequence starting with number 9:
9, 27, 140, 11, 36, 6, 2, 1

Given a number N, we have to print the Juggler Sequence for this number as the first term of the sequence.

Examples:

Input: N = 9
Output: 9, 27, 140, 11, 36, 6, 2, 1
We start with 9 and use above formula to get next terms.

Input: N = 6
Output: 6, 2, 1

Iterative approach: We have already seen the iterative approach in Set 1 of this problem.

Recursive approach: In this approach, we will recursively traverse starting from N. Follow the steps below for each recursive step

Output the value of N
If N has reached 1 end the recursion
Otherwise, follow the formula based on the number being odd or even and call the recursive function on the newly derived number.

Below is the implementation of the approach:

C++

// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Recursive function to print
// the juggler sequence
void jum_sequence(int N)
{
 
    cout << N << " ";
 
    if (N <= 1)
        return;
    else if (N % 2 == 0)
    {
        N = floor(sqrt(N));
        jum_sequence(N);
    }
    else
    {
        N = floor(N * sqrt(N));
        jum_sequence(N);
    }
}
 
// Driver code
int main()
{
   
    // Juggler sequence starting with 10
    jum_sequence(10);
    return 0;
}
 
// This code is contributed by Potta Lokesh

Java

// Java code for the above approach
class GFG
{
   
    // Recursive function to print
    // the juggler sequence
    public static void jum_sequence(int N) {
 
        System.out.print(N + " ");
 
        if (N <= 1)
            return;
        else if (N % 2 == 0) {
            N = (int) (Math.floor(Math.sqrt(N)));
            jum_sequence(N);
        } else {
            N = (int) Math.floor(N * Math.sqrt(N));
            jum_sequence(N);
        }
    }
 
    // Driver code
    public static void main(String args[]) {
 
        // Juggler sequence starting with 10
        jum_sequence(10);
    }
}
 
// This code is contributed by Saurabh Jaiswal

Python3

# Python code to implement the above approach
 
# Recursive function to print
# the juggler sequence
def jum_sequence(N):
     
    print(N, end =" ")
 
    if (N == 1):
        return
    elif N & 1 == 0:
        N = int(pow(N, 0.5))
        jum_sequence(N)
    else:
        N = int(pow(N, 1.5))
        jum_sequence(N)
 
 
# Juggler sequence starting with 10
jum_sequence(10)

C#

// C# code for the above approach
using System;
 
class GFG{
 
// Recursive function to print
// the juggler sequence
public static void jum_sequence(int N)
{
    Console.Write(N + " ");
 
    if (N <= 1)
        return;
    else if (N % 2 == 0) 
    {
        N = (int)(Math.Floor(Math.Sqrt(N)));
        jum_sequence(N);
    } 
    else
    {
        N = (int)Math.Floor(N * Math.Sqrt(N));
        jum_sequence(N);
    }
}
 
// Driver code
public static void Main() 
{
     
    // Juggler sequence starting with 10
    jum_sequence(10);
}
}
 
// This code is contributed by Saurabh Jaiswal

Javascript

<script>
// Javascript code for the above approach
 
// Recursive function to print
// the juggler sequence
function jum_sequence(N){
 
    document.write(N +" ");
 
    if (N <= 1)
        return;
    else if (N % 2 == 0)
    {
        N = Math.floor(Math.sqrt(N));
        jum_sequence(N);
    }
    else
    {
        N = Math.floor(N * Math.sqrt(N));
        jum_sequence(N);
    }
}
 
// Driver code
// Juggler sequence starting with 10
 
jum_sequence(10);
     
// This code is contributed by gfgking
</script>

Output:

10 3 5 11 36 6 2 1

Time Complexity: O(N)
Auxiliary Space: O(1)

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