Juggler Sequence | Set 2 (Using Recursion)

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 : 

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)