# 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 using namespace std; // Recursive function to print// the juggler sequencevoid 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 codeint main(){       // Juggler sequence starting with 10    jum_sequence(10);    return 0;} // This code is contributed by Potta Lokesh

## Java

 // Java code for the above approachclass 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 sequencedef 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 10jum_sequence(10)

## C#

 // C# code for the above approachusing System; class GFG{ // Recursive function to print// the juggler sequencepublic 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 codepublic static void Main() {         // Juggler sequence starting with 10    jum_sequence(10);}} // This code is contributed by Saurabh Jaiswal

## Javascript

 

Output:
10 3 5 11 36 6 2 1

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

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