Juggler Sequence - GeeksforGeeks

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

Input: 6
Output: 6, 2, 1

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

// C++ implementation of Juggler Sequence 
#include <bits/stdc++.h> 
using namespace std; 
  
// This function prints the juggler Sequence 
void printJuggler(int n) 
{ 
    int a = n; 
  
    // print the first term 
    cout << a << " "; 
  
    // calculate terms until  
    // last term is not 1 
    while (a != 1) 
    { 
        int b = 0; 
  
        // Check if previous term  
        // is even or odd 
        if (a % 2 == 0) 
  
            // calculate next term 
            b = floor(sqrt(a)); 
  
        else
  
            // for odd previous term  
            // calculate next term 
            b = floor(sqrt(a) *  
                      sqrt(a) * sqrt(a)); 
  
        cout << b << " "; 
        a = b; 
    } 
} 
  
// Driver Code 
int main() 
{ 
    printJuggler(3); 
    cout <<"\n"; 
    printJuggler(9); 
    return 0; 
} 
  
// This code is contributed by shubhamsingh10 

C

// C implementation of Juggler Sequence 
#include<stdio.h> 
#include<math.h> 
  
// This function prints the juggler Sequence 
void printJuggler(int n) 
{ 
    int a = n; 
  
    // print the first term 
    printf("%d ", a); 
  
    // calculate terms until last term is not 1 
    while (a != 1) 
    { 
        int b = 0; 
  
        // Check if previous term is even or odd 
        if (a%2 == 0) 
  
            // calculate next term 
            b  = floor(sqrt(a)); 
  
        else
  
            // for odd previous term calculate 
            // next term 
            b = floor(sqrt(a)*sqrt(a)*sqrt(a)); 
  
        printf("%d ", b); 
        a = b; 
    } 
} 
  
//driver program to test above function 
int main() 
{ 
    printJuggler(3); 
    printf("\n"); 
    printJuggler(9); 
    return 0; 
} 

Java

// Java implementation of Juggler Sequence 
import java.io.*; 
import java.math.*; 
  
class GFG { 
       
    // This function prints the juggler Sequence 
    static void printJuggler(int n) 
    { 
        int a = n; 
    
       // print the first term 
       System.out.print(a+" "); 
    
      // calculate terms until last term is not 1 
       while (a != 1) 
       { 
          int b = 0; 
     
          // Check if previous term is even or odd 
          if (a%2 == 0) 
     
             // calculate next term 
                b  = (int)Math.floor(Math.sqrt(a)); 
    
          else
    
            // for odd previous term calculate 
            // next term 
                b =(int) Math.floor(Math.sqrt(a) * 
                               Math.sqrt(a) * Math.sqrt(a)); 
    
          System.out.print( b+" "); 
          a = b; 
        } 
    } 
  
// Driver program to test above function 
public static void main (String[] args) { 
    printJuggler(3); 
    System.out.println(); 
    printJuggler(9); 
    } 
} 
   
//This code is contributed by Nikita Tiwari. 

Python

import math 
  
#This function prints the juggler Sequence 
def printJuggler(n) : 
    a = n 
      
    # print the first term 
    print a, 
      
    # calculate terms until last term is not 1 
    while (a != 1) : 
        b = 0
          
        # Check if previous term is even or odd 
        if (a%2 == 0) : 
              
            # calculate next term 
            b  = (int)(math.floor(math.sqrt(a))) 
   
        else : 
            # for odd previous term calculate 
            # next term 
            b = (int) (math.floor(math.sqrt(a)*math.sqrt(a)*
                                         math.sqrt(a))) 
   
        print b, 
        a = b 
  
printJuggler(3) 
print
printJuggler(9) 
  
# This code is contributed by Nikita Tiwari. 

C#

// C# implementation of Juggler Sequence 
using System; 
  
class GFG { 
      
    // This function prints the juggler Sequence 
    static void printJuggler(int n) 
    { 
        int a = n; 
  
    // print the first term 
    Console.Write(a+" "); 
  
    // calculate terms until last term is not 1 
    while (a != 1) 
    { 
        int b = 0; 
      
        // Check if previous term is even or odd 
        if (a%2 == 0) 
      
            // calculate next term 
                b = (int)Math.Floor(Math.Sqrt(a)); 
  
        else
  
            // for odd previous term calculate 
            // next term 
                b =(int) Math.Floor(Math.Sqrt(a) * 
                     Math.Sqrt(a) * Math.Sqrt(a)); 
  
        Console.Write( b+" "); 
        a = b; 
        } 
    } 
  
// Driver Code 
public static void Main () { 
    printJuggler(3); 
    Console.WriteLine(); 
    printJuggler(9); 
    } 
} 
  
// This code is contributed by Nitin Mittal 

PHP

<?php 
// PHP implementation of  
// Juggler Sequence 
  
// function prints the 
// juggler Sequence 
function printJuggler($n) 
{ 
    $a = $n; 
  
    // print the first term 
    echo($a . " "); 
  
    // calculate terms until  
    // last term is not 1 
    while ($a != 1) 
    { 
        $b = 0; 
  
        // Check if previous 
        // term is even or odd 
        if ($a % 2 == 0) 
  
            // calculate next term 
            $b = floor(sqrt($a)); 
  
        else
  
            // for odd previous term 
            // calculate next term 
            $b = floor(sqrt($a) * sqrt($a) * 
                                  sqrt($a)); 
  
        echo($b . " "); 
        $a = $b; 
    } 
} 
  
// Driver Code 
printJuggler(3); 
echo("\n"); 
printJuggler(9); 
  
// This code is contributed by Ajit. 
?> 

Output:

3 5 11 36 6 2 1 
9 27 140 11 36 6 2 1

Important Points:

The terms in Juggler Sequence first increases to a peak value and then starts decreasing.
The last term in Juggler Sequence is always 1.

Reference:
https://en.wikipedia.org/wiki/Juggler_sequence
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My Personal Notes

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

C

Java

Python

C#

PHP

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