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: 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

## 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.

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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