# Program to find N-th term of series 3, 5, 33, 35, 53….

Last Updated : 11 Aug, 2022

Given a series of numbers composed only of digits 3 and 5. The first few numbers in the series are:

3, 5, 33, 35, 53, 55, …..

Given a number N. The task is to find the n-th number in the given series.
Examples

Input : N = 2
Output : 5

Input : N = 5
Output : 53

The idea is based on the fact that the value of the last digit alternates in the series. For example, if the last digit of ith number is 3, then the last digit of (i-1)th and (i+1)th numbers must be 5.
Create an array of size (n+1) and push 3 and 5(These two are always first two elements of series) to it. For more elements check,

1) If i is odd,
arr[i] = arr[i/2]*10 + 3;
2) If it is even,
arr[i] = arr[(i/2)-1]*10 + 5;
At last return arr[n].

Below is the implementation of the above idea:

## C++

 // C++ program to find n-th number in a series // made of digits 3 and 5   #include using namespace std;   // Function to find n-th number in series // made of 3 and 5 int printNthElement(int n) {     // create an array of size (n+1)     int arr[n + 1];     arr[1] = 3;     arr[2] = 5;       for (int i = 3; i <= n; i++) {         // If i is odd         if (i % 2 != 0)             arr[i] = arr[i / 2] * 10 + 3;         else             arr[i] = arr[(i / 2) - 1] * 10 + 5;     }     return arr[n]; }   // Driver code int main() {     int n = 6;       cout << printNthElement(n);       return 0; }

## C

 // C program to find n-th number in a series // made of digits 3 and 5 #include   // Function to find n-th number in series // made of 3 and 5 int printNthElement(int n) {     // create an array of size (n+1)     int arr[n + 1];     arr[1] = 3;     arr[2] = 5;       for (int i = 3; i <= n; i++) {         // If i is odd         if (i % 2 != 0)             arr[i] = arr[i / 2] * 10 + 3;         else             arr[i] = arr[(i / 2) - 1] * 10 + 5;     }     return arr[n]; }   // Driver code int main() {     int n = 6;       printf("%d",printNthElement(n));       return 0; }   // This code is contributed by kothavvsaakash.

## Java

 // Java program to find n-th number in a series // made of digits 3 and 5   class FindNth {     // Function to find n-th number in series     // made of 3 and 5     static int printNthElement(int n)     {         // create an array of size (n+1)         int arr[] = new int[n + 1];         arr[1] = 3;         arr[2] = 5;           for (int i = 3; i <= n; i++) {             // If i is odd             if (i % 2 != 0)                 arr[i] = arr[i / 2] * 10 + 3;             else                 arr[i] = arr[(i / 2) - 1] * 10 + 5;         }         return arr[n];     }       // main function     public static void main(String[] args)     {         int n = 6;           System.out.println(printNthElement(n));     } }

## Python3

 # Python3 program to find n-th number  # in a series made of digits 3 and 5     # Return n-th number in series made  # of 3 and 5 def printNthElement(n) :             # create an array of size (n + 1)     arr =[0] * (n + 1);     arr[1] = 3     arr[2] = 5         for i in range(3, n + 1) :         # If i is odd         if (i % 2 != 0) :             arr[i] = arr[i // 2] * 10 + 3         else :             arr[i] = arr[(i // 2) - 1] * 10 + 5             return arr[n]         # Driver code n = 6 print(printNthElement(n))

## C#

 // C# program to find n-th number // in a series made of digits 3 and 5 using System;   class GFG { // Function to find n-th number // in series made of 3 and 5 static int printNthElement(int n) {     // create an array of size (n+1)     int [] arr = new int[n + 1];     arr[1] = 3;     arr[2] = 5;       for (int i = 3; i <= n; i++)     {         // If i is odd         if (i % 2 != 0)             arr[i] = arr[i / 2] * 10 + 3;         else             arr[i] = arr[(i / 2) - 1] * 10 + 5;     }     return arr[n]; }   // Driver Code static void Main() {     int n = 6;       Console.WriteLine(printNthElement(n)); } }   // This code is contributed by ANKITRAI1



## Javascript



Output:

55

Time complexity: O(n) as using a loop

Auxiliary space: O(n)