Find Nth term of the series 0, 1, 1, 2, 5, 29, 841…

Given a positive integer N, the task is to find the Nth term in the series 0, 1, 1, 2, 5, 29, 841…

Examples:

Input: N = 6
Output: 29
Explanation: The 6th term of the given series is 29.

Input: N = 1
Output: 1

Input: N = 8
Output: 750797

Approach: The given problem is a basic maths-based problem where the A_i = A_i-1² + A_i-2². Therefore, create variables a = 0 and b = 1. Iterate using the variable i in the range [2, N], and for each i, calculate the i^th term and update the value of a and b to i – 1^th and i – 2^th term respectively.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find N-th term
// of the given series
int getNthTerm(int N)
{
    if (N < 3)
        return N - 1;

    // Initialize Variables repre-
    // senting 1st and 2nd term
    long int a = 0, b = 1;

    // Loop to iterate through the
    // range [3, N] using variable i
    for (int i = 3; i <= N; i++) {

        // pow((i - 2)th term, 2) +
        // pow((i - 1)th term, 2)
        long int c = a * a + b * b;

        // Update a and b
        a = b;
        b = c;
    }

    // Return Answer
    return b;
}

// Driver Code
int main()
{
    int N = 8;
    cout << getNthTerm(N);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{

  // Function to find N-th term
  // of the given series
  static int getNthTerm(int N)
  {
    if (N < 3)
      return N - 1;

    // Initialize Variables repre-
    // senting 1st and 2nd term
    int a = 0, b = 1;

    // Loop to iterate through the
    // range [3, N] using variable i
    for (int i = 3; i <= N; i++) {

      // Math.pow((i - 2)th term, 2) +
      // Math.pow((i - 1)th term, 2)
      int c = a * a + b * b;

      // Update a and b
      a = b;
      b = c;
    }

    // Return Answer
    return b;
  }

  // Driver Code
  public static void main(String[] args)
  {
    int N = 8;
    System.out.print(getNthTerm(N));
  }
}

// This code is contributed by 29AjayKumar

Python3

# Python code for the above approach

# Function to find N-th term
# of the given series
def getNthTerm(N):
    if (N < 3):
        return N - 1

    # Initialize Variables repre-
    # senting 1st and 2nd term
    a = 0
    b = 1

    # Loop to iterate through the
    # range [3, N] using variable i
    for i in range(3, N + 1):

        # pow((i - 2)th term, 2) +
        # pow((i - 1)th term, 2)
        c = a * a + b * b

        # Update a and b
        a = b
        b = c

    # Return Answer
    return b

# Driver Code
N = 8
print(getNthTerm(N))

# This code is contributed by Saurabh Jaiswal

C#

// C# program for the above approach
using System;
class GFG
{

// Function to find N-th term
// of the given series
static int getNthTerm(int N)
{
    if (N < 3)
        return N - 1;

    // Initialize Variables repre-
    // senting 1st and 2nd term
    long a = 0, b = 1;

    // Loop to iterate through the
    // range [3, N] using variable i
    for (int i = 3; i <= N; i++) {

        // pow((i - 2)th term, 2) +
        // pow((i - 1)th term, 2)
        long c = a * a + b * b;

        // Update a and b
        a = b;
        b = c;
    }

    // Return Answer
    return (int)b;
}

// Driver Code
public static void Main()
{
    int N = 8;
    Console.Write(getNthTerm(N));

}
}

// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
  // JavaScript code for the above approach


  // Function to find N-th term
  // of the given series
  function getNthTerm(N) {
    if (N < 3)
      return N - 1;

    // Initialize Variables repre-
    // senting 1st and 2nd term
    let a = 0, b = 1;

    // Loop to iterate through the
    // range [3, N] using variable i
    for (let i = 3; i <= N; i++) {

      // pow((i - 2)th term, 2) +
      // pow((i - 1)th term, 2)
      let c = a * a + b * b;

      // Update a and b
      a = b;
      b = c;
    }

    // Return Answer
    return b;
  }

  // Driver Code

  let N = 8;
  document.write(getNthTerm(N));


// This code is contributed by Potta Lokesh
</script>

Output

750797

Time complexity: O(N)
Auxiliary space: O(1)