Nth Fibonacci Number

Last Updated : 03 Oct, 2023

Given a number n, print n-th Fibonacci Number.

The Fibonacci numbers are the numbers in the following integer sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ……..

Examples:

Input : n = 1

Output : 1

Input : n = 9

Output : 34

Input : n = 10

Output : 55

[/Tex]with seed values and $F_0 = 0$ and $F_1 = 1$ .

C++

// Fibonacci Series using Space Optimized Method
#include <bits/stdc++.h>
using namespace std;
 
int fib(int n)
{
    int a = 0, b = 1, c, i;
    if (n == 0)
        return a;
    for (i = 2; i <= n; i++) {
        c = a + b;
        a = b;
        b = c;
    }
    return b;
}
 
// Driver code
int main()
{
    int n = 9;
 
    cout << fib(n);
    return 0;
}
 
// This code is contributed by Code_Mech

C

// Fibonacci Series using Space Optimized Method
#include <stdio.h>
int fib(int n)
{
    int a = 0, b = 1, c, i;
    if (n == 0)
        return a;
    for (i = 2; i <= n; i++) {
        c = a + b;
        a = b;
        b = c;
    }
    return b;
}
 
int main()
{
    int n = 9;
    printf("%d", fib(n));
    getchar();
    return 0;
}

Java

// Java program for Fibonacci Series using Space
// Optimized Method
public class fibonacci {
    static int fib(int n)
    {
        int a = 0, b = 1, c;
        if (n == 0)
            return a;
        for (int i = 2; i <= n; i++) {
            c = a + b;
            a = b;
            b = c;
        }
        return b;
    }
 
    public static void main(String args[])
    {
        int n = 9;
        System.out.println(fib(n));
    }
};
 
// This code is contributed by Mihir Joshi

Python3

# Function for nth fibonacci number - Space Optimisation
# Taking 1st two fibonacci numbers as 0 and 1
 
 
def fibonacci(n):
    a = 0
    b = 1
    if n < 0:
        print("Incorrect input")
    elif n == 0:
        return a
    elif n == 1:
        return b
    else:
        for i in range(2, n+1):
            c = a + b
            a = b
            b = c
        return b
 
# Driver Program
 
 
print(fibonacci(9))
 
# This code is contributed by Saket Modi

C#

// C# program for Fibonacci Series
// using Space Optimized Method
using System;
 
namespace Fib {
public class GFG {
    static int Fib(int n)
    {
        int a = 0, b = 1, c = 0;
 
        // To return the first Fibonacci number
        if (n == 0)
            return a;
 
        for (int i = 2; i <= n; i++) {
            c = a + b;
            a = b;
            b = c;
        }
 
        return b;
    }
 
    // Driver function
    public static void Main(string[] args)
    {
 
        int n = 9;
        Console.Write("{0} ", Fib(n));
    }
}
}
 
// This code is contributed by Sam007.

Javascript

<script>
 
// Javascript program for Fibonacci Series using Space Optimized Method
 
function fib(n)
{
    let a = 0, b = 1, c, i;
    if( n == 0)
        return a;
    for(i = 2; i <= n; i++)
    {
    c = a + b;
    a = b;
    b = c;
    }
    return b;
}
 
// Driver code
 
    let n = 9;
     
    document.write(fib(n));
 
// This code is contributed by Mayank Tyagi
 
</script>

PHP

<?php
// PHP program for Fibonacci Series 
// using Space Optimized Method
 
function fib( $n)
{
    $a = 0; 
    $b = 1; 
    $c;
    $i;
    if( $n == 0)
        return $a;
    for($i = 2; $i <= $n; $i++)
    {
        $c = $a + $b;
        $a = $b;
        $b = $c;
    }
    return $b;
}
 
// Driver Code
$n = 9;
echo fib($n);
 
// This code is contributed by anuj_67.
?>

Output

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

Recursion Approach to Find and Print Nth Fibonacci Numbers:

In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation: $F_{n} = F_{n-1} + F_{n-2}$ with seed values and $F_0 = 0$ and $F_1 = 1$ .

The Nth Fibonacci Number can be found using the recurrence relation shown above:

if n = 0, then return 0.
If n = 1, then it should return 1.
For n > 1, it should return F_n-1 + F_n-2

Below is the implementation of the above approach:

C++

// Fibonacci Series using Recursion
#include <bits/stdc++.h>
using namespace std;
 
int fib(int n)
{
    if (n <= 1)
        return n;
    return fib(n - 1) + fib(n - 2);
}
 
int main()
{
    int n = 9;
    cout << n << "th Fibonacci Number: " << fib(n);
    return 0;
}
 
// This code is contributed
// by Akanksha Rai

C

// Fibonacci Series using Recursion
#include <stdio.h>
int fib(int n)
{
    if (n <= 1)
        return n;
    return fib(n - 1) + fib(n - 2);
}
 
int main()
{
    int n = 9;
    printf("%dth Fibonacci Number: %d", n, fib(n));
    return 0;
}

Java

// Fibonacci Series using Recursion
import java.io.*;
class fibonacci {
    static int fib(int n)
    {
        if (n <= 1)
            return n;
        return fib(n - 1) + fib(n - 2);
    }
 
    public static void main(String args[])
    {
        int n = 9;
        System.out.println(
            n + "th Fibonacci Number: " + fib(n));
    }
}
/* This code is contributed by Rajat Mishra */

Python3

# Fibonacci series using recursion
def fibonacci(n):
    if n <= 1:
        return n
    return fibonacci(n-1) + fibonacci(n-2)
 
 
if __name__ == "__main__":
    n = 9
    print(n, "th Fibonacci Number: ")
    print(fibonacci(n))
 
 # This code is contributed by Manan Tyagi.

C#

// C# program for Fibonacci Series
// using Recursion
using System;
 
public class GFG {
    public static int Fib(int n)
    {
        if (n <= 1) {
            return n;
        }
        else {
            return Fib(n - 1) + Fib(n - 2);
        }
    }
 
    // driver code
    public static void Main(string[] args)
    {
        int n = 9;
        Console.Write(n + "th Fibonacci Number: " + Fib(n));
    }
}
 
// This code is contributed by Sam007

Javascript

// Javascript program for Fibonacci Series
// using Recursion
 
function Fib(n) {
  if (n <= 1) {
    return n;
  } else {
    return Fib(n - 1) + Fib(n - 2);
  }
}
 
// driver code
let n = 9;
console.log(n + "th Fibonacci Number: " + Fib(n));

PHP

<?php
// PHP program for Fibonacci Series
// using Recursion
 
function Fib($n)
{
    if ($n <= 1) {
        return $n;
    }
    else {
        return Fib($n - 1) + Fib($n - 2);
    }
}
 
// driver code
$n = 9;
echo $n . "th Fibonacci Number: " . Fib($n);
 
// This code is contributed by Sam007
?>

Output

Time Complexity: Exponential, as every function calls two other functions.
Auxiliary space complexity: O(n), as the maximum depth of the recursion tree is n.

Dynamic Programming Approach to Find and Print Nth Fibonacci Numbers:

Consider the Recursion Tree for the 5th Fibonacci Number from the above approach:

                          fib(5)   
                     /                \
               fib(4)                fib(3)   
             /        \              /       \ 
         fib(3)      fib(2)         fib(2)   fib(1)
        /    \       /    \        /      \
  fib(2)   fib(1)  fib(1) fib(0) fib(1) fib(0)
  /     \
fib(1) fib(0)

If you see, the same method call is being done multiple times for the same value. This can be optimized with the help of Dynamic Programming. We can avoid the repeated work done in the Recursion approach by storing the Fibonacci numbers calculated so far.

Dynamic Programming Approach to Find and Print Nth Fibonacci Numbers:

Dynamic Programming Approach to Find and Print Nth Fibonacci Numbers:

Below is the implementation of the above approach:

C++

// C++ program for Fibonacci Series
// using Dynamic Programming
#include <bits/stdc++.h>
using namespace std;
 
class GFG {
 
public:
    int fib(int n)
    {
 
        // Declare an array to store
        // Fibonacci numbers.
        // 1 extra to handle
        // case, n = 0
        int f[n + 2];
        int i;
 
        // 0th and 1st number of the
        // series are 0 and 1
        f[0] = 0;
        f[1] = 1;
 
        for (i = 2; i <= n; i++) {
 
            // Add the previous 2 numbers
            // in the series and store it
            f[i] = f[i - 1] + f[i - 2];
        }
        return f[n];
    }
};
 
// Driver code
int main()
{
    GFG g;
    int n = 9;
 
    cout << g.fib(n);
    return 0;
}
 
// This code is contributed by SoumikMondal

C

// Fibonacci Series using Dynamic Programming
#include <stdio.h>
 
int fib(int n)
{
    /* Declare an array to store Fibonacci numbers. */
    int f[n + 2]; // 1 extra to handle case, n = 0
    int i;
 
    /* 0th and 1st number of the series are 0 and 1*/
    f[0] = 0;
    f[1] = 1;
 
    for (i = 2; i <= n; i++) {
        /* Add the previous 2 numbers in the series
           and store it */
        f[i] = f[i - 1] + f[i - 2];
    }
 
    return f[n];
}
 
int main()
{
    int n = 9;
    printf("%d", fib(n));
    getchar();
    return 0;
}

Java

// Fibonacci Series using Dynamic Programming
public class fibonacci {
    static int fib(int n)
    {
        /* Declare an array to store Fibonacci numbers. */
        int f[]
            = new int[n
                      + 2]; // 1 extra to handle case, n = 0
        int i;
 
        /* 0th and 1st number of the series are 0 and 1*/
        f[0] = 0;
        f[1] = 1;
 
        for (i = 2; i <= n; i++) {
            /* Add the previous 2 numbers in the series
              and store it */
            f[i] = f[i - 1] + f[i - 2];
        }
 
        return f[n];
    }
 
    public static void main(String args[])
    {
        int n = 9;
        System.out.println(fib(n));
    }
};
/* This code is contributed by Rajat Mishra */

Python3

# Fibonacci Series using Dynamic Programming
def fibonacci(n):
 
    # Taking 1st two fibonacci numbers as 0 and 1
    f = [0, 1]
 
    for i in range(2, n+1):
        f.append(f[i-1] + f[i-2])
    return f[n]
 
 
print(fibonacci(9))

C#

// C# program for Fibonacci Series
// using Dynamic Programming
using System;
class fibonacci {
 
    static int fib(int n)
    {
 
        // Declare an array to
        // store Fibonacci numbers.
        // 1 extra to handle
        // case, n = 0
        int[] f = new int[n + 2];
        int i;
 
        /* 0th and 1st number of the
           series are 0 and 1 */
        f[0] = 0;
        f[1] = 1;
 
        for (i = 2; i <= n; i++) {
            /* Add the previous 2 numbers
               in the series and store it */
            f[i] = f[i - 1] + f[i - 2];
        }
 
        return f[n];
    }
 
    // Driver Code
    public static void Main()
    {
        int n = 9;
        Console.WriteLine(fib(n));
    }
}
 
// This code is contributed by anuj_67.

Javascript

<script>
 
// Fibonacci Series using Dynamic Programming
 
    function  fib(n)
    {
        /* Declare an array to store Fibonacci numbers. */
        let f = new Array(n+2); // 1 extra to handle case, n = 0
        let i;
        /* 0th and 1st number of the series are 0 and 1*/
        f[0] = 0;
        f[1] = 1;
        for (i = 2; i <= n; i++)
        {
            /* Add the previous 2 numbers in the series
            and store it */
            f[i] = f[i-1] + f[i-2];
        }
        return f[n];
    }
    let n=9;
    document.write(fib(n));
     
    // This code is contributed by avanitrachhadiya2155
     
</script>

PHP

<?php
//Fibonacci Series using Dynamic
// Programming
 
function fib( $n)
{
     
    /* Declare an array to store
    Fibonacci numbers. */
     
    // 1 extra to handle case,
    // n = 0
    $f = array();
    $i;
     
    /* 0th and 1st number of the
    series are 0 and 1*/
    $f[0] = 0;
    $f[1] = 1;
     
    for ($i = 2; $i <= $n; $i++)
    {
         
        /* Add the previous 2
        numbers in the series
        and store it */
        $f[$i] = $f[$i-1] + $f[$i-2];
    }
     
    return $f[$n];
}
 
$n = 9;
echo fib($n);
 
// This code is contributed by
// anuj_67. 
?>

Output

Time complexity: O(n) for given n
Auxiliary space: O(n)

Nth Power of Matrix Approach to Find and Print Nth Fibonacci Numbers

This approach relies on the fact that if we n times multiply the matrix M = {{1,1},{1,0}} to itself (in other words calculate power(M, n)), then we get the (n+1)th Fibonacci number as the element at row and column (0, 0) in the resultant matrix.

If n is even then k = n/2:
Therefore Nth Fibonacci Number = F(n) = [2*F(k-1) + F(k)]*F(k)
If n is odd then k = (n + 1)/2:
Therefore Nth Fibonacci Number = F(n) = F(k)*F(k) + F(k-1)*F(k-1)

How does this formula work?
The formula can be derived from the matrix equation.

$\begin{bmatrix}1 & 1 \\1 & 0 \end{bmatrix}^n = \begin{bmatrix}F_{n+1} & F_n \\F_n & F_{n-1} \end{bmatrix}$

Taking determinant on both sides, we get (-1)ⁿ = F_n+1F_n-1 – F_n²

Moreover, since AⁿA^m = A^n+m for any square matrix A, the following identities can be derived (they are obtained from two different coefficients of the matrix product)

F_mF_n + F_m-1F_n-1 = F_{m+n-1 —————————(1)}

By putting n = n+1 in equation(1),

F_mF_n+1 + F_m-1F_n = F_{m+n ————————–(2)}

Putting m = n in equation(1).

F_2n-1 = F_n² + F_n-1²

Putting m = n in equation(2)

F_2n = (F_n-1 + F_n+1)F_n = (2F_n-1 + F_n)F_n ——–

( By putting Fn+1 = Fn + Fn-1 )

To get the formula to be proved, we simply need to do the following

If n is even, we can put k = n/2
If n is odd, we can put k = (n+1)/2

Below is the implementation of the above approach

C++

// Fibonacci Series using Optimized Method
#include <bits/stdc++.h>
using namespace std;
 
void multiply(int F[2][2], int M[2][2]);
void power(int F[2][2], int n);
 
// Function that returns nth Fibonacci number
int fib(int n)
{
    int F[2][2] = { { 1, 1 }, { 1, 0 } };
    if (n == 0)
        return 0;
    power(F, n - 1);
 
    return F[0][0];
}
 
// Optimized version of power() in method 4
void power(int F[2][2], int n)
{
    if (n == 0 || n == 1)
        return;
    int M[2][2] = { { 1, 1 }, { 1, 0 } };
 
    power(F, n / 2);
    multiply(F, F);
 
    if (n % 2 != 0)
        multiply(F, M);
}
 
void multiply(int F[2][2], int M[2][2])
{
    int x = F[0][0] * M[0][0] + F[0][1] * M[1][0];
    int y = F[0][0] * M[0][1] + F[0][1] * M[1][1];
    int z = F[1][0] * M[0][0] + F[1][1] * M[1][0];
    int w = F[1][0] * M[0][1] + F[1][1] * M[1][1];
 
    F[0][0] = x;
    F[0][1] = y;
    F[1][0] = z;
    F[1][1] = w;
}
 
// Driver code
int main()
{
    int n = 9;
 
    cout << fib(9);
    getchar();
 
    return 0;
}
 
// This code is contributed by Nidhi_biet

C

#include <stdio.h>
 
void multiply(int F[2][2], int M[2][2]);
 
void power(int F[2][2], int n);
 
/* function that returns nth Fibonacci number */
int fib(int n)
{
    int F[2][2] = { { 1, 1 }, { 1, 0 } };
    if (n == 0)
        return 0;
    power(F, n - 1);
    return F[0][0];
}
 
/* Optimized version of power() in method 4 */
void power(int F[2][2], int n)
{
    if (n == 0 || n == 1)
        return;
    int M[2][2] = { { 1, 1 }, { 1, 0 } };
 
    power(F, n / 2);
    multiply(F, F);
 
    if (n % 2 != 0)
        multiply(F, M);
}
 
void multiply(int F[2][2], int M[2][2])
{
    int x = F[0][0] * M[0][0] + F[0][1] * M[1][0];
    int y = F[0][0] * M[0][1] + F[0][1] * M[1][1];
    int z = F[1][0] * M[0][0] + F[1][1] * M[1][0];
    int w = F[1][0] * M[0][1] + F[1][1] * M[1][1];
 
    F[0][0] = x;
    F[0][1] = y;
    F[1][0] = z;
    F[1][1] = w;
}
 
/* Driver program to test above function */
int main()
{
    int n = 9;
    printf("%d", fib(9));
    getchar();
    return 0;
}

Java

// Fibonacci Series using Optimized Method
public class fibonacci {
    /* function that returns nth Fibonacci number */
    static int fib(int n)
    {
        int F[][] = new int[][] { { 1, 1 }, { 1, 0 } };
        if (n == 0)
            return 0;
        power(F, n - 1);
 
        return F[0][0];
    }
 
    static void multiply(int F[][], int M[][])
    {
        int x = F[0][0] * M[0][0] + F[0][1] * M[1][0];
        int y = F[0][0] * M[0][1] + F[0][1] * M[1][1];
        int z = F[1][0] * M[0][0] + F[1][1] * M[1][0];
        int w = F[1][0] * M[0][1] + F[1][1] * M[1][1];
 
        F[0][0] = x;
        F[0][1] = y;
        F[1][0] = z;
        F[1][1] = w;
    }
 
    /* Optimized version of power() in method 4 */
    static void power(int F[][], int n)
    {
        if (n == 0 || n == 1)
            return;
        int M[][] = new int[][] { { 1, 1 }, { 1, 0 } };
 
        power(F, n / 2);
        multiply(F, F);
 
        if (n % 2 != 0)
            multiply(F, M);
    }
 
    /* Driver program to test above function */
    public static void main(String args[])
    {
        int n = 9;
        System.out.println(fib(n));
    }
};
/* This code is contributed by Rajat Mishra */

Python3

# Fibonacci Series using
# Optimized Method
 
# function that returns nth
# Fibonacci number
 
 
def fib(n):
 
    F = [[1, 1],
         [1, 0]]
    if (n == 0):
        return 0
    power(F, n - 1)
 
    return F[0][0]
 
 
def multiply(F, M):
 
    x = (F[0][0] * M[0][0] +
         F[0][1] * M[1][0])
    y = (F[0][0] * M[0][1] +
         F[0][1] * M[1][1])
    z = (F[1][0] * M[0][0] +
         F[1][1] * M[1][0])
    w = (F[1][0] * M[0][1] +
         F[1][1] * M[1][1])
 
    F[0][0] = x
    F[0][1] = y
    F[1][0] = z
    F[1][1] = w
 
# Optimized version of
# power() in method 4
 
 
def power(F, n):
 
    if(n == 0 or n == 1):
        return
    M = [[1, 1],
         [1, 0]]
 
    power(F, n // 2)
    multiply(F, F)
 
    if (n % 2 != 0):
        multiply(F, M)
 
 
# Driver Code
if __name__ == "__main__":
    n = 9
    print(fib(n))
 
# This code is contributed
# by ChitraNayal

C#

// Fibonacci Series using
// Optimized Method
using System;
 
class GFG {
    /* function that returns
    nth Fibonacci number */
    static int fib(int n)
    {
        int[, ] F = new int[, ] { { 1, 1 }, { 1, 0 } };
        if (n == 0)
            return 0;
        power(F, n - 1);
 
        return F[0, 0];
    }
 
    static void multiply(int[, ] F, int[, ] M)
    {
        int x = F[0, 0] * M[0, 0] + F[0, 1] * M[1, 0];
        int y = F[0, 0] * M[0, 1] + F[0, 1] * M[1, 1];
        int z = F[1, 0] * M[0, 0] + F[1, 1] * M[1, 0];
        int w = F[1, 0] * M[0, 1] + F[1, 1] * M[1, 1];
 
        F[0, 0] = x;
        F[0, 1] = y;
        F[1, 0] = z;
        F[1, 1] = w;
    }
 
    /* Optimized version of
    power() in method 4 */
    static void power(int[, ] F, int n)
    {
        if (n == 0 || n == 1)
            return;
        int[, ] M = new int[, ] { { 1, 1 }, { 1, 0 } };
 
        power(F, n / 2);
        multiply(F, F);
 
        if (n % 2 != 0)
            multiply(F, M);
    }
 
    // Driver Code
    public static void Main()
    {
        int n = 9;
        Console.Write(fib(n));
    }
}
 
// This code is contributed
// by ChitraNayal

Javascript

<script>
 
// Fibonacci Series using Optimized Method
 
// Function that returns nth Fibonacci number 
function fib(n)
{
    var F = [ [ 1, 1 ], [ 1, 0 ] ];
    if (n == 0)
        return 0;
         
    power(F, n - 1);
 
    return F[0][0];
}
 
function multiply(F, M)
{
    var x = F[0][0] * M[0][0] + F[0][1] * M[1][0];
    var y = F[0][0] * M[0][1] + F[0][1] * M[1][1];
    var z = F[1][0] * M[0][0] + F[1][1] * M[1][0];
    var w = F[1][0] * M[0][1] + F[1][1] * M[1][1];
 
    F[0][0] = x;
    F[0][1] = y;
    F[1][0] = z;
    F[1][1] = w;
}
 
// Optimized version of power() in method 4 */
function power(F, n) 
{
    if (n == 0 || n == 1)
        return;
         
    var M = [ [ 1, 1 ], [ 1, 0 ] ];
 
    power(F, n / 2);
    multiply(F, F);
 
    if (n % 2 != 0)
        multiply(F, M);
}
 
// Driver code
var n = 9;
 
document.write(fib(n));
 
// This code is contributed by gauravrajput1
 
</script>

Output

Time Complexity: O(Log n)
Auxiliary Space: O(Log n) if we consider the function call stack size, otherwise O(1).

Related Articles:
Large Fibonacci Numbers in Java

GeeksforGeeks

Improve

Previous Article

How to check if a given number is Fibonacci number?

C++ Program For Fibonacci Numbers

Similar Reads

Check if a M-th fibonacci number divides N-th fibonacci number

Given two numbers M and N, the task is to check if the M-th and N-th Fibonacci numbers perfectly divide each other or not.Examples: Input: M = 3, N = 6 Output: Yes F(3) = 2, F(6) = 8 and F(6) % F(3) = 0 Input: M = 2, N = 9 Output: Yes A naive approach will be to find the N-th and M-th Fibonacci numbers and check if they are perfectly divisible or n

Check if sum of Fibonacci elements in an Array is a Fibonacci number or not

Given an array arr[] containing N elements, the task is to check if the sum of Fibonacci elements of the array is a fibonacci number or not.Examples: Input: arr[] = {2, 3, 7, 11} Output: Yes Explanation: As there are two Fibonacci numbers in the array i.e. 2 and 3. So, the sum of Fibonacci numbers is 2 + 3 = 5 and 5 is also a Fibonacci number. Inpu

C/C++ Program for nth multiple of a number in Fibonacci Series

Given two integers n and k. Find position the n'th multiple of K in the Fibonacci series. Examples: Input : k = 2, n = 3 Output : 9 3'rd multiple of 2 in Fibonacci Series is 34 which appears at position 9. Input : k = 4, n = 5 Output : 30 5'th multiple of 4 in Fibonacci Series is 832040 which appears at position 30. An Efficient Solution is based o

Find nth Fibonacci number using Golden ratio

Fibonacci series = 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ........Different methods to find nth Fibonacci number are already discussed. Another simple way of finding nth Fibonacci number is using golden ratio as Fibonacci numbers maintain approximate golden ratio till infinite. Golden ratio: [Tex]\varphi ={\frac {1+{\sqrt {5}}}{2}}=1.6180339887\ldots [/T

Nth XOR Fibonacci number

Given three integers a, b and N where a and b are the first two terms of the XOR Fibonacci series and the task is to find the Nth term. The Nth term of the XOR Fibonacci series is defined as F(N) = F(N - 1) ^ F(N - 2) where ^ is the bitwise XOR.Examples: Input: a = 1, b = 2, N = 5 Output: 3 F(0) = 1 F(1) = 2 F(2) = 1 ^ 2 = 3 F(3) = 2 ^ 3 = 1 F(4) =

Nth Fibonacci number using Pell's equation

Given an integer N, the task is to find the Nth Fibonacci number. Examples: Input: N = 13 Output: 144 Input: N = 19 Output: 2584 Approach: The Nth Fibonacci number can be found using the roots of the pell's equation. Pells equation is generally of the form (x2) - n(y2) = |1|. Here, consider y2 = x, n = 1. Also, taken positive (+1) in the right-hand

Program to find Nth odd Fibonacci Number

Given an integer N. The task is to find the Nth odd Fibonacci number.The odd number fibonacci series is as: 1, 1, 3, 5, 13, 21, 55, 89, 233, 377, 987, 1597.............and so on.Note: In the above series we have omitted even terms from the general fibonacci sequence. Examples: Input: N = 3 Output: 3 Input: N = 4 Output: 5 Approach: On observing car

Find Nth Fibonacci Number using Binet's Formula

Given a number n, print n-th Fibonacci Number, using Binet's Formula. Examples: Input: n = 5 Output: 1 Input: n = 9 Output: 34 What is Binet's Formula?Binet's Formula states that: If [Tex]F_n[/Tex] is the [Tex]n_{th}[/Tex] Fibonacci number, then[Tex] F_{n} = \frac{1}{\sqrt{5}}\left ( \left ( \frac{1+\sqrt{5}}{2} \right )^{n} - \left ( \frac{1-\sqrt

Fast Doubling method to find the Nth Fibonacci number

Given an integer N, the task is to find the N-th Fibonacci numbers.Examples: Input: N = 3 Output: 2 Explanation: F(1) = 1, F(2) = 1 F(3) = F(1) + F(2) = 2 Input: N = 6 Output: 8 Approach: The Matrix Exponentiation Method is already discussed before. The Doubling Method can be seen as an improvement to the matrix exponentiation method to find the N-

Program to find last two digits of Nth Fibonacci number

Given a number ‘n’, write a function that prints the last two digits of n-th (‘n’ can also be a large number) Fibonacci number.Examples: Input : n = 65 Output : 65 Input : n = 365 Output : 65 Recommended: Please solve it on “PRACTICE” first, before moving on to the solution. A simple solution is to find n-th Fibonacci number and print its last two

Article Tags :

Practice Tags :