Lucas Numbers

Lucas numbers are similar to Fibonacci numbers. Lucas numbers are also defined as the sum of its two immediately previous terms. But here the first two terms are 2 and 1 whereas in Fibonacci numbers the first two terms are 0 and 1 respectively.
Mathematically, Lucas Numbers may be defined as:
{\displaystyle L_{n}:={\begin{cases}2&{\text{if }}n=0;\\1&{\text{if }}n=1;\\L_{n-1}+L_{n-2}&{\text{if }}n>1.\\\end{cases}}}
The Lucas numbers are in the following integer sequence:
2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123 ..............
Write a function int lucas(int n) n as an argument and returns the nth Lucas number.
Examples :

Input : 3
Output : 4

Input : 7
Output : 29

Try It Yourself

Recursive Solution - O(2^n) Time and O(n) Space

Below is a recursive implementation based on a simple recursive formula.

C++

// Recursive C/C++ program 
// to find n'th Lucas number 
#include <stdio.h> 

// recursive function 
int lucas(int n) 
{ 
    // Base cases 
    if (n == 0) 
        return 2; 
    if (n == 1) 
        return 1; 

    // recurrence relation 
    return lucas(n - 1) + 
        lucas(n - 2); 
} 

// Driver Code 
int main() 
{ 
    int n = 9; 
    printf("%d", lucas(n)); 
    return 0; 
}

Java

// Recursive Java program to
// find n'th Lucas number

class GFG 
{

    // recursive function
    public static int lucas(int n)
    {

        // Base cases
        if (n == 0)
            return 2;
        if (n == 1)
            return 1;

        // recurrence relation
        return lucas(n - 1) + 
               lucas(n - 2);
    }

    // Driver Code
    public static void main(String args[])
    {
        int n = 9;
        System.out.println(lucas(n));
    }
}
// This code is contributed 
// by Nikita Tiwari.

Python

# Python3 program to
# find n'th Lucas number

# recursive function
def lucas(n):
  
    # Base cases
    if n == 0:
        return 2;
    if n == 1:
        return 1;
  
    # recurrence relation
    return lucas(n - 1) + lucas(n - 2);
    
# Driver code to test above methods
n = 9;
print(lucas(n));

# This code is contributed by phasing17

// Recursive C# program to
// find n'th Lucas number
using System;

class GFG {

    // recursive function
    public static int lucas(int n)
    {

        // Base cases
        if (n == 0)
            return 2;
        if (n == 1)
            return 1;

        // recurrence relation
        return lucas(n - 1) + lucas(n - 2);
    }

    // Driver program
    public static void Main()
    {

        int n = 9;

        Console.WriteLine(lucas(n));
    }
}

// This code is contributed by vt_m.

JavaScript

<script>

// Javascript program to
// find n'th Lucas number

    // recursive function
    function lucas(n)
    {
  
        // Base cases
        if (n == 0)
            return 2;
        if (n == 1)
            return 1;
  
        // recurrence relation
        return lucas(n - 1) + 
               lucas(n - 2);
    }

// Driver code to test above methods

        let n = 9;
        document.write(lucas(n));
 
 // This code is contributed by avijitmondal1998.
</script>

PHP

<?php
// Recursive php program to
// find n'th Lucas number

// recursive function
function lucas($n)
{
    
// Base cases 
if ($n == 0) 
    return 2;
if ($n == 1) 
    return 1;

// recurrence relation 
return lucas($n - 1) + 
       lucas($n - 2); 
}

// Driver Code
$n = 9;
echo lucas($n);

// This code is contributed by ajit.
?>

Output :

Iterative Solution - O(n) Time and O(1) Space

The time complexity of the above implementation is exponential. We can optimize it to work in O(n) time using iteration.

C++

// Iterative C/C++ program 
// to find n'th Lucas Number 
#include <stdio.h> 

// Iterative function 
int lucas(int n) 
{ 
    // declaring base values 
    // for positions 0 and 1 
    int a = 2, b = 1, c, i; 

    if (n == 0) 
        return a; 

    // generating number 
    for (i = 2; i <= n; i++) 
    { 
        c = a + b; 
        a = b; 
        b = c; 
    } 
    return b; 
} 

// Driver Code 
int main() 
{ 
    int n = 9; 
    printf("%d", lucas(n)); 
    return 0; 
}