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Lucas Numbers

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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


 

Recommended Practice


Method 1 (Recursive Solution) 
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.

                    

Python3

# 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

                    

C#

// 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.

                    

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.
?>

                    

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>

                    

Output : 
 

76


Method 2 (Iterative Solution) 
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; 

                    

Java

// Iterative Java program to
// find n'th Lucas Number
class GFG 
{
    // Iterative function
    static 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
    public static void main(String args[])
    {
        int n = 9;
        System.out.println(lucas(n));
    }
}
  
// This code is contributed 
// by Nikita tiwari.

                    

Python3

# Iterative Python 3 program 
# to find n'th Lucas Number
  
# Iterative function
def lucas(n) :
  
    # declaring base values
    # for positions 0 and 1
    a = 2
    b = 1
      
    if (n == 0) :
        return a
   
    # generating number
    for i in range(2, n + 1) :
        c = a + b
        a = b
        b = c
      
    return b
      
   
# Driver Code
n = 9
print(lucas(n))
  
# This code is contributed
# by Nikita tiwari.

                    

C#

// Iterative C# program to
// find n'th Lucas Number
using System;
  
class GFG {
  
    // Iterative function
    static 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
    public static void Main()
    {
        int n = 9;
  
        Console.WriteLine(lucas(n));
    }
}
  
// This code is contributed by vt_m.

                    

PHP

<?php
// Iterative php program 
// to find n'th Lucas Number
  
function lucas($n)
{
    // declaring base values
    // for positions 0 and 1
    $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
$n = 9;
echo lucas($n);
  
// This code is contributed by ajit
?>

                    

Javascript

<script>
    // Iterative Javascript program to find n'th Lucas Number
      
    // Iterative function
    function lucas(n)
    {
   
        // declaring base values
        // for positions 0 and 1
        let 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;
    }
      
    let n = 9;
   
      document.write(lucas(n));
          
</script>

                    

Output : 
 

76

Time complexity: O(n) since using a for loop

Space complexity: O(1) since using constant variables, since no extra space has been taken.


References: 
https://en.wikipedia.org/wiki/Lucas_number


 



Last Updated : 11 Sep, 2023
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