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 argument and returns the n’th Lucas number.

Examples :

Input : 3
Output : 4

Input : 7
Output : 29

Method 1 ( Recursive Solution )
Below is recursive implementation based on simple recursive formula.

C/C++

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// 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;
}

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Java

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// 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.

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Python3

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# Recursive Python 3 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
n = 9
print(lucas(n))
  
# This code is contributed by Nikita Tiwari.

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C#

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// 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.

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PHP

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<?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.
?>

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Output :

76

Method 2 ( Iterative Solution )
The time complexity of above implementation is exponential. We can optimize it to work in O(n) time using iteration.

C/C++

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// 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;
}

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Java

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// 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.

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Python3

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# 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.

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C#

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// 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.

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PHP

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<?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
?>

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Output :

76

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

This article is contributed by Harsh Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : jit_t, nidhi_biet