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Lucas Numbers
• Difficulty Level : Easy
• Last Updated : 29 Jul, 2019

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: 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


## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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

## C/C++

 // Recursive C/C++ program   // to find n'th Lucas number  #include     // 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

 # 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.

## 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

 

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++

 // Iterative C/C++ program  // to find n'th Lucas Number  #include     // 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

 

Output :

76


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

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