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: 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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Improved By : jit_t, nidhi_biet