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 an argument and returns the nth Lucas number.
Examples :
Input : 3
Output : 4
Input : 7
Output : 29
Method 1 (Recursive Solution)
Below is a recursive implementation based on a simple recursive formula.
C++
#include <stdio.h>
int lucas(int n)
{
if (n == 0)
return 2;
if (n == 1)
return 1;
return lucas(n - 1) +
lucas(n - 2);
}
int main()
{
int n = 9;
printf("%d", lucas(n));
return 0;
}
|
Java
class GFG
{
public static int lucas(int n)
{
if (n == 0)
return 2;
if (n == 1)
return 1;
return lucas(n - 1) +
lucas(n - 2);
}
public static void main(String args[])
{
int n = 9;
System.out.println(lucas(n));
}
}
|
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#
using System;
class GFG {
public static int lucas(int n)
{
if (n == 0)
return 2;
if (n == 1)
return 1;
return lucas(n - 1) + lucas(n - 2);
}
public static void Main()
{
int n = 9;
Console.WriteLine(lucas(n));
}
}
|
PHP
<?php
function lucas($n)
{
if ($n == 0)
return 2;
if ($n == 1)
return 1;
return lucas($n - 1) +
lucas($n - 2);
}
$n = 9;
echo lucas($n);
?>
|
Javascript
<script>
function lucas(n)
{
if (n == 0)
return 2;
if (n == 1)
return 1;
return lucas(n - 1) +
lucas(n - 2);
}
let n = 9;
document.write(lucas(n));
</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++
#include <stdio.h>
int lucas(int n)
{
int a = 2, b = 1, c, i;
if (n == 0)
return a;
for (i = 2; i <= n; i++)
{
c = a + b;
a = b;
b = c;
}
return b;
}
int main()
{
int n = 9;
printf("%d", lucas(n));
return 0;
}
|
Java
class GFG
{
static int lucas(int n)
{
int a = 2, b = 1, c, i;
if (n == 0)
return a;
for (i = 2; i <= n; i++)
{
c = a + b;
a = b;
b = c;
}
return b;
}
public static void main(String args[])
{
int n = 9;
System.out.println(lucas(n));
}
}
|
Python3
def lucas(n) :
a = 2
b = 1
if (n == 0) :
return a
for i in range(2, n + 1) :
c = a + b
a = b
b = c
return b
n = 9
print(lucas(n))
|
C#
using System;
class GFG {
static int lucas(int n)
{
int a = 2, b = 1, c, i;
if (n == 0)
return a;
for (i = 2; i <= n; i++) {
c = a + b;
a = b;
b = c;
}
return b;
}
public static void Main()
{
int n = 9;
Console.WriteLine(lucas(n));
}
}
|
PHP
<?php
function lucas($n)
{
$a = 2; $b = 1; $c; $i;
if ($n == 0)
return $a;
for ($i = 2; $i <= $n; $i++)
{
$c = $a + $b;
$a = $b;
$b = $c;
}
return $b;
}
$n = 9;
echo lucas($n);
?>
|
Javascript
<script>
function lucas(n)
{
let a = 2, b = 1, c, i;
if (n == 0)
return a;
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
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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