Hexagonal Number

Given an integer n, the task is to find the n’th hexagonal number . The n’th hexagonal number Hn is the number of distinct dots in a pattern of dots consisting of the outlines of regular hexagons with sides up to n dots, when the hexagons are overlaid so that they share one vertex.{Source : wiki }

Input : n = 2
Output : 6
Input : n = 5
Output : 45
Input : n = 7
Output : 91

In general, a polygonal number (triangular number, square number, etc) is a number represented as dots or pebbles arranged in the shape of a regular polygon. The first few pentagonal numbers are: 1, 5, 12, etc.

If s is the number of sides in a polygon, the formula for the nth s-gonal number P (s, n) is

nth s-gonal number P(s, n) = (s - 2)n(n-1)/2 + n
If we put s = 6, we get
n'th Hexagonal number Hn = 2(n*n)-n
= n(2n - 1) C/C++ // C program for above approach
#include <stdio.h>
#include <stdlib.h>
// Finding the nth Hexagonal Number
int hexagonalNum(int n)
{
return n*(2*n - 1);
}
// Driver program to test above function
int main()
{
int n = 10;
printf("10th Hexagonal Number is = %d",
hexagonalNum(n));
return 0;
} Java // Java program for above approach
class Hexagonal
{
int hexagonalNum(int n)
{
return n*(2*n - 1);
}
}
public class GeeksCode
{
public static void main(String[] args)
{
Hexagonal obj = new Hexagonal();
int n = 10;
System.out.printf("10th Hexagonal number is = "
+ obj.hexagonalNum(n));
}
} python # Python program for finding pentagonal numbers
def hexagonalNum( n ):
return n*(2*n - 1)
# Driver code
n = 10
print "10th Hexagonal Number is = ", hexagonalNum(n)

Output:

10th Hexagonal Number is = 190

Reference: https://en.wikipedia.org/wiki/Hexagonal_number

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