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 H_{n}= 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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