Icosikaiheptagonal Number
Last Updated :
23 Jun, 2021
Given a number N, the task is to find Nth icosikaiheptagonal number.
An icosikaiheptagonal number is a class of figurate numbers. It has 27 – sided polygon called icosikaiheptagon. The N-th icosikaiheptagonal number count’s the 27 number of dots and all other dots are surrounding with a common sharing corner and make a pattern. The first few icosikaiheptagonol numbers are 1, 27, 78, 154 …
Examples:
Input: N = 2
Output: 27
Explanation:
The second icosikaiheptagonol number is 27.
Input: N = 3
Output: 78
Approach: The N-th icosikaiheptagonal number is given by the formula:
- Nth term of s sided polygon =
- Therefore Nth term of 27 sided polygon is
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int icosikaiheptagonalNum( int n)
{
return (25 * n * n - 23 * n) / 2;
}
int main()
{
int n = 3;
cout << "3rd icosikaiheptagonal Number is "
<< icosikaiheptagonalNum(n);
return 0;
}
|
Java
class GFG{
static int icosikaiheptagonalNum( int n)
{
return ( 25 * n * n - 23 * n) / 2 ;
}
public static void main(String[] args)
{
int n = 3 ;
System.out.print( "3rd icosikaiheptagonal Number is " +
icosikaiheptagonalNum(n));
}
}
|
Python3
def icosikaiheptagonalNum(n):
return ( 25 * n * n - 23 * n) / / 2 ;
n = 3 ;
print ( "3rd icosikaiheptagonal Number is " ,
icosikaiheptagonalNum(n));
|
C#
using System;
class GFG{
static int icosikaiheptagonal( int n)
{
return (25 * n * n - 23 * n) / 2;
}
public static void Main(String[] args)
{
int n = 3;
Console.Write( "3rd icosikaiheptagonal Number is " +
icosikaiheptagonal(n));
}
}
|
Javascript
<script>
function icosikaiheptagonalNum( n)
{
return (25 * n * n - 23 * n) / 2;
}
let n = 3;
document.write( "3rd icosikaiheptagonal Number is " + icosikaiheptagonalNum(n));
</script>
|
Output: 3rd icosikaiheptagonal Number is 78
Time Complexity: O(1)
Auxiliary Space: O(1)
Reference: http://www.2dcurves.com/line/linep.html
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