n’th Pentagonal Number
Given an integer n, find the nth Pentagonal number. The first three pentagonal numbers are 1, 5, and 12 (Please see the below diagram).
The n’th pentagonal number Pn is the number of distinct dots in a pattern of dots consisting of the outlines of regular pentagons with sides up to n dots when the pentagons are overlaid so that they share one vertex [Source Wiki]
Examples :
Input: n = 1 Output: 1 Input: n = 2 Output: 5 Input: n = 3 Output: 12
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 = 5, we get n'th Pentagonal number Pn = 3*n*(n-1)/2 + n
Examples:
Pentagonal Number
Below are the implementations of the above idea in different programming languages.
C++
// C++ program for above approach #include<bits/stdc++.h> using namespace std; // Finding the nth pentagonal number int pentagonalNum( int n) { return (3 * n * n - n) / 2; } // Driver code int main() { int n = 10; cout << "10th Pentagonal Number is = " << pentagonalNum(n); return 0; } // This code is contributed by Code_Mech |
C
// C program for above approach #include <stdio.h> #include <stdlib.h> // Finding the nth Pentagonal Number int pentagonalNum( int n) { return (3*n*n - n)/2; } // Driver program to test above function int main() { int n = 10; printf ( "10th Pentagonal Number is = %d \n \n" , pentagonalNum(n)); return 0; } |
Java
// Java program for above approach class Pentagonal { int pentagonalNum( int n) { return ( 3 *n*n - n)/ 2 ; } } public class BeginnerCode { public static void main(String[] args) { Pentagonal obj = new Pentagonal(); int n = 10 ; System.out.printf( "10th petagonal number is = " + obj.pentagonalNum(n)); } } |
Python3
# Python program for finding pentagonal numbers def pentagonalNum( n ): return ( 3 * n * n - n) / 2 #Script Begins n = 10 print ( "10th Pentagonal Number is = " , pentagonalNum(n)) #Scripts Ends |
C#
// C# program for above approach using System; class GFG { static int pentagonalNum( int n) { return (3 * n * n - n) / 2; } public static void Main() { int n = 10; Console.WriteLine( "10th petagonal" + " number is = " + pentagonalNum(n)); } } // This code is contributed by vt_m. |
PHP
<?php // PHP program for above approach // Finding the nth Pentagonal Number function pentagonalNum( $n ) { return (3 * $n * $n - $n ) / 2; } // Driver Code $n = 10; echo "10th Pentagonal Number is = " , pentagonalNum( $n ); // This code is contributed by ajit ?> |
Javascript
<script> // Javascript program for above approach function pentagonalNum(n) { return (3 * n * n - n) / 2; } // Driver code to test above methods let n = 10; document.write( "10th petagonal" + " number is = " + pentagonalNum(n)); // This code is contributed by avijitmondal1998. </script> |
10th Pentagonal Number is = 145
Time Complexity: O(1) // since no loop or recursion is used the algorithm takes up constant time to perform the operations
Auxiliary Space: O(1) // since no extra array or data structure is used so the space taken by the algorithm is constant
Another Approach:
The formula indicates that the n-th pentagonal number depends quadratically on n. Therefore, try to find the positive integral root of N = P(n) equation.
P(n) = nth pentagonal number
N = Given Number
Solve for n:
P(n) = N
or (3*n*n – n)/2 = N
or 3*n*n – n – 2*N = 0 … (i)
The positive root of equation (i)
n = (1 + sqrt(24N+1))/6
After obtaining n, check if it is an integer or not. n is an integer if n – floor(n) is 0.
C++
// C++ Program to check a // pentagonal number #include <bits/stdc++.h> using namespace std; // Function to determine if // N is pentagonal or not. bool isPentagonal( int N) { // Get positive root of // equation P(n) = N. float n = (1 + sqrt (24*N + 1))/6; // Check if n is an integral // value of not. To get the // floor of n, type cast to int. return (n - ( int ) n) == 0; } // Driver Code int main() { int N = 145; if (isPentagonal(N)) cout << N << " is pentagonal " << endl; else cout << N << " is not pentagonal" << endl; return 0; } |
Java
// Java Program to check a // pentagonal number import java.util.*; public class Main { // Function to determine if // N is pentagonal or not. public static boolean isPentagonal( int N) { // Get positive root of // equation P(n) = N. float n = ( 1 + ( float )Math.sqrt( 24 * N + 1 )) / 6 ; // Check if n is an integral // value of not. To get the // floor of n, type cast to int. return (n - ( int )n) == 0 ; } // Driver Code public static void main(String[] args) { int N = 145 ; if (isPentagonal(N)) System.out.println(N + " is pentagonal " ); else System.out.println(N + " is not pentagonal" ); } } |
Python3
import math # Function to determine if N is pentagonal or not. def isPentagonal(N): # Get positive root of equation P(n) = N n = ( 1 + math.sqrt( 24 * N + 1 )) / 6 # Check if n is an integral value or not. # To get the floor of n, use the int() function return (n - int (n)) = = 0 # Driver code if __name__ = = "__main__" : N = 145 if isPentagonal(N): print (N, "is pentagonal" ) else : print (N, "is not pentagonal" ) |
C#
using System; public class Program { // Function to determine if // N is pentagonal or not. public static bool IsPentagonal( int n) { // Get positive root of equation P(n) = N. float x = (1 + MathF.Sqrt(24 * n + 1)) / 6; // Check if x is an integral value or not return MathF.Floor(x) == x; } public static void Main() { int n = 145; if (IsPentagonal(n)) Console.WriteLine(n + " is pentagonal" ); else Console.WriteLine(n + " is not pentagonal" ); // Pause the console so that we can see the output Console.ReadLine(); } } // This code is contributed by divyansh2212 |
Javascript
// js equivalent // import math functions function isPentagonal(N) { // Get positive root of equation P(n) = N let n = (1 + Math.sqrt(24 * N + 1)) / 6; // Check if n is an integral value or not. // To get the floor of n, use the Math.floor() function return (n - Math.floor(n)) === 0; } // Driver code let N = 145; if (isPentagonal(N)) { console.log(`${N} is pentagonal`); } else { console.log(`${N} is not pentagonal`); } |
145 is pentagonal
Time Complexity: O(log n) //the inbuilt sqrt function takes logarithmic time to execute
Auxiliary Space: O(1) // since no extra array or data structure is used so the space taken by the algorithm is constant
Reference:
https://en.wikipedia.org/wiki/Polygonal_number