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Pentagonal Pyramidal Number

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Given a number n, find the nth pentagonal pyramidal number.
A Pentagonal Pyramidal Number belongs to the figurate number class. It is the number of objects in a pyramid with a pentagonal base. The nth pentagonal pyramidal number is equal to sum of first n pentagonal numbers
Examples: 
 

Input : n = 3 
Output : 18

Input : n = 7
Output : 196

 

 

Method 1: (Naive Approach) : 
This approach is simple. It says to add all the pentagonal numbers up to n (by running loop) to get nth Pentagonal pyramidal number. 
 
Below is the implementation of this approach:

C++




// CPP Program to get nth Pentagonal
// pyramidal number.
#include <bits/stdc++.h>
using namespace std;
 
// function to get nth Pentagonal
// pyramidal number.
int pentagon_pyramidal(int n)
{
    int sum = 0;
 
    // Running loop from 1 to n
    for (int i = 1; i <= n; i++) {
 
        // get nth pentagonal number
        int p = (3 * i * i - i) / 2;
 
        // add to sum
        sum = sum + p;
    }
    return sum;
}
 
// Driver Program
int main()
{
    int n = 4;
    cout << pentagon_pyramidal(n) << endl;
    return 0;
}


Java




// Java Program to get nth
// Pentagonal pyramidal number.
import java.io.*;
 
class GFG
{
 
// function to get nth
// Pentagonal pyramidal number.
static int pentagon_pyramidal(int n)
{
    int sum = 0;
 
    // Running loop from 1 to n
    for (int i = 1; i <= n; i++)
    {
 
        // get nth pentagonal number
        int p = (3 * i * i - i) / 2;
 
        // add to sum
        sum = sum + p;
    }
    return sum;
}
 
// Driver Code
public static void main (String[] args)
{
    int n = 4;
    System.out.println(pentagon_pyramidal(n));
}
}
 
// This code is contributed by anuj_67.


Python3




# Python3 Program to get nth Pentagonal
# pyramidal number.
   
# function to get nth Pentagonal
# pyramidal number.
def pentagon_pyramidal(n):
    sum = 0
 
    # Running loop from 1 to n
    for i in range(1, n + 1):
   
        # get nth pentagonal number
        p = ( 3 * i * i - i ) / 2
 
        # add to sum
        sum = sum + p      
  
    return sum
 
   
# Driver Program
n = 4
print(int(pentagon_pyramidal(n)))


C#




// C# Program to get nth
// Pentagonal pyramidal number.
using System;
 
class GFG
{
     
// function to get nth
// Pentagonal pyramidal number.
static int pentagon_pyramidal(int n)
{
    int sum = 0;
 
    // Running loop from 1 to n
    for (int i = 1; i <= n; i++)
    {
 
        // get nth pentagonal number
        int p = (3 * i *
                 i - i) / 2;
 
        // add to sum
        sum = sum + p;
    }
    return sum;
}
 
// Driver Code
static public void Main ()
{
    int n = 4;
    Console.WriteLine(pentagon_pyramidal(n));
}
}
 
// This code is contributed by ajit.


PHP




<?php
// PHP Program to get nth
// Pentagonal pyramidal number.
 
// function to get nth
// Pentagonal pyramidal number.
function pentagon_pyramidal($n)
{
    $sum = 0;
 
    // Running loop from 1 to n
    for ($i = 1; $i <= $n; $i++)
    {
 
        // get nth pentagonal number
        $p = (3 * $i *
                  $i - $i) / 2;
 
        // add to sum
        $sum = $sum + $p;
    }
    return $sum;
}
 
// Driver Code
$n = 4;
echo pentagon_pyramidal($n);
 
// This code is contributed by m_kit
?>


Javascript




<script>
// javascript Program to get nth
// Pentagonal pyramidal number.
 
// function to get nth
// Pentagonal pyramidal number.
function pentagon_pyramidal(n)
{
    var sum = 0;
 
    // Running loop from 1 to n
    for (i = 1; i <= n; i++)
    {
 
        // get nth pentagonal number
        var p = (3 * i * i - i) / 2;
 
        // add to sum
        sum = sum + p;
    }
    return sum;
}
 
// Driver Code
var n = 4;
document.write(pentagon_pyramidal(n));
 
// This code is contributed by Amit Katiyar
</script>


Output : 
 

40

Time Complexity: O(n) 
Auxiliary space: O(1)
 
Method 2: (Efficient Approach) : 
In this approach, we use formula to get nth Pentagonal pyramidal number in O(1) time. 
nth Pentagonal pyramidal number = n2 (n + 1) / 2 

Below is the implementation of this approach: 

C++




// CPP Program to get nth Pentagonal
// pyramidal number.
#include <bits/stdc++.h>
using namespace std;
 
// function to get nth Pentagonal
// pyramidal number.
int pentagon_pyramidal(int n)
{
    return n * n * (n + 1) / 2;
}
 
// Driver Program
int main()
{
    int n = 4;
    cout << pentagon_pyramidal(n) << endl;
    return 0;
}


Java




// Java Program to get nth
// Pentagonal pyramidal number.
import java.io.*;
 
class GFG
{
     
// function to get nth
// Pentagonal pyramidal number.
static int pentagon_pyramidal(int n)
{
    return n * n *
          (n + 1) / 2;
}
 
// Driver Code
public static void main (String[] args)
{
    int n = 4;
    System.out.println(pentagon_pyramidal(n));
}
}
 
// This code is contributed by ajit


Python3




# Python3 Program to get nth Pentagonal
# pyramidal number.
   
# function to get nth Pentagonal
# pyramidal number.
def pentagon_pyramidal(n):    
    return n * n * (n + 1) / 2
 
   
# Driver Program
n = 4
print(int(pentagon_pyramidal(n)))


C#




// C# Program to get nth
// Pentagonal pyramidal number.
using System;
 
class GFG
{
     
// function to get nth
// Pentagonal pyramidal number.
static int pentagon_pyramidal(int n)
{
    return n * n *
          (n + 1) / 2;
}
 
// Driver Code
static public void Main ()
{
    int n = 4;
    Console.WriteLine(
            pentagon_pyramidal(n));
}
}
 
// This code is contributed
// by ajit


PHP




<?php
// PHP Program to get
// nth Pentagonal
// pyramidal number.
 
// function to get
// nth Pentagonal
// pyramidal number.
 
function pentagon_pyramidal($n)
{
    return $n * $n *
          ($n + 1) / 2;
}
 
// Driver Code
$n = 4;
echo pentagon_pyramidal($n);
     
// This code is contributed
// by akt_mit
?>


Javascript




<script>
// javascript Program to get nth
// Pentagonal pyramidal number.
 
     
// function to get nth
// Pentagonal pyramidal number.
function pentagon_pyramidal(n)
{
    return n * n *
          (n + 1) / 2;
}
 
// Driver Code
var n = 4;
document.write(pentagon_pyramidal(n));
 
// This code is contributed by Princi Singh
</script>


Output

40

Time Complexity: O(1) 
Space complexity: O(1)



Last Updated : 23 Sep, 2022
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