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Balanced Prime
• Difficulty Level : Medium
• Last Updated : 20 Dec, 2018

In number theory, a Balanced Prime is a prime number with equal-sized prime gaps above and below it, so that it is equal to the arithmetic mean of the nearest primes above and below. Or to put it algebraically, given a prime number pn, where n is its index in the ordered set of prime numbers, First few balanced prime are 5, 53, 157, 173……

Given a positive integer N. The task is to print Nth balanced prime number.
Examples:

Input : n = 2
Output : 53

Input : n = 3
Output : 157


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is to generate prime numbers using Sieve of Eratosthenes and store it in an array. Now iterate over the array to check whether it is balanced prime or not and keep counting the balanced prime. Once you reach the nth prime, return it.

Below is the implementation of this approach:

## C++

 // CPP Program to find Nth Balanced Prime#include#define MAX 501using namespace std;  // Return the Nth balanced prime.int balancedprime(int n){    // Sieve of Eratosthenes    bool prime[MAX+1];    memset(prime, true, sizeof(prime));      for (int p = 2; p*p <= MAX; p++)    {        // If prime[p] is not changed, then it is a prime        if (prime[p] == true)        {            // Update all multiples of p            for (int i = p*2; i <= MAX; i += p)                prime[i] = false;        }    }      // storing all primes    vector<int> v;    for (int p = 3; p <= MAX; p += 2)       if (prime[p])          v.push_back(p);                int count = 0;          // Finding the Nth balanced Prime    for (int i = 1; i < v.size(); i++)    {        if (v[i] == (v[i+1] + v[i - 1])/2)          count++;                    if (count == n)          return v[i];    }}  // Driven Programint main(){  int n = 4;    cout << balancedprime(n) << endl;  return 0;}

## Java

 // Java Program to find Nth Balanced Primeimport java.util.*;  public class GFG{    static int MAX = 501;      // Return the Nth balanced prime.    public static int balancedprime(int n)    {                  // Sieve of Eratosthenes        boolean[] prime = new boolean[MAX+1];                  for(int k = 0 ; k < MAX+1; k++)            prime[k] = true;          for (int p = 2; p*p <= MAX; p++)        {                          // If prime[p] is not changed,            // then it is a prime            if (prime[p] == true)            {                                  // Update all multiples of p                for (int i = p*2; i <= MAX;                                     i += p)                    prime[i] = false;            }        }          // storing all primes        Vector v =                      new Vector();        for (int p = 3; p <= MAX; p += 2)            if (prime[p])                v.add(p);          int count = 0;          // Finding the Nth balanced Prime        for (int i = 1; i < v.size(); i++)        {            if ((int)v.get(i) == ((int)v.get(i+1)                           + (int)v.get(i-1))/2)                count++;                          if (count == n)                return (int) v.get(i);        }          return 1;    }          // Driven Program    public static void main(String[] args)    {        int n = 4;        System.out.print(balancedprime(n));    }}  // This code is contributed by Prasad Kshirsagar

## Python3

 # Python3 code to find Nth Balanced Prime  MAX = 501  # Return the Nth balanced prime.def balancedprime( n ):      # Sieve of Eratosthenes    prime = [True]*(MAX+1)    p=2    while p * p <= MAX:          # If prime[p] is not changed,         # then it is a prime        if prime[p] == True:                      # Update all multiples of p            i = p * 2            while i <= MAX:                prime[i] = False                i = i + p        p = p +1              # storing all primes    v = list()    p = 3    while p <= MAX:        if prime[p]:            v.append(p)        p = p + 2              count = 0          # Finding the Nth balanced Prime    i=1    for i in range(len(v)):        if v[i] == (v[i+1] + v[i - 1])/2:            count += 1                      if count == n:            return v[i]   # Driven Programn = 4print(balancedprime(n))   # This code is contributed by "Sharad_Bhardwaj".

## PHP

 

## C#

 // C# Program to find Nth Balanced Primeusing System;using System.Collections.Generic;public class GFG{    static int MAX = 501;      // Return the Nth balanced prime.    public static int balancedprime(int n)    {                  // Sieve of Eratosthenes        bool[] prime = new bool[MAX+1];                  for(int k = 0 ; k < MAX+1; k++)            prime[k] = true;          for (int p = 2; p*p <= MAX; p++)        {                          // If prime[p] is not changed,            // then it is a prime            if (prime[p] == true)            {                                  // Update all multiples of p                for (int i = p*2; i <= MAX;                                    i += p)                    prime[i] = false;            }        }          // storing all primes        List<int> v = new List<int>();        for (int p = 3; p <= MAX; p += 2)            if (prime[p])                v.Add(p);          int c = 0;        // Finding the Nth balanced Prime        for (int i = 1; i < v.Count-1; i++)        {            if ((int)v[i]==(int)(v[i+1]+v[i-1])/2)            c++;            if (c == n)                return (int) v[i];        }          return 1;    }          // Driven Program    public static void Main()    {        int n = 4;        Console.WriteLine(balancedprime(n));    }}  // This code is contributed by mits

Output:
173


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