# Ramanujan Prime

The Nth Ramanujan prime is the least integer Rn for which

where π(x) is a prime-counting function

Note that the integer Rn is necessarily a prime number: π(x) – π(x/2) and, hence, π(x) must increase by obtaining another prime at x = Rn. Since π(x) – π(x/2) can increase by at most 1,

Range of Rn is (2n log(2n), 4n log(4n)).

Ramanujan primes:

2, 11, 17, 29, 41, 47, 59, 67, 71, 97

For a given N, the task is to print first N Ramanujan primes

Examples:

Input : N = 5
Output : 2, 11, 17, 29, 41

Input : N = 10
Output : 2, 11, 17, 29, 41, 47, 59, 67, 71, 97

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

Approach:
Let us divide our solution into parts,
First, we will use sieve of Eratosthenes to get all the primes less than 10^6

Now we will have to find the value of π(x), π(x) is the count of primes which are less than or equal to x. Primes are stored in increasing order. So we can perform a binary search to find all the primes less than x, which can be done in O(log n).

Now we have the range Rn lies between : 2n log(2n) < Rn < 4n log(4n) So, we will take the upper bound and iterate from upper bound to the lower bound until π(i) – π(i/2) < n, i+1 is the nth Ramanujan prime.

Below is the implementation of the above approach :

## C++

 // CPP program to find Ramanujan numbers  #include  using namespace std;  #define MAX 1000000     // FUnction to return a vector of primes  vector<int> addPrimes()  {      int n = MAX;         // Create a boolean array "prime[0..n]" and initialize      // all entries it as true. A value in prime[i] will      // finally be false if i is Not a prime, else true.      bool prime[n + 1];      memset(prime, true, sizeof(prime));         for (int p = 2; p * p <= n; p++)       {          // If prime[p] is not changed, then it is a prime          if (prime[p] == true)           {              // Update all multiples of p greater than or              // equal to the square of it              // numbers which are multiple of p and are              // less than p^2 are already been marked.              for (int i = p * p; i <= n; i += p)                  prime[i] = false;          }      }         vector<int> ans;      // Print all prime numbers      for (int p = 2; p <= n; p++)          if (prime[p])              ans.push_back(p);         return ans;  }  // Function to find number of primes   // less than or equal to x  int pi(int x, vector<int> v)  {      int l = 0, r = v.size() - 1, m, in = -1;         // Binary search to find out number of      // primes less than or equal to x      while (l <= r) {          m = (l + r) / 2;          if (v[m] <= x) {              in = m;              l = m + 1;          }          else {              r = m - 1;          }      }      return in + 1;  }     // Function to find the nth ramanujan prime  int Ramanujan(int n, vector<int> v)  {      // For n>=1, a(n)<4*n*log(4n)      int upperbound = 4 * n * (log(4 * n) / log(2));         // We start from upperbound and find where      // pi(i)-pi(i/2) v = addPrimes();         for (int i = 1; i <= n; i++) {          cout << Ramanujan(i, v);          if(i!=n)              cout << ", ";      }  }     // Driver code  int main()  {      int n = 10;         Ramanujan_Numbers(n);         return 0;  }

## Java

 // Java program to find Ramanujan numbers  import java.util.*;  class GFG   {         static int MAX = 1000000;     // FUnction to return a vector of primes  static Vector addPrimes()  {      int n = MAX;         // Create a boolean array "prime[0..n]" and       // initialize all entries it as true.       // A value in prime[i] will finally be false       // if i is Not a prime, else true.      boolean []prime= new boolean[n + 1];      Arrays.fill(prime, true);         for (int p = 2; p * p <= n; p++)       {          // If prime[p] is not changed,          // then it is a prime          if (prime[p] == true)           {              // Update all multiples of p greater than or              // equal to the square of it              // numbers which are multiple of p and are              // less than p^2 are already been marked.              for (int i = p * p; i <= n; i += p)                  prime[i] = false;          }      }         Vector ans = new Vector();             // Print all prime numbers      for (int p = 2; p <= n; p++)          if (prime[p])              ans.add(p);         return ans;  }     // Function to find number of primes   // less than or equal to x  static int pi(int x, Vector v)  {      int l = 0, r = v.size() - 1, m, in = -1;         // Binary search to find out number of      // primes less than or equal to x      while (l <= r)       {          m = (l + r) / 2;          if (v.get(m) <= x)          {              in = m;              l = m + 1;          }          else          {              r = m - 1;          }      }      return in + 1;  }     // Function to find the nth ramanujan prime  static int Ramanujan(int n, Vector v)  {      // For n>=1, a(n)<4*n*log(4n)      int upperbound = (int) (4 * n * (Math.log(4 * n) /                                        Math.log(2)));         // We start from upperbound and find where      // pi(i)-pi(i/2) v = addPrimes();         for (int i = 1; i <= n; i++)       {          System.out.print(Ramanujan(i, v));          if(i != n)              System.out.print(", ");      }  }     // Driver code  public static void main(String[] args)   {      int n = 10;         Ramanujan_Numbers(n);  }  }     // This code is contributed by 29AjayKumar

## Python3

 # Python3 program to find Ramanujan numbers  from math import log, ceil  MAX = 1000000    # FUnction to return a vector of primes  def addPrimes():         n = MAX        # Create a boolean array "prime[0..n]"       # and initialize all entries it as true.       # A value in prime[i] will finally be      # false if i is Not a prime, else true.      prime = [True for i in range(n + 1)]         for p in range(2, n + 1):          if p * p > n:              break                        # If prime[p] is not changed,           # then it is a prime          if (prime[p] == True):                 # Update all multiples of p               # greater than or equal to the               # square of it. numbers which are               # multiple of p and are less than p^2              # are already been marked.              for i in range(2 * p, n + 1, p):                  prime[i] = False        ans = []             # Print all prime numbers      for p in range(2, n + 1):          if (prime[p]):              ans.append(p)         return ans         # Function to find number of primes  # less than or equal to x  def pi(x, v):         l, r = 0, len(v) - 1        # Binary search to find out number of      # primes less than or equal to x      m, i = 0, -1     while (l <= r):          m = (l + r) // 2         if (v[m] <= x):              i = m              l = m + 1         else:              r = m - 1        return i + 1    # Function to find the nth ramanujan prime  def Ramanujan(n, v):         # For n>=1, a(n)<4*n*log(4n)      upperbound = ceil(4 * n * (log(4 * n) / log(2)))         # We start from upperbound and find where      # pi(i)-pi(i/2)

## C#

 // C# program to find Ramanujan numbers  using System;  using System.Collections.Generic;      class GFG   {  static int MAX = 1000000;     // FUnction to return a vector of primes  static List<int> addPrimes()  {      int n = MAX;         // Create a boolean array "prime[0..n]" and       // initialize all entries it as true.       // A value in prime[i] will finally be false       // if i is Not a prime, else true.      Boolean []prime = new Boolean[n + 1];      for(int i = 0; i < n + 1; i++)          prime[i] = true;         for (int p = 2; p * p <= n; p++)       {          // If prime[p] is not changed,          // then it is a prime          if (prime[p] == true)           {              // Update all multiples of p greater than               // or equal to the square of it              // numbers which are multiple of p and are              // less than p^2 are already been marked.              for (int i = p * p; i <= n; i += p)                  prime[i] = false;          }      }         List<int> ans = new List<int>();             // Print all prime numbers      for (int p = 2; p <= n; p++)          if (prime[p])              ans.Add(p);         return ans;  }     // Function to find number of primes   // less than or equal to x  static int pi(int x, List<int> v)  {      int l = 0, r = v.Count - 1, m, i = -1;         // Binary search to find out number of      // primes less than or equal to x      while (l <= r)       {          m = (l + r) / 2;          if (v[m] <= x)          {              i = m;              l = m + 1;          }          else         {              r = m - 1;          }      }      return i + 1;  }     // Function to find the nth ramanujan prime  static int Ramanujan(int n, List<int> v)  {      // For n>=1, a(n)<4*n*log(4n)      int upperbound = (int) (4 * n * (Math.Log(4 * n) /                                        Math.Log(2)));         // We start from upperbound and find where      // pi(i)-pi(i/2) v = addPrimes();         for (int i = 1; i <= n; i++)       {          Console.Write(Ramanujan(i, v));          if(i != n)              Console.Write(", ");      }  }     // Driver code  public static void Main(String[] args)   {      int n = 10;         Ramanujan_Numbers(n);  }  }     // This code is contributed by 29AjayKumar 

Output:

2, 11, 17, 29, 41, 47, 59, 67, 71, 97
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