Ramanujan Prime

The Nth Ramanujan prime is the least integer R_n for which

where ?(x) is a prime-counting function
Note that the integer Rn is necessarily a prime number: ?(x) – ?(x/2) and, hence, ?(x) must increase by obtaining another prime at x = R_n. Since ?(x) – ?(x/2) can increase by at most 1, 
Range of R_n 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 

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 ?(x), ?(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 R_n lies between : 2n log(2n) < R_n < 4n log(4n) So, we will take the upper bound and iterate from upper bound to the lower bound until ?(i) – ?(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 <bits/stdc++.h>
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)<n the previous number being
    // the nth ramanujan prime
    for (int i = upperbound;; i--) {
        if (pi(i, v) - pi(i / 2, v) < n)
            return 1 + i;
    }
}

// Function to find first n Ramanujan numbers
void Ramanujan_Numbers(int n)
{
    int c = 1;

    // Get the prime numbers
    vector<int> 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<Integer> 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<Integer> ans = new Vector<Integer>();

    // 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<Integer> 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<Integer> 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)<n the previous number being
    // the nth ramanujan prime
    for (int i = upperbound;; i--)
    {
        if (pi(i, v) - pi(i / 2, v) < n)
            return 1 + i;
    }
}

// Function to find first n Ramanujan numbers
static void Ramanujan_Numbers(int n)
{
    int c = 1;

    // Get the prime numbers
    Vector<Integer> 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)<n the previous number being
    # the nth ramanujan prime
    for i in range(upperbound, -1, -1):
        if (pi(i, v) - pi(i / 2, v) < n):
            return 1 + i

# Function to find first n Ramanujan numbers
def Ramanujan_Numbers(n):
    c = 1

    # Get the prime numbers
    v = addPrimes()

    for i in range(1, n + 1):
        print(Ramanujan(i, v), end = "")
        if(i != n):
            print(end = ", ")

# Driver code
n = 10

Ramanujan_Numbers(n)

# This code is contributed
# by Mohit Kumar

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)<n the previous number being
    // the nth ramanujan prime
    for (int i = upperbound;; i--)
    {
        if (pi(i, v) - pi(i / 2, v) < n)
            return 1 + i;
    }
}

// Function to find first n Ramanujan numbers
static void Ramanujan_Numbers(int n)
{
    int c = 1;

    // Get the prime numbers
    List<int> 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

Javascript

<script>
// Javascript program to find Ramanujan numbers

const MAX = 1000000;

// FUnction to return a vector of primes
function addPrimes()
{
    let 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.
    let prime = new Array(n + 1).fill(true);

    for (let 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 (let i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }

    let ans = [];
    // Print all prime numbers
    for (let p = 2; p <= n; p++)
        if (prime[p])
            ans.push(p);

    return ans;
}
// Function to find number of primes
// less than or equal to x
function pi(x, v)
{
    let l = 0, r = v.length - 1, m, inn = -1;

    // Binary search to find out number of
    // primes less than or equal to x
    while (l <= r) {
        m = parseInt((l + r) / 2);
        if (v[m] <= x) {
            inn = m;
            l = m + 1;
        }
        else {
            r = m - 1;
        }
    }
    return inn + 1;
}

// Function to find the nth ramanujan prime
function Ramanujan(n, v)
{
    // For n>=1, a(n)<4*n*log(4n)
    let upperbound = 4 * n * parseInt((Math.log(4 * n) / Math.log(2)));

    // We start from upperbound and find where
    // pi(i)-pi(i/2)<n the previous number being
    // the nth ramanujan prime
    for (let i = upperbound;; i--) {
        if (pi(i, v) - pi(parseInt(i / 2), v) < n)
            return 1 + i;
    }
}

// Function to find first n Ramanujan numbers
function Ramanujan_Numbers(n)
{
    let c = 1;

    // Get the prime numbers
    let v = addPrimes();

    for (let i = 1; i <= n; i++) {
        document.write(Ramanujan(i, v));
        if(i!=n)
            document.write(", ");
    }
}

// Driver code
    let n = 10;

    Ramanujan_Numbers(n);

</script>

Output:

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

Time Complexity : O(nlogn)

Auxiliary Space: O(MAX)