Count of ways to represent N as sum of a prime number and twice of a square
Last Updated :
15 Jul, 2025
Given an integer N, the task is to count the number of ways so that N can be written as the sum of a prime number and twice of a square, i.e.
N = 2*A^{2} + P
, where P can be any prime number and A is any positive integer.
Note:
N <= 10^{6}
Examples:
Input: N = 9
Output: 1
Explanation:
9 can be represented as sum of prime number and twice a square in only one way -
N = 9 = 7 + 2*(1^{2})
Input: N = 15
Output: 2
Explanation:
15 can be represented as sum of prime number and twice a square in two ways -
N = 15 = 7 + 2 * (2^{2}) N = 15 = 13 + 2 * (1^{2})
Approach: The idea is to use Sieve of Eratosthenes to find all the primes and then for each prime number check for every possible number starting from 1. If any prime number and twice a square is equal to the given number then increment the count of the number of ways by 1.
Below is the implementation of the above approach:
C++
// C++ implementation to count the
// number of ways a number can be
// written as sum of prime number
// and twice a square
#include <bits/stdc++.h>
using namespace std;
long long int n = 500000 - 2;
vector<long long int> v;
// Function to mark all the
// prime numbers using sieve
void sieveoferanthones()
{
bool prime[n + 1];
// Initially all the numbers
// are marked as prime
memset(prime, true, sizeof(prime));
// Loop to mark the prime numbers
// upto the Square root of N
for (long long int i = 2; i <= sqrt(n); i++) {
if (prime[i])
for (long long int j = i * i; j <= n; j += i) {
prime[j] = false;
}
}
// Loop to store the prime
// numbers in an array
for (long long int i = 2; i < n; i++) {
if (prime[i])
v.push_back(i);
}
}
// Function to find the number
// ways to represent a number
// as the sum of prime number and
// square of a number
void numberOfWays(long long int n)
{
long long int count = 0;
// Loop to iterate over all the
// possible prime numbers
for (long long int j = 1; 2 * (pow(j, 2)) < n; j++) {
for (long long int i = 1; v[i] + 2 <= n; i++) {
// Increment the count if
// the given number is a
// valid number
if (n == v[i] + (2 * (pow(j, 2))))
count++;
}
}
cout << count << endl;
}
// Driver Code
int main()
{
sieveoferanthones();
long long int n = 9;
// Function Call
numberOfWays(n);
return 0;
}
// Java implementation to count the
// number of ways a number can be
// written as sum of prime number
// and twice a square
import java.util.*;
class GFG {
static int n = 500000 - 2;
static Vector<Integer> v = new Vector<>();
// Function to mark all the
// prime numbers using sieve
static void sieveoferanthones()
{
boolean[] prime = new boolean[n + 1];
// Initially all the numbers
// are marked as prime
Arrays.fill(prime, true);
// Loop to mark the prime numbers
// upto the Square root of N
for (int i = 2; i <= Math.sqrt(n); i++) {
if (prime[i])
for (int j = i * i; j <= n; j += i) {
prime[j] = false;
}
}
// Loop to store the prime
// numbers in an array
for (int i = 2; i < n; i++) {
if (prime[i])
v.add(i);
}
}
// Function to find the number
// ways to represent a number
// as the sum of prime number and
// square of a number
static void numberOfWays(int n)
{
int count = 0;
// Loop to iterate over all the
// possible prime numbers
for (int j = 1; 2 * (Math.pow(j, 2)) < n; j++) {
for (int i = 1; v.get(i) + 2 <= n; i++) {
// Increment the count if
// the given number is a
// valid number
if (n == v.get(i) + (2 * (Math.pow(j, 2))))
count++;
}
}
System.out.print(count + "\n");
}
// Driver Code
public static void main(String[] args)
{
sieveoferanthones();
int n = 9;
// Function Call
numberOfWays(n);
}
}
// This code is contributed by Princi Singh
# Python3 implementation to count the
# number of ways a number can be
# written as sum of prime number
# and twice a square
import math
n = 500000 - 2
v = []
# Function to mark all the
# prime numbers using sieve
def sieveoferanthones():
prime = [1] * (n + 1)
# Loop to mark the prime numbers
# upto the Square root of N
for i in range(2, int(math.sqrt(n)) + 1):
if (prime[i] != 0):
for j in range(i * i, n + 1, i):
prime[j] = False
# Loop to store the prime
# numbers in an array
for i in range(2, n):
if (prime[i] != 0):
v.append(i)
# Function to find the number
# ways to represent a number
# as the sum of prime number and
# square of a number
def numberOfWays(n):
count = 0
# Loop to iterate over all the
# possible prime numbers
j = 1
while (2 * (pow(j, 2)) < n):
i = 1
while (v[i] + 2 <= n):
# Increment the count if
# the given number is a
# valid number
if (n == v[i] +
(2 * (math.pow(j, 2)))):
count += 1
i += 1
j += 1
print(count)
# Driver Code
sieveoferanthones()
n = 9
# Function call
numberOfWays(n)
# This code is contributed by sanjoy_62
// C# implementation to count the
// number of ways a number can be
// written as sum of prime number
// and twice a square
using System;
using System.Collections;
using System.Collections.Generic;
class GFG {
static int n = 500000 - 2;
static ArrayList v = new ArrayList();
// Function to mark all the
// prime numbers using sieve
static void sieveoferanthones()
{
bool[] prime = new bool[n + 1];
// Initially all the numbers
// are marked as prime
Array.Fill(prime, true);
// Loop to mark the prime numbers
// upto the Square root of N
for (int i = 2; i <= (int)Math.Sqrt(n); i++) {
if (prime[i]) {
for (int j = i * i; j <= n; j += i) {
prime[j] = false;
}
}
}
// Loop to store the prime
// numbers in an array
for (int i = 2; i < n; i++) {
if (prime[i])
v.Add(i);
}
}
// Function to find the number
// ways to represent a number
// as the sum of prime number and
// square of a number
static void numberOfWays(int n)
{
int count = 0;
// Loop to iterate over all the
// possible prime numbers
for (int j = 1; 2 * (Math.Pow(j, 2)) < n; j++) {
for (int i = 1; (int)v[i] + 2 <= n; i++) {
// Increment the count if
// the given number is a
// valid number
if (n == (int)v[i] + (2 * (Math.Pow(j, 2))))
count++;
}
}
Console.Write(count);
}
// Driver Code
public static void Main(string[] args)
{
sieveoferanthones();
int n = 9;
// Function call
numberOfWays(n);
}
}
// This code is contributed by rutvik_56
<script>
// JavaScript implementation to count the
// number of ways a number can be
// written as sum of prime number
// and twice a square
let n = 500000 - 2;
let v = [];
// Function to mark all the
// prime numbers using sieve
function sieveoferanthones()
{
let prime = Array.from({length: n+1},
(_, i) => true);
// Loop to mark the prime numbers
// upto the Square root of N
for (let i = 2;
i <= Math.sqrt(n); i++)
{
if (prime[i])
for (let j = i * i;
j <= n; j += i)
{
prime[j] = false;
}
}
// Loop to store the prime
// numbers in an array
for (let i = 2; i < n; i++)
{
if (prime[i])
v.push(i);
}
}
// Function to find the number
// ways to represent a number
// as the sum of prime number and
// square of a number
function numberOfWays(n)
{
let count = 0;
// Loop to iterate over all the
// possible prime numbers
for (let j = 1; 2 *
(Math.pow(j, 2)) < n; j++)
{
for (let i = 1; v[i] +
2 <= n; i++)
{
// Increment the count if
// the given number is a
// valid number
if (n == v[i] +
(2 * (Math.pow(j, 2))))
count++;
}
}
document.write(count + "<br/>");
}
// Driver Code
sieveoferanthones();
let N = 9;
// Function Call
numberOfWays(N);
</script>
Time Complexity: O(n2)
Auxiliary Space: O(n)
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