Consecutive Prime Sum Program in Java

The sum of consecutive prime numbers refers to the total obtained by adding up a series of prime numbers that follow each other in sequence.

To find consecutive prime numbers that sum up to a given value in Java, we can use a sliding window approach. Some prime numbers can be expressed as a sum of other consecutive prime numbers.

For example

5 = 2 + 3,
17 = 2 + 3 + 5 + 7,
41 = 2 + 3 + 5 + 7 + 11 + 13.
The task is to find out how many prime numbers that satisfy this property are present in the range 3 to N subject to a constraint that summation should always start with number 2.

Input: First line contains a number N.

Output: Print the total number of all such prime numbers which are less or equal to N.

The time complexity of the code is O(n² * b + k).

The actual time complexity may vary depending on the values of n and b.

ConsecutiveSum.java

import java.util.Scanner;
 class ConsecutiveSum {
    static int prime(int b) {
        int j,cnt;
        cnt=1;
        for (j = 2; j <= b/2; j++) {
            if(b%j==0)
                cnt=0;
        }
        if(cnt==0)
        return 1;
        else
            return 0;
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int i,j,n=0,cnt,c=0,sum=0,count=0,k=0;
        Main t = new Main();
        int[] a = new int[25];
        System.out.println("Enter no");
        n = sc.nextInt();
        for (i = 2; i <=n ; i++) {
            cnt=1;
            for (j = 2; j <= n/2; j++) {
                if(i%j==0)
                    cnt=0;
            }
            if(cnt==1) {
                a[k]=i;
                k++;
            }
        }
        for (i = 0; i < k; i++) {
            sum=sum+a[i];
            c=t.prime(sum);
            if(c==1)
                count++;
        }
        System.out.println(count);
    }
}

Output:

Enter no
10
4

Time complexity: O(sqrt(N))

Space complexity: O(1)

Let's see another approach for the same.

ConsecutivePrimeSum.java

public class ConsecutivePrimeSum {
    public static int consutiveNumberSum(int N) {
        int ans = 0, n = 1, s = 1;
        for (; s <= N; s += ++n) {
            if ((N - s) % n == 0) ++ans;
        }       
 return ans;
    }
    public static void main(String[] args) {
        int results = consutiveNumberSum(50);
        System.out.println(results);
    }
}

Output:

50
3

Another solution:

Assume N = (x+1)+(x+2)+...+(x+k), where x >= 0, k >= 1.

We have 2N = k(2x+k+1), which has two factors, k and 2x+k+1, one of which is odd number and another is even. So, the question is how many ways to factor 2N into one even and one odd number.

Assume 2N = 2^t * M, where M is odd. If we factor M = a * b, then multiply 2^t to one of them will yield the even number. So the question now becomes how many ways to factor the odd part of N.

If N = 3*3*3*5*5, how many ways to factor N into two numbers?

For one number, we can pick 0 to 3 3s, and 0 to 2 5s, so we have 4 * 3 = 12 options.

Here we can see the answer is multiplication of 1 + {count of a factor}.

But what if there is a factor greater than sqrt(N)? In this case, it has to be a prime number and there will be only one of it. So, if eventually N > 1, we multiply 2 to answer.

ConsecutivePrimeSum.java

public class ConsecutivePrimeSum {
    public static int consutiveNumberSum(int N) {
        int consecutiveNumbersSum(int N) {
        while ((N & 1) == 0) N >>= 1;
        int ans = 1, d = 3;
        while (d * d <= N) {
            int e = 0;
            for (; N % d == 0; N /= d, ++e);
            ans *= e + 1;
            d += 2;
        }
        if (N > 1) ans <<= 1;
        return ans;
    }
    public static void main(String[] args) {
        int results = consutiveNumberSum(50);
        System.out.println(results);
    }
}

Output:

Time complexity: O(sqrt(N)

Space complexity: O(1)

Next TopicCount distinct XOR values among pairs using numbers in range 1 to N

← prev next →