Find a sequence of distinct integers whose Harmonic Mean is N

Given an integer N, the task is to find a sequence of distinct integers whose Harmonic Mean is N itself. In case such a sequence doesn’t exist, print -1.

Examples:

Input: N = 5
Output: {3, 4, 5, 6, 20}
Explanation: Given sequence is the correct answer because 5/(1/3​+1/4+1/5​+1/6​+1/20)​=5. Note that there are other possible sequences as well. For example – {2, 6, 12, 20, 5}.

Input: N=3
Output: {2, 3, 6}

Approach: To solve the problem follow the below observations:

Observations:

• For N = 1, the sequence {1} satisfies the conditions.
• For N = 2, there is no possible sequence satisfying the given conditions. So, we’ll consider the case for N > 2.
• We know that 1/X(X+1) = 1/X – 1/(X+1). Using this fact, we can see that the sequence {1*2, 2*3, …, (N-1)*N, N} would satisfy the required conditions. (Reason: {1*2, 2*3, …, (N-1)*N, N} = {2, 6, 12, 20…..(N-1)*N, N}.
• The harmonic mean of this sequence:
  • N/(1/(1*2) + 1/(2*3) + 1/(3*4) +…+ 1/(N-1)*N) + 1/N)
  • N/(1-1/2 + 1/2-1/3 + 1/3-1/4) +…+ 1/(N-1)-1/N + 1/N)
  • N/1 = N
• However, for the cases when N can be represented as N=X*(X-1), for some integer X, it can be seen that the above observation fails as the sequence would contain a duplicate.

For example, for N=6 (6=2*3), the sequence would be {2, 6, 12, 20, 30, 6}. Here 6 is repeated, but we want all distinct integers in the sequence.

• Suppose the sequence from observation 1 is {A1, A2, A3….AN}. To avoid repeating numbers when N is even (as N = X*(X-1) will be possible only when N is even), we’ll print the sequence {2, 2*A1, 2*A2,….,2*A(N-2), 2*(N-1)}.

The following steps can be used to solve the problem :

• For N = 1, print 1.
• For N = 2, print -1.
• For odd N, print the sequence {1*2, 2*3, …, (N-1)*N, N}.
• For even N, print the sequence {2, 2*(1*2, 2*3, …, (N-2)(N-1), N-1}.

Following is the code based on the above approach:

C++

// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find a sequence of
// distinct integers whose
// Harmonic Mean is N
void printSequence(int N)
{

    // Sequence is not possible for N=2
    if (N == 2) {
        cout << -1;
        return;
    }

    // For N=1, the sequence is {1} itself
    if (N == 1) {
        cout << 1;
        return;
    }

    // If N is odd
    if (N % 2) {

        // Sequence would be
        // 1*2, 2*3,...,(N-1)*N, N
        for (int i = 1; i <= N - 1; i++) {
            cout << i * (i + 1) << " ";
        }
        cout << N << "\n";
    }

    // If N is even
    else {

        // Sequence would be
        // 2, 2*(1*2, 2*3,...,(N-2)(N-1), N-1)
        cout << 2 << " ";
        for (int i = 1; i <= N - 2; i++) {
            cout << 2 * i * (i + 1) << " ";
        }
        cout << 2 * (N - 1) << "\n";
    }
}

// Driver Code
int main()
{
    int N = 5;

    // Function Call
    printSequence(N);
}

Java

// Java Implementation
import java.util.*;

public class GFG {
    // Function to find a sequence of
    // distinct integers whose
    // Harmonic Mean is N
    static void printSequence(int N) {

        // Sequence is not possible for N=2
        if (N == 2) {
            System.out.println(-1);
            return;
        }

        // For N=1, the sequence is {1} itself
        if (N == 1) {
            System.out.println(1);
            return;
        }

        // If N is odd
        if (N % 2 != 0) {

            // Sequence would be
            // 1*2, 2*3,...,(N-1)*N, N
            for (int i = 1; i <= N - 1; i++) {
                System.out.print(i * (i + 1) + " ");
            }
            System.out.println(N);
        }

        // If N is even
        else {

            // Sequence would be
            // 2, 2*(1*2, 2*3,...,(N-2)(N-1), N-1)
            System.out.print(2 + " ");
            for (int i = 1; i <= N - 2; i++) {
                System.out.print(2 * i * (i + 1) + " ");
            }
            System.out.println(2 * (N - 1));
        }
    }

    // Driver Code
    public static void main(String[] args) {
        int N = 5;

        // Function Call
        printSequence(N);
    }
}

//this code is contributed by uttamdp_10

Python3

#Python Implementation

# Function to find a sequence of
# distinct integers whose
# Harmonic Mean is N
def print_sequence(N):
    # Sequence is not possible for N=2
    if N == 2:
        print(-1)
        return

    # For N=1, the sequence is {1} itself
    if N == 1:
        print(1)
        return

    # If N is odd
    if N % 2 != 0:
        # Sequence would be
        # 1*2, 2*3,...,(N-1)*N, N
        for i in range(1, N):
            print(i * (i + 1), end=" ")
        print(N)

    # If N is even
    else:
        # Sequence would be
        # 2, 2*(1*2, 2*3,...,(N-2)(N-1), N-1)
        print(2, end=" ")
        for i in range(1, N - 1):
            print(2 * i * (i + 1), end=" ")
        print(2 * (N - 1))

# Driver Code
if __name__ == "__main__":
    N = 5

    # Function Call
    print_sequence(N)

 # this code is contributed by uttamdp_10

C#

// C# code for the above approach
using System;

public class GFG {
    // Function to find a sequence of
    // distinct integers whose
    // Harmonic Mean is N
    static void PrintSequence(int N)
    {
        // Sequence is not possible for N=2
        if (N == 2) {
            Console.WriteLine(-1);
            return;
        }

        // For N=1, the sequence is {1} itself
        if (N == 1) {
            Console.WriteLine(1);
            return;
        }

        // If N is odd
        if (N % 2 != 0) {
            // Sequence would be
            // 1*2, 2*3,...,(N-1)*N, N
            for (int i = 1; i <= N - 1; i++) {
                Console.Write(i * (i + 1) + " ");
            }
            Console.WriteLine(N);
        }

        // If N is even
        else {
            // Sequence would be
            // 2, 2*(1*2, 2*3,...,(N-2)(N-1), N-1)
            Console.Write(2 + " ");
            for (int i = 1; i <= N - 2; i++) {
                Console.Write(2 * i * (i + 1) + " ");
            }
            Console.WriteLine(2 * (N - 1));
        }
    }

    // Driver Code
    static public void Main()
    {
        int N = 5;

        // Function Call
        PrintSequence(N);
    }
}

Javascript

// JavaScript code for the above approach

// Function to find a sequence of
// distinct integers whose
// Harmonic Mean is N
function printSequence(N) {

    // Sequence is not possible for N=2
    if (N === 2) {
        console.log(-1);
        return;
    }

    // For N=1, the sequence is [1] itself
    if (N === 1) {
        console.log(1);
        return;
    }

    // If N is odd
    if (N % 2) {

        // Sequence would be
        // 1*2, 2*3,...,(N-1)*N, N
        for (let i = 1; i <= N - 1; i++) {
            process.stdout.write(i * (i + 1) + " ");
        }
        console.log(N);
    }

    // If N is even
    else {

        // Sequence would be
        // 2, 2*(1*2, 2*3,...,(N-2)(N-1), N-1)
        process.stdout.write(2 + " ");
        for (let i = 1; i <= N - 2; i++) {
            process.stdout.write(2 * i * (i + 1) + " ");
        }
        console.log(2 * (N - 1));
    }
}

// Driver Code
const N = 5;

// Function Call
printSequence(N);

// This code is contributed by prasad264

Output

2 6 12 20 5

Time Complexity: O(N)
Auxiliary Space: O(1)