Check if a number can be expressed as a sum of consecutive numbers

Given a number n, the task is to check whether it can be expressed as a sum of two or more consecutive numbers or not. 
Examples: 

Input  : n = 10 
Output : true
It can be expressed as sum of two consecutive
numbers 1 + 2 + 3 + 4.

Input  : n = 16  
Output : false
It cannot be expressed as sum of two consecutive
numbers.

Input  : n = 5  
Output : true
2 + 3 = 5

There is a direct and quick method to solve this. If a number is a power of two, then it cannot be expressed as a sum of consecutive numbers otherwise Yes.
The idea is based on below two facts. 
1) Sum of any two consecutive numbers is odd as one of them has to be even and the other odd. 
2) 2ⁿ = 2^n-1 + 2^n-1
If we take a closer look at 1) and 2), we can get the intuition behind the fact.
Below is the implementation of the above idea. 

C++

// C++ program to check if a number can
// be expressed as sum of consecutive numbers
#include<bits/stdc++.h>
using namespace std;

// This function returns true if n can be
// expressed sum of consecutive.
bool canBeSumofConsec(unsigned int n)
{
    // We basically return true if n is a
    // power of two
    return ((n&(n-1)) && n);
}

// Driver code
int main()
{
    unsigned int n = 15;
    canBeSumofConsec(n)? cout << "true" :
                         cout << "false";
    return 0;
}

Java

// Java program to check if a number can
// be expressed as sum of consecutive numbers

class Test
{
    // This function returns true if n can be
    // expressed sum of consecutive.
    static boolean canBeSumofConsec(int n)
    {
        // We basically return true if n is a
        // power of two
        return (((n&(n-1))!=0) && n!=0);
    }

    // Driver method
    public static void main(String[] args)
    {
        int n = 15;
        System.out.println(canBeSumofConsec(n) ? "true" : "false");
    }
}

Python3

# Python 3 program to check if a number can
# be expressed as sum of consecutive numbers


# This function returns true if n
# can be expressed sum of consecutive.
def canBeSumofConsec(n) :

    # We basically return true if n is a
    # power of two
    return ((n&(n-1)) and n)


# Driver code
n = 15
if(canBeSumofConsec(n)) :
    print("true")
else :
    print("false")

# This code is contributed by Nikita Tiwari.

C#

// C# program to check if a number can be
// expressed as sum of consecutive numbers
using System;

class Test
{
    // This function returns true if n
    // can be expressed sum of consecutive.
    static bool canBeSumofConsec(int n)
    {
        // We basically return true if n is a
        // power of two
        return (((n & (n - 1)) != 0) && n != 0);
    }

    // Driver Code
    public static void Main()
    {
        int n = 15;
        Console.Write(canBeSumofConsec(n) ? "True" : "False");
    }
}

// This code is contributed by Nitin Mittal.

PHP

<?php
// php program to check if a number
// can be expressed as sum of
// consecutive numbers

// This function returns true if n
// can be expressed sum of consecutive.
function canBeSumofConsec($n)
{

    // We basically return true if n is a
    // power of two
    return (($n & ($n - 1)) && $n);
}

// Driver code
    $n = 15;
    if(canBeSumofConsec($n))
        echo "true" ;
    else
        echo "false";

// This code is contributed by
// nitin mittal.
?>

Javascript

<script>

// Javascript program to check if a number can
// be expressed as sum of consecutive numbers


    // This function returns true if n can be
    // expressed sum of consecutive.
    function canBeSumofConsec(n)
    {
        // We basically return true if n is a
        // power of two
        return (((n&(n-1))!=0) && n!=0);
    }


// function call

    let n = 15;
    document.write(canBeSumofConsec(n) ? "true" : "false");

</script>

Output: 

True

Time Complexity: O(1)

Auxiliary Space: O(1)

Another Approach :

Let number chosen to represent N as a sum of consecutive numbers be X + 1, X + 2, X + 3 …. Y 

Sum of these chosen numbers = Sum of first Y natural numbers – Sum of first X natural number 

Sum of first Y natural number =
Sum of first X natural number =
We know that,
N = Sum of first Y natural number – Sum of first X natural number




Let Y – X = a, Y + X + 1 = b
Y + X + 1 > Y – X, b > a
,
2N = a * b
It means that a and b are factor of 2N, we know that X and Y are integers so,
1. b – a – 1 => multiple of 2 (Even number)
2. b + a + 1 => multiple of 2 (Even number)

Both conditions must be satisfied

From 1 and 2 we can say that either one of them (a, b) should be Odd and another one Even

So if the number (2N) has only odd factors (can’t be possible as it is an even number (2N not N) ) or only even factors we can’t represent it as a sum of any consecutive natural numbers

So now, we have to now only check whether it has an odd factor or not

1. If the number (2N not N) does not have any odd factor (contains only even factor means can be represented as) then we can’t represent it as a sum of consecutive number

2. If the number (2N not N) has an odd factor then we can represent it as a sum of a consecutive number

After this we have to only check whether we can represent (2N as ) or not

• if Yes then answer is false or 0
• if No then answer is true or 1

Below is the implementation of the above idea : 

C++14

#include <bits/stdc++.h>
using namespace std;

long long int canBeSumofConsec(long long int n)
{
    // Updating n with 2n
    n = 2 * n;
    // (n & (n - 1)) => Checking whether we can write 2n as 2^k
    // if yes (can't represent 2n as 2^k) then answer 1
    // if no (can represent 2n as 2^k) then answer 0
    return ((n & (n - 1)) != 0);
}

int main()
{
    long long int n = 10;
    cout<<canBeSumofConsec(n)<<"\n";
}

C

#include <stdio.h>

long long int canBeSumofConsec(long long int n)
{
    // Updating n with 2n
    n = 2 * n;
    // (n & (n - 1)) => Checking whether we can write 2n as 2^k
    // if yes (can't represent 2n as 2^k) then answer 1
    // if no (can represent 2n as 2^k) then answer 0
    return ((n & (n - 1)) != 0);
}

int main()
{
    long long int n = 10;
    printf("%lld", canBeSumofConsec(n));
}

Java

import java.util.*;
class GFG{

 static int canBeSumofConsec( int n)
{
    // Updating n with 2n
    n = 2 * n;

    // (n & (n - 1)) => Checking whether we can write 2n as 2^k
    // if yes (can't represent 2n as 2^k) then answer 1
    // if no (can represent 2n as 2^k) then answer 0
    return ((n & (n - 1)) != 0)?1:0;
}

public static void main(String[] args)
{
    int n = 10;
    System.out.print(canBeSumofConsec(n)+"\n");
}
}

// This code is contributed by umadevi9616

Python3

def canBeSumofConsec(n):

    # Updating n with 2n
    n = 2 * n;

    # (n & (n - 1)) => Checking whether we can write 2n as 2^k
    # if yes (can't represent 2n as 2^k) then answer 1
    # if no (can represent 2n as 2^k) then answer 0
    if((n & (n - 1)) != 0):
        return 1;
    else:
        return 0;

if __name__ == '__main__':
    n = 10;
    print(canBeSumofConsec(n));

# This code is contributed by umadevi9616

C#

using System;

public class GFG {

    static int canBeSumofConsec(int n)
    {

        // Updating n with 2n
        n = 2 * n;

        // (n & (n - 1)) => Checking whether we can write 2n as 2^k
        // if yes (can't represent 2n as 2^k) then answer 1
        // if no (can represent 2n as 2^k) then answer 0
        return ((n & (n - 1)) != 0) ? 1 : 0;
    }

    public static void Main(String[] args) {
        int n = 10;
        Console.Write(canBeSumofConsec(n) + "\n");
    }
}

// This code is contributed by umadevi9616

Javascript

<script>

    function canBeSumofConsec(n) {
        // Updating n with 2n
        n = 2 * n;

        // (n & (n - 1)) => Checking whether we can write 2n as 2^k
        // if yes (can't represent 2n as 2^k) then answer 1
        // if no (can represent 2n as 2^k) then answer 0
        return ((n & (n - 1)) != 0) ? 1 : 0;
    }

        var n = 10;
        document.write(canBeSumofConsec(n) + "\n");

// This code is contributed by umadevi9616
</script>

Output

1

Time Complexity: O(1)

Auxiliary Space: O(1)

Reference:
http://www.cut-the-knot.org/arithmetic/UnpropertyOfPowersOf2.shtml
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