Positive integers have so many gorgeous features. 
Some of them could be expressed as a sum of two or more consecutive positive numbers.
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# Consider an Example :

* `10` , could be expressed as a sum of `1 + 2 + 3 + 4 `.
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# Task

**_Given_** *Positive integer*, N , **_Return_** true if it could be expressed as a sum of two or more consecutive positive numbers , OtherWise return false .
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# Notes 

~~~if-not:clojure,csharp,java
* Guaranteed constraint  :  **_2  ≤  N  ≤  (2^32) -1_** .
~~~
~~~if:clojure,csharp,java
* Guaranteed constraint  :  **_2  ≤  N  ≤  (2^31) -1_** .
~~~
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# Input >> Output Examples:

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# [Playing with Numbers Series](https://www.codewars.com/collections/playing-with-numbers)

# [Playing With Lists/Arrays Series](https://www.codewars.com/collections/playing-with-lists-slash-arrays)

# [Bizarre Sorting-katas](https://www.codewars.com/collections/bizarre-sorting-katas)

# [For More Enjoyable Katas](http://www.codewars.com/users/MrZizoScream/authored)
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## ALL translations are welcomed

## Enjoy Learning !!
# Zizou

Difficulty: introductory

Starter code:

def consecutive_ducks(n):
	

Solutions:
from math import log2

def consecutive_ducks(n):
    return not log2(n).is_integer()

def consecutive_ducks(n):
    while n % 2 == 0:
        n //= 2
    return n > 1

def consecutive_ducks(n):
    '''
    find x + (x + 1) = n thus x exists if (n-1) % 2 == 0
    if not, find x + (x + 1) + (x + 2) = n
    4*x + 1+2+3+4
    5*x + 1+2+3+4+5
    '''
    sum = 0
    for i in range(1,n//2):
        sum += i
        if sum > n:
            continue
        if (n-sum) % (i+1) == 0:
            return True
    return False
        
        
    


def consecutive_ducks(n):
    return bool(n & (n - 1))

import math
def consecutive_ducks(n):
    return math.ceil(math.log(n,2))!=math.floor(math.log(n,2))


def consecutive_ducks(n):
    return f"{n:b}".count("1") > 1

def consecutive_ducks(n):
    return "1" in bin(n)[3:]

def consecutive_ducks(n):
    return n != 1 << n.bit_length()-1

def consecutive_ducks(n):
    return (((n & (n - 1)) != 0) and n != 0)

def consecutive_ducks(n):
    while n>1:
        n=n/2
        if str(n).split('.')[1]!='0':
            return True
    return False