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: EASY
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):
	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