##Task:

You have to write a function **pattern** which creates the following pattern upto n number of rows. 

* If the Argument is 0 or a Negative Integer then it should return "" i.e. empty string.

* If any odd number is passed as argument then the pattern should last upto the largest even number which is smaller than the passed odd number.

* If the argument is 1 then also it should return "".

##Examples:

pattern(8):

    22
    4444
    666666
    88888888
    
pattern(5):
 
    22
    4444



```Note: There are no spaces in the pattern```

```Hint: Use \n in string to jump to next line```

Source: codewars | Difficulty: EASY
Tags: String algorithms

Fundamentals

Starter code: def pattern(n):
	

Solutions:
def pattern(n):
	return '\n'.join([str(i) * i for i in range(2, n + 1, 2)])


def pattern(n):
	result = []
	if n == 1:
		return ''
	for i in range(n + 1):
		if i % 2 == 0:
			result.append(str(i) * i + '\n')
	return ''.join(result).strip()


def pattern(n):
	if n % 2 != 0:
		n = n - 1
	return '' if n <= 1 else '\n'.join([str(i) * i for i in range(1, n + 1) if i % 2 == 0])


pattern = lambda n: '\n'.join([str(x) * x for x in range(2, n + 1, 2)])


def pattern(n):
	return '\n'.join([str(a) * a for a in range(2, n + 1, 2)])


def pattern(n):
	return '\n'.join((str(x) * x for x in range(2, n + 1, 2)))


pattern = lambda n: '\n'.join((i * str(i) for i in range(2, n + 1, 2)))