For every positive integer N, there exists a unique sequence starting with 1 and ending with N and such that every number in the sequence is either the double of the preceeding number or the double plus 1. 

For example, given N = 13, the sequence is [1, 3, 6, 13], because . . . :
```
 3 =  2*1 +1
 6 =  2*3
 13 = 2*6 +1
```

Write a function that returns this sequence given a number N. Try generating the elements of the resulting list in ascending order, i.e., without resorting to a list reversal or prependig the elements to a list.

Difficulty: EASY
Starter code: def climb(n):
	

Solutions:
def climb(n):
	return [1] if n == 1 else climb(int(n / 2)) + [n]


def climb(n):
	res = []
	cur = 1
	mask = 1 << max(0, n.bit_length() - 2)
	while cur <= n:
		res.append(cur)
		cur = 2 * cur + (1 if n & mask != 0 else 0)
		mask >>= 1
	return res


def climb(n):
	result = [1]
	for x in '{:b}'.format(n)[1:]:
		result.append(result[-1] * 2 + (x == '1'))
	return result


def climb(n):
	return [n >> i for i in range(len(f'{n:b}') - 1, -1, -1)]


from math import log
from collections import deque

def climb(n):
	return [n // 2 ** i for i in range(int(log(n, 2)) + 1)][::-1]

def climb(n):
	seq = deque([n])
	while n > 1:
		n //= 2
		seq.appendleft(n)
	return list(seq)


def climb(n):
	return [n >> n.bit_length() - i - 1 for i in range(n.bit_length())]


def climb(n):
	return list(climb_iterator(n))

def climb_iterator(n):
	cursor = 0
	for digit in bin(n)[2:]:
		cursor = 2 * cursor + int(digit)
		yield cursor


def climb(n):
	res = [n]
	while res[-1] != 1:
		res.append(res[-1] // 2)
	return res[::-1]


def climb(n):
	arr = []
	while n:
		arr.append(n)
		n //= 2
	return arr[::-1]