def collatz(n):
	steps = 1
	#Because the number itself is the first in the sequence.
	while n!=1:
		if n%2:
			n = (3*n+1)/2
			steps+=2
		else:
			n /= 2
			steps+=1
	
	return steps


for n in range(1, 10**9):
	'''
	This is the main loop that tests every number.
	The last number it looks up for is the second
	number in the range() function - 1, in the 
	example it would be 10^9-1
	'''
	if collatz(n)>n:
		print str(n)+',',
		
print