Your reasoning is pretty confusing, and I don't see ANY application for this, but here is how I understand your problem...

The reason is because you aren't actually returning the values, you're simply returning the index + 3, which is wrong to begin with. What you're trying to do is point to a new index of the array based on its value and return the index if it contains a value greater than 0.

You need to reference the index you want, check its value, then return the index which contains a value.

a = [0, 0, 1, 0, 0, 1, 0]
for i, v in enumerate(a):
   if i == 0:
      print(i)
      next
   if v == 0:
      next
   else
      print(i)

But let's be honest, this is extremely ugly and un-pythonic. Let's simply check for whether a[i] contains a value, and if so, return the index...

for i, v in enumerate(a):
   if v or i == 0:
      print(i)

The purpose of if v or i == 0 is to check if v has a value, if so, print the index. OR if we are looking at the first element of i.

If you want to EXPLICITLY move the index by two, you must set your index at the start and use a while loop, since enumerate can't help you here and doesn't actually let you move the indicies...

a = [0, 0, 1, 0, 0, 1, 0]
i = 0
while i < len(a) - 1:   # -1 to avoid out of bounds error
   print(i)
   if a[i + 2] == 0:
      i += 2
   elif a[i + 2] == 1:
      i += 1
print(i)            # Final print statement for the last index at the end of the while loop

I want to impress upon you the fact that this solution does NOT scale with different or larger lists, which is why it isn't recommended. By simply checking for whether a value does or doesn't exist, you guarantee your accuracy.

You should simply return the index based upon whether or not it contains a value. Even for very large lists, this will be extremely fast, and will always scale, even for values greater than 1.

The only other reason I would see you would want to do this differently is if you're doing string-search algorithms.