Sum of elements on the list recursive

[perfringo]

I deduct that if your 'iteration version works flawlessly' then objective is:

[1,0,2,0,0,3,1,3,2,1] --> [13, 12, 12, 10, 10, 10, 7, 6, 3, 1]

I think that your solution is too complicated.

Lists support sum(), so:

>>> sum([1, 2])
3

Lists also support slicing:

>>> a = [1, 2, 3, 4, 5]
>>> a[1:]
[2, 3, 4, 5]
>>> sum(a[1:])
14

Combining those two things into list comprehension and taking into account noisefloor comment about Python antipattern you can write simply:

>>> a = [1,0,2,0,0,3,1,3,2,1] 
>>> [sum(a[i:]) for i, v in enumerate(a)]
[13, 12, 12, 10, 10, 10, 7, 6, 3, 1]

Iterative solution should give you pretty good idea how to proceed with recursive solution.

EDIT: how Python core developers do it

There is built-in module itertools which have function accumulate (Make an iterator that returns accumulated sums).

There is rough equivalent provided:

def accumulate(iterable, func=operator.add):
    'Return running totals'
    # accumulate([1,2,3,4,5]) --> 1 3 6 10 15
    # accumulate([1,2,3,4,5], operator.mul) --> 1 2 6 24 120
    it = iter(iterable)
    try:
        total = next(it)
    except StopIteration:
        return
    yield total
    for element in it:
        total = func(total, element)
        yield total