logical deduction - Self referential puzzles - follow-up

https://puzzling.stackexchange.com/questions/23554/self-referential-puzzles made me wonder. How many solutions are there to the below box when multiple digits are allowed? I don't know the answer myself, so a good proof of your answer is appreciated. We're counting characters, so 11 is considered as two 1's and not as one 11.

\begin{array}{|cc|} \hline \text{The number of 0's in this box is} & \text{_______} \\ \text{The number of 1's in this box is} & \text{_______} \\ \text{The number of 2's in this box is} & \text{_______} \\ \text{The number of 3's in this box is} & \text{_______} \\ \text{The number of 4's in this box is} & \text{_______} \\ \text{The number of 5's in this box is} & \text{_______} \\ \text{The number of 6's in this box is} & \text{_______} \\ \text{The number of 7's in this box is} & \text{_______} \\ \text{The number of 8's in this box is} & \text{_______} \\ \text{The number of 9's in this box is} & \text{_______} \\ \hline \end{array}

Answer

(assuming numbers with leading zeros, ex "05" are forbidden)

Part 1. Establishing an upper boundary for digit size

Suppose there exists a solution where the largest blank is filled in with three digits.
Then the box must contain at least 108 digits, so one blank must contain at least 9 digits. But the largest blank contains only three digits, which is a contradiction.
So there can't be a solution whose largest blank is three digits.

Suppose there exists a solution where the largest blank is filled in with four digits.
Then the box must contain at least 1008 digits, so one blank must contain at least 99 digits. But the largest blank contains only four digits, which is a contradiction.
So there can't be a solution whose largest blank is four digits.

...

This pattern of contradiction continues upwards forever. So, if a solution exists, its largest blank must have no more than two digits.

Part 2. Establishing an upper boundary for number of two-digit slots

Suppose there exists a solution where exactly two blanks are filled with two digits.
The total number of digits in the entire box is 10 + 8*1 + 2*2 = 22.
The sum of the numbers in each blank must exactly equal the total number of digits in the entire box.
But the minimum theoretical sum of all numbers in each blank is 1*8 + 2*10 = 28
So there can't be a solution where exactly two blanks are filled with two digts.

Suppose there exists a solution where exactly three blanks are filled with two digits.
The total number of digits in the entire box is 10 + 7*1 + 3*2 = 23.
The sum of the numbers in each blank must exactly equal the total number of digits in the entire box.
But the minimum theoretical sum of all numbers in each blank is 1*7 + 3*10 = 37.
So there can't be a solution where exactly three blanks are filled with two digts.

...

This pattern of contradictions continues upwards forever. So, if a solution exists, it can have no more than one blank slot with two digits.

In the previous question, the solution that has zero two-digit slots was already found, so the remainder of this post will search for a solution containing exactly one blank slot with two digits.

Part 3. Determining which slot has two digits

Suppose there exists a solution where exactly one blank is filled in with two digits.
the total number of digits in the entire box = 10 + (9*1) + (1*2) = 21.
The sum of the numbers in each blank must exactly equal the total number of digits in the entire box.

Due to the existing labels, the number in each blank must be at least 1.
No single-digit blank may contain a digit larger than 3, because then the sum of the numbers in each blank would exceed 21.
The two-digit blank may not exceed 12, because then the sum of the numbers in each blank would exceed 21.

Because none of them appear in any blanks, The blanks in slots 4, 5, 6, 7, 8, and 9 are all "1".
The digit in blank slot 1 must be at least 7.
If blank slot 1 is not the two-digit slot, the minimum theoretical sum of numbers in each blank is 7+10+(1*8) = 25. But this is larger than 21, so blank slot 1 must contain two digits.

Part 4. Finding the value of the two digit slot

Blank slot 1 can't contain "12", because once you account for the "1" in "12" and the "1" in the label, you need to fit ten "1"s in nine remaining slots.
Blank slot 1 can't contain "10", because blank slot 0 will be at least 2, and the slot corresponding to 0's number will be at least 2, and you'll need to fit eight "1"s in the seven remaining slots.

So blank slot 1 must contain "11".

Part 5. Finding the remaining values

Accounting for the "11" in the two-digit slot, and the "1" in the label, eight slots must contain "1". Exactly one blank slot contains a single digit other than 1.
The blank slot not containing a 1 must be equal to 21 - (11 + 8*1) = 2.

That will be the second "2" to appear in the box, so it must go in slot 2.

So the only multi-digit solution is [1, 11, 2, 1, 1, 1, 1, 1, 1, 1].

Appendix. Brute-forcing all one digit solutions

The digits in the one-digit solution add up to 20, and no slot contains a 0. Using this information, you can iterate through all possible configurations and find the self-referential ones. Sample Python implementation:

from collections import Counter
import functools

def iter_ways(x, nums):
    if x == 0:
        return

    if nums == 1:
        if 1 <= x<= 9:
            yield [x]
        return
    for i in range(1,10):
        if i <= x:
            for rest in iter_ways(x-i, nums-1):
                yield [i] + rest

def valid(seq):

    c = Counter()
    for i in range(10):
        c[i] = seq[i]
    digit_str = "0123456789" + "".join(map(str, seq))
    digits = map(int, digit_str)
    return c == Counter(digits)

for seq in iter_ways(20,10):
    if valid(seq):
        print seq

Result:

[1, 7, 3, 2, 1, 1, 1, 2, 1, 1]

So the only one-digit solution is [1, 7, 3, 2, 1, 1, 1, 2, 1, 1].