1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3

[d, d+1, d+2, d+3]

1 2 3 4 5 6 7 8 9

[d, x, d+1, x]

[x,y,d,x,y,d+1]

>>> import itertools
>>> g = (digit for number in itertools.count(1) for digit in map(int, str(number)))
>>> digits = 2, 2, 3, 2, 4
>>> i=9+(22-10)*2 + 1; list(itertools.islice(g, i, i+len(digits)))
[2, 2, 3, 2, 4]

def find_min_index(number):
    digits = get_digits(number)
    # find minimal n and start,end such that
    # digits = get_digits(n)[start:] + get_digits(n+1) +...+ get_digits(n+m)[:end]
    # brute force approach
    for ndigits_in_n in range(1, len(digits)):  # m=ceil(len(digits)/ndigits_in_n)
        for start in range(ndigits_in_n):
            i = ndigits_in_n - start  # where (n+1) starts in *digits*
            # is10pow = (10**ndigits_in_n == (n+1))
            is10pow = (digits[i:i + ndigits_in_n + 1] == [1] + [0] * ndigits_in_n)
            # get_digits(n+1) == digits[i:i+ndigits_in_n+is10pow]
            n = digits2number(digits[i:i+ndigits_in_n+is10pow]) - 1
            if matched_n(digits, n, start):
                return get_index(n) + start
    return get_index(number) # m == 0

>>> find_min_index(22324)
34 # верно

def get_digits(number):
    """
    >>> get_digits(22324)
    [2, 2, 3, 2, 4]
    """
    return list(map(int, str(number)))  # NOTE: O(k**2) in CPython

import functools

def digits2number(digits):
    """
    >>> digits2number([1,2,3])
    123
    """
    return functools.reduce(lambda number, digit: 10 * number + digit, digits)