[Medium] String Decompression Solutions Counter

Given an input c, outputs the number of distinct lists of strings lst such that:

1. ''.join(lst) == c
2. for every string s in lst, s consists of an arbitrary character followed by one or more characters from '0123456789'

Sure hope I didn't mess this up, because I think the fundamental idea is quite elegant! Should run successfully for all "reasonable" inputs (as in, the numeric output fits in a uint64 and the input isn't ridiculously long). Fundamental algorithm is O(n) if you assume all arithmetic is O(1). (If not, then I don't know what the time complexity is, and I don't feel like working it out.)

from functools import cache
from itertools import pairwise
from math import prod

@cache
def fibonacci(n: int) -> int:
    if n == 0:
        return 0
    if n == 1:
        return 1
    return fibonacci(n - 1) + fibonacci(n - 2)

def main(compressed: str) -> int:
    is_fragment_start = [i == 0 or c not in '0123456789' for i, c in enumerate(compressed)]
    fragment_start_positions = [i for i, s in enumerate(is_fragment_start) if s]
    fragment_lengths = [stop - start for start, stop in pairwise(fragment_start_positions + [len(compressed)])]
    return prod(fibonacci(fragment_length - 1) for fragment_length in fragment_lengths)

if __name__ == '__main__':
    from argparse import ArgumentParser
    parser = ArgumentParser()
    parser.add_argument('compressed')
    print(main(parser.parse_args().compressed))

Idea: a00010 -> [a000, 10] -> [length 4, length 2] -> F(4) * F(2)
01a102b0305 -> [01, a102, b0305] -> [length 2, length 4, length 5] -> F(2) * F(4) * F(5)
where F(n) = fibonacci(n - 1) is the number of ways to partition a string of length n into a list of strings of length ≥2.

F(2) = 1 = fibonacci(1), F(3) = 1 = fibonacci(2), and F(n) = F(n - 2) + F(n - 1), so F is indeed just an offset version of the Fibonacci sequence.
To see why F(n) = F(n - 2) + F(n - 1), here are the ways to split up 'abcde': ['ab'] + (split up 'cde'), ['abc'] + (split up 'de'), and ['abcde'], corresponding to F(5) = F(3) + F(2) + 1.
And the ways to split up 'abcdef': ['ab'] + (split up 'cdef'), ['abc'] + (split up 'def'), ['abcd'] + (split up 'ef'), and ['abcdef'], corresponding to F(6) = F(4) + F(3) + F(2) + 1 = F(4) + F(5) = F(6 - 2) + F(6 - 1).
The same logic generalizes to all n >= 4.