Consider on the natural number the lexicographic ordering on the prime factorization:
If $m = p_1^{a_1}\cdots p_r^{a_r},n = q_1^{b_1}\cdots q_s^{b_s}$ then we define:

$$m \vartriangleleft n :\iff [(p_1,a_1),(p_2,a_2),\cdots,(p_r,a_r)] \prec [(q_1,b_1),(q_2,b_2),\cdots,(q_s,b_s)]$$

where the right hand side $\prec$ is the lexicographic ordering of the two lists, where $(p,a) \prec (q,b) :\iff p < q  \text{ or } ( p=q \text{ and } a < b)$ and the primes in the factorization list are sorted by usual absolute value: $p_i < p_{i+1}$.

Example:

1) For instance for $n=1,\cdots , 10 $ we get the following sorting:

$$1, 2, 6, 10, 4, 8, 3, 9, 5, 7$$

2) 

Examples :

sorted by absolute value:  $[2, 7, 11, 30, 60, 121]$

lexicographically sorted:  $[2, 30, 60, 7, 11, 121]$

    [(2, 1)] = [(2, 1)]
    [(2, 1)] < [(7, 1)]
    [(2, 1)] < [(11, 1)]
    [(2, 1)] < [(2, 1), (3, 1), (5, 1)]
    [(2, 1)] < [(2, 2), (3, 1), (5, 1)]
    [(2, 1)] < [(11, 2)]
    [(7, 1)] > [(2, 1)]
    [(7, 1)] = [(7, 1)]
    [(7, 1)] < [(11, 1)]
    [(7, 1)] > [(2, 1), (3, 1), (5, 1)]
    [(7, 1)] > [(2, 2), (3, 1), (5, 1)]
    [(7, 1)] < [(11, 2)]
    [(11, 1)] > [(2, 1)]
    [(11, 1)] > [(7, 1)]
    [(11, 1)] = [(11, 1)]
    [(11, 1)] > [(2, 1), (3, 1), (5, 1)]
    [(11, 1)] > [(2, 2), (3, 1), (5, 1)]
    [(11, 1)] < [(11, 2)]
    [(2, 1), (3, 1), (5, 1)] > [(2, 1)]
    [(2, 1), (3, 1), (5, 1)] < [(7, 1)]
    [(2, 1), (3, 1), (5, 1)] < [(11, 1)]
    [(2, 1), (3, 1), (5, 1)] = [(2, 1), (3, 1), (5, 1)]
    [(2, 1), (3, 1), (5, 1)] < [(2, 2), (3, 1), (5, 1)]
    [(2, 1), (3, 1), (5, 1)] < [(11, 2)]
    [(2, 2), (3, 1), (5, 1)] > [(2, 1)]
    [(2, 2), (3, 1), (5, 1)] < [(7, 1)]
    [(2, 2), (3, 1), (5, 1)] < [(11, 1)]
    [(2, 2), (3, 1), (5, 1)] > [(2, 1), (3, 1), (5, 1)]
    [(2, 2), (3, 1), (5, 1)] = [(2, 2), (3, 1), (5, 1)]
    [(2, 2), (3, 1), (5, 1)] < [(11, 2)]
    [(11, 2)] > [(2, 1)]
    [(11, 2)] > [(7, 1)]
    [(11, 2)] > [(11, 1)]
    [(11, 2)] > [(2, 1), (3, 1), (5, 1)]
    [(11, 2)] > [(2, 2), (3, 1), (5, 1)]
    [(11, 2)] = [(11, 2)]

I used this sorting to visualize the following Jaccard similarty kernel, where $\Omega$ counts the prime divisors with multiplicity:

$$\frac{\Omega(\gcd(a,b))}{\Omega(\operatorname{lcm}(a,b))}$$

where the entries $n=1,\ldots, N=1400$ of the matrix are sorted by the lexicographic ordering above:

[![prime_factorization_fractal_with_lexicographi_sorting][1]][1]

**Q: While there seems to be some visual fractal pattern, I am asking myself, how does one quantify / define this fractal pattern?**

One idea would be to let $N \rightarrow \infty$ and make the matrix larger, so that one can zoom in but I am not sure how to formalize this idea.

**Edit**:
Here is another image for the kernel $\omega(\gcd(a,b))$ whose Gram matrix of size $n \times n$ has rank $\pi(n)$, where $\omega$ counts the distinct prime divisors of $n$ and $\pi$ is the prime counting function:

[![prime_factorization_fractal_lexicographic_sorting_omega_kernel][2]][2]

Counting the size of the blocks on the main diagonal, we find the [following OEIS sequence][3]:

$$0, 1, 1, 1, 2, 1, 2, 1, 1, 3, 1, 1, 3, 1, 1, 1, 4, 1, 1, 1, 4, 2, 1, 1, 5, 2, 1, 1, 5, 2, 1, 1, 1, 6, 2, 1, 1, 1, 6, 2, 1, 1, 1, 1, 7, 2, 1, 1, 1, 1, 7, 3, 1, 1, 1, 1, 8, 3, 1, 1, 1, 1, 8, 3, 1, 1, 1, 1, 1, 9, 3, 1, 1, 1, 1, 1, 9, 3, 1, 1, 1, 1, 1, 1, 10, 3, 1, 1, 1, 1, 1, 1, 10, 4, 1, 1, 1, 1, 1, 1, 11, 4, 1, 1, 1, 1, 1, 1, 11, 4, 1, 1, 1, 1, 1, 1, 1$$

[SageMath-script][4]


  [1]: https://i.sstatic.net/EUioQ.jpg
  [2]: https://i.sstatic.net/SQXeY.png
  [3]: https://oeis.org/A249809
  [4]: https://sagecell.sagemath.org/?z=eJytVF1v2jAUfa_U_-DlyU49ILTTpK7ew6Ttkb3sDVXISQy1YsfGNl3593P8EaDQTpOGEME35557fO-xW7YGSrINhT26v74C_mOY25keCNZDbbhkq5Y_c6uM9RB0fXV91fqcnxdz7E7CZyp21HHVwx5rBNbKAA14D97m2hgqYddhY17RSeoMf4HLZddBiutIRgcyYx7Dok6LkWtNG3cmawm3-FjWFkWq7ZDt1TjWnm8VG_bMjGXkBxWWocdcQTAo8VhBa0BiUYliZLvNkR6dqPDQB_924IlMjZKamhM6PvID2rdAfiB9enNE9LEawZJcAsxO6t5UQ8G1URKsd33jlBIWcKn9vkEj9cqpVcf2A4a9UKkFs34DqSvLOf6Mq8p_y3KO5-Vt-cn_luFPbLrvW-9g8T2n3hdoDEYSUO8Bra0SO8fAMAZ2Dwqcax3Qgr3wRnkz6CfeUCH2ScSAznpyFgZeMjmoh6mXKLpq9EnGpx6NlnkVD5MMKsLgKMLFQ5GGEYzHvAVAQUKMElKnwFevLGTUsW7qqvetFsoJXk8a-QXkHZ6EJy11tJ147RbG5GgKQ9sz_2qjLhh09HzXVUHlwUObpg0RQqq_uSMe5JyApnEtGhnX4FBkflwkZavT7AP49gJ4XmZgWd5OIS3rEtKbk7S7C2nVdOjJiJ4ekaS-dZ13rBeYR78ItwLtNwxWeF6NZzVMYZFOpdwLbp1PHB4woxc3FUoAO5wDmIwX0e_aLpTw84X5PrMW4UYJZWpqyC-zY7iRVJPimz8CtsDUata4lRluJVKhiX1Sv2GrOalms7Cxha9_N5sNvNlCg_h3hf-r6P8ieIH-AHGz2dk=&lang=sage&interacts=eJyLjgUAARUAuQ==