We can do better for RSA and bigger number. Feel free to comment

    We take the number to be factorized,
    N, and its base 4 representation, N4,
    expand them in powers of (2 pi) weighted
    by the sum of adjacent and errant digits.

    Then we take the ratio of N4/N10 or N4/N

Then trying to follow through the next part of the example you seem to have apparently arbitrary correction factors, so I have no idea what you mean by "errant digits", nor how you get these weird expansions in powers of (2.pi). I'm sure you must have some process for doing this, but it's not clear from what you have here.

And despite what I said earlier[0] you still haven't just taken an existing, difficult example and given us the factorisation. You say you can do one of the RSA numbers, and yet you haven't.

In an earlier comment[1] you said:

    Now factorising Ras 1536.  Will
    post the results when done.

Let me say that people in the past have made amazing advances that have then been ignored by everyone, only later to be found to be exactly right. You may be right, and you may have made an amazing discovery. But you might not, and it's up to you to communicate clearly what you've done. That is, to communicate exactly what you've done.

Demonstrate all of the process, in detail, written clearly, for two small examples. Factor 10^4+1 for me using your method, then factor (10^7-1)/9. Then show your working in parallel for a large example, perhaps 40 digits. Finally, factor a large number such as RSA 1536 and present your answer.

I have to say that I'll also listen more if you spell my name correctly.

[0] https://news.ycombinator.com/item?id=9549682

[1] https://news.ycombinator.com/item?id=9499462