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  1. #11
    Originally Posted by Garnabby View Post
    Garnabby got up, worked on some files, throughout the day, but, had time on the end to work on a few more numerals, in particular, those to do with the main terms for the quarks, or, as I like to call 'em, the partial charges. Ha.

    At first glance, the numerals, 16,410, with 73,784, weren't so apparent as with the number of chemical elements of the periodic table, at 291, which translates to (1,029 + 1) for the other universe. But, although a bit more doodling revealed that 16,410 = [-1,000 + 9,000 + (29^2 * 10)] ---> 19291 ---> 192_291, and, that 73,785 = {1 + [2^3 * (2 + 9220 + 1]} ---> 1229221 ---> 1029_9201, Garnabby didn't get around to further relating the second group of numerals, to the first, namely in terms of 291*1030 = 299,730 = (-1,000 + 30,073*10) ---> 137_731.

    Maybe, later on, today, given the clue that between the two realms of the full, and partial charges, the thus numerals come out as 685, and 1,950, respectively, of which 686*3,465 = 2,376,990 = (-1,000,000 - 10 + 3377000) ---> 00113377 . Then the numerals of the partial charges, which work inward, should have their 137's from within, given that the numerals are apart, and, next, together, in between.
    Firstly, I edited a couple of mistakes in the quote above.

    Secondly, I was right, in that the numerals for the quarks go, to a version of 137_731, from within.

    For the quarks, their numbers of elements give (16,410 * 73,784) = 1,210,795,440 = [(100899623)^2 - (100899617)^2] = {[-1,000 + 100,000,000 + 300*3,000 + (700 - 77)]^2 - [(0^0 - 7 - 77 + 700) + 300*3,000 + 100,000,000 - 1,000]^2} ---> 1/133777_7777331\1 ---> 1/137_731\1 . Note that 133777 is an additive pattern, by one 1, two 3's, and three 7's, but, 133777 is multiplicative pattern, by one 1, two 3's, and four 7's.

    Things went from the atomic side, of 137_731, to the quark-side, of 1/137_731\1, and, with 00113377 in between. Note again, the in-between spot must be of the form(x^x + y^y + ...), and, so, it becomes (0^0 + 1^1 + 3^3 + 7^7) = 823,572 = 2*3*22877*6 = 2 * (√9 * {22900 - 6^0 * [antilog(0^0)]^2 + 77} * 6) ---> 9229_6776__6776_9229, another way to express the in-between part, but, which brings it, too, back to the 2's, and 9's, of the chemical, and, then, quark-, periodic tables. The 6776-parts invert to the 9229-parts, with the 6's rotating to 9's, and 7's rotating to 2's.

    I omitted a few of the complicated (but spot-on in the theory) details, such as how the zeroes end up in front of the 00113377-part, and, how the in-between part has also a 1/137 form, but, suffice it to say how the tables above are related. When I have a bit more time, I will show how the tables, themselves, are formulated.
    Last edited by Garnabby; 03-17-2024 at 08:44 PM.
    Every one /everyone knows it all; yet, no thing /nothing is truly known by any one /anyone. Similarly, the suckers think that they win, but, the house always wins, unless to hand out an even worse beating.

    https://youtu.be/OxgmMbSZ99w

    Garnabby + OppsIdidItAgain + ThomasClines (or TomasHClines) + TheGrimReaper + LMR + OneHitWonder (or 1HitWonder, 1Hit1der) + Bill Yung ---> GOTTLOB1, or GOTTLOB = Praise to God! And, MHF.

    Blog at https://garnabby.blogspot.com/

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