## Monday, July 30, 2018

### How its possible 111,111,111 x 111,111,111 = 12,345,678,987,654,321

I'm not sure how you knew to ask me this question, but since you did I will provide a little feedback. Ironically, I was recently working on these very same configurations of the intrinsic set of palindromic sequences in an attempt to develop potential bedrock deep learning methodologies.

The sequences using 1’s are extremely important. However, to get a clearer comprehension, we might want to look at this from a big picture standpoint. By looking at the transitional dynamics. Below is a formula representation of the internal count of ones versus zeroes that can be seen if we did the multiplication by hand.

x2∑1→(x−1)×(x−1)=2xx2∑1→(x−1)×(x−1)=2x

It may seem trivial to analyze these usually hidden variables, however sometimes guiding patterns can be found in even the most unusual and unexpected places. If I find any kind of symmetry, I go into bloodhound mode, and in that state, no pattern is safe…

Looking at the set of palindromic numbers generated by the increasing multiplication using only ones, we find something very interesting indeed.

Note that the symmetry holds for the first 20. After that, we must adjust the logic if we want to keep the symmetry. First, notice that if we take the palindromic numbers as rows and sum those rows; the result is equal to an n2→n(n+1)n2→n(n+1) situation where;

the natural numbers of n equate to an exact representation of the increasing ones that were multiplied to generate the palindromic sequences we're now summing…

Example:

[11×11=121][11×11=121] while the sum of

[1+2+1=4][1+2+1=4] where

[4≡2×2][4≡2×2] and

[(2×2)≡lens(11)×lens(11)][(2×2)≡lens(11)×lens(11)]

Think about that for a second…

Realize that this remains true for twenty terms. The only reason the logic falters after that is because after the twentieth term, the palindromic numbers begin to generate double digit numbers which breaks the symmetry…

In order to keep the symmetry going, we simply have to stop using the ones and continue on using the equivalent n representations. That and we must from that point on consider the palindromic numbers as having separate elements while introducing each new term as a distinct value. I know that process seems weird but what I haven't yet mentioned is the special use of such a (now infinite) set of rows.

Actually, its better I stop here for now because the steps that follow are related to a different process pertaining to prime numbers.

Anyway, hopefully a bit more light now shines on some of the palindromic numbers and the logic that those numbers subscribe too…

There is always something interesting indeed when analyzing the logic surrounding symmetries…

*Edit:

I want to express gratitude for the upvotes and support. I noticed that many young minds have read this writing, which is great. Although, I now feel responsible to clarify a few things. First, the mathematics used here, although correct in process, are incorrect in terms of the conventions used to express notations, especially regarding the way I chose to express the summations. I am not aware of any convention that expresses the addition version of the factorial methods that are expressed with multiplication via the ! Symbol.

I therefore created a makeshift notation using ∑1→n∑1→n in an attempt to achieve said expressions. The Sum to n method is expressing an intention to add 1+2+3+…n≡∑1→n1+2+3+…n≡∑1→n.

In other words, I modified the notation method to express my purposes. However, that's not the right way to learn these things. We cannot just invent things and expect others to understand our meaning. So please do not follow or let these examples influence you to learn the language of mathematics without respect for existing notations. The point of mathematical notation is to communicate on agreed terms. That being said, I'm adding a few more formulas related to the observations above.

(∑1→(x−1)x2)2x=(x−1)(∑1→(x−1)x2)2x=(x−1)

(2[∑1→(x−1)]x2)x=(x−1)(2[∑1→(x−1)]x2)x=(x−1)

(2[∑1→(x−1)]x2)−1(x−1)=x(2[∑1→(x−1)]x2)