Tasty Bread! Thanks Baker!
>>2506 Made this for ya'
Hello PMA! Can you update me on your ideas for how we can use a=1b=C, CC, AA, BB, etc to jump records? Also if anyone else has working formulas or programs for using these to make the jumps between records, jump in too!
On a separate note, what's the latest on mirroring x, lads?
Lol! Thanks for the deep laugh, Topol! That was just what the doctor ordered.
Nice work last night on n and na, Anon! I also think (e,1) is the key. How do we USE (e,1) to get what we need in higher values of n? Working to solve that over here, still haven't made the breakthrough. Hopefully we'll find it soon.
Hello Isee! Thanks for all your work, anon. We are getting close. Let's keep going!
Hello PMA! Thanks for the clear explanation and great output as always. So the left column is used in this example to search for factors of c=65? Cool. Then you can change your search to whatever factors you want. Pulling them out will be the next step?
>>2603
Hmmm... Quiz, Ok!
Q1: (1,1) and (1,2).
At (1,1) we have a,b factors of 1,5 and c=5
At (2,2) we have a,b factors of 1,3 and c=3
Q2: Seems like a repeat of Q1?
Yes, this is very useful for analyzing factor trees. Not sure what you're hinting at. Have you found something new to report, Anon?
Thanks Topol for the fun as usual, and thanks Hobo for updating your project. Sorry we don't have any cool new visual output for you guys! The vibe here has been off too. We're missing MA, CA, Teach, Baker, AA, and the BruteForce anons! I know I just got over being sick for almost three days, so maybe a few people are under the weather.
In the attached pic, note the value for var e, remainder. It's freaking huge!!! Baker, I know you love running RSA thru your shit. Is it normal to get huge e values like that?
Thanks for the quick reply, anon! Is this Baker?
Interesting! So from the algebra approach, and starting only from C we'd need to use your ideas to factor backwards to one or more factorizations of C.
To find the n value for a=1,b=c I took the basic formulas and just solved for n. Anyone able to verify that a cell exists at this location?
Hey lads, thanks for your replies. So the first part is just the (e,1) factorization we already know. The newer highlighted part is just setting ab to a=1 b=c and then using the basic equation (you weren't being a downer, PMA) to solve for n when ab are set to 1,c.
What I should have taken more time to be clear about is that starting from only C, this is a way to generate two valid cells for C, one at: {1,1,d,x,a,b} which we know using f, and a new one at {1,BigN,d,BigX,1,c}. This gives us an upper bound for n and x. All other factorizations for c should be in lesser n values.
At a conceptual level, I guess I'm trying to take our knowledge in (e,1) and figure out how to expand it. This 1,c bigN may have no value at all, just keeping the mental oven baking over here.
Baker mentioned that he had been able to get the f formula to generate factors for the RSA numbers, although they were in n=1. Baker, any chance you could post some examples of RSA numbers that you've been able to factor in n=1?Thanks!
Teach, nice to see you! I'm following all your ideas here. Var f seems likely to hold some more goodies for us to discover.
>How does this pattern change as n grows?
>How do we use the na cell in (e,1) to locate our axb for n>1?
Hello MA! Hope to see you over the weekend!
Sure Baker. Also, thanks for clearing up a question I had. Only two factorizations for semiprime c? 1c and ab. perfect. Here's the formula.
x= floor(SQRT( abs(f)))
then use x to create a and b.
It works for ALL (e,1), but only for (e,1). However, within row 1 it's very powerful. Just start from C, derive d,e,f,x, then calculate a and b. Boom, done. Try it out. grab your grid, pick a random element, multiply a and b to get c, then feed your c var into the calculator. It will solve any c value in row 1, and with basic math. Somehow f is always related to x. Like a "golden ratio" but different. It's pretty incredible.
Cool, thanks for looking into it Baker.
Thanks for contributing, Anon!
Hello lads! Here's what I'm working on: took the formulas and built a sheet that calculates all vars and displays them visually for the difference of squares equation. Looking for ways to solve for x when n>1. Row one is solved, as you anons know. Seeing patterns in n>2, but still nothing yet.
I'm honing in on x being the key. X increases in a linear pattern. Can we link X to the c^2 patterns? I'm wondering if the stable growth of x can be tied to the exponential growth of C somehow. Maybe this ties into the -x patterns? Just keeping my head down and looking for patterns over here.
Thanks Baker! Here's the spreadsheet, anons. Couple of quick notes:
-
The formula for x works for all (e,1) but not yet for higher values of n. Replace the x value if you're inspecting an element that has n>1. It will auto fill the visual and re-calc all values.
-
The third box on the right is actually just the top left box (a * (x+n)) moved down to more easily see a*b = c in rectangular form. This is where b = a+2x+2n comes from. I left both boxes for accuracy, but you only use one.
-
Hope you lads find it helpful for visualizing the problem at hand!
https://anonfile.com/H4Qawed4be/DiffOfSquares.xls
https://anonfile.com/J1Qcw5dab4/DiffOfSquares.xlsx
What's up Topol!! Nice to see you. Thanks for the inspiration as always!
>I'm not going to give up this easily. I don't have anything new, I've been working for the past few days and after work I have other commitments, but I still read over crumbs and think every day.
Thanks IseePatterns! I think we're all feeling pretty much the same. Let's keep going. I'm seriously falling asleep thinking about the grid at night, and going over crumbs every day too. The solution will present itself if we keep adding to and writing down the rules like VQC asked us to.