Alright anons, I'm gonna swing for the fences like Babe Ruth.

(1,c)

(1,RSAc)

{e,n,d,x,1,RSAc}

solve for d and e.

n= (a+b)/2-d

x= d-a = d-1 = RSAx for 1,c

RSAt= (x+2)/2 for even, (x+1)/2 for odd.

Here's our starting position for any RSAc.

Now, use (n-1)*a to find the transform for the (e,1) position.

Then find the prime solution by finding the equivalent t value in (e,1).

I know it's a big swing, just looking for that home run. Thoughts?

>>3124

Hey PMA! That formula works in n=1. For n>1 it is a+2x+2n. (e,1) is unique. As VQC said, (e,1) is the Row To Rule Them All.

Lads, sincere apologies for all my UID's. I keep getting kicked offline by my VPN.

Hey PMA! can you look this over and give thoughts? I posted this a few days ago. Posting again for your analysis.

>>3042

"starting record at c=145

(1,61,6) = {1:61:12:11:1:145}

na transform to (1,1,6)

(1,1,6) = {1:1:72:11:61:85}

prime solution at a=5,b=29

(1,5,4) = {1:5:12:7:5:29}

na transform to (1,1,4)

(1,1,4) = {1:1:32:7:25:41}

Thinking out loud here…

For c record (1,c)

[t] is the same in (e,1) and (e,n)

x[t] is the same in (e,1) and (e.n)

d[t] in (e,1) is original d (12) + (61-1)*1 = transformed d = 72

a[t] = na = (161)1 = 61

For prime record:

[t] is the same in (e,1) and (e,n)

x[t] is the same in (e,1) and (e,n)

d[t] in (e,1) is original d (12) + (n-1)*a

transformed d is 12+ (5-1)*5 = 32

a[t] = na = 5*5 for transformed a

is there supposed to be a (n-1)*a connection for d[t]?"

Thoughts, anons?

>>3119

I just love math. RSA whatever. I also really love VQC's blend of spirituality, math, and truth. He attracted us all here with his crazy shit. I honestly have been super happy working with you all on this math challenge. This part of my self (math love) has been laying dormant for many years, and now it's wide awake.If all I get out of this is to commune with you fine Anons all around the world working on math, fuck it, I'm in 100%. Our small part in the #GreatAwakening will become huge when we unlock this. Remember, this has already been solved by other minds. We can solve it too. This is a crowdsourced effort to share this with the world. Don't lose sight, Anons! Persistence wins the day.

Future generations will know that we did our duty for truth and justice. Grab your balls, faggots. For our kids and future generations, we need transparency and NO MORE LIES. That's really what we're working for here.

>>3134

Ok, PMA. Let's work. Can you give another (1,c) example that we can work on now? Pen, paper, and calculator handy over here. AlgebraAnons ready to work.

RATM for your listening pleasure:

https://youtu.be/fI677jYfKz0

take any (1,c)

derive d and e.

Use n=(a+b)/2-d to find n for (1,c)

Derive x,b

we now have {e,n,d,x,a,b} for our starting (1,c) position.

Now we transform.

>>3139

Fuck yeah. "All your base are belong to us." I like hidden in plain sight. Let's find it, Anons.

>>3131

So the (e,1) record will always have equal t and x values as the (e,n) record? That makes it easy to find, right? We need to examine more (1,c) records to confirm this pattern.

So what's the connection between (1,c) and prime records? That's what we're working on now. Prime records will always have lower x and t values than (1,c). Can we make a pattern connection between the c and prime records?

>>3142

Well that's the trick. How do we make (e,1) do the work for us?

For the examples shown, I'm seeing that the prime solution lies in c[t]-2 = prime[t]. However, that pattern may not hold as we move up.

>>3144

Baker, did you already figure it out? :)

>>3146

Chris said some of the patterns are easier to see in binary. You know this already, but posting it for everyone to see.