So obviously he's not doing what he said he would again. I've been having a look at the RoD thing. Over the next few posts, I'm going to post every cell containing a and b values that multiply together to produce a given c value for the sequence of c values starting from RSA2048 replacing c with d recursively until reaching the lowest possible cell (obviously excluding RSA2048, since we don't know its a and b). It's a lot of numbers, so give me a bit. He said look at big semiprimes to see the pattern, so why not use the biggest semiprime we’ve dealt with so far?

This is the sequence of c values where each time we replace c with d (excluding RSA2048 itself):

158732191050391204174482508661063007579358463444809715795726627753579970080749948404278643259568101132671402056190021464753419480472816840646168575222628934671405739213477439533870489791038973166834068736234020361664820266987726919453356824138007381985796493621233035112849373047484148339095287142097834807844, 12598896421924866518220704506279920262503827873668266660431014021495031828185862096976477460760574959532030199266536614252274764167453116999041838483272614, 112244805768128382102848738141332806509123753930790945631219843869937264197591, 335029559543823509028205969685172744250, 18303812705112110303, 4278295537, 65408, 255, 15, 3, 1, 1

Let's start at the bottom because I'm waiting for this webtool to factor the second d (I somehow doubt I'll get to the first d, but who knows). To begin with, I accidentally added an extra 1. The lowest cell we find is c = 1. Above that, c = 3, then above that c = 15, then c = 255, then c = 65408, then c = 4278295537, then c = 18303812705112110303, then c = 335029559543823509028205969685172744250, then c = 112244805768128382102848738141332806509123753930790945631219843869937264197591, then c = 12598896421924866518220704506279920262503827873668266660431014021495031828185862096976477460760574959532030199266536614252274764167453116999041838483272614 (and I'm probably not going to do the original d since it's quite large, and then the original c we want to factor is obviously too large itself). I'll do 65408 etc in the next post because that one is even so it has a lot of factors.

c = 1, factors are 1

(0, 0, 1) = {0:0:1:0:1:1} f = -3

c = 3, factors are 1 and 3

(3, 2, 0) = {3:2:0:-1:1:3}, f = 4 (which has a negative x value and t = 0, so it doesn't exist)

c = 15, factors are 1, 3, 5 and 15

(6, 5, 2) = {6:5:3:2:1:15}, f = -1

(6, 1, 1) = {6:1:3:0:3:5}, f = -1

c = 255, factors are 1, 3, 5, 15, 17, 51, 85 and 255

(30, 113, 8) = {30:113:15:14:1:255}, f = -1

(30, 29, 7) = {30:29:15:12:3:85}, f = -1

(30, 13, 6) = {30:13:15:10:5:51}, f = -1

(30, 1, 1) = {30:1:15:0:15:17}, f = 1

>>7425

Reading this the first time I was so confused and thought someone read "I'm up to RSA #9" as "I'm new and I'm reading the old threads" and thought they'd introduce me, but then why would they give that second clue?

…is this VQC giving us another crumb without his trip? They have no posting history.