Okay guys, I think found something interesting. I was just messing around with that brute force algorithm, comparing variables. I’m not sure what this means, but it’s a thing.

The t where a[t]=a(BigN-1) in (-f,1) and the t where a[t]=c*BigN in (e,1) are BigN or BigN-1 apart, depending on the parity of e.

If e is even, BigN + (the t where a[t]=a(BigN-1) in (-f,1)) - 1 = the t where a[t] = c*BigN in (e,1)

If e is odd, BigN + the t where a[t]=a(BigN-1) in (-f,1) = the t where a[t] = c*BigN in (e,1)

Here are some examples. First with odd e:

>c=209, e=13, BigN=91

>a[t]=a(BigN-1) at (-16,1,7) = {-16:1:104:14:90:120}, t=7

>a[t]=c*BigN at (13,1,98) = {13:1:19214:195:19019:19411}, t=98

>91+7=98

>c=403, e=3, BigN=182

>a[t]=a(BigN-1) at (-38,1,10) = {-38:1:201:20:181:223}, t=10

>a[t]=c*BigN at (3,1,192) = {3:1:73729:383:73346:74114}, t=192

>182+10=192

Now with even e:

>c=559, e=30, BigN=257

>a[t]=a(BigN-1) at (-17,1,12) = {-17:1:279:23:256:304}, t=12

>a[t]=c*BigN at (30,1,268) = {30:1:144199:536:143663:144737}, t=268

>257+12-1=268

>c=551, e=22, BigN=253

>a[t]=a(BigN-1) at (-25,1,12) = {-25:1:275:23:252:300}, t=12

>a[t]=c*BigN at (22,1,264) = {22:1:139931:528:139403:140461}, t=264

>253+12-1=264

I’ve tested this on over 20 semiprimes at it holds true for all of them. I would have to rewrite my entire algorithm in order to accommodate for non-semiprimes, so I’ve only tested it on semiprimes.