Ok, I ran the elements out. Here's what I found.
(-111,1,6) {-111:1:16:11:5:29} a and b pair that matches our (prime) a and b
(-119,1,6) {-119:1:12:11:1:25} a=1 and b=25
Hello Lads, here's my responses to VQC's latest questions. Happy 4th of July to you all!
>When does c first appear at a[t]?
For c287 It appears in (-f,1) at a[12], and is the b(n-1) value. If we move to (31,1,13), the related element is at a[13] and is our bn value of 328
328 - 287 (c) = 41 = b
328 / 41 = 8 = n
>When is the second time it appears?
287 also appears in (-2,1,11) in the b column at t[11]. It’s related record is (31,1,12) b = 328
You can perform the same calculation as above, 328 - 287
>When is the first time a squared appears?
It appears in (-2,1,5) at t[5], a =49
It is also the a(n-1) value, since the corresponding an record in (31,15) is a = 56, which is 7 * 8 = 56
so a^2 is also a(n-1).
What is the factor it is multiplied by?
7 * 7 =49
7 * (8-1) = 49
So the two factors are the same, but n = a+1 ?? That's correct for this c287 example, but we'll need to test more.
>What is the first time b squared appears?
hmm. 41^2 = 1681
Can you guise help me find more in (-2,1) or (31,1) ?
b^2 appears at
(-2,1,29) = {-2:1:1623:56:1567:1681}
>The second?
(-2,1,30) = {-2:1;1739:58:1681:1799}
>When a appears as "an" it appears another time.
yes, it appears in (31,1) at t[5] a = 56 and also at t[4] b = 56
It also appears at (-2,1) t[5] as a(n-1) = 49
It also appears as our x value at the b^2 record above, (-2,1,29) = {-2:1:1623:56:1567:1681}
>When does n first appear?
at (-2,1,1) a = 1 * (8-1) = 1 * (n-1) = 7
at (31,1,1) b = 2 * 8 = 2 * n = 16
so n first appears in t[1] for both (-2,1) and (31,1)
>What are the rules?
Solving n from c is easy if you’re a faggot?
The na transform record is our starting point in looking for an, ab, a(n-1), b(n-1)
an will always occur in (e,1)
a(n-1) wil always occur in (-f,1)
they share the same [t] value, which is less than the na transform [t]
the bn record will be [t+n] apart from the an record
the b(n-1) record will be [t+n-1] apart from the a(n-1) record
>In column -f too?
In column (-f,1) everything seems to be based on factors of (n-1), but other than that its patterns are the same.
>Any one of these patterns well understood between -f and e is the step you are looking for.
Thanks Senpai! Looking now.
>Godspeed anons.
Blessings to you as well.
>Happy Independence Day.
Thanks! Happy 4th of July to all Anons here. Glad to be with you all.
"for I have sworn upon the altar of god eternal hostility against every form of tyranny over the mind of man.” Thomas Jefferson