>>8736

So are you proposing that we create the BigN (1,c * small prime) elements?

Like so for c145:

{1:61:12:11:1:145} c' = 1 * 145

{35:198:20:19:1:435} c' = 3 * 145

{49:337:26:25:1:725} c' = 5 * 145

{54:539:31:30:1:1015} c' = 7 * 145

{74:759:39:38:1:1595} c' = 11 * 145

{36:900:43:42:1:1885} c' = 13 * 145

{59:1042:46:45:1:2175} c' = 15 * 145

now we've reached 1 * 3 * 5 = 15 * c' , so the small prime product is greater than d, so we have the primes we need. (I think)

This gives all the BigN values for the c' = c * small prime elements.

>Lets say you take the first fifteen primes and multiple c to make c'.

>You would focus on the column with e',d' and c'.

>There is a very fast (way?) to do this.

>It would not make enough sense if we didn't build to this.

So now we go to (e,1) in each of the BigN columns?

I'm finding prime b = 29 for145 everywhere in the respective (e,1) columns for the BigN values.

>>8736

>>8740

Continuing the c145 exploration:

Here's a good one in (49,1)

the first element (a,b) values are (an) and (prime b) !!

>>8741

Could be a fluke, but trying to work through the new hints.

In essence, the (c * small prime) calcs give us new (e,n) columns to explore, and row (e,1) for each of these new BigN's contains the factors we're looking for in the first few elements.

Can I please get some eyes on this to double check my work?

>>8742

Here's the BigN for (49,n)

{49:337:26:25:1:725} c' = 5 * 145 = 725

First element in (49,1) is

{49:1:26:1:25:29} c = 725

(an) * (b) = 725 = 5 * 5 * 29

25 * 29 = 725 = 145 * 5

For each (c * small prime) you get a new BigN and new e value.

And then we check (e,1) for the factors

And since we can calc a[1] for any e value, maybe a direct (or nearly direct) calc ???

Lookup indeed if this turns out to be legit.

Lol, calc'd every element by hand with pencil, paper, and my trusty TI-89.

Love this math game.