>>6695

Thanks baker!

>>6736

Morning VQC.

The formulas posted work to construct infinite elements in a chain and where n=1 will generate all records.

For n>1, there will be gaps.

This is equivalent to creating elements by ent where t=t+n.

>>6759

>>6767

These d[t] and a[t] rules are for (0,1).

>>6785

Definitely just getting started.

>>6787

Thanks, Teach.

That repeating a[t] pattern for perfect squares is interesting.

Also appears that difference in x values between those records is 2*(sqrt(n)).

>>6787

Also, the a[t] pattern repeats where n=a[t] from (e,1).

(e,1) a[t] = 2, 8, 18, 32, 50, etc.

((0,2), (0,8), (0,18), (0,32)…), a[t] are the perfect squares 1,4,9,16…

>>6793

Teach.

First pic is the pattern you pointed out in (0,n) where n is a perfect square. Examples show the first few records for n=1,4,9,16,81,169. The a values repeat. 0,2,8,18,32,etc.

Second pic, is the pattern mentioned above >>6792 where the a[t] value from (0,1) becomes the n value in (0,n), and the perfect squares move to a[t].

Think this is justing show how the n and a values are interchangeable in certain circumstances.

>>6801

>interchangeable in certain circumstances.

*all circumstances.

>>6834

>>6836

>are there only 2 sequences at n > 1

Answering my own question. No, there can be more than 2 sequences.

Example for c287 at (31,128) attached shows 4 sequences incrementing by t+n starting from t=8, 57, 72, and 121.

Not sure of the pattern yet to determine the number of sequences within each n.

Just posting a work in progress understanding of the RoD algorithm.

Attached pics are for c145, c287, c551, and c6107, and include the (-f,n-1) and (e,n) records, and factor tree.

Aside from the sqrt(2d), sqrt(e), and sqrt(f) values - still not quite sure how these apply - the only new piece is the (e,1) record at x=f or x=f-1. >>7047

Initial tests indicate that the x record will always be at f-1. Which essentially means that we are finding a record at (e,1) where x+n = f.