>>6079

>How does row one (n=1) relate and determine the patterns?

Have spent some time looking into n=1 for odd x+n.

First pic attached is for n=1 at x+n=11.

Because n, x, and f are fixed, these records can be generated for all valid combinations of d and e, where d starts at zero and carries on indefinitely. In other words, every d value exists.

Notice also that the records start at e = c = (1-f), and at some point have a c=0 value where d=x.

Taking this a bit further, the second pic is for n=1, x+n between 1 and 83, and a constant d value of 78, corresponding to the d value for the c6107 example, where f=134 the solution n=36 and x+n=83.

A couple of interesting records appeared:

1) at (36,1) where e equals the solution n with very close c and f values.

2) at (-6084,1) where d=78, x=78, c=0, and is very close to the x+n solution.

Not quite sure yet how this integrates into a possible solution.

>>6273

Attached pic is a first attempt at rendering animated even squares for x+n=16, and includes records between n=2 and n=15.

This example represents the nn + (2d-1)(n-1) + (n-1) + (f-1) breakdown with (n-1)+(f-1) occupying the middle.

>>6275

Think I've ironed out the kinks in the even x+n rendering.

Latest x+n=16 example attached (the top right and bottom left square were reversed).

Also included is shading for the different groups of "triangles" within the middle (n-1)+(f-1) portion.

>>6290

VQC. Running out of ways to say thank you. These words… saved.

>>6291

>>6292

WE, will look deeper into these hints.

>>6292

>The gifs are key and there are patterns that you're just about to look at to compare with the even squares.

Pics attached are for x+n=77 and x+n=78, both with fixed n=13 in the positive e space.

No conclusions yet, but the n-1 distribution in the middle, and the pattern of the yellow squares are interesting.

>>6324

Excellent connection, 3D.

This started a further analysis in the (-e,n) space and what we thought would be a way to find the largest factor record for any even c number. ( mentioned previously >>6325 )

The idea was that starting from a (1,c) entry record, we could jump directly to (-f, 1) at d+1 and find a connection. This turned out to be true, but only for a handful of cases, irrespective of x+n parity or f mod 8.

Unfortunately, I think all we've confirmed is that (-f, n-1) is indeed a mirror, and those entries at (-f, 1) have corresponding entries at (e, 2) anyway.

Attached are sample outputs for c145, c6107, and c112220 that explore factor records in the (e,n), (-f, n-1), and (e,n) at negative x spaces.

>>6328

Pic attached are key records for the c145 test case showing (e,n), (-f,n-1) and their respective na and nb records.

Differences in the x+n values are:

e na x+n = 12

-f na x+n = 13

e nb x+n = 134

-f nb x+n = 133

Also noticed for the -f nb record:

(d[t] - d)/(n-1) = (8832 - 12)/(61-1) = 147

when using the original d and n values.

>>6346

>>6347

This is an excellent explanation. Just fantastic.

>>6328

>"three are in row 1, column zero and the side by side diagonal cells from the origin"

Have been thinking about VQC's comment about 3 solutions, and went in search of the diagonal cells from the origin.

Aside from the n=0 special cases, there are 4 quadrants to the grid.

(e, n)

(e, n) at -x

(-f, n-1)

(-f, n-1) at -x

For c145, the respective entries in each quadrant are:

en (1,61,6) = {1:61:12:11:1:145} = 145; f=24; (f%8)=0; (x+n)=72; u=36; (d+n)=73

enx (1,61,-5) = {1:61:-10:-11:1:101} = 101; f=-20; (f%8)=4; (x+n)=50; u=25; (d+n)=51

fn (-24,60,7) = {-24:60:13:12:1:145} = 145; f=51; (f%8)=3; (x+n)=72; u=36; (d+n)=73

fnx (-24,60,-5) = {-24:60:-11:-12:1:97} = 97; f=3; (f%8)=3; (x+n)=48; u=24; (d+n)=49

And images for each are attached.

In the (e,n) -x image, notice that the entire image is comprised of only nn and n-1 portions, due to the negative d value.

Confirmed similar behavior for other negative x records when compared to their positive e counterparts.

>>6328

Pic attached is current work in progress, related to the idea originally posted >>6245 and VQC's subsequent hint in the (-f, n-1) mirror.

For a small example c287, the output starts with the factor records, then moves to the (-f, n-1) mirror, and then to the nb record at (-2,1,136).

Then it shows the factors of a = 36449, just as a confirmation that this nb record indeed contains the a=7 and b=41 values from the prime solution.

Finally, the result of an n,d,f query is shown within the nb x+n = 271 for our solution d=16 value.

Note the nb starting f value of 73441 compared to the query results f value of 73154.

In this case, the difference just happens to be our starting c value, or dd+e.

If we can correctly determine the f value we are looking for, we will be able to calculate our prime solution directly.

>>6384

Figured out the grid coordinates and how to navigate between any of the 8 related records.

Examples attached are for c145, c303, c6107 and the c6107 prime solution.

They show records in the 4 quadrants for +/- x.

>>6395

>>6401

Initial wip for c145, comparing (e,n) and (-f,n-1) records. Related records are highlighted.

>>6470

>>6494

Found something interesting while analyzing (e,1) and (-f,1) records that seemed a bit too coincidental not to share.

Pic attached is for c329832 and shows three groups of records.

The first group includes all the factor records in (e,n).

The second group shows these records nb transformed to (e,1).

The third group shows these records nb transformed to (-f,1).

For the records tagged with an "x", (356,1,223) in (e,1) and (-793,1,222) in (-f,1), there appears to be a way to use the their d and a values to navigate to the next factor record at (356,1,732).

(356,1,223).d = 99190

(-793,1,222).a = 97728

99190 - 97728 = 1462

Where 1462 is the x value in (356,1,732).

So perhaps a way to find these factor records using the n=1 columns is:

(e,1) d[t] - (-f,1) a[t] = (e,1) x[t].

The records marked with a "y" can be used the same way.

(356,1,1241).d = 3077858

(-793,1,1240).a = 3072324

3077858 - 3072324 = 5534

Which is the x value for (356,1,2768).

>>6557

For records in (e,n) and mirrored to (-f,n-1), the following relationship applies to their na transform records:

(e,1) d[t] - (-f,1) a[t] = (e,n) d[t]

For example:

(e,n)

(1,61,6) = {1:61:12:11:1:145} = 145

(-f,n-1)

(-24,60,7) = {-24:60:13:12:1:145} = 145

(e,1) na

(1,1,6) = {1:1:72:11:61:85} = 5185

(-f,1) na

(-24,1,7) = {-24:1:72:12:60:86} = 5160

(e,1) d - (-f,1) a = (e,n) d

72 - 60 = 12

>>6572

Thanks.

The f % 8 analysis, lead to a very efficient and accurate way of iterating through valid n, d, f combinations within any odd x+n.

The original code posted in the previous thread >>5632 has been changed just slightly to enable iterative searching to work where n=1, but otherwise those patterns work very well, and have been the source for the animated images posted here and various other queries.

There were a couple of problems that I ran into and was not able to solve:

1) Identify patterns that could yield more than n+4 jumps when searching for factor records.

2) Identify patterns that consistently applied to even x+n.

The pic attached is the latest groupings for f % 8 based on odd and even x+n. Made a mistake earlier in thinking only 0,1,2,4,5,6 were valid values when in fact that rule only applied to odd x+n.

That being said, I believe you're asking if the f % 8 rules can be applied to vertical movements within a given (e,n), or if there is any consistency in jumping between records.

In the case of (3,6,5), you don't need the f % 8 rules to jump to the next record in the sequence.

This can be accomplished by incrementing t in steps of n, which is effectively increasing x by 2n.

Therefore your sequence would be (3,6) where t=5, t=5+6, t=5+12, t=5+18, etc.

I understand this pattern works everywhere.

>>6561

>>6579

Based on VQC's post about the gcd and p+1-t factors in the Grid Patterns thread, I went back through some older code to review factor searching in (e,1).

My notes indicate there were two ways to search for factor records, would appreciate a double check on this:

new t = m*p+1 - t

and

new t = m*p + t

where m is a multiple and t represents the starting t.

The pics attached start at the na records for c145 in (e,1) and (-f,1) and iterate through m, creating both alternative factor records along the way, and represent factor matches for the known a=5 and b=29 values.

In the gcd(a[t], a[p]) column, a[t] is the starting na value, and a[p] is from the current record.

Each of these gcd values represent a valid n in their respective spaces, and would enable a reverse na transform back to their original starting position.

For example, (-24,1,9), with a gcd result of 4, can reverse to (-24,4,9).

(-24,1,9) = {-24:1:132:16:116:150}

(-24,4,9) = {-24:4:45:16:29:69}

Initial tests indicate that sometimes the gcd values of n appear in (e,1), and sometimes in (-f,1). Perhaps related to different parities of n or x+n.

>>6631

Much appreciated!

>>6619

Attached is one more example of factor records in (e,1) for c145 and p=5. This time starting at the (1,1,67) nb record and includes iterations into negative x.

A column has been added for "valid n(s)", that show all factors for the gcd result.

The second picture shows the grid entries at t=67 for all of these valid n values.

Just confirming how the (e,1) records indeed point to multiple (e,n) records.

>>6629

>This has always been about (you).

>>6630

>That n-1 is a factor of col -f and n is a factor of e will show you a shortcut to the triangle solution.

>>6632

>Do not be afraid of negative x values in row 1 for -f and e.

>>6647

>Have you discovered ant new insights from the negative x values?

Pics attached are for c145 and c287 and include related entries into negative x.

Records are grouped by (-f,1), (-f,n-1), (e,1), and (e,n), sorted in each group by e,n,t, and the label on the left indicates the "origin" for each.

Have been thinking about the (you) comment, and triangles, and stumbled across something interesting related to "u" in (e,1) and (-f,1).

Final spreadsheet image attached shows a comparison between nb and nb -x records for odd/even x+n in (e,1) and (-f,1).

The differences between d, b, and f values can be calculated in terms of u, using the following formulas:

odd x+n

new d = d - 4u

new b = b - 8u

new f = f - 8u

even x+n

new d = d - (4u-2)

new b = b - (8u-4)

new f = f - (8u-4)

Think this is somehow relevant as a "triangle solution" relies on being able to calculate the correct (x+n)(x+n) from u.

>>6654

Lot to cover in these crumbs.

>When is the first time a squared appears?

>What is the factor it is multiplied by?

Not sure if I'm interpreting these two questions correctly, but here is a pattern that I noticed that holds across a number of test cases.

In (e,1), there is a record where a = n*(a^2), that points to an (e,n) record where a = a^2 for the first time.

Examples:

c6107

a=31,b=197

The first time a mod 31^2 appears in (23,1):

(23,1,132) = {23:1:34859:263:34596:35124} = 1215149904; f=69696; (f%8)=0; (x+n)=264; u=132; (d+n)=34860

34596 / 961 = 36

And the first time a = a^2 is where n=36:

(23,36,132) = {23:36:1224:263:961:1559} = 1498199; f=2426; (f%8)=2; (x+n)=299; u=149; (d+n)=1260

c551

a=19, b=29

First time a mod 19^2 appears in (22,1):

(22,1,136) = {22:1:36731:270:36461:37003} = 1349166383; f=73441; (f%8)=1; (x+n)=271; u=135; (d+n)=36732

36461 / 361 = 101

First time a = a^2 is where n=101:

(22,101,136) = {22:101:631:270:361:1103} = 398183; f=1241; (f%8)=1; (x+n)=371; u=185; (d+n)=732

c493,

a=17, b=29

First time a mod 17^2 appears in (9,1)

(9,1,88) = {9:1:15492:175:15317:15669} = 240002073; f=30976; (f%8)=0; (x+n)=176; u=88; (d+n)=15493

15317 / 289 = 53

First time a = a^2 is where n=53

(9,53,88) = {9:53:464:175:289:745} = 215305; f=920; (f%8)=0; (x+n)=228; u=114; (d+n)=517

Not sure yet if a similar pattern exists in the (-f,1) space.

>>6643

Following up on an earlier post for the valid n values from the c145 nb record at (1,1,67).

The attached pic shows each of those records into (e,n), (e,1), (-f,n-1), and (-f,1) for positive and negative x.

Figured there would be some difference in these records, especially on the (-f,1) side, that would give additional insight into determining the valid n values for (1,1,67).