PMA !!y5/EVb5KZI ID: 7e5bcd May 9, 2018, 11:07 p.m. No.5907   🗄️.is 🔗kun   >>5932 >>5934 >>5936

>>5887

>The final steps involve finding a base using (f-1)/8 for each of the eight triangles (some a bit bigger than others by a unit in length) which is larger than (n-1)/8 BUT smaller than (x+n)/8.

 

Pics attached are an idea of how this could look for the c6701 and c27707 examples.

 

Starting with the (n-1)+(f-1) portion in the middle, a new red square outline has been added (f-1)/8 units from the center.

 

For the c27707, the light green squares around that red outline represent the u1=(f-1)/8 + 1 value of 23. And the light green squares represent the u2=(f-1)/8 value of 22.

 

The c6107 example computes to u1=17 and u2=16.

 

Attached are also rough calculations showing successful matches in 7 steps.

PMA !!y5/EVb5KZI ID: 7e5bcd May 11, 2018, 8:04 p.m. No.5960   🗄️.is 🔗kun   >>5961 >>5964 >>5967

>>5950

Some additional examples for (x+n)=83 for different f mod 8 and n starting values.

 

Looks like the earlier formula for f can be generalized for e movements within the same x+n.

 

e = e + 2n

f = f - 2(n-1)

 

So there's our (n-1) again.

 

Also looks like there might be some relationship between f mod 8 and (n-1)+(f-1) mod 8, but not certain of that yet.

 

>>5959

>make an equation for finding the range of n and (x+n) for a given c if that would help (it sounds like it would to me but I don't tend to get feedback about these things)

I also looked for the formula for max n in a given x+n. But ran into a problem where that max n could be greater than the n we were searching for during iteration, so tabled the idea.

 

The starting n and increments are known - I've posted that code snippet previously. It would be nice to know that max n value. If not for iteration purposes, then perhaps for another use down the road.

PMA !!y5/EVb5KZI ID: 7e5bcd May 13, 2018, 12:03 a.m. No.5976   🗄️.is 🔗kun   >>5985 >>5989 >>6003 >>6026

>>5942

Pics attached are visuals for the e = e + 2n movement that VQC referenced previously.

 

Starting from a new test record for c86747, these images show the e movement and affects on f moving 2 records to the right as well as 2 records left.

 

For reference, the starting prime solution is:

 

(311,12,36) = {311:12:294:71:223:389} = 86747; f=278; (x+n)=83; u=41; (d+n)=306

 

One image shows the regular formula for nn + 2d(n-1) + f - 1 with the (f-1) portion in the center.

 

The other image breaks the 2d(n-1) into another (n-1) piece and combines with (f-1).

 

All images are partial views of the x+n=83 square.

PMA !!y5/EVb5KZI ID: 7e5bcd May 13, 2018, 5:04 p.m. No.5989   🗄️.is 🔗kun

>>5976

>>5985

Topol, thanks for stitching these together.

 

>>5977

I think the most compelling of the two views is the (n-1)+(f-1) example, which shows a single green square moving around the top portion.

 

These gifs show only 5 iterations, but many more examples for the same x+n=83 can be found by decreasing n from 12 to 2.

 

I believe the following repeating patterns apply to (n-1)+(f-1) mod 8, but would appreciate someone else verifying:

 

f mod 8 = 0

Starting from n=2

(n-1)+(f-1) mod 8 pattern: 0, 2, 4, 6

 

f mod 8 = 1

Starting from n=3

(n-1)+(f-1) mod 8 pattern: 2, 4, 6, 0

 

f mod 8 = 2

Starting from n=2

(n-1)+(f-1) mod 8 pattern: 2, 4, 6, 0

 

f mod 8 = 4

Starting from n=2

(n-1)+(f-1) mod 8 pattern: 4, 6, 0, 2

 

f mod 8 = 5

Starting from n=3

(n-1)+(f-1) mod 8 pattern: 6, 4, 2, 0

 

f mod 8 = 6

Starting from n=2

(n-1)+(f-1) mod 8 pattern: 6, 0, 2, 4

 

These mod patterns apply only within the same f mod 8 value. Keep in mind that the e = e + 2n movement changes the f mod 8 value for each record.

 

So for the c86747 record, 2 left and 2 right posted, the (n-1)+(f-1) mod 8 values are 4, 6, 0, 2, 4.

PMA !!y5/EVb5KZI ID: 7e5bcd May 14, 2018, 10:32 p.m. No.6003   🗄️.is 🔗kun   >>6004 >>6022

>>5991

Still playing with the odd x+n square and triangles, but managed to take things a bit further.

 

Pics attached show a bit of a progression:

 

1) Initial layout of 8 triangles around a center square.

2) Iterating the 8 triangles sequentially and filling in numerical values.

3) Dynamically generating the n, d, f view for c=6107. (This corresponds to the nn + 2d(n-1) + f - 1 layout from >>5976.)

PMA !!y5/EVb5KZI ID: 7e5bcd May 15, 2018, 11:26 p.m. No.6026   🗄️.is 🔗kun   >>6027 >>6034

Now able to generate odd x+n squares dynamically to image format.

 

Attached are animated examples again for x+n=83 where n=11 or 12 in the same two styles mentioned previously at >>5976.

 

These records are not associated by e=e+2n movement. Currently just testing for any layout adjustments.