ProgramMathAnon !dSvrkhSLR6 ID: 6a1cc2 Jan. 21, 2018, 9:14 p.m. No.3188   🗄️.is 🔗kun   >>3189

>>3184

Just tested your {144:0:17:12:5:29} record.

 

It fails this check:

c == ( d * d ) + e

 

If you look at the original data in the grid, the t value in the (-e, 0) space is unreliable. See pic of records in (-16,0) with t = 2.

 

I don't believe you can generate these records using t as an input parameter.

 

>>3187

>Something interesting is that it shows 17, and 17^2-12^2 = 145

That is interesting. Will look into also.

ProgramMathAnon !dSvrkhSLR6 ID: 6a1cc2 Jan. 21, 2018, 9:25 p.m. No.3190   🗄️.is 🔗kun   >>3191

>>3189

>You can't rely on d^2+e

Not sure about that. I've been generating records all over the place (including the left half) without any failures just using that IsValid check I posted before. Happy to post it again.

 

The 17^2 - 12^2 = 145 comment is a good observation. Have confirmed on other records.

 

This makes sense, because normally (d+n)^2 - (x+n)^2 = c, and we are dealing with n = 0. So everything here is a perfect square.

ProgramMathAnon !dSvrkhSLR6 ID: 6a1cc2 Jan. 21, 2018, 9:39 p.m. No.3192   🗄️.is 🔗kun

>>3186

MA - we're still searching.

 

Can't summarize everyone's work, but VQC did drop some hints again about (n-1)*a transforms from (e,n) into (e,1).

 

He mentioned that an offset can be found to solve the problem by analyzing (d[t] - d)/(n-1) factors and comparing them with a[t]/n.

>>2990

 

>>3134

I've posted an example for c=145 that shows my limited understanding of this.

 

There was also a recent reminder about records in the (-e, 0) space that also contain the prime results we are looking for.

 

Patterns and patterns all over the place. But nothing yet that links any c records to prime solutions.

ProgramMathAnon !dSvrkhSLR6 ID: 6a1cc2 Jan. 21, 2018, 9:54 p.m. No.3193   🗄️.is 🔗kun   >>3194 >>3207

Just posting some additional tests of the (-e,0) space.

 

This includes records for both c and p (prime solution), and the (-d^2, 0) record they both share.

 

Interesting that the c value from (-d^2) less the original e equals the e value in the (-e,0) record.

ProgramMathAnon !dSvrkhSLR6 ID: 6a1cc2 Jan. 21, 2018, 10:19 p.m. No.3196   🗄️.is 🔗kun

>>3194

And corrected method is:

 

public static TheEndRecord CreateForZeroN( TheEndRecord ter ) {

 

BigInteger a = ter.a;

BigInteger b = ter.b;

BigInteger n = 0;

BigInteger x = (b - a) / 2;

BigInteger d = b - x; // this was posted incorrectly earlier.

BigInteger e = -(x * x);

 

return new TheEndRecord( e, n, d, x, a, b );

 

}

ProgramMathAnon !dSvrkhSLR6 ID: 6a1cc2 Jan. 23, 2018, 8:16 p.m. No.3240   🗄️.is 🔗kun

Here is another example of what doesn't quite work in different test cases, but works for c=901.

 

Might be considered a follow up to my post at >>3093 but using the (e,1) na records only to solve.

 

Starting na record for c=901

(1,1,15) = {1:1:450:29:421:481}

 

Destination na record for prime solution

(1,1,7) = {1:1:98:13:85:113}

 

New d formula:

 

newD = d - ( (d[t] diff) * (x+1) - (a[t] diff) * (x-1) )

 

the d[t] diff value can be calculated from (x^2 - f)

the a[t] diff value is d[t] diff - 4

 

In this example:

 

d[t] diff: 120

a[t] diff: 116

x+1 = 30

x-1 = 28

 

98 = 450 - (120 * 30 - 116 * 28)

ProgramMathAnon !dSvrkhSLR6 ID: 6a1cc2 Jan. 24, 2018, 6:32 p.m. No.3305   🗄️.is 🔗kun   >>3308 >>3313

>>3296

Thank you for clarifying.

 

>>3300

I looked into f records previously. But not with regards to the d[t]-d pattern. Will revisit.

 

Regarding the (-e,0) discussion earlier, attached pic are various test cases showing c and prime solution and their corresponding (-e,0) records.

 

In some cases, there is a defined relationship between the 2.