Ready to go!
Don't know if this will help anyone, but starting to focus a bit more on the size of the small and large squares.
>The cell at (2,1) is identical to the cell at (0,1) with one added to each value a,b,d while the values of n and x remain the same.
>The cell at (3,1) is identical to the cell at (1,1) with one added to each value a,b,d while the values of n and x remain the same.
Enumerating e=0, n=1, for values of t from 1 to 10.
(0,1,1) = {0:1:0:0:0:2} = 0; (x+n)=1x1=1; (d+n)=1x1=1
(0,1,2) = {0:1:4:2:2:8} = 16; (x+n)=3x3=9; (d+n)=5x5=25
(0,1,3) = {0:1:12:4:8:18} = 144; (x+n)=5x5=25; (d+n)=13x13=169
(0,1,4) = {0:1:24:6:18:32} = 576; (x+n)=7x7=49; (d+n)=25x25=625
(0,1,5) = {0:1:40:8:32:50} = 1600; (x+n)=9x9=81; (d+n)=41x41=1681
(0,1,6) = {0:1:60:10:50:72} = 3600; (x+n)=11x11=121; (d+n)=61x61=3721
(0,1,7) = {0:1:84:12:72:98} = 7056; (x+n)=13x13=169; (d+n)=85x85=7225
(0,1,8) = {0:1:112:14:98:128} = 12544; (x+n)=15x15=225; (d+n)=113x113=12769
(0,1,9) = {0:1:144:16:128:162} = 20736; (x+n)=17x17=289; (d+n)=145x145=21025
(0,1,10) = {0:1:180:18:162:200} = 32400; (x+n)=19x19=361; (d+n)=181x181=32761
Enumerating e=2, n=1, for values of t from 1 to 10.
(2,1,1) = {2:1:1:0:1:3} = 3; (x+n)=1x1=1; (d+n)=2x2=4
(2,1,2) = {2:1:5:2:3:9} = 27; (x+n)=3x3=9; (d+n)=6x6=36
(2,1,3) = {2:1:13:4:9:19} = 171; (x+n)=5x5=25; (d+n)=14x14=196
(2,1,4) = {2:1:25:6:19:33} = 627; (x+n)=7x7=49; (d+n)=26x26=676
(2,1,5) = {2:1:41:8:33:51} = 1683; (x+n)=9x9=81; (d+n)=42x42=1764
(2,1,6) = {2:1:61:10:51:73} = 3723; (x+n)=11x11=121; (d+n)=62x62=3844
(2,1,7) = {2:1:85:12:73:99} = 7227; (x+n)=13x13=169; (d+n)=86x86=7396
(2,1,8) = {2:1:113:14:99:129} = 12771; (x+n)=15x15=225; (d+n)=114x114=12996
(2,1,9) = {2:1:145:16:129:163} = 21027; (x+n)=17x17=289; (d+n)=146x146=21316
(2,1,10) = {2:1:181:18:163:201} = 32763; (x+n)=19x19=361; (d+n)=182x182=33124
Enumerating e=1, n=1, for values of t from 1 to 10.
(1,1,1) = {1:1:2:1:1:5} = 5; (x+n)=2x2=4; (d+n)=3x3=9
(1,1,2) = {1:1:8:3:5:13} = 65; (x+n)=4x4=16; (d+n)=9x9=81
(1,1,3) = {1:1:18:5:13:25} = 325; (x+n)=6x6=36; (d+n)=19x19=361
(1,1,4) = {1:1:32:7:25:41} = 1025; (x+n)=8x8=64; (d+n)=33x33=1089
(1,1,5) = {1:1:50:9:41:61} = 2501; (x+n)=10x10=100; (d+n)=51x51=2601
(1,1,6) = {1:1:72:11:61:85} = 5185; (x+n)=12x12=144; (d+n)=73x73=5329
(1,1,7) = {1:1:98:13:85:113} = 9605; (x+n)=14x14=196; (d+n)=99x99=9801
(1,1,8) = {1:1:128:15:113:145} = 16385; (x+n)=16x16=256; (d+n)=129x129=16641
(1,1,9) = {1:1:162:17:145:181} = 26245; (x+n)=18x18=324; (d+n)=163x163=26569
(1,1,10) = {1:1:200:19:181:221} = 40001; (x+n)=20x20=400; (d+n)=201x201=40401
Enumerating e=3, n=1, for values of t from 1 to 10.
(3,1,1) = {3:1:3:1:2:6} = 12; (x+n)=2x2=4; (d+n)=4x4=16
(3,1,2) = {3:1:9:3:6:14} = 84; (x+n)=4x4=16; (d+n)=10x10=100
(3,1,3) = {3:1:19:5:14:26} = 364; (x+n)=6x6=36; (d+n)=20x20=400
(3,1,4) = {3:1:33:7:26:42} = 1092; (x+n)=8x8=64; (d+n)=34x34=1156
(3,1,5) = {3:1:51:9:42:62} = 2604; (x+n)=10x10=100; (d+n)=52x52=2704
(3,1,6) = {3:1:73:11:62:86} = 5332; (x+n)=12x12=144; (d+n)=74x74=5476
(3,1,7) = {3:1:99:13:86:114} = 9804; (x+n)=14x14=196; (d+n)=100x100=10000
(3,1,8) = {3:1:129:15:114:146} = 16644; (x+n)=16x16=256; (d+n)=130x130=16900
(3,1,9) = {3:1:163:17:146:182} = 26572; (x+n)=18x18=324; (d+n)=164x164=26896
(3,1,10) = {3:1:201:19:182:222} = 40404; (x+n)=20x20=400; (d+n)=202x202=40804
oops. PMA here.
yes.
>We will explore that for any row n, the values of a,b and d will increase by one in a cell that is twice n to the right.
>If a cell has values at (e,n) then there will be values in a cell at (e+2n,n)
>Please explore that pattern.
Started looking into enumerating (e,n) cells other than n = 1, and ran into a problem. t increments irregularly. But there is a pattern. Perhaps someone else can assist in narrowing the formula down.
(1,5,2) = {1:5:4:3:1:17} = 17; (x+n)=8x8=64; (d+n)=9x9=81
(1,5,4) = {1:5:12:7:5:29} = 145; (x+n)=12x12=144; (d+n)=17x17=289 x + 4
(1,5,7) = {1:5:30:13:17:53} = 901; (x+n)=18x18=324; (d+n)=35x35=1225 x + 6
(1,5,9) = {1:5:46:17:29:73} = 2117; (x+n)=22x22=484; (d+n)=51x51=2601 x + 4
(1,5,12) = {1:5:76:23:53:109} = 5777; (x+n)=28x28=784; (d+n)=81x81=6561 x + 6
(1,13,3) = {1:13:6:5:1:37} = 37; (x+n)=18x18=324; (d+n)=19x19=361
(1,13,11) = {1:13:38:21:17:85} = 1445; (x+n)=34x34=1156; (d+n)=51x51=2601 x + 16
(1:13:16) = {1:13:68:31:37:125} = 4625; (x+n)=44x44=1936; (d+n)=81x81=6561 x + 10
(1:13:24) = {1:13:132:47:85:205} = 17425; (x+n)=60x60=3600; (d+n)=145x145=21025 x + 16
(1,17,7) = {1:17:18:13:5:65} = 325; (x+n)=30x30=900; (d+n)=35x35=1225
(1,17,11) = {1:17:34:21:13:89} = 1157; (x+n)=38x38=1444; (d+n)=51x51=2601 x + 8
(1,17,24) = {1:17:112:47:65:193} = 12545; (x+n)=64x64=4096; (d+n)=129x129=16641 x + 26
(1,17,28) = {1:17:144:55:89:233} = 20737; (x+n)=72x72=5184; (d+n)=161x161=25921 x + 8
Pattern: x changes by 2n over 2 records.
for (1,5), the total change is 10. x + 4 for the change from (1,5,2) to (1,5,4), then x + 6 from (1,5,4) to (1,5,7)
for (1,13), the total change is 26. x + 16 for the change from (1,13,3) to (1,13,11), then x + 10 from (1,13,11) to (1,13,16)
for (1,17), the total change is 34. x + 8 for the change from (1,17,7) to (1,17,11), then x + 26 from (1,17,11) to (1,17,24)
>We will explore that for any row n, the values of a,b and d will increase by one in a cell that is twice n to the right.
>If a cell has values at (e,n) then there will be values in a cell at (e+2n,n)
Exploring relationship between (e,n) and (e + 2n, n).
Starting with record (1,5,4) = {1:5:12:7:5:29} = 145; (x+n)=12x12=144; (d+n)=17x17=289
Record at e=1 + 2 * 5: (11,5,4) = {11:5:13:7:6:30} = 180; (x+n)=12x12=144; (d+n)=18x18=324
Record at e=1 + 4 * 5: (21,5,4) = {21:5:14:7:7:31} = 217; (x+n)=12x12=144; (d+n)=19x19=361
Record at e=1 + 6 * 5: (31,5,4) = {31:5:15:7:8:32} = 256; (x+n)=12x12=144; (d+n)=20x20=400
Record at e=1 + 8 * 5: (41,5,4) = {41:5:16:7:9:33} = 297; (x+n)=12x12=144; (d+n)=21x21=441
Record at e=1 + 10 * 5: (51,5,4) = {51:5:17:7:10:34} = 340; (x+n)=12x12=144; (d+n)=22x22=484
Questions are this stage:
- How do we determine valid n's for a given e?
- What is the x relationship between records at n 1. Looks like 2n defines the total x change between 2 records, but the breakdown is strange.
- how do we determine valid factors for a given c?
I don't think we have enough information yet to determine a from c,d,e.
Just using the formula to create an entry from (e,n,t) we can effectively create any record in the first row.
I think the problem is figuring out the jumps to the correct n levels, and then how to iterate based on x.
took a closer look at your pastebin.
Just an example. My theory is that the sequence adds up to 2n, then repeats.
Originally thought it was repeating every 2 records, but based on the following, it could be in groups of 2, 3 or even 6.
3 records 54 2x27.
[0,27] =18,18,18,18,18,18,18
2 records repeat to total 54.
[2,27] =22,10,44,10,44
[5,27] =7,40,14,40,14
[8,27] =10,34,20,34,20
[11,27] =23,8,46,8,46
[14,27] =16,22,32,22,32
[17,27] =19,16,38,16,38
[20,27] =14,26,28,26,28
[23,27] =25,4,50,4,50
[26,27] =26,2,52,2,52,2
[27,27] =27,18,18,18,18,18,18
[29,27] =49,10,44,10
[32,27] =20,14,40,14,40
[35,27] =17,20,34,20,34
6 records repeat to total 54.
[18,27] =12,12,6,12,6,12,6,12,6,12,6,12,6,12
Makes me wonder if this is the result of another equation as opposed to something that can be calculated directly.
>t+p, t+2p, t+3p,..
>and
>p+1-t, 2p+1-t, 3p+1-t,..
Put together a sample that shows how to step through the different factor methods.
The depth variable is the number of iterations.
———————————
TestPrimeResult for a=5, b=157.
P(e,n) =(1,365,14) = {1:365:28:27:1:785} = 785; (x+n)^2=392x392=153664; (d+n)^2=393x393=154449.
P(e,1) a=n =(1,1,14) = {1:1:392:27:365:421} = 153665; (x+n)^2=28x28=784; (d+n)^2=393x393=154449.
Factors by Method 1: newT = depth * p + t;
(1,1,2) = {1:1:8:3:5:13} = 65; (x+n)^2=4x4=16; (d+n)^2=9x9=81
(1,1,7) = {1:1:98:13:85:113} = 9605; (x+n)^2=14x14=196; (d+n)^2=99x99=9801
(1,1,12) = {1:1:288:23:265:313} = 82945; (x+n)^2=24x24=576; (d+n)^2=289x289=83521
Factors by Method 2: newT = depth * p + 1 - t;
(1,1,-1) = {1:1:2:-3:5:1} = 5; (x+n)^2=-2x-2=4; (d+n)^2=3x3=9
(1,1,4) = {1:1:32:7:25:41} = 1025; (x+n)^2=8x8=64; (d+n)^2=33x33=1089
(1,1,9) = {1:1:162:17:145:181} = 26245; (x+n)^2=18x18=324; (d+n)^2=163x163=26569
(1,1,14) = {1:1:392:27:365:421} = 153665; (x+n)^2=28x28=784; (d+n)^2=393x393=154449
p(e,n) answer =(1,53,12) = {1:53:28:23:5:157} = 785; (x+n)^2=76x76=5776; (d+n)^2=81x81=6561
>Since xx+e = 2na, every cell in row (e,1) contains a catalog of all values of na for a remainder e within the value a for those elements.
However, I found an example that doesn't quite fit the factor iteration pattern:
———————————
TestPrimeResult for a=17, b=53.
P(e,n) =(1,421,15) = {1:421:30:29:1:901} = 901; (x+n)^2=450x450=202500; (d+n)^2=451x451=203401.
P(e,1) a=n =(1,1,15) = {1:1:450:29:421:481} = 202501; (x+n)^2=30x30=900; (d+n)^2=451x451=203401.
All records where t < 15
(1,1,14) = {1:1:392:27:365:421} = 153665; (x+n)^2=28x28=784; (d+n)^2=393x393=154449;
(1,1,13) = {1:1:338:25:313:365} = 114245; (x+n)^2=26x26=676; (d+n)^2=339x339=114921;
(1,1,12) = {1:1:288:23:265:313} = 82945; (x+n)^2=24x24=576; (d+n)^2=289x289=83521;
(1,1,11) = {1:1:242:21:221:265} = 58565; (x+n)^2=22x22=484; (d+n)^2=243x243=59049;
(1,1,10) = {1:1:200:19:181:221} = 40001; (x+n)^2=20x20=400; (d+n)^2=201x201=40401;
(1,1,9) = {1:1:162:17:145:181} = 26245; (x+n)^2=18x18=324; (d+n)^2=163x163=26569;
(1,1,8) = {1:1:128:15:113:145} = 16385; (x+n)^2=16x16=256; (d+n)^2=129x129=16641;
(1,1,7) = {1:1:98:13:85:113} = 9605; (x+n)^2=14x14=196; (d+n)^2=99x99=9801;
(1,1,6) = {1:1:72:11:61:85} = 5185; (x+n)^2=12x12=144; (d+n)^2=73x73=5329;
(1,1,5) = {1:1:50:9:41:61} = 2501; (x+n)^2=10x10=100; (d+n)^2=51x51=2601;
(1,1,4) = {1:1:32:7:25:41} = 1025; (x+n)^2=8x8=64; (d+n)^2=33x33=1089;
(1,1,3) = {1:1:18:5:13:25} = 325; (x+n)^2=6x6=36; (d+n)^2=19x19=361;
(1,1,2) = {1:1:8:3:5:13} = 65; (x+n)^2=4x4=16; (d+n)^2=9x9=81;
(1,1,1) = {1:1:2:1:1:5} = 5; (x+n)^2=2x2=4; (d+n)^2=3x3=9;
p(e,n) answer =(1,5,7) = {1:5:30:13:17:53} = 901; (x+n)^2=18x18=324; (d+n)^2=35x35=1225
The factor of a = 17 doesn't exist in (1,1). Not sure how iterating factors is going help solve the prime problem.
There is a direct relationship between the t and x values. Given an e, you can compute easily from one to the other.
VQC has sent us down a path of using factors to navigate the (e,1) space.
I view the t values as normalized data for iteration purposes. X is more closely aligned to the dimensions of the squares.
>What exactly is the big square, and what exactly is the small square?
The entire space we are working in is calculated from the Difference of 2 squares.
c = (d+n)^2 - (x+n)^2.
i.e. c = big square - small square.
I've adjusted my test output to indicate these square values, as I believe they better describe the relationship between the grid entries than the actual keys.
I wrote a confusing post yesterday, that I thought would lead us to the prime solution.
Following is another attempt at my thinking:
Step 1: Starting from c = 758 and a = 1, we can create:
(1,365,14) = {1:365:28:27:1:785} = 785;
small square = (x+n)^2=392x392 = 153664.
large square = (d+n)^2=393x393 = 154449.
Step 2: substituting n for a and using the same x, we can create:
(1,1,14) = {1:1:392:27:365:421} = 153665
small square = (x+n)^2=28x28=784. (this value is the same in step 2)
large square = (d+n)^2=393x393=154449.
Notice that the small square here of 784 = the c value in the first step minus e. I believe this is always true when create records at the same x by substituting n for a.
Step 3: I created the correct record for e where a and b are the prime values we are looking for, and then switched the n.
(1,5,12) = {1:5:76:23:53:109} = 5777
small square = (x+n)^2=28x28=784; (same value from step 2)
large square = (d+n)^2=81x81=6561. (this value is the same in step 4).
Step 4: substitute n for a and using the same x for the record from step 3, we can create:
From (1,53,12) = {1:53:28:23:5:157} = 785;
small square = (x+n)^2=76x76=5776;
large square = (d+n)^2=81x81=6561. (same value from step 3)
For step 4, the prime solution record, we already know e, d, and c. We just need one more variable to solve this problem.
Based on the movement in the small and large squares, I thought that if we could figure out how to create the record in Step 3, then we would have this solved.
For the record in step 3, we know e, and now we know (x+n).
This may still be a wild goose chase.
I don't know about triangles. But the relationship between records seems to hold.
Here is one more example for c = 20413.
TestABSquared a=137, b=149
P(0,n) for c^2 =(0,208324872,10207) = {0:208324872:20413:20412:1:416690569} = 416690569
P(0,n) a^2, b^2 =(0,72,823) = {0:72:20413:1644:18769:22201} = 416690569
n division =208324872/72 = 2893401;
x diff =20412-1644 = 18768;
xdiffBy24 =18768/24 = 782;
p(e,n) a*b =(249,1,3) = {249:1:142:5:137:149} = 20413
Next step is to figure out movement in the (0,n) column.
not yet. will look into.
I think I'm on the same page now.
65=5x13
1 x c =(1,25,4) = {1:25:8:7:1:65} = 65
a x b =(1,1,2) = {1:1:8:3:5:13} = 65
c x c =(0,0,1) = {0:0:65:0:65:65} = 4225
a x cb =(0,360,31) = {0:360:65:60:5:845} = 4225
b x ca =(0,104,27) = {0:104:65:52:13:325} = 4225
1 x cc =(0,2048,33) = {0:2048:65:64:1:4225} = 4225
I changed the order, and added in the aa x bb record. I think that's the one we're looking for.
Square root of a, Square root of b gets us the solution.
65=5x13
1 x c =(1,25,4) = {1:25:8:7:1:65} = 65
a x b *=(1,1,2) = {1:1:8:3:5:13} = 65
c x c =(0,0,1) = {0:0:65:0:65:65} = 4225
aa x bb =(0,32,21) = {0:32:65:40:25:169} = 4225
b x ca =(0,104,27) = {0:104:65:52:13:325} = 4225
a x cb =(0,360,31) = {0:360:65:60:5:845} = 4225
1 x cc =(0,2048,33) = {0:2048:65:64:1:4225} = 4225
Here are all my samples. * indicates important record
Really is beautiful.
65=5x13
1 x c *=(1,25,4) = {1:25:8:7:1:65} = 65
a x b *=(1,1,2) = {1:1:8:3:5:13} = 65
c x c =(0,0,1) = {0:0:65:0:65:65} = 4225
aa x bb *=(0,32,21) = {0:32:65:40:25:169} = 4225
b x ca =(0,104,27) = {0:104:65:52:13:325} = 4225
a x cb =(0,360,31) = {0:360:65:60:5:845} = 4225
1 x cc *=(0,2048,33) = {0:2048:65:64:1:4225} = 4225
145=5x29
1 x c *=(1,61,6) = {1:61:12:11:1:145} = 145
a x b *=(1,5,4) = {1:5:12:7:5:29} = 145
c x c =(0,0,1) = {0:0:145:0:145:145} = 21025
aa x bb *=(0,288,61) = {0:288:145:120:25:841} = 21025
b x ca =(0,232,59) = {0:232:145:116:29:725} = 21025
a x cb =(0,1960,71) = {0:1960:145:140:5:4205} = 21025
1 x cc *=(0,10368,73) = {0:10368:145:144:1:21025} = 21025
785=5x157
1 x c *=(1,365,14) = {1:365:28:27:1:785} = 785
a x b *=(1,53,12) = {1:53:28:23:5:157} = 785
c x c =(0,0,1) = {0:0:785:0:785:785} = 616225
aa x bb *=(0,11552,381) = {0:11552:785:760:25:24649} = 616225
b x ca =(0,1256,315) = {0:1256:785:628:157:3925} = 616225
a x cb =(0,60840,391) = {0:60840:785:780:5:123245} = 616225
1 x cc *=(0,307328,393) = {0:307328:785:784:1:616225} = 616225
901=17x53
1 x c *=(1,421,15) = {1:421:30:29:1:901} = 901
a x b *=(1,5,7) = {1:5:30:13:17:53} = 901
c x c =(0,0,1) = {0:0:901:0:901:901} = 811801
aa x bb *=(0,648,307) = {0:648:901:612:289:2809} = 811801
b x ca =(0,6784,425) = {0:6784:901:848:53:15317} = 811801
a x cb =(0,22984,443) = {0:22984:901:884:17:47753} = 811801
1 x cc *=(0,405000,451) = {0:405000:901:900:1:811801} = 811801
6107=31x197
1 x c *=(23,2976,39) = {23:2976:78:77:1:6107} = 6107
a x b *=(23,36,24) = {23:36:78:47:31:197} = 6107
c x c =(0,0,1) = {0:0:6107:0:6107:6107} = 37295449
aa x bb *=(0,13778,2574) = {0:13778:6107:5146:961:38809} = 37295449
b x ca =(0,88650,2956) = {0:88650:6107:5910:197:189317} = 37295449
a x cb =(0,595448,3039) = {0:595448:6107:6076:31:1203079} = 37295449
1 x cc *=(0,18641618,3054) = {0:18641618:6107:6106:1:37295449} = 37295449
20413=137x149
1 x c *=(249,10065,71) = {249:10065:142:141:1:20413} = 20413
a x b *=(249,1,3) = {249:1:142:5:137:149} = 20413
c x c =(0,0,1) = {0:0:20413:0:20413:20413} = 416690569
aa x bb *=(0,72,823) = {0:72:20413:1644:18769:22201} = 416690569
b x ca =(0,1377952,10133) = {0:1377952:20413:20264:149:2796581} = 416690569
a x cb =(0,1500424,10139) = {0:1500424:20413:20276:137:3041537} = 416690569
1 x cc *=(0,208324872,10207) = {0:208324872:20413:20412:1:416690569} = 416690569
then we need a formula!
no idea. I was looking for relationships between the records, and that number came out.