20210108, 20:21  #111  
Nov 2016
2820_{10} Posts 
Quote:
This prime is likely the thirdlargest "base 13 minimal prime (start with base+1)" (there is a larger probable prime 80_{32017}111, and there is an unsolved family in base 13: 9{5}) Last fiddled with by sweety439 on 20210108 at 20:23 

20210108, 20:37  #112 
"Carlos Pinho"
Oct 2011
Milton Keynes, UK
4886_{10} Posts 
Have you thought about writing a book about all of this exciting stuff?

20210108, 21:11  #113 
"Curtis"
Feb 2005
Riverside, CA
3·1,579 Posts 
14 posts on a single topic in 36 hours. Are you trying to get banned? It's working.

20210109, 10:34  #114 
Nov 2016
2^{2}·3·5·47 Posts 
I have a pdf file for the proof (not complete, continue updating), I will complete the proof for bases 7, 9, 12

20210109, 10:38  #115 
Nov 2016
2^{2}×3×5×47 Posts 
It is conjectured that for all simple families x{y}z cannot be proved as only contain composites (for numbers > base) in one of these four ways:
** Periodic sequence p of prime divisors with p(n)  (xyyy...yyyz with n y's) ** Algebraic factors (e.g. differenceofsquares factorization, differenceofcubes factorization, sumofcubes factorization, differenceof5thpowers factorization, sumof5thpowers factorization, Aurifeuillian factorization of x^4+4*y^4, etc.) of x{y}z ** The combine of the above two ways (like the case of {B}9B in base 12) ** Reduced to (b^(r*n+s)+1)/gcd(b+1,2), and r*n+s can never be power of 2 (like the case of 8{0}1 in base 128) Then x{y}z contain primes (for numbers > base). Last fiddled with by sweety439 on 20210405 at 16:12 
20210109, 10:53  #116 
Nov 2016
2820_{10} Posts 
The simple families x{y}z (where x and z are strings of base b digits, y is base b digit) in base b are of the form (a*b^n+c)/gcd(a+c,b1) (where a>=1, c != 0, gcd(a,c) = 1, gcd(b,c) = 1), this number has algebra factors if and only if:
either * there is an integer r>1 such that both a*b^n and c are perfect rth powers or * a*b^n*c is of the form 4*m^4 with integer m If (a*b^n+c)/gcd(a+c,b1) (where a>=1, c != 0, gcd(a,c) = 1, gcd(b,c) = 1) has algebra factors, then it must be composite, the only exception is when it is either GFN (generalized Fermat number) base b or GRU (generalized repunit number) base b, in these two cases this number may be prime, the only condition is the n is power of 2 if it is GFN, and the n is prime if it is GRU GFNs and GRUs are the only simple families x{y}z (where x and z are strings of base b digits, y is base b digit) in base b which are also cyclotomic numbers (i.e. numbers of the form Phi(n,b)/gcd(Phi(n,b),n), where Phi is cyclotomic polynomial) or Zsigmondy numbers Zs(n,b,1) (see Zsigmondy's theorem) GFNs and GRUs in bases 2<=b<=36: Code:
base GFN family GRU family 2 1{0}1 {1} 3 {1}2 {1} 4 1{0}1 1{3}, {2}3 5 {2}3 {1} 6 1{0}1 {1} 7 {3}4 {1} 8 2{0}1, 4{0}1 1{7}, 3{7} 9 {4}5 1{4}, {6}7 10 1{0}1 {1} 11 {5}6 {1} 12 1{0}1 {1} 13 {6}7 {1} 14 1{0}1 {1} 15 {7}8 {1} 16 1{0}1 1{F}, 7{F}, {A}B, 2{A}B 17 {8}9 {1} 18 1{0}1 {1} 19 {9}A {1} 20 1{0}1 {1} 21 {A}B {1} 22 1{0}1 {1} 23 {B}C {1} 24 1{0}1 {1} 25 {C}D 1{6}, {K}L 26 1{0}1 {1} 27 1{D}E, 4{D}E 1{D}, 4{D} 28 1{0}1 {1} 29 {E}F {1} 30 1{0}1 {1} 31 {F}G {1} 32 2{0}1, 4{0}1, 8{0}1, G{0}1 1{V}, 3{V}, 7{V}, F{V} 33 {G}H {1} 34 1{0}1 {1} 35 {H}I {1} 36 1{0}1 1{7}, {U}V Code:
base GFN family GRU family 4 {1} 8 1{0}1 {1} 9 {1} 16 {1}, 1{5}, {C}D 25 {1} 27 {D}E {1} 32 1{0}1 {1} 36 {1} Last fiddled with by sweety439 on 20210326 at 13:23 
20210109, 21:18  #117 
Nov 2016
2^{2}·3·5·47 Posts 
These bases 2<=b<=1024 have unsolved families which are GFNs:
{31, 32, 37, 38, 50, 55, 62, 63, 67, 68, 77, 83, 86, 89, 91, 92, 93, 97, 98, 99, 104, 107, 109, 117, 122, 123, 125, 127, 128, 133, 135, 137, 143, 144, 147, 149, 151, 155, 161, 168, 177, 179, 182, 183, 186, 189, 193, 197, 200, 202, 207, 211, 212, 213, 214, 215, 216, 217, 218, 223, 225, 227, 233, 235, 241, 244, 246, 247, 249, 252, 255, 257, 258, 263, 265, 269, 273, 277, 281, 283, 285, 286, 287, 291, 293, 294, 297, 298, 302, 303, 304, 307, 308, 311, 319, 322, 324, 327, 338, 343, 344, 347, 351, 354, 355, 356, 357, 359, 361, 362, 367, 368, 369, 377, 380, 381, 383, 385, 387, 389, 390, 393, 394, 397, 398, 401, 402, 404, 407, 410, 411, 413, 416, 417, 421, 422, 423, 424, 437, 439, 443, 446, 447, 450, 454, 457, 458, 465, 467, 468, 469, 473, 475, 480, 481, 482, 483, 484, 489, 493, 495, 497, 500, 509, 511, 512, 514, 515, 518, 524, 528, 530, 533, 534, 538, 541, 547, 549, 552, 555, 558, 563, 564, 572, 574, 578, 580, 590, 591, 593, 597, 601, 602, 603, 604, 608, 611, 615, 619, 620, 621, 622, 625, 626, 627, 629, 632, 633, 635, 637, 638, 645, 647, 648, 650, 651, 653, 655, 659, 662, 663, 666, 667, 668, 670, 671, 673, 675, 678, 679, 683, 684, 687, 691, 692, 693, 694, 697, 698, 706, 707, 709, 712, 717, 720, 722, 724, 731, 733, 734, 735, 737, 741, 743, 744, 746, 749, 752, 753, 754, 755, 757, 759, 762, 765, 766, 767, 770, 771, 773, 775, 777, 783, 785, 787, 792, 793, 794, 797, 801, 802, 806, 807, 809, 812, 813, 814, 817, 818, 823, 825, 836, 840, 842, 844, 848, 849, 851, 853, 854, 865, 867, 868, 870, 872, 873, 877, 878, 887, 888, 889, 893, 896, 897, 899, 902, 903, 904, 907, 908, 911, 915, 922, 923, 924, 926, 927, 932, 933, 937, 938, 939, 941, 942, 943, 944, 945, 947, 948, 953, 954, 957, 958, 961, 964, 967, 968, 974, 975, 977, 978, 980, 983, 987, 988, 993, 994, 998, 999, 1000, 1002, 1003, 1005, 1006, 1009, 1014, 1016, 1017, 1024} Such families are: * 4:{0}:1, 16:{0}:1 for b = 32 * 12:{62}:63 for b = 125 * 16:{0}:1 for b = 128 * 36:{0}:1 for b = 216 * 24:{171}:172 for b = 343 * 2:{0}:1, 4:{0}:1, 16:{0}:1, 32:{0}:1, 256:{0}:1 for b = 512 * 10:{0}:1, 100:{0}:1 for b = 1000 * 4:{0}:1, 16:{0}:1, 256:{0}:1 for b = 1024 * 1:{0}:1 for other even bases b * {((b1)/2)}:((b+1)/2) for other odd bases b These bases 2<=b<=1024 have unsolved families which are GRUs: {185, 243, 269, 281, 380, 384, 385, 394, 452, 465, 511, 574, 601, 631, 632, 636, 711, 713, 759, 771, 795, 861, 866, 881, 938, 948, 951, 956, 963, 1005, 1015} Such families are: * 40:{121} for b = 243 * {1} for other bases b Last fiddled with by sweety439 on 20210325 at 10:46 
20210109, 21:44  #118 
"Sam"
Nov 2016
2·163 Posts 
Here's a suggestion:
Instead of constantly posting search limits and reservations that can be done in minutes, use a doc or pdf instead. Attach or provide the link in a single post. If you need to edit the link or update the attachment, edit the post instead of creating a new post. It saves time and space! The updates you see on CRUS or other prime search projects take at least several months, if not years to complete. Now your searches and updates on the other hand, can be done in minutes. Why post something that is so trivial that anyone who wants to do it can do it in such a short amount of time? 
20210110, 12:09  #119  
Nov 2016
5404_{8} Posts 
Quote:
Well, the proofs for base 2, 3, 4 are really trivial, but they are part of the project, I want to store these proofs, and the pdf file was made recently 

20210110, 19:27  #120 
Nov 2016
2^{2}·3·5·47 Posts 
Now, we proved the set of minimal primes (start with b+1, which is equivalent to start with b, if b is composite) of base b=12:
Code:
11 15 17 1B 25 27 31 35 37 3B 45 4B 51 57 5B 61 67 6B 75 81 85 87 8B 91 95 A7 AB B5 B7 221 241 2A1 2B1 2BB 401 421 447 471 497 565 655 665 701 70B 721 747 771 77B 797 7A1 7BB 907 90B 9BB A41 B21 B2B 2001 200B 202B 222B 229B 292B 299B 4441 4707 4777 6A05 6AA5 729B 7441 7B41 929B 9777 992B 9947 997B 9997 A0A1 A201 A605 A6A5 AA65 B001 B0B1 BB01 BB41 600A5 7999B 9999B AAAA1 B04A1 B0B9B BAA01 BAAA1 BB09B BBBB1 44AAA1 A00065 BBBAA1 AAA0001 B00099B AA000001 BBBBBB99B B0000000000000000000000000009B 400000000000000000000000000000000000000077 
20210110, 19:28  #121 
Nov 2016
2^{2}×3×5×47 Posts 
There are totally 106 minimal primes (start with 2 digits) in base 12, there are 77 such primes in base 10

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