These Sierpinski bases (up to 1024) cannot be proven with current knowledge and technology, since they have either GFN (for even bases) or half GFN (odd bases) remain: (half GFN is much worse, since for these (probable) primes, the divisor gcd(k+-1,b-1) is not 1 (it is 2), and when n is large (for all numbers of the form (k*b^n+-1)/gcd(k+-1,b-1) whose gcd(k+-1,b-1) is not 1) the known primality tests for such a number are too inefficient to run. In this case one must resort to a probable primality test such as a Miller–Rabin test, unless a divisor of the number can be found)

[CODE]
b k
2 65536
6 1296
10 100
12 12
15 225
18 18
22 22
31 1
32 4
36 1296
37 37
38 1
40 1600
42 42
50 1
52 52
55 1
58 58
60 60
62 1
63 1
66 4356
67 1
68 1
70 70
72 72
77 1
78 78
83 1
86 1
89 1
91 1
92 1
93 93
97 1
98 1
99 1
104 1
107 1
108 108
109 1
117 117
122 1
123 1
124 15376
126 15876
127 1
128 16
135 1
136 136
137 1
138 138
143 1
144 1
147 1
148 148
149 1
151 1
155 1
161 1
166 166
168 1
178 178
179 1
180 1049760000
182 1
183 1
186 1
189 1
192 192
193 193
196 196
197 1
200 1
202 1
207 1
210 1944810000
211 1
212 1
214 1
215 1
216 36
217 217
218 1
222 222
223 1
225 225
226 226
227 1
232 232
233 1
235 1
241 1
243 27
244 1
246 1
247 1
249 1
252 1
255 1
257 1
258 1
262 262
263 1
265 1
268 268
269 1
273 273
280 78400
281 1
282 282
283 1
285 1
286 1
287 1
291 1
293 1
294 1
298 1
302 1
303 1
304 1
307 1
308 1
310 310
311 1
316 316
319 1
322 1
324 1
327 1
336 336
338 1
343 49
344 1
346 346
347 1
351 1
354 1
355 1
356 1
357 357
358 358
359 1
361 361
362 1
366 366
367 1
368 1
369 1
372 372
377 1
380 1
381 381
383 1
385 385
387 1
388 388
389 1
390 1
393 393
394 1
397 397
398 1
401 1
402 1
404 1
407 1
408 408
410 1
411 1
413 1
416 1
417 1
418 418
420 176400
422 1
423 1
424 1
437 1
438 438
439 1
443 1
446 1
447 1
450 1
454 1
457 457
458 1
460 460
462 462
465 465
467 1
468 1
469 1
473 1
475 1
480 1
481 481
482 1
483 1
484 1
486 486
489 1
493 1
495 1
497 1
500 1
509 1
511 1
512 2, 4, 16
514 1
515 1
518 1
522 522
524 1
528 1
530 1
533 1
534 1
538 1
541 541
546 546
547 1
549 1
552 1
555 1
558 1
563 1
564 1
570 324900
572 1
574 1
578 1
580 1
586 586
590 1
591 1
593 1
597 1
600 129600000000
601 1
602 1
603 1
604 1
606 606
608 1
611 1
612 612
615 1
618 618
619 1
620 1
621 621
622 1
626 1
627 1
629 1
630 630
632 1
633 633
635 1
637 1
638 1
645 1
647 1
648 1
650 1
651 1
652 652
653 1
655 1
658 658
659 1
660 660
662 1
663 1
666 1
667 1
668 1
670 1
671 1
672 672
675 1
678 1
679 1
683 1
684 1
687 1
691 1
692 1
694 1
698 1
706 1
707 1
708 708
709 1
712 1
717 717
720 1
722 1
724 1
731 1
734 1
735 1
737 1
741 1
743 1
744 1
746 1
749 1
752 1
753 1
754 1
755 1
756 756
759 1
762 1
765 765
766 1
767 1
770 1
771 1
773 1
775 1
777 777
783 1
785 1
787 1
792 1
793 793
794 1
796 796
797 1
801 801
802 1
806 1
807 1
809 1
812 1
813 1
814 1
817 817
818 1
820 820
822 822
823 1
825 1
836 1
838 838
840 1
842 1
844 1
848 1
849 1
851 1
852 852
853 1
854 1
858 858
865 865
867 1
868 1
870 1
872 1
873 1
878 1
880 880
882 882
886 886
887 1
888 1
889 1
893 1
896 1
897 897
899 1
902 1
903 1
904 1
907 1
908 1
910 828100
911 1
915 1
922 1
923 1
924 1
926 1
927 1
932 1
933 933
937 1
938 1
939 1
941 1
942 1
943 1
944 1
945 1
947 1
948 1
953 1
954 1
958 1
961 1
964 1
966 870780120336
967 1
968 1
970 970
974 1
975 1
977 1
978 1
980 1
983 1
987 1
988 1
993 1
994 1
998 1
999 1
1000 10
1002 1
1003 1
1005 1005
1006 1
1008 1008
1009 1
1012 1012
1014 1
1016 1
1017 1017
1020 1020
1024 4, 16
[/CODE]

We don't know the conjectures yet on bases 66, 120, 156, 180, 210, 240, 280, 330, 358, 420, 456, 462, 546, ... and Sierp 190, 540 and Riesel 276, 486. They are definitely k > 1M. They're just too unmanageable with today's computing power and knowledge. That said, if people are open to the idea, I might suggest searching some of those up to k=100K at some point to get future generations of prime searchers started on those huge efforts. To me, the main thing is collecting data for us and for future generations to analyze.

There are upper bounds of these conjectures: (the lower bounds of these conjectures are all 1000001, since all k<=1M are checked)

S66: 21314443 (if not this number, then must be == 4 mod 5 or == 12 mod 13)
S120: 374876369 (if not this number, then must be == 6 mod 7 or == 16 mod 17)
S156: 18406311208 (if not this number, then must be == 4 mod 5 or == 30 mod 31)
S180: 1679679 (if not this number, then must be == 178 mod 179)
S190: 3146151 (if not this number, then must be == 2 mod 3 or == 6 mod 7)
S210: 147840103 (if not this number, then must be == 10 mod 11 or == 18 mod 19)
S240: 1722187 (if not this number, then must be == 238 mod 239)
S280: 82035074042274 (if not this number, then must be == 2 mod 3 or == 30 mod 31)
S330: 16636723 (if not this number, then must be == 6 mod 7 or == 46 mod 47)
S358: 27478218 (if not this number, then must be == 2 mod 3 or == 6 mod 7 or == 16 mod 17)
S420: 2288555 (if not this number, then must be == 418 mod 419)
S456: 14836963 (if not this number, then must be == 4 mod 5 or == 6 mod 7 or == 12 mod 13)
S462: 6880642 (if not this number, then must be == 460 mod 461)
S540: 1091739 (if not this number, then must be == 6 mod 7 or == 10 mod 11)
S546: 45119296 (if not this number, then must be == 4 mod 5 or == 108 mod 109)

R66: 101954772 (if not this number, then must be == 1 mod 5 or == 1 mod 13)
R120: 166616308 (if not this number, then must be == 1 mod 7 or == 1 mod 17)
R156: 2113322677 (if not this number, then must be == 1 mod 5 or == 1 mod 31)
R180: 7674582 (if not this number, then must be == 1 mod 179)
R210: 80176412 (if not this number, then must be == 1 mod 11 or == 1 mod 19)
R240: 2952972 (if not this number, then must be == 1 mod 239)
R276: 1552307 (if not this number, then must be == 1 mod 5 or == 1 mod 11)
R280: 513613045571841 (if not this number, then must be == 1 mod 3 or == 1 mod 31)
R330: 16527822 (if not this number, then must be == 1 mod 7 or == 1 mod 47)
R358: 27606383 (if not this number, then must be == 1 mod 3 or == 1 mod 7 or == 1 mod 17)
R420: 6548233 (if not this number, then must be == 1 mod 419)
R456: 76303920 (if not this number, then must be == 1 mod 5 or == 1 mod 7 or == 1 mod 13)
R462: 2924772 (if not this number, then must be == 1 mod 461)
R486: 1525283 (if not this number, then must be == 1 mod 5 or == 1 mod 97)

Any Riesel k that is a perfect square will always be lower weight than average because the even-n have algebraic factors.

A Sierpinski form (k*b^n+1)/gcd(k+1,b-1) has algebra factors if and only if k*b^n is perfect odd power [URL="https://oeis.org/A070265"]A070265[/URL] or of the form 4*m^4 [URL="https://oeis.org/A141046"]A101046[/URL]

A Riesel form (k*b^n-1)/gcd(k-1,b-1) has algebra factors if and only if k*b^n is perfect power [URL="https://oeis.org/A001597"]A001597[/URL]

Situations in which srsieve will make that statement:

1. For any k on any Riesel base that is a perfect square, when sieving, all n-values that are divisible by 2 can be manually removed.
2. For any k on any Sierpinski base that is of the form 4*m^4, all n-values that are divisible by 4 can be manually removed.
3. For any k on any (Riesel or Sierpinski) base that is a perfect cube, all n-values that are divisible by 3 can be manually removed.
4. (etc.) for k's that are perfect 5th powers, 7th powers, 11th powers, or any prime power where p=power, any n's divisible by p can be manually removed.

Taken to an extreme, for k=2048, which is 2^11, you could manually eliminate all n-values that are divisible by 11.

A special case is k=1:

* For Riesel base, all n-values that are not prime can be manually removed.
* For Sierpinski base, all n-values that are not power of 2 can be manually removed.

To put the above in a different way: You can only eliminate the k if manually removing all of these n-values leaves you with no n's remaining, which would have been the case had you attempted to sieve 900*67^n-1. Had you sieved it, you would have ended up with a sieve file with very few EVEN n-values remaining and zero ODD n-values remaining. Once you manually removed the n's divisible by 2, you would have had nothing left. That means that the k-value can be removed from conjecture testing because it has partial algebraic factors that combine with a numeric factor to make a full "covering set" of factors.

Analysis on both k=125 and 729 shows that they should remain because there is no factor or factors that eliminate the n's that the algebraic factors do not.

When sieving, as per the above, on k=125, you can manually remove all n-values divisible by 3. On k=729, that is one of the few that you can eliminate n-values that are divisible by 2 -or- that are divisible by 3. In effect, you're only left with n-values that are n==(1 or 5 mod 6). If you manually remove those n's, you'll stop getting that message from srsieve.

k=729 would normally be extremely low weight except for the fact that it is divisible by 3, which eliminates any possibility of a factor of 3 for all n-values. That increases the weight to something like a k that is a perfect square.

A k-value that would be extremely low-weight is one that is a perfect square and cube but is not divisible by 2 or 3. I think the lowest one of that nature would be 5^6=15625, which would only be an issue on very few bases.

Exclusions:

k's that are multiples of the base where (k+-1)/gcd(k+-1,b-1) (+ for Sierp, - for Riesel) is not prime.

I'll give some examples for S22:

k=44, 154, 220, 242, 264, 374, 440 would be excluded from testing because 45/3, 155, 221, 243/3, 265, 375/3, 441/21 are not primes.

k=22, 66, 88, 110, 132, 176, 198, 286, 308, 330, 352, 396, 418 would be INcluded in testing because 23, 67, 89, 111/3, 133/7, 177/3, 199, 287/7, 309/3, 331, 353, 397, 419 are primes.

The exclusions for multiples of the base are the same on all bases. Check for (k-1)/gcd(k+-1,b-1) being prime on the Riesel side and check for (k+1)/gcd(k+-1,b-1) being prime on the Sierp side. If it's prime, include it; if it's not prime, exclude it.

So far, there is only two Riesel k and base <= 128 where algebraic factors on even-n combine with a covering set of MORE than one factor on odd-n to eliminate a k. That is 1369*30^n-1 and (400*88^n-1)/3

(101*73^2146-1)/4 is (probable) prime!!!

R73 has only k=79 remain, reserve it to n=10K

[QUOTE=sweety439;549922](101*73^2146-1)/4 is (probable) prime!!!

R73 has only k=79 remain, reserve it to n=10K[/QUOTE]

[URL="https://docs.google.com/document/d/e/2PACX-1vR3ubC0-QCspbRFX-YnLRxPLm_Y448tN8BkLjyIeKsVYbb7EOrLYhw5tw6T-LV6SPlXLCncZ4OqEDGd/pub"]https://docs.google.com/document/d/e/2PACX-1vR3ubC0-QCspbRFX-YnLRxPLm_Y448tN8BkLjyIeKsVYbb7EOrLYhw5tw6T-LV6SPlXLCncZ4OqEDGd/pub[/URL]

Newest status of Riesel problems

Checking whether a k-value makes a full covering set with algebraic factors not always very easy. The way I do it is to look for patterns in the factors of the various n-values for specific k-values. If there are algebraic factors, it's most common for them to be in a pattern of f*(f+2), i.e.:
11*13
179*181
etc.

In other cases there may be a consistent steady increase in the differences of their factors, which is especially tricky to find but indicates the existence of algebraic factors.

e.g. for the case R15 k=47
n-value : factors
1 : 2^5 · 11
2 : 17 · 311
3 : 2^4 · 4957
4 : 31 · 38377
6 : 11 · 43 · 565919
8 : 199 · 1627 · 186019
10 : 17 · 61 · 13067776451
12 : 37 · 82406457849451
20 : 15061 · 236863181 · 2190492030407

Analysis:
For n=1 & 3 (and all odd n), all values are divisible by 2 so we only consider even n's.
For n=4, the two prime factors does not close.
For n=6 & 10, multiplying the 2 lower prime factors together does not come close to the higher prime factor so little chance of algebraic factors.
For n=12, the large lowest prime factor that bears no relation to the other prime factor means that there is unlikely to be a pattern to the occurrences of large prime factors so there must be a prime at some point.

R33 k=257:
n-value : factors
1 : 5 · 53
2 : 2 · 4373
3 : 397 · 727
4 : 2^2 · 2381107
5 : 5^3 · 7 · 359207
7 : 11027 · 31040117
15 : 13337 · 706661 · 51076716238627
19 : 38231 · 14932493857679888742000509
For n=15 & 19 same explanation as R15 k=47

R36 k=1555:
n-value : factors
1 : 11 · 727
2 : 31 · 37 · 251
3 : 67 · 154691
4 : 37 · 127 · 271 · 293
7 : 4943 · 3521755879
9 : 59 · 382386761790283
For n=7 & 9 same explanation as R15 k=47


The prime factors for n=12, n=15, and n=7 respectively make it clear to me that these k-values should all yield primes at some point so you are correct to include them as remaining.

The higher-math folks may be able to chime in and answer why there are an abnormally large # of k's that are perfect squares that end up remaining even though they don't have known algebraic factors for most bases. IMHO, it's because there ARE algebraic factors for a subset of the universe of n-values on them but not for all of the n-values. Hence they are frequently lower weight than the other k's but NOT zero weight and so should eventually yield a prime.

[QUOTE=sweety439;549958]Checking whether a k-value makes a full covering set with algebraic factors not always very easy. The way I do it is to look for patterns in the factors of the various n-values for specific k-values. If there are algebraic factors, it's most common for them to be in a pattern of f*(f+2), i.e.:
11*13
179*181
etc.

In other cases there may be a consistent steady increase in the differences of their factors, which is especially tricky to find but indicates the existence of algebraic factors.

e.g. for the case R15 k=47
n-value : factors
1 : 2^5 · 11
2 : 17 · 311
3 : 2^4 · 4957
4 : 31 · 38377
6 : 11 · 43 · 565919
8 : 199 · 1627 · 186019
10 : 17 · 61 · 13067776451
12 : 37 · 82406457849451
20 : 15061 · 236863181 · 2190492030407

Analysis:
For n=1 & 3 (and all odd n), all values are divisible by 2 so we only consider even n's.
For n=4, the two prime factors does not close.
For n=6 & 10, multiplying the 2 lower prime factors together does not come close to the higher prime factor so little chance of algebraic factors.
For n=12, the large lowest prime factor that bears no relation to the other prime factor means that there is unlikely to be a pattern to the occurrences of large prime factors so there must be a prime at some point.

R33 k=257:
n-value : factors
1 : 5 · 53
2 : 2 · 4373
3 : 397 · 727
4 : 2^2 · 2381107
5 : 5^3 · 7 · 359207
7 : 11027 · 31040117
15 : 13337 · 706661 · 51076716238627
19 : 38231 · 14932493857679888742000509
For n=15 & 19 same explanation as R15 k=47

R36 k=1555:
n-value : factors
1 : 11 · 727
2 : 31 · 37 · 251
3 : 67 · 154691
4 : 37 · 127 · 271 · 293
7 : 4943 · 3521755879
9 : 59 · 382386761790283
For n=7 & 9 same explanation as R15 k=47


The prime factors for n=12, n=15, and n=7 respectively make it clear to me that these k-values should all yield primes at some point so you are correct to include them as remaining.

The higher-math folks may be able to chime in and answer why there are an abnormally large # of k's that are perfect squares that end up remaining even though they don't have known algebraic factors for most bases. IMHO, it's because there ARE algebraic factors for a subset of the universe of n-values on them but not for all of the n-values. Hence they are frequently lower weight than the other k's but NOT zero weight and so should eventually yield a prime.[/QUOTE]

In fact, (k*b^n+1)/gcd(k+1,b-1) has algebra factors if and only if k*b^n is either perfect odd power or of the form 4*m^4, and (k*b^n-1)/gcd(k-1,b-1) has algebra factors if and only if k*b^n is perfect power, thus forms like (257*33^n-1)/32 cannot have algebra factors, since there is no n such that 257*33^n is perfect power (after all, the exponent of the prime 257 in the prime factorization of 257*33^n is 1 for all n, but any prime factor in the prime factorization of a perfect power cannot have exponent 1).

[QUOTE=sweety439;549922](101*73^2146-1)/4 is (probable) prime!!!

R73 has only k=79 remain, reserve it to n=10K[/QUOTE]

(79*73^9339-1)/6 is (probable) prime!!!

R73 is proven!!!