[QUOTE=Uncwilly;550359]Why do you insist on quoting whole posts all the time?[/QUOTE]

Why shouldn't I do this?

[QUOTE=sweety439;550364]Also these cases:

S15 k=343:

since 343 is cube, all n divisible by 3 have algebra factors, and we only want to know whether it has a covering set of primes for all n not divisible by 3, if so, then this k makes a full covering set with algebraic factors and be excluded from the conjecture; if not, then this k does not make a full covering set with algebraic factors and be included from the conjecture.

n-value : factors
1 : 31 · 83
2 : 2^2 · 11 · 877
4 : 2^2 · 809 · 2683
5 : 811 · 160583
7 : 11^2 · 242168453
11 : 31 · 101 · 25357 · 18684739
13 : 397 · 1281101 · 656261029
17 : 11 · 27479311 · 55900668804553
29 : 53 · 197741 · 209188613429183386499227445981
35 : 1337724923 · 18667724069720862256321575167267431
43 : 20943991 · 3055827403675875709696160949928034201885723243
61 : 23539 · (a 61-digit prime)

and it does not appear to be any covering set of primes, so there must be a prime at some point.

S61 k=324:

since 324 is of the form 4*m^4, all n divisible by 4 have algebra factors, and we only want to know whether it has a covering set of primes for all n not divisible by 4, if so, then this k makes a full covering set with algebraic factors and be excluded from the conjecture; if not, then this k does not make a full covering set with algebraic factors and be included from the conjecture.

n-value : factors
1 : 59 · 67
2 : 41 · 5881
3 : 13 · 1131413
5 : 5 · 7 · 1563709723
6 : 13 · 256809250661
7 : 23 · 1255679 · 7051433
13 : 191 · 7860337 · 27268229 · 256289843
14 : 1540873 · 1698953 · 244480646906833
31 : 1888149043321 · 441337391577139 · 1721840403480692512106884569347
34 : 10601 · 174221 · (a 54-digit prime)

and it does not appear to be any covering set of primes, so there must be a prime at some point.[/QUOTE]

However, there are cases that cannot have prime:

R88 k=400:

since 400 is square, all even n have algebra factors, and we only want to know whether it has a covering set of primes for all odd n, if so, then this k makes a full covering set with algebraic factors and be excluded from the conjecture; if not, then this k does not make a full covering set with algebraic factors and be included from the conjecture.

n-value : factors
1 : 3 · 3911
3 : 7 · 101 · 128519
5 : 13 · 12119 · 4466239
7 : 3^3 · 3799333 · 53118563
9 : 7 · 167 · 36096764394509957
11 : 13 · 25136498347515268468841
13 : 3 · 1877 · 1156907 · 5976181 · 64998862429
15 : 7^2 · 239 · 6079 · 483551 · 1173283 · 485185295929
17 : 13 · 7417 · 1573883316708285469700513209073

and there is covering set {3, 7, 13} (n == 1 mod 6: factor of 3; n == 3 mod 6: factor of 7; n == 5 mod 6: factor of 13), thus for R88, k=400 proven composite by partial algebra factors.

R10 k=343:

since 343 is cube, all n divisible by 3 have algebra factors, and we only want to know whether it has a covering set of primes for all n not divisible by 3, if so, then this k makes a full covering set with algebraic factors and be excluded from the conjecture; if not, then this k does not make a full covering set with algebraic factors and be included from the conjecture.

n-value : factors
1 : 3 · 127
2 : 37 · 103
4 : 3 · 127037
5 : 17 · 37 · 73 · 83
7 : 3^2 · 42345679
8 : 37 · 113 · 613 · 1487
10 : 3 · 2399 · 52954163
11 : 37 · 103003003003

and there is covering set {3, 37} (n == 1 mod 3: factor of 3; n == 2 mod 3: factor of 37), thus for R10, k=343 proven composite by partial algebra factors.

For Sierpinski side:

Case gcd(k+1,b-1) = 1 is the same as the Sierpinski side of CRUS

Case k = 1, b even is the same as finding the smallest generalized Fermat prime base b

Case k = 1, b odd is the same as finding the smallest half generalized Fermat prime base b

Case k = b-2 is the same as finding the smallest prime of the form ((b-2)*b^n+1)/(b-1)

For Riesel side:

Case gcd(k-1,b-1) = 1 is the same as the Riesel side of CRUS

Case k = 1 is the same as finding the smallest generalized repunit prime base b

Case k = b-1 is the same as finding the smallest Williams prime base b (primes of the form (b-1)*b^n-1, some authors use base (b-1) instead of base b for (b-1)*b^n-1, but I don't think this is good, since the "base" should be the number with an exponent)

[QUOTE=sweety439;549757]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][/QUOTE]

Since a large probable prime n can be proven to be prime if and only if at least one of n-1 and n+1 can be trivially written into a product.

Thus, if n is large, a probable prime (k*b^n+-1)/gcd(k+-1,b-1) can be proven to be prime if and only if gcd(k+-1,b-1) = 1.

[QUOTE=sweety439;550375]Since a large probable prime n can be proven to be prime if and only if at least one of n-1 and n+1 can be trivially written into a product.

Thus, if n is large, a probable prime (k*b^n+-1)/gcd(k+-1,b-1) can be proven to be prime if and only if gcd(k+-1,b-1) = 1.[/QUOTE]

If gcd(k+-1,b-1) = 1 (+ for Sierpinski, - for Riesel), then this is completely the same as the Sierpinski/Riesel problem in CRUS for this (k,b) combo, these Sierpinski/Riesel problems just extend the Sierpinski/Riesel problems in CRUS to the k such that gcd(k+-1,b-1) (+ for Sierpinski, - for Riesel) is not 1, since in this case, k*b^n+-1 is always divisible by gcd(k+-1,b-1), thus we should take out this factor and find the smallest n such that (k*b^n+-1)/gcd(k+-1,b-1) is prime.

[QUOTE=sweety439;550375]Since a large probable prime n can be proven to be prime if and only if at least one of n-1 and n+1 can be [B]trivially written into a product[/B].

Thus, if n is large, a probable prime (k*b^n+-1)/gcd(k+-1,b-1) can be proven to be prime if and only if gcd(k+-1,b-1) = 1.[/QUOTE]

That is incorrect. What if we only know 12.5% of the factorization of n^2-1 (CHG proof), and thus, niether n+1, nor n-1 has to be trivially written as a product ? What if the factors found were non-trivial?

Also, have you considered where

[URL="https://primes.utm.edu/prove/prove3_3.html"]n^2+n+1, n^2-n+1, n^2+1,[/URL]

are partially factored?

I could keep going you know.

[QUOTE=sweety439;550297]If

k is square and k == -1 mod p and b == -1 mod p for some odd prime p (this situation only exists for p == 1 mod 4)

or

k is square and k == 2^(r-1)+1 mod 2^r and b == 2^(r-1)+1 mod 2^r for some r >= 2 (this situation only exists for r >= 4)

Then this k proven composite by partial algebraic factors (has algebraic factors (difference of two squares) for even n and divisible by a prime (p or 2, respectively) for odd n)[/QUOTE]

Thus, the second case (i.e. has algebra factors for even n and divisible by 2 for odd n) only exists for bases b == 1 mod 8

[QUOTE=sweety439;550378]Thus, the second case (i.e. has algebra factors for even n and divisible by 2 for odd n) only exists for bases b == 1 mod 8[/QUOTE]

and such k's are square numbers k such that the highest power of 2 dividing (k-1) is the same as the highest power of 2 dividing (b-1)

(not consider the case that b is square, since any square k for any square Riesel base b proven composite by full algebra factors)

These bases b have many small Sierpinski/Riesel numbers k:

[CODE]
base Sierpinski numbers k Riesel numbers k
5 == 7, 11 mod 24 == 13, 17 mod 24
8 == 47, 79, 83, 181 mod 195 == 14, 112, 116, 148 mod 195
9 == 31, 39 mod 80 == 41, 49 mod 80
11 == 5, 7 mod 12 == 5, 7 mod 12
12 == 521, 597, 1143, 1509 mod 1885 == 376, 742, 1288, 1364 mod 1885
13 == 15, 27 mod 56 == 29, 41 mod 56
14 == 4, 11 mod 15 == 4, 11 mod 15
16 == 38, 194, 524, 608, 647, 719 mod 819 == 100, 172, 211, 295, 625, 781 mod 819
17 == 31, 47 mod 96 == 49, 65 mod 96
18 == 398, 512, 571, 989 mod 1235 == 246, 664, 723, 837 mod 1235
19 == 9, 11 mod 20 == 9, 11 mod 20
20 == 8, 13 mod 21 == 8, 13 mod 21
21 == 23, 43 mod 88 == 45, 65 mod 88
23 == 5, 7 mod 12 == 5, 7 mod 12
25 == 79, 103 mod 208 == 105, 129 mod 208
27 == 13, 15 mod 28 == 13, 15 mod 28
29 == 4, 11 mod 15 or == 7, 11 mod 24 or == 19, 31 mod 40 == 4, 11 mod 15 or == 13, 17 mod 24 or == 9, 21 mod 40
32 == 10, 23 mod 33 == 10, 23 mod 33
[/CODE]

There are several reasons not to quote an entire post:
1. It causes the reader to have to scroll past the giant block to get to fresh content.
- - If someone is trying to catch up on a thread it slows them down.
2. It does not help the reader know particularly what bit of the quote that is being referred to.
3. If it is the immediately preceding post, a quote is not needed at all.
4. If someone is searching the forum to find something, excessive quoting produces extra hits that are chaff and not wheat.
5. It slows forum operation and adds extra unneeded volume to the database.

An example of an effective quote:

[QUOTE=In a previous post][FONT="Comic Sans MS"][I]Here are 5 examples of dogs I like[/I][/FONT][/QUOTE]
[I][FONT="Comic Sans MS"]I also now like these 2 more:
Dog 6
Dog 7[/FONT][/I]

Notice that the limited quote tells the user what is being referred to. The concept of Hypertext was that things could be referenced without placing an entire work in the document. This is done all of the time in documents, a small bit is quoted and there are footnotes to the original work.

Found an error of S81: (34*81^734+1)/gcd(34+1,81-1) is prime

Double checking S81....