A Sierpinski/Riesellike problem
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For the original Sierpinski/Riesel problem, it is finding and proving the smallest k such that k*b^n+1 is composite for all integer n>=1 and gcd(k+1, b1) = 1. Now, I extend to the k's such that gcd(k+1, b1) is not 1. Of course, all numbers of the form k*b^n+1 is divisible by gcd(k+1, b1). Thus, I must take out this factor and find and prove the smallest k such that (k*b^n+1)/gcd(k+1, b1) is composite for all integer n>=1. (of course, in the base 2 case, this is completely the same as the original Sierpinski/Riesel problem)
In the original Sierpinski/Riesel problems, kvalues with all nvalues have a single trivial factor are excluded from the conjectures. However, in these problems, we take out this trivial factor, thus all kvalues are included from the conjectures. (thus, in this problems, the divisor of k*b^n+1 is the largest trivial factor of k*b^n+1, which equals gcd(k+1, b1)) The research is form [URL]http://mersenneforum.org/showthread.php?t=21832[/URL]. The Riesel case are also researched in [URL]https://www.rosehulman.edu/~rickert/Compositeseq/[/URL]. The strong (extended) Sierpinski problem base 4 is proven, with the conjectured k=419. Also, the strong (extended) Riesel problem base 10 is proven, with the conjectured k=334. For the strong (extended) Riesel problem base 3, for k<=500, I cannot find a prime for k = {119, 313, 357}. For k=291, the prime is the same as the k=97 prime: (97*3^31311)/2, since 291 = 97*3, and since 357 = 119*3, the prime for k=357 is the same as the prime for k=119, but both are unknown. Edit: According to the link, (313*3^247611)/2 is a probable prime. Extended Sierpinski problem base b: Finding and proving the smallest integer k>=1 such that (k*b^n+1)/gcd(k+1, b1) is not prime for all integer n>=1. Extended Riesel problem base b: Finding and proving the smallest integer k>=1 such that (k*b^n1)/gcd(k1, b1) is not prime for all integer n>=1. The last text file is the list of the conjectured smallest strong Sierpinski/Riesel number number to base b for b = 2 to b = 12. 
[QUOTE]For the strong (extended) Riesel problem base 3, for k<=500, I cannot find a prime for k = {119, 313, 357}. For k=291, the prime is the same as the k=97 prime: (97*3^31311)/2, since 291 = 97*3, and since 357 = 119*3, the prime for k=357 is the same as the prime for k=119, but both are unknown.
Edit: According to the link, (313*3^247611)/2 is a probable prime. [/QUOTE]I would suggest keeping this in the "And now for something completely different" subforum. It does not apply to CRUS. If you want some smaller (not top10) primes that CRUS has found I have already said that I'll give them to you. Someone put a lot of effort into the page that you linked to but there is a lot of inconsistency in how the kvalues are being shown both in that link and in your post here for base 3 append 1. It creates a lot of confusion over what needs to be searched. First suggestion: When showing which k's need a prime use the kvalue in this form: k*3^n+(3^n1)/2 Reasoning for the suggestion. For a more complex base/append combination, such as base 9 append 7, it would not be clear what k is being searched unless you use the form: k*9^n+(9^n1)*7/8 Second suggestion: Do not show k=3m as remaining. It is redundant. Using that format, what you are asking for are PRPs for k= {59, 156}. Here they are: 59*3^8972+(3^89721)/2 is PRP 156*3^24761+(3^247611)/2 is PRP The author of the site that you linked to did not do a good job of consistency in how he displayed his k's remaining and primes. K's should properly be reduced to their lowest form lest they end up as higher than the actual conjecture. For instance many of his k's and PRP's for base 3 append 1 are k==(1 mod 3). Such k's can be reduced to (k1)/3 while adding 1 to the nvalue. (Many can be reduced multiple times.) Here are two related examples that would have helped you on both of the above PRP's: 1. k=4819 with a PRP at n=8968 is the same as and should be shown as k=59 with a PRP at n=8972. 2. k=4225 with a PRP at n=24758 is the same as and should be shown as k=156 with a PRP at n=24761. Had he shown these PRP's in correct reduced format you would have realized that you already had the PRPs that you were looking for. This goes to the crux of the problem when running projects or asking others for help with your efforts: It is important that you are very clear and consistent else people get confused and become disinterested in helping. [B]Back to the page that you linked to: Note that he searched all base 3 append 1 k's to n=50K. I double checked him to n=25K. k=806 is the smallest k remaining at n=50K[/B]. His results are correct except that his k's are shown in the inconsistent format demonstrated above. For example he shows that the following "single k's" are remaining: (806, 2419) (915, 2746) In other words the second number is the first number * 3 + 1, which will result in the same prime, so he is avoiding searching both k's. That is a good thing BUT... he is missing that k=5383 should show: (1794, 5383) Someone double checking the base (like me) could easily conclude that k=1794 is remaining and think that he had missed it. In other words k==(1 mod 3) needs to be reduced for consistency and the same applies to the primes that he shows as demonstrated above. The linked page is extremely old and inconsistent. You should consider doing some of your own double checking on it and check for other inconsistencies or errors. 
:goodposting: It shows you put some time and effort into that reply! But... What is the English equivalent of "[URL="https://translate.google.com/#ro/en/a%20strica%20orzul%20pe%20g%C3%A2%C8%99te"]a strica orzul pe gâște[/URL]"? (rhetoric question, no answer needed) (edit: [URL="http://www.englishtest.net/forum/ftopic86928.html"]found it[/URL]! the linked thread is nice to read, my answer is in posts #7, #10, etc in that thread)
[QUOTE=gd_barnes;449175] Had he shown these PRP's in correct reduced format you would have realized that you already had the PRPs that you were looking for. <snip> The linked page is extremely old and inconsistent. You should consider doing some of your own double checking on it and check for other inconsistencies or errors.[/QUOTE] +1 
[QUOTE=gd_barnes;449175]I would suggest keeping this in the "And now for something completely different" subforum. It does not apply to CRUS. If you want some smaller (not top10) primes that CRUS has found I have already said that I'll give them to you.
Someone put a lot of effort into the page that you linked to but there is a lot of inconsistency in how the kvalues are being shown both in that link and in your post here for base 3 append 1. It creates a lot of confusion over what needs to be searched. First suggestion: When showing which k's need a prime use the kvalue in this form: k*3^n+(3^n1)/2 Reasoning for the suggestion. For a more complex base/append combination, such as base 9 append 7, it would not be clear what k is being searched unless you use the form: k*9^n+(9^n1)*7/8 Second suggestion: Do not show k=3m as remaining. It is redundant. Using that format, what you are asking for are PRPs for k= {59, 156}. Here they are: 59*3^8972+(3^89721)/2 is PRP 156*3^24761+(3^247611)/2 is PRP The author of the site that you linked to did not do a good job of consistency in how he displayed his k's remaining and primes. K's should properly be reduced to their lowest form lest they end up as higher than the actual conjecture. For instance many of his k's and PRP's for base 3 append 1 are k==(1 mod 3). Such k's can be reduced to (k1)/3 while adding 1 to the nvalue. (Many can be reduced multiple times.) Here are two related examples that would have helped you on both of the above PRP's: 1. k=4819 with a PRP at n=8968 is the same as and should be shown as k=59 with a PRP at n=8972. 2. k=4225 with a PRP at n=24758 is the same as and should be shown as k=156 with a PRP at n=24761. Had he shown these PRP's in correct reduced format you would have realized that you already had the PRPs that you were looking for. This goes to the crux of the problem when running projects or asking others for help with your efforts: It is important that you are very clear and consistent else people get confused and become disinterested in helping. [B]Back to the page that you linked to: Note that he searched all base 3 append 1 k's to n=50K. I double checked him to n=25K. k=806 is the smallest k remaining at n=50K[/B]. His results are correct except that his k's are shown in the inconsistent format demonstrated above. For example he shows that the following "single k's" are remaining: (806, 2419) (915, 2746) In other words the second number is the first number * 3 + 1, which will result in the same prime, so he is avoiding searching both k's. That is a good thing BUT... he is missing that k=5383 should show: (1794, 5383) Someone double checking the base (like me) could easily conclude that k=1794 is remaining and think that he had missed it. In other words k==(1 mod 3) needs to be reduced for consistency and the same applies to the primes that he shows as demonstrated above. The linked page is extremely old and inconsistent. You should consider doing some of your own double checking on it and check for other inconsistencies or errors.[/QUOTE] Unfortunately, it is only for the extendRiesel problem, not the extendSierpinski problem. 
In these conjectures, the form is (k*b^n+1)/gcd(k+1,b1). Thus, for example, for the extended Riesel problem base 38 and k=1, the form is (1*38^n1)/gcd(11, 381) = (1*38^n1)/37. Thus, the corresponding prime is not 1*38^11, it is (1*38^31)/37.

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R7 (extended) has only one k remain: 197. (its conjecture is k=457)
Some (probable) primes not in the text file are given in the link: (159*7^48961)/2 (313*7^59071)/6 (367*7^151181)/6 (429*7^38151)/2 Besides, according to the link, (197*7^n1)/2 is checked to n=15000 with no (probable) prime found. Thus, this conjecture is [I]nearly[/I] proven. The extended S7 conjecture is easily to proven, with k=209 the smallest strong Sierpinski number to base 7. The extended R4, S5, R5, S8, R8, S9, R9, S11, R11 and R12 conjectures are trivial and very easily to proven. 
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The conjecture of S10 (extended) is k=989, with a covering set {3, 7, 11, 13}, period=6.
I tested the conjecture and it has only 3 k's remain: 100, 269, and 804. However, k=100 is a GFN, for k=804, it is a prime given by the CRUS: 804*10^5470+1. Thus, only k=269 is remain, the form is (269*10^n+1)/9. Also, the conjecture of R10 (extended) is k=334, with a covering set {3, 7, 13, 37}, period=6. This conjecture can be easily to proven. 
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The conjecture of S12 (extended) is proven, since only GFNs (k=12 and k=144) are remain.
The conjecture of R12 (extended) is easily to proven. Thus, the only harder conjecture is the extended conjecture of S2, R2, S3, R3, S6, and R6. (the base 2 extended Sierpinski/Riesel conjecture is completely the same as the original base 2 Sierpinski/Riesel conjecture) 
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Update the extended S6/R6 files. (only up to k=500)

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Update the extend Sierpinski/Riesel conjectures base 5, 8, 9, 11 files.
All these conjectures are easily to prove. 
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Can you find the smallest [I]conjectured[/I] strong Sierpinski number in base 3 and 6?

How about 'yet another, unrelated to the original Sierpinski/Riesel, problem'
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[QUOTE=sweety439;449149]For the original Sierpinski/Riesel problem, it is finding and proving ...
Now, I extend to the k's such that gcd(k+1, b1) is not 1. The [B]strong[/B] (extended) Sierpinski problem ...[/QUOTE] Staying away from words that have an established meaning in the community is probably a good idea before you actually can talk the language of the community. How do you know that you are not actually meaning 'weak' when you are saying 'strong'? Or rather, how do you know that things you are talking about are altogether related? Because they aren't! If one proved Goldbach, the would have also proven weak Goldbach. If one proved weak Goldbach (and they did!), then nothing happened to Goldbach. If you (well, let's imagine) proved "The [B]strong[/B] Sierpinski problem", then ... <care to fill the blanks?> ... [SPOILER]Nothing would happen to the "normal" Sierpinski problem, of course![/SPOILER] Origin: [url]https://xkcd.com/1310/[/url] 
[QUOTE=Batalov;449245]Staying away from words that have an established meaning in the community is probably a good idea before you actually can talk the language of the community. How do you know that you are not actually meaning 'weak' when you are saying 'strong'?
Or rather, how do you know that things you are talking about are altogether related? Because they aren't! If one proved Goldbach, the would have also proven weak Goldbach. If one proved weak Goldbach (and they did!), then nothing happened to Goldbach. If you (well, let's imagine) proved "The [B]strong[/B] Sierpinski problem", then ... <care to fill the blanks?> ... [SPOILER]Nothing would happen to the "normal" Sierpinski problem, of course![/SPOILER] Origin: [URL]https://xkcd.com/1310/[/URL][/QUOTE] Since this is the [B]extending [/B]of the original Sierpinski/Riesel problem. If you found all (probable) primes of the form (k*b^n+1)/gcd(k+1, b1) for all k, then of course, you also found all primes of the form k*b^n+1 for all k such that gcd(k+1, b1) = 1, and the latter is the original Sierpinski/Riesel conjecture. However, although the conjecture [B][I]smallest [/I][/B]k does need to be the same, if we also include the k's > conjectured smallest k, then this conjecture covers the original conjecture. For example, for R10, the conjecture smallest Riesel k is 10176, but the conjecture smallest strong Riesel k is 334. Of course, this conjecture does not cover the original conjecture. However, if we also include the k's > conjectured smallest k (334), if we include all k's < 10176 which are not proven composite for all exponent n, then this conjecture covers the original R10 conjecture. 
This is only my extending of the Sierpinski/Riesel problem. Now, I know that it does not apply to CRUS since the correspond primes are often only probable primes, i.e. not proved primes.
Thanks. Now, I know this should be in "and now for something completely different". 
[QUOTE=sweety439;449225]Can you find the smallest [I]conjectured[/I] strong Sierpinski number in base 3 and 6?[/QUOTE]
base 3: conjectured smallest strong Sierpinski k=11047, cover set: {2, 5, 7, 13, 73}, period=12 For the number (11047*3^n+1)/2: If n=0 (mod 2), then this number is divisible by 2. If n=3 (mod 4), then this number is divisible by 5. If n=1 (mod 12), then this number is divisible by 73. If n=5 (mod 12), then this number is divisible by 13. If n=9 (mod 12), then this number is divisible by 7. 
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base 6: conjectured smallest strong Sierpinski k=174308, cover set: {7, 13, 31, 37, 97}, period=12.
This k is the same as the conjectured smallest original base 6 Sierpinski k. Thus, the strong Sierpinski conjecture base 6 covers the original Sierpinski conjecture in the same base. Update the text file for all conjectured smallest strong Sierpinski/Riesel number base 2 to 12. 
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Due to the CRUS, if we only consider the k's with cover set (i.e. not with full or partial algebra factors), then the conjectured k's for S8, R4, R9 and R12 should be larger, they are in this text file.
Of course, there are k's can be proven composite with full or partial algebra factors, e.g. square k's in R4 and R9, cube k's in S8, and k=25, 27, 64, 300, 324 in R12, these k's are excluded from the conjectures. The conjectured k should be: S8: 1 > 47 R4: 9 > 361 R9: 1 > 41 R12: 25 > 376 
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Update the extend S8, R4, R9, R12 files. (with the CRUS definition, i.e. exclude the k's can be proven composite with full or partial algebra factors)
All of the four conjectures are proven. 
For base 2, the strong Sierpinski/Riesel problem is completely the same as the original Sierpinski/Riesel problem, since gcd(k+1,21) = 1 for all k.

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I continued to search bases 13 to 24.
This is the text file for the conjectured smallest strong Sierpinski/Riesel k. I will find the conjectured smallest strong Sierpinski/Riesel k for base 15 and 24. (The two bases are harder) 
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I continued to search bases 25 to 64.
This is the text file for the conjectured smallest strong Sierpinski/Riesel k. For these bases, I still found no k with a cover set. (but of course, there are infinitely many such k's) S15, S24, S40, S42, S52, S60, S63. R15, R24, R30, R36, R40, R42, R48, R52, R60. All such bases have been tested to at least k=1000 (k=1000 is the test limit for all bases b>=25 except S28, R28 and S36) Edit: For R28, I found k=3769, cover: {5, 29, 157} period=4. I will find (probable) primes for these problems in next few weeks. 
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Add the conjectured k's I found for S63, R28, R30 and R48.
Now, I still found no k with a cover set only for these bases <= 64. S15, S24, S40, S42, S52, S60. R15, R24, R36, R40, R42, R52, R60. 
Found more conjectured k for the extended Sierpinski/Riesel problems:
S24: 30651 S42: 13372 S60: 16957 R24: 32336 R42: 15137 R60: 20558 The six conjectured k's are the same as the conjectured k's for the original Sierpinski/Riesel problem. Thus, for base 24, base 42 and base 60 (but not for all bases), the extend Sierpinski/Riesel problem covers the original Sierpinski/Riesel problem. 
Found the conjectured k for R36: 33791.
(33791*36^n1)/5 has a cover set: {13, 31, 43, 97}. Now, I still found no k with a cover set only for these bases <= 64: S15, S40, S52. R15, R40, R52. 
Found more conjectured k for the extended Sierpinski/Riesel problems:
S15: 673029 cover: {2, 17, 113, 1489} period=8 S40: 47723 cover: {3, 7, 41, 223} period=6 S52: 28674 cover: {5, 53, 541} period=4 R15: 622403 cover: {2, 17, 113, 1489} period=8 R40: 25462 cover: {3, 7, 41, 223} period=6 R52: 25015 cover: {3, 7, 53, 379} period=6 Now, the list of the conjectured smallest strong (extended) Sierpinski/Riesel number for bases 2<=b<=64 is completed!!! :fish1: 
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Update the complete text file for the conjectured smallest strong (extended) k to all bases 2<=b<=64.

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Now, I am running the extended Sierpinski/Riesel conjectures for 13<=b<=24. Since the conjectured k for base 15, 22 and 24 (on both sides) are larger, I only run other bases.

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All extended Sierpinski conjectures I ran are proven. (S18 is proven since only GFNs (18*18^n+1 and 324*18^n+1) are remain)
Define of GFNs: Only exist for extended Sierpinski conjectures ((k*b^n+1)/gcd(k+1,b1)). gcd(k+1,b1)=1. k is a rational power of b. Thus, 100*10^n+1, 18*18^n+1 and 4*32^n+1 are GFNs, but 4*155^n+1, (25*5^n+1)/2 and (7*49^n+1)/8 are not. 
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Also running extended Riesel conjectures.

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All extended Riesel conjectures I ran are proven except R17, R17 has only k=29 remain.
Can someone find a prime of the form (29*17^n1)/4? 
The extended Sierpinski/Riesel conjectures for bases 2<=b<=24 with only one k remain:
R7, k=197 ((197*7^n1)/2) S10, k=269 ((269*10^n+1)/9) R17, k=29 ((29*17^n1)/4) Can you find the smallest n? 
List of the status for the extended Sierpinski/Riesel conjectures to bases 2<=b<=24: (the number of remain k does not contain the k excluded from testing, i.e. k's that is multiple of b and (k+1)/gcd(k+1, b1) are composite, and also not contain GFN's)
S2: conjectured k=78557, 5 k's remain (21181, 22699, 24747, 55459, 67607) S3: conjectured k=11047, not completely started. S4: conjectured k=419, proven. S5: conjectured k=7, proven. S6: conjectured k=174308, not completely started. S7: conjectured k=209, proven. S8: conjectured k=47, proven. S9: conjectured k=31, proven. S10: conjectured k=989, only k=269 remain. S11: conjectured k=5, proven. S12: conjectured k=521, proven. S13: conjectured k=15, proven. S14: conjectured k=4, proven. S15: conjectured k=673029, not completely started. S16: conjectured k=38, proven. S17: conjectured k=31, proven. S18: conjectured k=398, proven. S19: conjectured k=9, proven. S20: conjectured k=8, proven. S21: conjectured k=23, proven. S22: conjectured k=2253, not completely started. S23: conjectured k=5, proven. S24: conjectured k=30651, not completely started. R2: conjectured k=509203, 52 k's remain (2293, 9221, 23669, 31859, 38473, 46663, 67117, 74699, 81041, 93839, 97139, 107347, 121889, 129007, 143047, 146561, 161669, 192971, 206039, 206231, 215443, 226153, 234343, 245561, 250027, 273809, 315929, 319511, 324011, 325123, 327671, 336839, 342847, 344759, 351134, 362609, 363343, 364903, 365159, 368411, 371893, 384539, 386801, 397027, 409753, 444637, 470173, 474491, 477583, 478214, 485557, 494743) R3: conjectured k=12119, 15 k's remain (1613, 1831, 1937, 3131, 3589, 5755, 6787, 7477, 7627, 7939, 8713, 8777, 9811, 10651, 11597) R4: conjectured k=361, proven. R5: conjectured k=13, proven. R6: conjectured k=84687, 13 k's remain (1597, 2626, 6236, 9491, 37031, 49771, 50686, 53941, 55061, 57926, 76761, 79801, 83411) R7: conjectured k=457, only k=197 remain. R8: conjectured k=14, proven. R9: conjectured k=41, proven. R10: conjectured k=334, proven. R11: conjectured k=5, proven. R12: conjectured k=376, proven. R13: conjectured k=29, proven. R14: conjectured k=4, proven. R15: conjectured k=622403, not completely started. R16: conjectured k=100, proven. R17: conjectured k=49, only k=29 remain. R18: conjectured k=246, proven. R19: conjectured k=9, proven. R20: conjectured k=8, proven. R21: conjectured k=45, proven. R22: conjectured k=2738, not completely started. R23: conjectured k=5, proven. R24: conjectured k=32336, not completely started. 
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Found the probable prime (29*17^49041)/4.
Extended R17 is proven!!! 
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The correct text file for extended R17 is here. (the above file only lists the k's < 44, we should lists all k's < 49)

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You can surely post 10 messages an hour. You have proven your capacity to type.
You come across as a graphomaniac. But do you think anybody is reading your inane stream of conscience? Why would they  if you are appear to be deaf to other people's messages? 
The (probable) primes with n > 1000 for the extended Sierpinski/Riesel problems (with bases b <= 32, except b = 2, 3, 6, 15, 22, 24, 28, 30) are:
S4: 186*4^10458+1 S7: (141*7^1044+1)/2 S10: 804*10^5470+1 S12: 404*12^714558+1 378*12^2388+1 S16: (23*16^1074+1)/3 S17: 10*17^1356+1 S18: 122*18^292318+1 381*18^24108+1 291*18^2415+1 S25: (61*25^3104+1)/2 S26: 32*26^318071+1 217*26^11454+1 95*26^1683+1 178*26^1154+1 R4: (106*4^45531)/3 74*4^12761 296*4^12751 ( = 74*4^12761, the two primes are the same, since 296 = 74 * 4) R7: (367*7^151181)/6 (313*7^59071)/6 (159*7^48961)/2 (429*7^38151)/2 (419*7^10521)/2 R12: (298*12^16761)/11 R17: 44*17^64881 (29*17^49041)/4 (13*17^11231)/4 R25: 86*25^10291 R26: 115*26^5202771 32*26^98121 (121*26^15091)/5 
[QUOTE=sweety439;449149]
Extend Sierpinski problem base b: Finding and proving smallest k such that (k*b^n+1)/gcd(k+1, b1) is composite for all integer n>1. Extend Riesel problem base b: Finding and proving smallest k such that (k*b^n1)/gcd(k1, b1) is composite for all integer n>1. [/QUOTE] It should not be "n>1", it should be "n>=1", we allow n=1 for (k*b^n+1)/gcd(k+1, b1) and (k*b^n1)/gcd(k1, b1), but we do not allow n=0. 
[QUOTE=gd_barnes;449175]
First suggestion: When showing which k's need a prime use the kvalue in this form: k*3^n+(3^n1)/2 Reasoning for the suggestion. For a more complex base/append combination, such as base 9 append 7, it would not be clear what k is being searched unless you use the form: k*9^n+(9^n1)*7/8 Second suggestion: Do not show k=3m as remaining. It is redundant. [/QUOTE] This is another problem, different from the "[URL="http://mersenneforum.org/showthread.php?t=21832"]Add repeated digits after a number until it is prime[/URL]" problem. First, although for the Riesel case this problem are special cases (when d divides b1) of that problem, but for the Sierpinski case it is not. Second, in that problem, if d does not divide b1, then it cannot be transfer to this problem. Thus, in this problem, we say "k=1617 is remaining for the extended Riesel base 3 problem", but in that problem, we say "k=806 is remaining for the base 3 d=1 problem". 
These problems are just my extending for the original Sierpinski/Riesel problems.

For (269*10^n+1)/9, tested to n=6000, still no (probable) prime found.

For the two bases that have only one k remaining:
S10, k=269: (269*10^n+1)/9 seems to be tested to n=10000 (I ran the program of 25 hours and 38 minutes!!!), still no (probable) prime found, base released. R7, k=197: See the link: [URL]https://www.rosehulman.edu/~rickert/Compositeseq/[/URL], (197*7^n1)/2 is already tested to n=15000 with no (probable) prime found. 
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Also tested extended S15, S22 and S24 to k=500.
For extended S15, there are 4 k's <= 500 remain: 219, 225, 341, 343. For extended S22, there are 5 k's <= 500 (excluding the GFNs: i.e. k=22 and k=484) remain: 173, 346, 383, 461, 464. For extended S24, there are 5 k's <= 500 remain: 319, 346, 381, 461, 486. However, all of them are correspond to CRUS primes, since all these k's satisfy that gcd(k+1,241)=1, and according to the CRUS page, the smallest remain k for the original S24 problem is k=656. Thus, there are in fact no k's <= 500 remain for extended S24. 
If the k satisfies that gcd(k+1, b1)=1, then the correspond prime of the extended Sierpinski/Riesel problem is the same as the correspond prime of the original Sierpinski/Riesel problem.
The extended Sierpinski/Riesel problem is just my extending of the original Sierpinski/Riesel problem to the k's such that gcd(k+1, b1) is not 1. Since k*b^n+1 is always divisible by gcd(k+1, b1), the division is necessary for there to be any chance of finding primes. 
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Also tested extended R15, R22 and R24 to k=500.
For extended R15, there are 6 k's <= 500 remain: 47, 203, 239, 407, 437, 451. For extended R22, there are 4 k's <= 500 (k=185 prime is given by the CRUS page) remain: 208, 211, 355, 436. For extended R24, there are 7 k's <= 500 remain: 69, 201, 339, 346, 364, 389, 461. However, the k's != 1 mod 23 are correspond to CRUS primes, according to the CRUS page, k=389 is the only k <= 500 remain for the original R24 problem. 
[QUOTE=sweety439;450039](269*10^n+1)/9 seems to be tested to n=10000 ([B][COLOR="Red"]I ran the program of 25 hours and 38 minutes!!![/COLOR][/B]), still no (probable) prime found, base released.
[/QUOTE] Mr Winston Wolf once said: "That's 30 minutes away. I'll be there in 10." In comparison, it takes you 25 hours and 38 minutes to do a job that takes 10 minutes. That doesn't bother you? Not at all? 
[QUOTE=Batalov;450051]Mr Winston Wolf once said: "That's 30 minutes away. I'll be there in 10."
In comparison, it takes you 25 hours and 38 minutes to do a job that takes 10 minutes. That doesn't bother you? Not at all?[/QUOTE] Hey now! How ruuuuuuude! :smile: All of my machines are more than 4 years old. I ran a 5+ year old version of PFGW (version 3.3.6) with no sieving (trialfactoring only) on my newest 4yearold machine with 5 other cores running at the same time and it took 18 minutes. So now I'm sad that you make fun of my old machines and software that take almost twice as long as you say it should. (lol) Anyway...a modern machine running a modern version of PFGW with some sieving done should likely take ~5 minutes. I don't think anything bothers Sweety. We can insult him all day long and he just keeps posting trivial results. 
[QUOTE=gd_barnes;450053]I ran a 5+ year old version of [COLOR="Red"]PFGW[/COLOR] (version 3.3.6) with no sieving (trialfactoring only) [/QUOTE]
Well for these PRPs with a denominator one should run only LLR or P95. They are a few times faster. And there are <300 tests for n<10,000 and each is much faster than a second, ...so 10 minutes was said only to match the Wolf quote. [QUOTE="Mr.Wolf"]I'm Winston Wolf. I solve problems.[/QUOTE] 
[QUOTE=Batalov;450059]Well for these PRPs with a denominator one should run only LLR or P95. They are a few times faster. And there are <300 tests for n<10,000 and each is much faster than a second, ...so 10 minutes was said only to match the Wolf quote.[/QUOTE]
Hum....How might one run LLR with a denominator? I've never tried it before and just now tried this: ABC ($a*10^$b$c)/9 // Sieved to 1000000000 with srsieve 269 47 +1 269 51 +1 269 84 +1 269 96 +1 269 105 +1 (etc.) LLR would not run. It runs fine without the denominator. I'm running LLR 3.8.13...not the latest version but it should be recent enough. 
[CODE]ABC ($a*$b^$c$d)/$e
269 10 47 +1 9 ...[/CODE] You will be very pleasantly surprised. Essentially the speed is the same as if without $e. 
[QUOTE=Batalov;450078][CODE]ABC ($a*$b^$c$d)/$e
269 10 47 +1 9 ...[/CODE]You will be very pleasantly surprised. Essentially the speed is the same as if without $e.[/QUOTE] Thanks! Not bad. About 8 mins on that same old machine with 5 other cores running, which includes about 1 minute to oversieve to 1G. If done efficiently on a modern machine...likely 3 mins. 
[QUOTE=gd_barnes;450067]
ABC ($a*10^$b$c)/9 // Sieved to 1000000000 with srsieve 269 47 +1 269 51 +1 269 84 +1 269 96 +1 269 105 +1 (etc.) [/QUOTE] What version of srsieve can do that sieving? :shock: 
[QUOTE=LaurV;450087]What version of srsieve can do that sieving? :shock:[/QUOTE]
Are you worried about the denominator 9? Just skip 3, and srsieve will take care of correctly sieving the numerator k*b^n+/c 
[QUOTE=LaurV;450087]What version of srsieve can do that sieving? :shock:[/QUOTE]
[QUOTE=axn;450089]Are you worried about the denominator 9? Just skip 3, and srsieve will take care of correctly sieving the numerator k*b^n+/c[/QUOTE] Just to clarify: I did not actually sieve ($a*10^$b$c)/9. I sieved $a*10^$b$c accomplishing exactly what axn said with p 4 P 1e9. After the sieving was done I then added the parens and denominator. Btw, this was a technique that I learned from Serge less than a week ago...one that I was not aware of because I've never searched repeatdigit type forms. A note about this: In this case the sieving still leaves terms with factors of 3^q after dividing by 9. When running PFGW I get around this by trial factoring to 1% with the f1 switch so it doesn't do a full primality test. It generally takes less than 1/10th of a second to trial factor to 1%. This begs a question: Does LLR look for teeny factors before doing a primality test? I could not tell with the short test that I ran on it here. If not I may stick with PFGW because it can quickly get rid of very small factors that are left when sieving a form with a denominator like this. 
Yes, I did exactly the same in the other thread of sweety, I just created a folder (called bp and copied his bases there, in a vector), filled it with sr_bxxx.pfgw files using pari:
[CODE]gp> \r bp\bases gp > (16:13:53) gp > v %1 = [184, 185, 200, 210, 269, 281, 306, 311, 326, 331, 371, 380, 384, 385, 394, 396, 452, 465, 485, 511, 522, 570, 574, 598, 601, 629, 631, 632, 636, 640, 649, 670, 684, 691, 693, 711, 713, 731, 752, 759, 771, 795, 820, 861, 866, 872, 881, 932, 938, 948, 951, 956, 963, 996, 1005, 1015] gp > for(i=1,#v,write("BP\\nrep_b"v[i]".pfgw","ABC $a*"v[i]"^$b$c // Sieved to "nextprime(v[i])" with srsieve");forprime(p=3,10^4,write("BP\\nrep_b"v[i]".pfgw","1 "p" 1"))) time = 1min, 4,259 ms. [/CODE] This just took a minute and the result was: (random sample) [CODE]ABC $a*636^$b$c // Sieved to 641 with srsieve 1 3 1 1 5 1 1 7 1 1 11 1 1 13 1 1 17 1 1 19 1 1 23 1 ...etc... [/CODE] This can be sieved with primes larger than 641 (inclusive) and a simple dos for command solves the problem: [CODE] for %p in (nrep_b*.pfgw) do srsieve S 2 m 1e8 w "%p" [/CODE] In minutes, all the files [U]with even base[/U] are reduced to less then half candidates, and files like "sr_bxxx.pfgw" are created from "nrep_bxxx.fgw". For the odd bases, srsieve gives an error that all terms are even, which I don't know how to avoid. Then a perl command will replace all headers: [CODE]for %p in (sr_*.pfgw) do perl i.bak p e "s/ABC \$a\*(\d*)\^\$b\$c/ABC (\$a*\1^\$b\$c)\/(\11)/g;" %p[/CODE] (this is heavy artillery, hehe) And of course, you will want pfgw to stop when a prime is found, therefore [CODE]for %p in (*.pfgw) do perl i.bak p e "s/Sieved to \d* with srsieve/{number_primes,\$a,1}/g;" %p [/CODE] Result: [CODE]ABC ($a*210^$b$c)/(2101) // {number_primes,$a,1} 1 7 1 1 11 1 1 17 1 1 31 1 1 103 1 1 107 1 1 157 1 1 173 1 1 227 1 1 257 1 1 269 1 1 331 1 1 347 1 1 353 1 1 383 1 1 397 1 ...etc... [/CODE] This feed to pfgw with another dos for command for all files. I know my tools, trust me, I was just wondering if there is a newer version of srsieve. Especially because the version which is marketed like 1.0.6 on the web, shows 1.0.5 when is executed (I downloaded from many places, and yes, all are the same). So I was more like wondering if there is another, [U]real[/U] version 1.0.6 Thanks anyhow... 
[QUOTE=LaurV;450092]For the odd bases, srsieve gives an error that all terms are even, which I don't know how to avoid.[/QUOTE]
There is no way around it except for forms that are c=1 or +1 and even the solution for those is not ideal. The program would have to be changed to avoid checking for the error. Both srsieve and sr2sieve have the same problem. For c=1 or +1 forms, you can use sr1sieve but it will only sieve where the factors are greater than the base so you'd have to figure a way to remove smaller factors or just do like I did and run PFGW with factoring set to 1% with f1. Also if you are sieving multiple k's at once, which I am frequently doing on some of these conjectures, you have to sieve them one k at a time...not a great solution. Would someone want to get ahold of Mark (rogue) and see if he can remove the error check from srsieve and sr2sieve where it says all terms are even? That becomes a major issue when sieving near repeat digit forms for odd bases. 
[QUOTE=sweety439;449953]The (probable) primes with n > 1000 for the completedstarted extended Sierpinski/Riesel problems (with bases b <= 24, except R2, R3 and R6) are:
S4: 186*4^10458+1 S7: (141*7^1044+1)/2 S10: 804*10^5470+1 S12: 404*12^714558+1 378*12^2388+1 S16: (23*16^1074+1)/3 S17: 10*17^1356+1 S18: 122*18^292318+1 381*18^24108+1 291*18^2415+1 R4: (106*4^45531)/3 74*4^12761 296*4^12751 ( = 74*4^12761, the two primes are the same, since 296 = 74 * 4) R7: (367*7^151181)/6 (313*7^59071)/6 (159*7^48961)/2 (429*7^38151)/2 (419*7^10521)/2 R12: (298*12^16761)/11 R17: (29*17^49041)/4 (13*17^11231)/4[/QUOTE] A prime is missing for R17: 44*17^64881. Thus, the list of R17 should be: 44*17^64881 (29*17^49041)/4 (13*17^11231)/4 
[QUOTE=Batalov;449245]Staying away from words that have an established meaning in the community is probably a good idea before you actually can talk the language of the community. How do you know that you are not actually meaning 'weak' when you are saying 'strong'?
Or rather, how do you know that things you are talking about are altogether related? Because they aren't! If one proved Goldbach, the would have also proven weak Goldbach. If one proved weak Goldbach (and they did!), then nothing happened to Goldbach. If you (well, let's imagine) proved "The [B]strong[/B] Sierpinski problem", then ... <care to fill the blanks?> ... [SPOILER]Nothing would happen to the "normal" Sierpinski problem, of course![/SPOILER] Origin: [URL]https://xkcd.com/1310/[/URL][/QUOTE] For example, in the most popular Riesel base 10 case, the smallest [I]original [/I]Riesel number base 10 is conjectured to be 10176, but the smallest [I]strong[/I] Riesel number is proven to be 334, since all numbers of the form (334*10^n1)/gcd(3341,101) are divisible by 3, 7, 13 or 37, all numbers of this form are composite. However, if we research all k's < 10176 (except the k's that are proven composite, these k's are 334, 343, 1585, 1882, 3340, 3430, 3664, 7327, 8425, 9208, k=343 and 3430 are proven composite by partial algebraic factors, other k's are proven composite by covering set), then the problem covers the original Riesel base 10 problem. This problem (the strong Riesel base 10 problem) has only 3 k's < 10176 remain: 2452, 4421 and 5428. (see the links [URL]https://www.rosehulman.edu/~rickert/Compositeseq/[/URL] and [URL="http://www.worldofnumbers.com/Appending%201s%20to%20n.txt"]http://www.worldofnumbers.com/Appending%201s%20to%20n.txt[/URL]) The top primes of this problem are: (with k < 10176, n > 10000) 7019 (881309) 8579 (373260) 6665 (60248) 1935 (51836) 1803 (45882) 1231 (37398) 6373 (37156) 1343 (29711) 6742 (22850) 505 (18470) 3499 (12689) 3356 (13323) 450 (11958) The test limits for the remain k's < 10176: 2452 (554K) (see [URL]https://www.rosehulman.edu/~rickert/Compositeseq/#b10d3[/URL]) 4421 (1.69M) (see [URL]http://www.noprimeleftbehind.net/crus/Rieselconjectures.htm[/URL]) 5428 (300K) (see [URL]http://www.worldofnumbers.com/Appending%201s%20to%20n.txt[/URL]) 
You copy for other resources very well :yucky:

In fact, I am interested in finding (probable) primes of the form (a*b^n+c)/gcd(a+c,b1), with integers a, b, c, a > 0, b > 1, c != 0, gcd(a,c)=1, gcd(b,c)=1.
Also see [URL]http://mersenneforum.org/showthread.php?t=21819[/URL] for the generalized minimal primes search for some triples (a, b, c) without known (probable) primes. 
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The old file is wrong for R6, the smallest extended Riesel number to base 6 should be 84687, not 84686.
Update correct text file. 
[QUOTE=sweety439;450281]In fact, I am interested in finding (probable) primes of the form (a*b^n+c)/gcd(a+c,b1), with integers a, b, c, a > 0, b > 1, c != 0, gcd(a,c)=1, gcd(b,c)=1.
Also see [URL]http://mersenneforum.org/showthread.php?t=21819[/URL] for the generalized minimal primes search for some triples (a, b, c) without known (probable) primes.[/QUOTE] In fact : you are very well in copying other stuff. 
[QUOTE=pepi37;450291]In fact : you are very well in copying other stuff.[/QUOTE]
He is quite amazing at doing that isn't he? Or is it she? Sweety, are you female? What is your native language? I (we) kind of feel like you don't quite understand our posts sometimes. 
Maybe she / he will understaned ban?

[QUOTE=gd_barnes;450305]He is quite amazing at doing that isn't he?
Or is it she? Sweety, are you female? What is your native language? I (we) kind of feel like you don't quite understand our posts sometimes.[/QUOTE] Oh, I know, you means I should not post my own research here? No, I am not female. 
[QUOTE=sweety439;450318]Oh, I know, you means I should not post my own research here?[/QUOTE]
You dont have your "own research" you have only copies from others. And yes, you should not post "your" irrelevant research here... 
[QUOTE=pepi37;450319]You dont have your "own research" you have only copies from others.
And yes, you should not post "your" irrelevant research here...[/QUOTE] I really download those programs, but my computer cannot run them. (my computer is windows 10) However, I have some programs, and found some (probable) primes, such as (29*17^49041)/4. The extended Sierpinski/Riesel problems are really my own research, and I proved the extended Riesel base 17 problem. 
[QUOTE=sweety439;450322]I really download those programs, but my computer cannot run them. (my computer is windows 10)
[/QUOTE] On Earth there is about 7.1 bilion peoples, and only you have computer that doesnot run those programs. You must be really special case of human! :ick: 
[QUOTE=sweety439;450322]I really download those programs, but my computer cannot run them. (my computer is windows 10)[/QUOTE]
my problem when I attempt is that I don't know enough about the file syntax or what I need to set up not that it won't run I also have windows 10. 
[QUOTE=sweety439;450322]I really download those programs, but my computer cannot run them. (my computer is windows 10)
However, I have some programs, and found some (probable) primes, such as (29*17^49041)/4. The extended Sierpinski/Riesel problems are really my own research, and I proved the extended Riesel base 17 problem.[/QUOTE] Point 1: My brain hurts! Pls stop it. All programms are working, even on Windows 10. Maybe you have an 32bit system (normaly you have 64bit, but...yeah)) If so, you should use the right programm (32bitfile). [CODE]srsievex86windows[/CODE]^32BitVersion of srsieve OR: Maybe your trying to use the Linux app instead off the windows app. Point 2: PRP´s are USELESS, you MUST prove them with PFGW. Please understand that.If not,your wasting your resources (and ours, too). 
[QUOTE=science_man_88;450328]my problem when I attempt is that I don't know enough about the file syntax or what I need to set up not that it won't run I also have windows 10.[/QUOTE]
Examble srsieve (cmd) [CODE] srsievex86_64windows S15.txt a nmin 2500 nmax 25000 vv pmax 100e6[/CODE] a= Print file to .abcdFormat nmin/max= Set nvalue pmax= Max sieve range vv= Print more informations/Additional 
[QUOTE=MisterBitcoin;450332]Point 1:
My brain hurts! Pls stop it. All programms are working, even on Windows 10. Maybe you have an 32bit system (normaly you have 64bit, but...yeah)) If so, you should use the right programm (32bitfile). [CODE]srsievex86windows[/CODE]^32BitVersion of srsieve OR: Maybe your trying to use the Linux app instead off the windows app. Point 2: PRP´s are USELESS, you MUST prove them with PFGW. Please understand that.If not,your wasting your resources (and ours, too).[/QUOTE] Point 2 is incorrect. PRPs are very useful. People find them on this forum quite frequently. They just cannot be reported to the top5000 primes site. But if they are greater than 30000 digits they can be reported to the top10000 PRPs site. 
[QUOTE=sweety439;449830]List of the status for the extended Sierpinski/Riesel conjectures to bases 2<=b<=24: (the number of remain k does not contain the k excluded from testing, i.e. k's that is multiple of b and (k+1)/gcd(k+1, b1) are composite, and also not contain GFN's)
S2: conjectured k=78557, 5 k's remain (21181, 22699, 24737, 55459, 67607) S3: conjectured k=11047, not completely started. S4: conjectured k=419, proven. S5: conjectured k=7, proven. S6: conjectured k=174308, not completely started. S7: conjectured k=209, proven. S8: conjectured k=47, proven. S9: conjectured k=31, proven. S10: conjectured k=989, only k=269 remain. S11: conjectured k=5, proven. S12: conjectured k=521, proven. S13: conjectured k=15, proven. S14: conjectured k=4, proven. S15: conjectured k=673029, not completely started. S16: conjectured k=38, proven. S17: conjectured k=31, proven. S18: conjectured k=398, proven. S19: conjectured k=9, proven. S20: conjectured k=8, proven. S21: conjectured k=23, proven. S22: conjectured k=2253, not completely started. S23: conjectured k=5, proven. S24: conjectured k=30651, not completely started. R2: conjectured k=509203, 52 k's remain (2293, 9221, 23669, 31859, 38473, 46663, 67117, 74699, 81041, 93839, 97139, 107347, 121889, 129007, 143047, 146561, 161669, 192971, 206039, 206231, 215443, 226153, 234343, 245561, 250027, 273809, 315929, 319511, 324011, 325123, 327671, 336839, 342847, 344759, 351134, 362609, 363343, 364903, 365159, 368411, 371893, 384539, 386801, 397027, 409753, 444637, 470173, 474491, 477583, 478214, 485557, 494743) R3: conjectured k=12119, 15 k's remain (1613, 1831, 1937, 3131, 3589, 5755, 6787, 7477, 7627, 7939, 8713, 8777, 9811, 10651, 11597) R4: conjectured k=361, proven. R5: conjectured k=13, proven. R6: conjectured k=84687, 13 k's remain (1597, 2626, 6236, 9491, 37031, 49771, 50686, 53941, 55061, 57926, 76761, 79801, 83411) R7: conjectured k=457, only k=197 remain. R8: conjectured k=14, proven. R9: conjectured k=41, proven. R10: conjectured k=334, proven. R11: conjectured k=5, proven. R12: conjectured k=376, proven. R13: conjectured k=29, proven. R14: conjectured k=4, proven. R15: conjectured k=622403, not completely started. R16: conjectured k=100, proven. R17: conjectured k=49, only k=29 remain. R18: conjectured k=246, proven. R19: conjectured k=9, proven. R20: conjectured k=8, proven. R21: conjectured k=45, proven. R22: conjectured k=2738, not completely started. R23: conjectured k=5, proven. R24: conjectured k=32336, not completely started.[/QUOTE] Is there any reserving for extended S3 and extended S6? Also extended S10 k=269 and extended R7 k=197. (The probable prime (29*17^49041)/4 was found by me. Thus, the extended R17 problem was proven) 
There is no k<=500 remain for extended S3 and extended S6. For extended S3, k=41, in 2015, I found the prime (41*3^4892+1)/2. (This number has been proven to be prime)
Also, some (probable) primes were not found by me, they are given by other websites: e.g. extended R4, k=106, the prime (106*4^45531)/3, is given by [URL]http://oeis.org/A177330[/URL], and extended R6, k=251, the prime (251*6^30081)/5, is given by [URL]http://oeis.org/A217377[/URL]. (Both of them have been proven to be prime) 
Note: k's that are multiple of b and where the number ((k+1)/gcd(k+1,b1) for extended Sierpinski conjectures, (k1)/gcd(k1,b1) for extended Riesel conjectures) is not prime are included in the conjectures but excluded from testing, since such k's will have the same prime as k / b.

Extend Sierpinski problem base b: Finding and proving the smallest k such that (k*b^n+1)/gcd(k+1, b1) is composite for all integer n>=1.
Extend Riesel problem base b: Finding and proving the smallest k such that (k*b^n1)/gcd(k1, b1) is composite for all integer n>=1. 
Another note: gcd(0,n) = n for all integer n, gcd(1,n) = 1 for all integer n.

k's that make a full covering set with all or partial algebraic factors for extended Sierpinski/Riesel bases 2<=b<=24:
S8: all k = m^3 S16: all k = 4*m^4 R4: all k = m^2 R8: all k = m^3 R9: all k = m^2 R12: all k = m^2 and m = 5 or 8 mod 13, and all k = 3*m^2 and m = 3 or 10 mod 13 R14: all k = m^2 and m = 2 or 3 mod 5, and all k = 14*m^2 and m = 2 or 3 mod 5 R16: all k = m^2 R19: all k = m^2 and m = 2 or 3 mod 5, and all k = 19*m^2 and m = 2 or 3 mod 5 R24: all k = m^2 and m = 2 or 3 mod 5, and all k = 6*m^2 and m = 1 or 4 mod 5 
According to [URL]http://oeis.org/A217377[/URL], for extended R6:
k=2626: the smallest n such that (2626*6^n1)/gcd(26261,61) is (probable) prime is 27871, that is, (2626*6^278711)/5 is a probable prime. k=6236: tested to 86.5K, no primes. k=9491: tested to 58K, no primes. 
Thus, 2626 can be removed form the list for the remain k's for extended R6.
There are only 12 remain k's for extended R6: (k<84687) (excluding the k's that are a multiple of 6 and where (k1)/gcd(k1,61) is not prime, since such k's will have the same prime as k / 6) 1597, 6236, 9491, 37031, 49771, 50686, 53941, 55061, 57926, 76761, 79801, 83411 
Congrats on having your own subforum, Sweety439!!!

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Since the conjectured k for extended R4 (k=361) is a square, I found and proved the first nonsquare such k: k=919.
The (probable) primes for n>1000 are: 74*4^12761 (106*4^45531)/3 296*4^12751 (it is the same as the prime for k=74) (373*4^25081)/3 (424*4^45521)/3 (it is the same as the prime for k=106) 659*4^4002581 674*4^58381 (751*4^66151)/3 Except (751*4^66151)/3, all of the numbers have been proven to be prime, but (751*4^66151)/3 is only a probable prime. The text file does not include square k's, since all of the numbers of the form (k*4^n1)/gcd(k1,41) for square k's have algebra factors. 
Some k's of the conjectures make GFNs or half GFNs.
Define of GFN: q^m*b^n+1 where b is the base, m>=0, and q is a root of the base. Define of half GFN: (q^m*b^n+1)/2 where b is the base, m>=0, and q is a root of the base. Since if b=q^r and q^m*b^n+1 (=q^(r*n+m)+1) or (q^m*b^n+1)/2 (=(q^(r*n+m)+1)/2) is prime, then r*n+m must be a power of 2. However, in my definition of extended Sierpinski/Riesel numbers, I do not exclude the GFNs and half GFNs in the conjectures, but I exclude the k's make a full covering set with all or partial algebraic factors. These numbers are GFNs: 65536*2^n+1 12*12^n+1 18*18^n+1 1296*6^n+1 100*10^n+1 4*32^n+1 1*38^n+1 1*50^n+1 2*512^n+1 36*216^n+1 These numbers are half GFNs: (225*15^n+1)/2 (1*31^n+1)/2 (1*55^n+1)/2 (37*37^n+1)/2 (625*5^n+1)/2 (27*243^n+1)/2 These numbers are neither GFNs nor half GFNs: 4*155^n+1 4*737^n+1 (2*16^n+1)/3 (7*49^n+1)/8 (1*24^n1)/23 ***(1*9^n1)/8 ***2500*16^n+1 ***4*625^n+1 ***(512*8^n1)/7 ***(64*16^n+1)/5 ***These forms can not have any primes, but it is because the full algebra factors. Thus, in the Riesel case, (k*b^n1)/gcd(k1,b1) is never a GFN or a half GFN, in the Sierpinski case, (k*b^n+1)/gcd(k+1,b1) is a GFN if and only if k is a power of a root of b and gcd(k+1,b1) = 1, (k*b^n+1)/gcd(k+1,b1) is a half GFN if and only if k is a power of a root of b and gcd(k+1,b1) = 2. 
Start bases 25 to 32 in both sides.

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For the Sierpinski side, all bases 25<=b<=32 are done except base 28, which has a higher conjectured k.
These are text files for the primes of extended Sierpinski base 25 to 27. 
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These are primes for the extended Sierpinski base 29 to 32.
The remain k's for extended Sierpinski base 25 to 32 are: (except base 28) Base 25: 61, 71. Base 26: 65, 155. (both are CRUS numbers) Base 27: none. Base 29: none. Base 30: 278, 588. (both are CRUS numbers) Base 31: 1, 5, (31), 43, 51, 73, 77, 107, 117, 149, (155), 181, 189, 191, 209. Base 32: 4. (it is a CRUS number, but since it is a GFN, it is excluded in CRUS conjectures) (numbers with "()" are k's included in the conjectures but excluded from testing, i.e. k's that are multiple of b and where (k+1)/gcd(k+1, b1) is not prime. In these examples, k=31 will have the same prime as k=1, and k=155 will have the same prime as k=5) 
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For the Riesel side, all bases 25<=b<=32 are done except base 28 and base 30, which have higher conjectured k's.
These are text files for the primes of extended Riesel base 25 to 27. 
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These are primes for the extended Sierpinski base 29, 31 and 32.
The remain k's for extended Sierpinski base 25 to 32 are: (except base 28 and base 30) Base 25: none. Base 26: 121. Base 27: none. Base 29: none. Base 31: 5, 19, 49, 51, 73, 97, 113, 123, 124. Base 32: none. Although 124 is a multiple of 31, but since (1241)/gcd(1241,311) = 41 is prime. Thus, 124 is not excluded from testing, we should find a prime of the form (124*31^n1)/3 with n>=1. All multiples of b need an n>=1 prime. 
Found the probable prime (61*25^3104+1)/2.
k=61 removed from S25. Now, S25 has only k=71 remain. 
Extended Sierpinski:
base conjectured k 2 78557 cover: {3, 5, 7, 13, 19, 37, 73} period=36 3 11047 cover: {2, 5, 7, 13, 73} period=12 4 419 cover: {3, 5, 7, 13} period=6 5 7 cover: {2, 3} period=2 6 174308 cover: {7, 13, 31, 37, 97} period=12 7 209 cover: {2, 3, 5, 13, 43} period=12 8 47 cover: {3, 5, 13} period=4 9 31 cover: {2, 5} period=2 10 989 cover: {3, 7, 11, 13} period=6 11 5 cover: {2, 3} period=2 12 521 cover: {5, 13, 29} period=4 13 15 cover: {2, 7} period=2 14 4 cover: {3, 5} period=2 15 673029 cover: {2, 17, 113, 1489} period=8 16 38 cover: {3, 7, 13} period=3 17 31 cover: {2, 3} period=2 18 398 cover: {5, 13, 19} period=4 19 9 cover: {2, 5} period=2 20 8 cover: {3, 7} period=2 21 23 cover: {2, 11} period=2 22 2253 cover: {5, 23, 97} period=4 23 5 cover: {2, 3} period=2 24 30651 cover: {5, 7, 13, 73, 79} period=12 25 79 cover: {2, 13} period=2 26 221 cover: {3, 7, 19, 37} period=6 27 13 cover: {2, 7} period=2 28 4554 cover: {5, 29, 157} period=4 29 4 cover: {3, 5} period=2 30 867 cover: {7, 13, 19, 31} period=6 31 239 cover: {2, 3, 7, 19} period=6 32 10 cover: {3, 11} period=2 33 511 cover: {2, 17} period=2 34 6 cover: {5, 7} period=2 35 5 cover: {2, 3} period=2 36 1886 cover: {13, 31, 37, 43} period=6 37 39 cover: {2, 19} period=2 38 14 cover: {3, 13} period=2 39 9 cover: {2, 5} period=2 40 47723 cover: {3, 7, 41, 223} period=6 41 8 cover: {3, 7} period=2 42 13372 cover: {5, 43, 353} period=4 43 21 cover: {2, 11} period=2 44 4 cover: {3, 5} period=2 45 47 cover: {2, 23} period=2 46 881 cover: {3, 7, 103} period=3 47 5 cover: {2, 3} period=2 48 1219 cover: {7, 13, 61, 181} period=6 49 31 cover: {2, 5} period=2 50 16 cover: {3, 17} period=2 51 25 cover: {2, 13} period=2 52 28674 cover: {5, 53, 541} period=4 53 7 cover: {2, 3} period=2 54 21 cover: {5, 11} period=2 55 13 cover: {2, 7} period=2 56 20 cover: {3, 19} period=2 57 47 cover: {2, 5, 13} period=4 58 488 cover: {3, 7, 163} period=3 59 4 cover: {3, 5} period=2 60 16957 cover: {13, 61, 277} period=4 61 63 cover: {2, 31} period=2 62 8 cover: {3, 7} period=2 63 1589 cover: {2, 5, 397} period=4 64 14 cover: {5, 13} period=2 Extended Riesel: base conjectured k 2 509203 cover: {3, 5, 7, 13, 17, 241} period=24 3 12119 cover: {2, 5, 7, 13, 73} period=12 4 361 cover: {3, 5, 7, 13} period=6 5 13 cover: {2, 3} period=2 6 84687 cover: {7, 13, 31, 37, 97} period=12 7 457 cover: {2, 3, 5, 13, 19} period=12 8 14 cover: {3, 5, 13} period=4 9 41 cover: {2, 5} period=2 10 334 cover: {3, 7, 13, 37} period=6 11 5 cover: {2, 3} period=2 12 376 cover: {5, 13, 29}, period=4 13 29 cover: {2, 7} period=2 14 4 cover: {3, 5} period=2 15 622403 cover: {2, 17, 113, 1489} period=8 16 100 cover: {3, 7, 13} period=3 17 49 cover: {2, 3} period=2 18 246 cover: {5, 13, 19} period=4 19 9 cover: {2, 5} period=2 20 8 cover: {3, 7} period=2 21 45 cover: {2, 11} period=2 22 2738 cover: {5, 23, 97} period=4 23 5 cover: {2, 3} period=2 24 32336 cover: {5, 7, 13, 73, 577} period=12 25 105 cover: {2, 13} period=2 26 149 cover: {3, 7, 31, 37} period=6 27 13 cover: {2, 7} period=2 28 3769 cover: {5, 29, 157} period=4 29 4 cover: {3, 5} period=2 30 4928 cover: {13, 19, 31, 67} period=6 31 145 cover: {2, 3, 7, 19} period=6 32 10 cover: {3, 11} period=2 33 545 cover: {2, 17} period=2 34 6 cover: {5, 7} period=2 35 5 cover: {2, 3} period=2 36 33791 cover: {13, 31, 43, 97} period=6 37 29 cover: {2, 5, 7, 13, 67} period=12 38 13 cover: {3, 5, 17} period=4 39 9 cover: {2, 5} period=2 40 25462 cover: {3, 7, 41, 223} period=6 41 8 cover: {3, 7} period=2 42 15137 cover: {5, 43, 353} period=4 43 21 cover: {2, 11} period=2 44 4 cover: {3, 5} period=2 45 93 cover: {2, 23} period=2 46 928 cover: {3, 7, 103} period=3 47 5 cover: {2, 3} period=2 48 3226 cover: {5, 7, 461} period=4 49 81 cover: {2, 5} period=2 50 16 cover: {3, 17} period=2 51 25 cover: {2, 13} period=2 52 25015 cover: {3, 7, 53, 379} period=6 53 13 cover: {2, 3} period=2 54 21 cover: {5, 11} period=2 55 13 cover: {2, 7} period=2 56 20 cover: {3, 19} period=2 57 144 cover: {5, 13, 29} period=4 58 547 cover: {3, 7, 163} period=3 59 4 cover: {3, 5} period=2 60 20558 cover: {13, 61, 277} period=4 61 125 cover: {2, 31} period=2 62 8 cover: {3, 7} period=2 63 857 cover: {2, 5, 397} period=4 64 14 cover: {5, 13} period=2 
S25, k=71 is likely tested to n=10000, no (probable) prime was found, base released.
Reserve R26, k=121. 
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S2: conjectured k=78557, 6 k's remain (21181, 22699, 24737, 55459, 65536, 67607)
S3: conjectured k=11047, not completely started. S4: conjectured k=419, proven. S5: conjectured k=7, proven. S6: conjectured k=174308, not completely started. S7: conjectured k=209, proven. S8: conjectured k=47, proven. S9: conjectured k=31, proven. S10: conjectured k=989, 2 k's remain (100 and 269). S11: conjectured k=5, proven. S12: conjectured k=521, only k=12 remain. S13: conjectured k=15, proven. S14: conjectured k=4, proven. S15: conjectured k=673029, not completely started. S16: conjectured k=38, proven. S17: conjectured k=31, proven. S18: conjectured k=398, only k=18 remain. S19: conjectured k=9, proven. S20: conjectured k=8, proven. S21: conjectured k=23, proven. S22: conjectured k=2253, not completely started. S23: conjectured k=5, proven. S24: conjectured k=30651, not completely started. S25: conjectured k=79, only k=71 remain. S26: conjectured k=221, 2 k's remain (65 and 155). S27: conjectured k=13, proven. S28: conjectured k=4554, not completely started. S29: conjectured k=4, proven. S30: conjectured k=867, 2 k's remain (278 and 588). S31: conjectured k=239, 13 k's remain (1, 5, 43, 51, 73, 77, 107, 117, 149, 181, 189, 191, 209). S32: conjectured k=10, only k=4 remain. R2: conjectured k=509203, 52 k's remain (2293, 9221, 23669, 31859, 38473, 46663, 67117, 74699, 81041, 93839, 97139, 107347, 121889, 129007, 143047, 146561, 161669, 192971, 206039, 206231, 215443, 226153, 234343, 245561, 250027, 273809, 315929, 319511, 324011, 325123, 327671, 336839, 342847, 344759, 351134, 362609, 363343, 364903, 365159, 368411, 371893, 384539, 386801, 397027, 409753, 444637, 470173, 474491, 477583, 478214, 485557, 494743) R3: conjectured k=12119, 15 k's remain (1613, 1831, 1937, 3131, 3589, 5755, 6787, 7477, 7627, 7939, 8713, 8777, 9811, 10651, 11597) R4: conjectured k=361, proven. R5: conjectured k=13, proven. R6: conjectured k=84687, 12 k's remain (1597, 6236, 9491, 37031, 49771, 50686, 53941, 55061, 57926, 76761, 79801, 83411) R7: conjectured k=457, only k=197 remain. R8: conjectured k=14, proven. R9: conjectured k=41, proven. R10: conjectured k=334, proven. R11: conjectured k=5, proven. R12: conjectured k=376, proven. R13: conjectured k=29, proven. R14: conjectured k=4, proven. R15: conjectured k=622403, not completely started. R16: conjectured k=100, proven. R17: conjectured k=49, proven. R18: conjectured k=246, proven. R19: conjectured k=9, proven. R20: conjectured k=8, proven. R21: conjectured k=45, proven. R22: conjectured k=2738, not completely started. R23: conjectured k=5, proven. R24: conjectured k=32336, not completely started. R25: conjectured k=105, proven. R26: conjectured k=149, only k=121 remain. R27: conjectured k=13, proven. R28: conjectured k=3769, not completely started. R29: conjectured k=4, proven. R30: conjectured k=4928, not completely started. R31: conjectured k=145, 9 k's remain (5, 19, 49, 51, 73, 97, 113, 123, 124). R32: conjectured k=10, proven. 
Found the probable prime (121*26^15091)/5.
Extended R26 is proven!!! 
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For R30, all k's not = 1 mod 29 are in CRUS. Thus, we only need to reserve the k's = 1 mod 29.
There is only one such k < 4928 remain: 1654. Compare with CRUS, there are 10 k's remain for extended R30: 659, 1024, 1580, 1654, 1936, 2293, 2916, 3719, 4372, 4897. 
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S3 done for k<=5000. (tested up to n=3000, this text file lists 0 if there is no prime for n<=3000)
For k<=500, there is already a text file. Thus, this text file is only for 501<=k<=5000. Reserve S3, 5001<=k<11047. (the conjectured k for extended S3 is 11047) 
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S3 is now completely done for all k<11047, tested up to n=3000, these text files list 0 if there is no prime for n<=3000.
These are the text files for 5001<=k<11047. Reserve R7 k=197 to 20K and S10 k=269 to 12K, use factordb. 
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The complete text file for extended S3, if you need it.
Added the n's for k=1107, 3321 and 9963, which have the same prime as k=41: (41*3^4892+1)/2. Extended S3 has 55 k's remain at n=3000. (including the k's without testing, e.g. 2463 (=821*3)) {621, 821, 823, 935, 1187, 1801, 1863, 2463, 2469, 2747, 2805, 2909, 3007, 3047, 3061, 3307, 3561, 4495, 5147, 5321, 5403, 5589, 5743, 5893, 6041, 6427, 6569, 6575, 6967, 7297, 7389, 7407, 7927, 8161, 8227, 8241, 8389, 8415, 8467, 8609, 8727, 8863, 8987, 9021, 9061, 9141, 9183, 9263, 9449, 9721, 9921, 10207, 10243, 10683, 10741} 
Reserve S28 (only need to test k's = 2 (mod 3), other k's are already in CRUS) and R28 (only need to test k's = 1 (mod 3), other k's are already in CRUS).
Both reserve to n=1000. 
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