20081209, 03:27  #122  
Jun 2003
Oxford, UK
2×953 Posts 
Quote:
show for integer x that (x1)*(2x1)^n1 is 0modx for n odd (x1)*(2x1)^n1==1modx*(1mod2x)^n1 == 1modx*(1modx)^n1 == 1modx^(n+1)1 if n even == 1modx1=2modx if n odd == 1modx1=0modx If this is the case then there is always Riesel algebraic factor partial cover provided through odd n for any odd base b when k=(b+1)/2 1 Is this a reasonable conjecture, I am really uncertain about this. Last fiddled with by robert44444uk on 20081209 at 03:56 

20081209, 04:34  #123  
Jun 2003
Oxford, UK
1906_{10} Posts 
Quote:
There is always Riesel factor x provided through odd n for any odd base b when k== 1modx Last fiddled with by robert44444uk on 20081209 at 04:36 

20081209, 14:02  #124 
I quite division it
"Chris"
Feb 2005
England
31·67 Posts 
Riesel base 68 is proven.
Confirmed primes: 5*68^135741 7*68^253951 I'll push Riesel base 101 higher and take Riesel base 93 to at least n=20k. 
20081209, 22:00  #125 
Jan 2005
111011111_{2} Posts 

20081209, 22:06  #126 
I quite division it
"Chris"
Feb 2005
England
31×67 Posts 

20081210, 04:24  #127  
May 2007
Kansas; USA
2801_{16} Posts 
Quote:
VERY nice! To find 2 primes out of 2 remaining k for n=10K25.4K is a most excellent find! We have our first proof where the final prime was n>25K in a long time! Gary 

20081211, 08:12  #128  
May 2007
Kansas; USA
7^{2}×11×19 Posts 
Quote:
To all: I understand a little bit about what Robert posted and should be able to grasp it all after studying it in detail a little later when I have more time. For the time being, the above that I stated is not really a proof at all, it essentially just restates the fact that all oddn are == (0 mod 37) in a different way. When I thought about it, it was kind of a stupid post by me. I'm glad the highermath folks didn't attack me. They could have easily. lol For the time being, here is a better analysis of how the modulo arithmetic works on this that provides what I believe to be a "generalized" proof of this specific base if you can really call it generalized for only one base. I'm calling it that because it proves ALL n for this base whereas my poor previous attempt at a proof did nothing but "prove" that the first 4 n's had a repeating pattern that may or may not repeat for higher nvalues. First, we'll start with the obvious: 73 == (36 mod 37) Use the above with variables as in: a == (b mod c) Using elementary modulo arithmetic, we have: To square a above, you would multiply (b mod c) by a, which would yield a^2 == (a*b mod c), where a*b reduces to some value in the range of 0 thru c1. Therefore: We have 73^2 == (36*73 mod 37) == (2628 mod 37) == (1 mod 37). Going one more exponent, you have 73^3 = 73^2*73 and since 73^2 == (1 mod 37), you have: 73^3 == (1*73 mod 37) == (73 mod 37) == (36 mod 37) Here, you can stop, because you have a repeated mod. The exponents of 1 and 3 are == (36 mod 37). You now know that all 73^n where n is odd will be == (36 mod 37) and where n is even will be == (1 mod 37). We know that the mods repeat every 2n for 73^n but we don't know if there is a trivial factor for the original form of 36*73^n1 at this point. Therefore: 73^1 == (36 mod 37); multiplying that by 36 and subtracting 1 gives: 36*73^11 == (36*361 mod 37) == (1295 mod 37) == (0 mod 37); therefore a factor of 37. 73^2 == (1 mod 37); multiplying that by 36 and subtracting 1 gives: 36*73^21 == (36*11 mod 37) == (35 mod 37); therefore a remainder of 35 when dividing by 37. Since 73^n is always == (36 mod 37) when n is odd and applying a multiplier of 36 and subtracting one always gives (0 mod 37), then the above proves that all odd n give a factor of 37. And taking it further for the conjecture: Since all evenn for any k that is a perfect square always yield algebraic factors on the Riesel side, then the k is eliminated from consideration. After grasping what Robert has done here a little better, I should avoid missing "basic" trivial factors that combine with algebraic factors to make a full covering set such as what I did on base 73. Gary Last fiddled with by gd_barnes on 20081211 at 21:18 

20081212, 09:26  #129 
Mar 2006
Germany
2×1,427 Posts 
Riesel Base 35
new PRP's
Code:
87064 8031 64760 8046 244466 8054 179312 8064 132574 8065 131290 8091 206962 8091 223562 8146 17468 8160 70526 8164 249296 8174 62008 8195 122156 8212 46120 8239 89338 8245 272306 8268 227564 8278 46288 8279 81772 8305 21346 8327 57848 8338 81532 8353 100504 8365 135062 8366 101648 8368 165608 8378 36214 8431 to Gary: the PRP's from post 149 are missing on your reservationpage for base 35. jerky job.... no time to update the pages... i know that too... Last fiddled with by kar_bon on 20081212 at 09:28 
20081212, 11:07  #130  
May 2007
Kansas; USA
7^{2}·11·19 Posts 
Quote:
Thanks for letting me know. You were doing too many posts for base 35! I'm bound to miss some of them. It wasn't a matter of time. I completely missed that post. I'm sure I would have noticed when I went to post these primes; wondering why there was a large nrange gap in primes. As I sit, I've been updating the Riesel pages for bases > 50 so I'll get these and post 149 for base 35 added in a little while. Gary Last fiddled with by gd_barnes on 20081212 at 11:42 

20081212, 11:12  #131 
May 2007
Kansas; USA
7^{2}×11×19 Posts 
To all:
I have my web pages updated for Riesel bases up to 80 now. What takes quite a while is correctly generalizing the algebraic factors. I had already found them and took them into account on the k's remaining that I posted previously. But to properly prove a conjecture, they need to be generalized, especially if we eventually go to proving the 2nd or 3rd conjectured kvalue in the future. Quite a few of the bases have them that allowed k's to be removed. The one that took the longest was Riesel base 54. It has 2 different sets of algebraic factors that are very similar to Riesel base 24. Both kinds were needed to eliminate k=4, 6, and 9. I've haven't uploaded the pages yet but will do so shortly. I'm beat now and will do some more updating Friday afternoon. Gary Last fiddled with by gd_barnes on 20081212 at 11:16 
20081212, 22:51  #132  
I quite division it
"Chris"
Feb 2005
England
100000011101_{2} Posts 
Quote:
Bases 93 and 100 are both at n>37k with no primes. :surprised It must be Max's turn to find one. I'll also take Riesel base 72 to at least n=20k. 

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