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2008-01-24, 18:15   #45
gd_barnes

May 2007
Kansas; USA

22×52×107 Posts

Quote:
 Originally Posted by jasong If no one objects, I'd like to reserve 16734*4^n-1. That's base=4, k=16734, Riesel numbers(-1). I believe the n-values that need to be tested start at n=100K. If there's a sieved file, I'd love to know about it. Also, if people would rather I sieve than LLR, I can do that to. I just ask that the digit length of the lowest untested value in the sieve file be no more than twice the digit length of any un-LLred value in a lower base. In that instance, I'd probably want to sieve a lower base.

Jasong is unreserving Riesel base 4 k=16734. He is working on another unrelated effort.

This also unreserves Riesel base 2 odd-n k=8367 and Riesel base 16 k=16734.

Gary

 2008-01-24, 18:17 #46 gd_barnes     May 2007 Kansas; USA 29CC16 Posts Jasong has unreserved Riesel base 4 k=16734 per a post that I just now put in the reservations/statuses thread. This also unreserves Riesel base 16 k=16734 and the Riesel base 2 odd-n conjecture of k=8367. I have changed the above post accordingly. Gary Last fiddled with by gd_barnes on 2008-01-24 at 18:21
2008-01-24, 20:12   #47
Jean Penné

May 2004
FRANCE

22·3·72 Posts
Unreserving k=19464

Quote:
 Originally Posted by gd_barnes Karsten and Jean, There are 3 more k's for Riesel base 2 ODD-n, base 4, and base 16 that I need to confirm statuses, reservations, and test limits. Here is what I show for k=13854, 16734, and 19464 and their coharts for Riesel base 2 odd-n k=6927 and 8367: Riesel base 2 odd-n: k=6927, test limit 261945; now unreserved k=8367, test limit 262045; now unreserved Riesel base 4: k=13854, test limit 130972; now unreserved k=16734; test limit 131022; now unreserved k=19464; test limit ??; reserved by Jean Riesel base 16: k=13854, test limit 100000; now unreserved (I tested this one separately to n=100000 since we had a sieved file for it. Odd n's would still need testing separately up to n=200000 base 4) k=16734, test limit 66511; now unreserved k=19464, test limit ??; reserved by Jean Can one or both of you verify that the above is correct? My main question is for Jean: What is your test limit on k=19464? There was an original post that stated n=93672 base 4. And then a later post that stated n=137000 base 2. Since the later post was a lower range, I've become confused. Note that all of this is coming about since I'm now incorporating the base 2 even-n and odd-n conjectures into my web pages. Thanks, Gary
Gary,

I reserved k=19464 because I believed falsely it might be tested for Riesel odd n's (but 19463 is prime), so, I wish to unreserve it.
I tested it up to n = 262032 base 2, then I created a base 4 sieve file up to n=1048576 (1024K) base 4. I will make it available to users by attaching it here. It is sieved up to 49.58 billions.

Regards,
Jean
Attached Files
 vk19464.zip (22.5 KB, 338 views)

 2008-01-24, 20:45 #48 Jean Penné     May 2004 FRANCE 22·3·72 Posts Riesel odd n's : 3 new k's eliminated. 3 new primes found : 130383 104123 260766*2^104122-1 is prime! Time : 71.272 sec. 131727 169621 263454*2^169620-1 is prime! Time : 79.613 sec. 135567 68325 271134*2^68324-1 is prime! Time : 23.265 sec. 20 k's remaining now... Also, for Sierpinski base 4 / odd n's, I completed the test of k = 9267 up to 2,000,000 base 2 (the very last n is 1999615), no prime... so, I am unreserving this k. Jean
 2008-01-25, 20:56 #49 gd_barnes     May 2007 Kansas; USA 1070010 Posts I'm reserving Riesel base 4 k=13854, 16734, and 19464 for further testing up to about n=250K base 4. I'll sieve to n=500K and leave my files for others if no primes are found on one or more of the k's. It will also be the same k's for Riesel base 16. Karsten, this also means I'll be reserving k=6927 and 8367 for the Riesel base 2 odd-n conjecture. Gary
 2008-01-25, 20:57 #50 gd_barnes     May 2007 Kansas; USA 22·52·107 Posts Riesel odd-n reservations Karsten, Part of a post from the reservations thread... I'm reserving Riesel base 4 k=13854, 16734, and 19464 for further testing up to about n=250K base 4. This also means I'll be reserving k=6927 and 8367 for the Riesel base 2 odd-n conjecture and taking them to n=500K base 2. Gary
2008-01-25, 22:10   #51
mdettweiler
A Sunny Moo

Aug 2007
USA (GMT-5)

141518 Posts

Quote:
 Originally Posted by gd_barnes I'm reserving Riesel base 4 k=13854, 16734, and 19464 for further testing up to about n=250K base 4. I'll sieve to n=500K and leave my files for others if no primes are found on one or more of the k's. It will also be the same k's for Riesel base 16. Karsten, this also means I'll be reserving k=6927 and 8367 for the Riesel base 2 odd-n conjecture. Gary
I thought that base 4 Riesel only covers the even n for base 2? Thus, wouldn't you be reserving it for the even-n conjecture?

2008-01-26, 10:06   #52
gd_barnes

May 2007
Kansas; USA

101001110011002 Posts

Quote:
 Originally Posted by Anonymous I thought that base 4 Riesel only covers the even n for base 2? Thus, wouldn't you be reserving it for the even-n conjecture?
No, it also covers odd-n IF the k base 4 is divisible by 2, i.e.

Let k=2m and let n=q

2m*4^q-1 = 2m*2^(2q)-1 = m*2^(2q+1)-1

Hence where the base is 4 and k=2m, then where the base is 2, n must be 2q+1; hence odd.

In this case, k=13854 and k=16734 base 4 equate to k=6927 and k=8367 base 2 odd-n.

It's surprising how tricky it has been to keep all of our reservations clean and consistent across all bases without stepping on one another; namely the bases that are powers of 2.

Edit: One more requirement...the k base 4 cannot be divisible by 4 in order to 'reduce' to a base 2 odd-n k. The even-n and odd-n conjectures have a requirement that the k must be odd and divisible by 3 or...as shown on my pages k==3 mod 6.

Gary

Last fiddled with by gd_barnes on 2008-01-26 at 10:12

2008-01-26, 16:20   #53
mdettweiler
A Sunny Moo

Aug 2007
USA (GMT-5)

3×2,083 Posts

Quote:
 Originally Posted by gd_barnes No, it also covers odd-n IF the k base 4 is divisible by 2, i.e. Let k=2m and let n=q 2m*4^q-1 = 2m*2^(2q)-1 = m*2^(2q+1)-1 Hence where the base is 4 and k=2m, then where the base is 2, n must be 2q+1; hence odd. In this case, k=13854 and k=16734 base 4 equate to k=6927 and k=8367 base 2 odd-n. It's surprising how tricky it has been to keep all of our reservations clean and consistent across all bases without stepping on one another; namely the bases that are powers of 2. Edit: One more requirement...the k base 4 cannot be divisible by 4 in order to 'reduce' to a base 2 odd-n k. The even-n and odd-n conjectures have a requirement that the k must be odd and divisible by 3 or...as shown on my pages k==3 mod 6. Gary
Okay, thanks for the explanation.

2008-01-27, 14:12   #54
tnerual

Oct 2006

7·37 Posts

Quote:
 Originally Posted by gd_barnes Go Tnerual go! 7707 k's...wow. Good luck! G
now at n=2414 ... 4960 k's remaining

stopping for 10 days ... hollidays

all computers switched back to llrnet CRUS1

Gary, can you tag sierp31 as reserved

see you next week

tnerual

 2008-01-28, 08:39 #55 gd_barnes     May 2007 Kansas; USA 22×52×107 Posts A big one drops for odd-n... From the 'regular' primes thread: Base 2 odd-n: 8367*2^313705-1 I'm still working on Base 2 odd-n k=6927. I'm currently at n=324K base 2. Gary

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