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#45 | |
May 2007
Kansas; USA
22×52×107 Posts |
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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 |
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#46 |
May 2007
Kansas; USA
29CC16 Posts |
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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 |
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#47 | |
May 2004
FRANCE
22·3·72 Posts |
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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 |
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#48 |
May 2004
FRANCE
22·3·72 Posts |
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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 |
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#49 |
May 2007
Kansas; USA
1070010 Posts |
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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 |
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#50 |
May 2007
Kansas; USA
22·52·107 Posts |
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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 |
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#51 | |
A Sunny Moo
Aug 2007
USA (GMT-5)
141518 Posts |
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#52 | |
May 2007
Kansas; USA
101001110011002 Posts |
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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 |
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#53 | |
A Sunny Moo
Aug 2007
USA (GMT-5)
3×2,083 Posts |
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#54 |
Oct 2006
7·37 Posts |
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#55 |
May 2007
Kansas; USA
22×52×107 Posts |
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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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Thread Tools | |
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Bases 501-1030 reservations/statuses/primes | KEP | Conjectures 'R Us | 4084 | 2022-05-15 06:34 |
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