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 2005-04-05, 21:53 #1 robert44444uk     Jun 2003 Oxford, UK 2,039 Posts Octoproths I am posting this again for those looking for a challenge - but this time in the Maths thread: Try to find the smallest and largest integer (maybe larger than you think) k*2^n+1 such that: k*2^n+1, k*2^n-1, k*2^(n+1)+1, k*2^(n+1)-1, 2^n+k, 2^n-k, 2^(n+1)+k, 2^(n+1)-k, all probable prime I have called these octoproths. On the large side of things, I have looked at n=32 and 1
 2005-04-06, 22:58 #2 robert44444uk     Jun 2003 Oxford, UK 2,039 Posts A bit bigger Maybe I am just being a bit shy, but I did not raise n too much but I devoted a whole 8 hours to sieving and came across a lot of larger values at n=40, namely: 3721055715 8781593205 9073509705 15541397325 20640857145 23820338205 27678219015 28483380255 29056766445 34144356345 38016547275 38354659875 43635963015 43838018925 48817120275 57779210775 58422340185 58443226395 59679412785 61308889305 62456594325 64989748725 76695589755 76715178345 79069804605 89329921695 92992163025 93765951525 97257415185 98114137575 99030787695 They are not all divisible by 105, but are all divisible by 15 I think this means that tonight my compuiter will sieve at the n=50 level Regards Robert Smith
 2005-04-07, 08:40 #3 axn     Jun 2003 123578 Posts k=925905105, n=64 925905105*2^64+1 925905105*2^64-1 925905105*2^(64+1)+1 925905105*2^(64+1)-1 2^64 + 925905105 2^64 - 925905105 2^(64+1) + 925905105 2^(64+1) - 925905105 All are prime!! Searched for 1 < k < 2^32
 2005-04-07, 10:31 #4 axn     Jun 2003 535910 Posts k=3539387145, n=65
 2005-04-07, 10:47 #5 Templus     Jun 2004 2·53 Posts How do you find these 'k'-s? Do you first sieve for a range 1 < k < 2^32 with a given 'n' for the form k*2^n+1, and after that you use that sieve as seed for a sieve for the form k*2^n-1? And use those results as seed for another sieve for another form, etcetera etcetera? Or is there are another (sieve-)program that can help? Quite interesting numbers though! Last fiddled with by Templus on 2005-04-07 at 10:47
 2005-04-07, 12:04 #6 axn     Jun 2003 23×233 Posts I used NewPGen BiTwin option which sieves for the 4 forms: k.2^n+/-1 and k.2^(n+1)+/-1 This reduces the billions of k's to about 10000. It is then run thru LLR, first for k.2^n+1, whose output is again LLR-ed for k.2^n-1, whose output is *again* LLR-ed for k.2^(n+1)+1, and finally (you guessed it!) LLR-ed for k.2^(n+1)-1 The resulting k's are then normalized (I sieve for even k's also) to odd k's. Then after a bit of script magic and Dario Alpern's batch factorization, I reduce the list to 2^n+k primes, then to 2^n-k primes, and finally the last two forms to come up with the solutions. Incidentally, just completed n=66 and no solutions Oh well. On to 67 !! Last fiddled with by axn on 2005-04-07 at 12:16
 2005-04-07, 13:00 #7 axn     Jun 2003 535910 Posts Nothing for n=67 either
 2005-04-07, 16:08 #8 robert44444uk     Jun 2003 Oxford, UK 2,039 Posts Another way I also use the bitwin option in NewPGen to reduce to an acceptable number of candidates. Sieving to 1 billion is fine, which is the minimum p which the large k range option NewPGen uses. My NewPgen splits the range into 120 or so smaller ranges and then batches all of the survivors together when each range is tested to 1 billion. At the next stage, I do not check the output for primes. Instead I alter the first line of the NewPgen output file to make this into an ABC file, and use the & option to check for the four other values of the octoproth. This file is then put through pfgw, with the -f100 option, which, for n=50 checks around 10,000 values of k a second. The output file for this run can be inspected to see if there are any values where all four possibilities are prp. It is quicker this way because the pfgw will give up testing the value of k as soon as it spots a composite. If you test the original output file then a lot of values will have to be tested for all four options as none of them, if they are composite, have factors of less than 1 billion. The final stage is to extract the few (maybe only ten values) where the output file shows the complete set is prp, and make this into a PFGW input file with the first line of the file the same as the output file from the bitwin output file. About half of the values will be octoproths. Therefore almost all of the work is in the sieving because a very large number of k must be looked at, for a given n. My computer takes 8 hours to sieve k=2 to k=10^11 up to p=1 billion. A further 2 hours takes the file to p=30 billion. I ran this range for n=50, and found about 8 octoproths. But it looks as if I have been overtaken by events. The bar has been raised! Regards Robert Smith
 2005-04-07, 18:01 #9 Templus     Jun 2004 2×53 Posts I tried this method for n=81 and 2 < k < 2^32. I made a sieve file which contains about 16000 lines, but I'm not sure about the ABC rule I have to construct. I now have: Code: ABC $a*2^$b+1 & $a*2^$b-1 & $a*2^($b+1)+1 & $a*2^($b+1)-1 & 2^$b+$a & 2^$b-$a & 2^($b+1)+$a & 2^($b+1)-$a Is this definition correct? When I run this code and I check the pfgw.log file I only see lines of the form k*2^n+1, k*2^n-1, k*2^(n+1)+1, k*2^(n+1)-1 but not of the form 2^n+k, 2^n-k, 2^(n+1)+k, 2^(n+1)-k. What am I doing wrong?
 2005-04-07, 21:35 #10 robert44444uk     Jun 2003 Oxford, UK 2,039 Posts Answer The line I use is (followed by first putput from the NewPGen file): ABC 2^$b+$a & 2^$b-$a & 2^($b+1)+$a & 2^($b+1)-$a 708435 40 Regards Robert Smith
2005-04-08, 03:48   #11
axn

Jun 2003

23×233 Posts

Quote:
 Originally Posted by robert44444uk My computer takes 8 hours to sieve k=2 to k=10^11 up to p=1 billion. A further 2 hours takes the file to p=30 billion.
How do you sieve with k as high as 10^11. I thought that NewPGen only supported upto k < 2^32 !

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