mersenneforum.org Near- and quasi-repunit PRPs
 Register FAQ Search Today's Posts Mark Forums Read

 2014-09-22, 18:39 #1 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 942810 Posts Near- and quasi-repunit PRPs M.Kamada has been collecting the near- and quasi-repunit primes and PRPs for many years. I've recently revisited this topic and remembered a few pitfalls that new searchers might fall in, so I decided to list them here: 1. Sieving is possible with sr*sieve, though a few tips are needed. First, because most forms have an implied denominator of 3 or 9, one needs to not sieve for p=3. For most forms srsieve works (with -p 5); start with srsieve to low limit, then use awk or similar to remove the 3-based composites. Then proceed with sr{1|2}sieve if possible (e.g. for +/-1 forms. 2. PRP testing is as fast as "normal" numbers (without denominator). For that, either use Prime95 (with worktodo entries like PRP={k},{b},{n},{c},"denominator") or LLR with ABC($a*$b^$c$d)/$e header! LLR runs a nice battery test: both SPRP and then a strong-Fermat, Lucas and Frobenius. 3. Use PFGW only for independent validation, because it uses general FFT and is many times slower. Example of application (711...11 series): 1. Find current limits at Kamada's page. 2. Sieve with srsieve ... -p 5 -P 2e6 "64*10^n-1" 3. Remove all entries where n != 1 (mod 6). This is because even n's (0,2,4) generate diff.of squares; n=3 (mod 6) generate diff.of cubes; and n=5 (mod 6) are divisible by additional 3's after being divided by 9. 4. Sieve with sr1sieve (or combine with some other siblings, e.g. I did the 133..33 series, and sieve with sr2sieve) 5. Prepare P95 or LLR input files with awk and run them. 6. ?????? 7. PROFIT! (64·10^762811-1)/9 is a (S-F-L-F)PRP. Similarly, (64·10^779465-1)/81 is a (non-repunit) SPRP, and (89·10^411590+1)/9 is a Plateau-and-Depression SPRP ("988...889"). All these are still in the PRP Top processing queue.  2014-09-22, 19:35 #2 Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 22×2,357 Posts P.S. We can imaginatively call the (64·10^779465-1)/81 a "rainbow" number, because it has all decimal digits (except 8) in order, over and over again. Code: 79012345679012345679012345679012345679012345679012345679012345679012345679012345679012345679012345679012345679012345679012345679012345679012345679012345679012345679012345679012345679012345679... 2014-10-25, 03:25 #3 davar55 May 2004 New York City 3·17·83 Posts Quote:  Originally Posted by Batalov P.S. We can imaginatively call the (64·10^779465-1)/81 a "rainbow" number, because it has all decimal digits (except 8) in order, over and over again. nice name  2014-10-27, 11:03 #4 Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 22·2,357 Posts A curio: "5" followed by 51111 "1"s is a PRP Here is a small but peculiar PRP. (self-describing, sort of) It is from the "511..11" near-repunit series, and in decimal expansion it has a "5" followed by 51111 "ones". Code: (46*10^51111-1)/9 is strong-Fermat, Lucas and Frobenius PRP! (P = 5, Q = 3, D = 13) Time : 88.194 sec.  2014-10-31, 00:29 #5 pepi37 Dec 2011 After milion nines:) 22·5·71 Posts (91*10^112098-1)/9 One 1 then one 0 then 1 until end :) Thanks Batalov for showing me way for sieving and llr-ing this type of PRP. Last fiddled with by pepi37 on 2014-10-31 at 00:30 Reason: edit PRP 2014-11-03, 22:28 #6 pepi37 Dec 2011 After milion nines:) 58C16 Posts Quote:  Originally Posted by pepi37 (91*10^112098-1)/9 One 1 then one 0 then 1 until end :) Thanks Batalov for showing me way for sieving and llr-ing this type of PRP. And new one: double in size :) (73*10^248145-1)/9  2019-06-04, 19:27 #7 sweety439 Nov 2016 B0316 Posts Found A200065(1033): (10^3376-1)/9-78 is prime!!! 2019-06-04, 20:08 #8 sweety439 Nov 2016 2,819 Posts Quote:  Originally Posted by sweety439 Found A200065(1033): (10^3376-1)/9-78 is prime!!! Another term is A200065(1079): (10^2553-1)/9-32 is prime!!!  2019-06-04, 20:11 #9 sweety439 Nov 2016 B0316 Posts Note that A200065(1073)=0, since (10^n-1)/9-38 is divisible by either 3 or 37 for n not divisible by 3, and has algebra factors (divisible by (10^(n/3)-7)/3, this number is >1 for n>3) for n divisible by 3. Last fiddled with by sweety439 on 2019-06-04 at 20:14  2019-09-12, 13:06 #10 Dylan14 "Dylan" Mar 2017 57710 Posts I'm looking to work on the 31111 series, and I want to make sure I have the procedure right before I start: 1. Since 31111 = 31w = (28*10^n-1)/9, there is no need to sieve p=3, and so we start with p = 5. Run srsieve to low limits (say, to 1e7) with the --newpgen flag, starting with n = 200001 (as it has been searched to 200000 from Kamada's website) to whatever max n (say, 500000). 2. Because by 2.4.1 of https://stdkmd.net/nrr/3/31111.htm, there is no need to test any n where n mod 3 = 0, so we can remove them. Not sure if it be worth removing the other factors shown in that section. 3. Continue sieve with sr1sieve up to desired sieve limit 4. Change the file header to an ABC header using srfile, and then add /9 at the end, so it would read Code: ABC (28*10^$a-1)/9 5. Use LLR to test and (hopefully) find PRP's. Let me know.
 2019-09-12, 13:31 #11 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 22×2,357 Posts ABC ($a*$b^$c$d)/\$e 28 10 n -1 9

 Similar Threads Thread Thread Starter Forum Replies Last Post sweety439 sweety439 231 2020-11-06 12:30 Bob Underwood Math 12 2020-10-11 20:01 schickel FactorDB 1 2015-08-03 02:50 Random Poster FactorDB 0 2012-07-24 10:53

All times are UTC. The time now is 20:52.

Wed May 12 20:52:55 UTC 2021 up 34 days, 15:33, 1 user, load averages: 2.32, 2.15, 2.12