20140922, 18:39  #1 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9428_{10} Posts 
Near and quasirepunit PRPs
M.Kamada has been collecting the near and quasirepunit 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 3based composites. Then proceed with sr{12}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 strongFermat, 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^n1" 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! (64·10^7628111)/9 is a (SFLF)PRP. Similarly, (64·10^7794651)/81 is a (nonrepunit) SPRP, and (89·10^411590+1)/9 is a PlateauandDepression SPRP ("988...889"). All these are still in the PRP Top processing queue. 
20140922, 19:35  #2 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}×2,357 Posts 
P.S. We can imaginatively call the (64·10^7794651)/81 a "rainbow" number, because it has all decimal digits (except 8) in order, over and over again.
Code:
79012345679012345679012345679012345679012345679012345679012345679012345679012345679012345679012345679012345679012345679012345679012345679012345679012345679012345679012345679012345679012345679... 
20141025, 03:25  #3 
May 2004
New York City
3·17·83 Posts 

20141027, 11:03  #4 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}·2,357 Posts 
A curio: "5" followed by 51111 "1"s is a PRP
Here is a small but peculiar PRP. (selfdescribing, sort of)
It is from the "511..11" nearrepunit series, and in decimal expansion it has a "5" followed by 51111 "ones". Code:
(46*10^511111)/9 is strongFermat, Lucas and Frobenius PRP! (P = 5, Q = 3, D = 13) Time : 88.194 sec. 
20141031, 00:29  #5 
Dec 2011
After milion nines:)
2^{2}·5·71 Posts 
(91*10^1120981)/9
One 1 then one 0 then 1 until end :) Thanks Batalov for showing me way for sieving and llring this type of PRP. Last fiddled with by pepi37 on 20141031 at 00:30 Reason: edit PRP 
20141103, 22:28  #6 
Dec 2011
After milion nines:)
58C_{16} Posts 

20190604, 20:11  #9 
Nov 2016
B03_{16} Posts 
Note that A200065(1073)=0, since (10^n1)/938 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 20190604 at 20:14 
20190912, 13:06  #10 
"Dylan"
Mar 2017
577_{10} 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^n1)/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^$a1)/9 Let me know. 
20190912, 13:31  #11 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}×2,357 Posts 
ABC ($a*$b^$c$d)/$e
28 10 n 1 9 
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