20151005, 21:10  #1  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}·2,393 Posts 
Smarandache prime(s)
Quote:
Instructions: 1a. Get PRPclient 5.4.0 for your platform from here. It contains all programs, bundled in. 1b. (optionally) Get PFGW from here if you already have prpclient but not PFGW 2. Edit prpclient.ini : add email, username, machineid, server and pfgwexe : Code:
//============= in prpclient.ini ============= email=youremail@somewhere.com userid=UserId machineid=PhenomII instanceid=1 //... server=SmarPRP:100:1:smarandache.ddns.net:1200 //... pfgwexe=pfgw64.exe The current typical test duration is 16 hours (on a Broadwell), 1024 hours on any modern CPU. The credit for the find will go to "A,B,PFGW", where A is the finder, B is me and PFGW is Mark+George and other authors of PFGW. Happy hunting! And read the whole thread, too (optional but important). . Last fiddled with by Batalov on 20161008 at 19:47 Reason: static ip added 

20151005, 21:20  #2 
Sep 2002
Database er0rr
3×1,291 Posts 
Are you going to take a punt at it, Serge

20151005, 21:39  #3 
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
well as far as I can tell only prime indexes need to be checked because for example 123456 = 1001 *123 +333 and both 123 and 333 divide 3 and so on for the other primes but this is likely well known already, in fact it sounds familiar ! not much else I can think of that's useful to help though. edit: doh I forgot that it expands the number of digits faster than doubling for doubling n for the most part.
Last fiddled with by science_man_88 on 20151005 at 21:41 
20151005, 21:54  #4  
I moo ablest echo power!
May 2013
2^{3}·223 Posts 
Quote:


20151005, 22:08  #5 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
I realize now that 1,12,123 .. means each number adds it's value mod 3 to the sum of digits and so the sum of digits is only not divisible by 3 when you are at an index that is 3x+1 and this is only odd if x is even so we get 6y+1 and then when you follow this up you'll see the sequence 1,7,13,19,25 (25 can't work as it ends in 5), mod 30 and then if any of the endings can't work with an even number in front of them I'd say try that to eliminate tests but once again I'm probably randomly ranting again.

20151005, 23:15  #6 
"Dana Jacobsen"
Feb 2011
Bangkok, TH
1110001101_{2} Posts 
sm88, a halfdecent primality test will find these cases (divisible by small values) quite easily, even at hundreds of thousands of digits. Filtering out values divisible by the first 10k or more small primes is also quite fast  a trivial amount of time compared to running a PRP test. I'd be leery of excluding values to test unless one was quite sure of the math.

20151005, 23:31  #7  
"Forget I exist"
Jul 2009
Dumbassville
10000011000000_{2} Posts 
Quote:


20151006, 01:45  #8  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9572_{10} Posts 
Quote:
I've previously taken a bite on upanddownconcatentated primes. And just like for them, I'd like to note that for every 10^{k} <= n < 10^{k+1}, there is an explicit summation formula. Additionally, Capt.Obvious reports that only n ≡ 1,3,7,9 (mod 10) need to be considered. And n>=77000. 

20151006, 02:12  #9 
Romulan Interpreter
"name field"
Jun 2011
Thailand
10011000111011_{2} Posts 
Sum of digits of any 3 consecutive numbers is divisible by 3, so only 2/3 of the 1,3,5,7 (mod 10) have to be considered
This is what sm88 tries to tell you hehe, and you don't get it... edit: pari goes quite fast to few thousands and stays there. Interesting that the "isprime" version is faster than other "probable prime" (modular) tests, because probably the builtin "isprime" filters the numbers which are obviously not primes (like even, etc). It can be seen how the "counter" progresses "in steps", jumping over big chunks of integers. Code:
gp> i=1; n=1; while(!isprime(n),print(i++); n=eval(concat(Str(n),Str(i)))) Last fiddled with by LaurV on 20151006 at 02:22 
20151006, 02:19  #10 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}×2,393 Posts 
By G*d, you are right!! I never get what sm88 tries to whatever... because...

20151006, 02:29  #11 
Romulan Interpreter
"name field"
Jun 2011
Thailand
9,787 Posts 
Bad Batalov! Bad! The
Just rewatched Jackie Brown, in which particular case Michael Keaton's name was the [STRIKE]Birdman[/STRIKE] Lord, as everyone most probably knows. It is a so much more mellow film :rolleyes: Last fiddled with by Batalov on 20151006 at 02:39 
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