20220817, 05:55  #12 
"Rashid Naimi"
Oct 2015
Remote to Here/There
4435_{8} Posts 
Just realized my error in the code checking valuation p instead of x.
Not getting any younger. 
20220817, 15:43  #13 
Jun 2012
Boulder, CO
5^{2}×17 Posts 
LLR2 does seem to be able to handle these:
Code:
Starting Fermat PRP test of M3636649/known_factors Using AVX512 FFT length 192K, Pass1=192, Pass2=1K, clm=1, 4 threads, a = 3 Iteration: 30000 / 3636649 [0.32%], ms/iter: 0.612, ETA: 00:36:48 
20220817, 20:40  #14  
"Rashid Naimi"
Oct 2015
Remote to Here/There
2,333 Posts 
Quote:
Code:
\\ EPP340A ExpPower1=2 subtrExpPBy =1 ExpPower2=ExpPower1subtrExpPBy a=2 forprime(q=2,19,{ p=(a^(q^ExpPower1)1)/(a^(q^ExpPower2)1); print("\n(",a,"^(",q,"^",ExpPower1,")1)/(",a,"^(",q,"^",ExpPower2,")1)"); if(ispseudoprime(p),print("** PRP ",p)); fordiv(p,x, if(x==1,next(1)); print("valuation = ",valuation(x1,q)); ); }) Code:
(2^(2^2)1)/(2^(2^1)1) ** PRP 5 valuation = 2 (2^(3^2)1)/(2^(3^1)1) ** PRP 73 valuation = 2 (2^(5^2)1)/(2^(5^1)1) valuation = 2 valuation = 2 valuation = 2 (2^(7^2)1)/(2^(7^1)1) ** PRP 4432676798593 valuation = 2 (2^(11^2)1)/(2^(11^1)1) valuation = 2 valuation = 2 valuation = 2 (2^(13^2)1)/(2^(13^1)1) valuation = 2 valuation = 2 valuation = 2 valuation = 2 valuation = 2 valuation = 2 valuation = 2 (2^(2^3)1)/(2^(2^2)1) ** PRP 17 valuation = 4 (2^(3^3)1)/(2^(3^2)1) ** PRP 262657 valuation = 3 (2^(5^3)1)/(2^(5^2)1) valuation = 3 valuation = 3 valuation = 3 (2^(7^3)1)/(2^(7^2)1) valuation = 3 valuation = 3 valuation = 4 valuation = 4 valuation = 3 valuation = 3 valuation = 5 valuation = 3 valuation = 5 valuation = 3 valuation = 3 valuation = 3 valuation = 4 valuation = 3 valuation = 3 (2^(2^4)1)/(2^(2^3)1) ** PRP 257 valuation = 8 (2^(3^4)1)/(2^(3^3)1) valuation = 4 valuation = 4 valuation = 4 valuation = 4 valuation = 5 valuation = 5 valuation = 4 

20220817, 21:27  #15  
Einyen
Dec 2003
Denmark
2×17×101 Posts 
Quote:
ABC (2^($a*$a)1)/(2^$a1) 59 result: (2^(59*59)1)/(2^591) is 3PRP! (0.0235s+0.0004s) Here are my results from back then, including more factors up to 10000: http://hoegge.dk/mersenne/M(p2)divM(p).txt Here is the remaining candidates 20005000: Code:
ABC (2^($a*$a)1)/(2^$a1) 2027 2029 2069 2081 2089 2099 2141 2143 2161 2179 2239 2243 2267 2269 2281 2287 2297 2311 2333 2341 2351 2377 2383 2423 2447 2459 2467 2477 2531 2539 2549 2551 2579 2617 2647 2677 2683 2689 2713 2719 2729 2731 2767 2789 2797 2801 2803 2833 2851 2857 2861 2897 2917 2927 2939 2963 3001 3011 3019 3119 3137 3169 3181 3187 3191 3203 3221 3251 3253 3271 3323 3331 3371 3407 3413 3433 3449 3457 3469 3491 3511 3571 3581 3631 3643 3697 3709 3719 3727 3779 3803 3823 3833 3881 3911 3919 3929 3931 3943 3967 3989 4007 4013 4019 4021 4027 4073 4091 4139 4153 4157 4159 4177 4231 4241 4243 4283 4297 4327 4337 4339 4349 4357 4363 4409 4423 4441 4447 4481 4513 4519 4523 4583 4591 4637 4651 4663 4703 4721 4723 4729 4733 4759 4793 4817 4831 4861 4877 4903 4919 4933 4967 4969 

20220818, 00:11  #16 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
23501_{8} Posts 
I would recommend squaring ($a*$a) externally, like this
ABC (2^$a1)/(2^$b1) 3481 59 And with this form, prefactor; and even better  prefactor each in its own workunit with option N k l f"9999{6962}" < insert 2p^{2} here Prefactoring preproject can be prepared with awk or perl and then pfgw (each in separate folder). And after pfgw ran for several hours, kill and check if pfgw.out has "has factor", then remove from candidate list. Then finish off with LLR as configured above (with /$e ) 
20220818, 05:18  #17 
Einyen
Dec 2003
Denmark
2·17·101 Posts 
You can trial factor 2^{p^2}1 more efficiently on GPU with a modified mfaktc that accepts p^2, or if on a CPU then a modified version of Ernst MFactor that accepts p^2.
If Pfgw is not used for prefactor is there any advantage of using this form? ABC (2^$a1)/(2^$b1) 3481 59 
20220818, 08:24  #18 
Einyen
Dec 2003
Denmark
110101101010_{2} Posts 
Factoring limits I factored remaining candidates up to back then with mfaktc (only looking for factors 2*k*p^2 + 1)
20002803 65 bits 28333989 66 bits 40075653 67 bits 56577993 68 bits 800910000 69 bits Last fiddled with by ATH on 20220818 at 08:25 
20220819, 01:27  #19 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2,333 Posts 
A false alarm with no small factors!
Code:
PFGW Version 3.8.3.64BIT.20161203.Win_Dev [GWNUM 28.6] (3^(19^4)1)/(3^(19^3)1) is 3PRP! (56.4782s+0.0033s) And one with no known factors: http://factordb.com/index.php?query=...23%5E2%291%29 Code:
PFGW Version 3.8.3.64BIT.20161203.Win_Dev [GWNUM 28.6] (3^(23^3)1)/(3^(23^2)1) is 3PRP! (0.3976s+0.0002s) Looks like will have to avoid any of the bases equal to the power of q: Code:
PFGW Version 3.8.3.64BIT.20161203.Win_Dev [GWNUM 28.6] ***WARNING! file CIM100E.txt may have already been fully processed. Primality testing (3^(23^4)1)/(3^(23^3)1) [N1/N+1, BrillhartLehmerSelfridge] Running N1 test using base 2 (3^(23^4)1)/(3^(23^3)1) is composite (426.7206s+0.0078s) Last fiddled with by a1call on 20220819 at 02:14 
20220819, 20:49  #20  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
13·773 Posts 
Quote:
Of course, (with cold statistician's eyes) the expected yield of this fishing expedition is approximately the same as for Motzkin primes, i.e. 0.000. 

20220820, 05:55  #21 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2,333 Posts 
Yes indeed. They are a general format of
(a^q^n1) / (a^q^(n1)1 for primes q Fermat numbers are a subset of this format and we know how easy it is to find Fermat primes. :smile Like Fermat numbers they seem to be easily factorable and compared to other integers of similar size they are, but they get so large so fast. I hope I didn’t encourage anyone to go on a wild goose chase. Additionally there seems to be many pseudoprimes in small bases. 
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