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Old 2019-12-15, 17:22   #23
Branger
 
Oct 2018

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For the next one I had a bit more luck and found it after only 20 SNFS factorizations.

10^167+6453 =

203214913448641292965085614133875784826110271627178496334164562386280018360230767193 *
492089868321958070178727516157409397743940386446977363088474243145123507652778037821
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Old 2020-05-01, 09:55   #24
Branger
 
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The next one took much longer, requiring 140 SNFS factorizations, but now I am happy to report that

10^167-38903 =

203295679518280624355545616168150860499969671339902409710914658195811040122874591267 *
491894369014408217255986821288848144293491232238922901468113805403238598987818347491
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Old 2020-08-30, 20:47   #25
Branger
 
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And an additional 90 SNFS factorizations show that

10^169 + 25831 =

1578640553322706420836164892965282526510795833878698113106432074544532020287270837641 *
6334564241968890714235608069337466422072649801566867783807088391771261962286575780591
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Old 2020-09-13, 14:26   #26
swishzzz
 
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Are these recently found brilliant numbers tracked anywhere? https://www.alpertron.com.ar/BRILLIANT.HTM doesn't seem to have anything above 155 digits.

Reserving 10^147 - n for n < 10000.
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Old 2020-09-13, 17:02   #27
alpertron
 
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At this moment I'm making changes to my calculator that factors and finds the roots of polynomials (you can see it at https://www.alpertron.com.ar/POLFACT.HTM). After that, I will update the page of brilliant numbers.

You can select whether you want to appear with your real name or with the username at this forum.

Thanks a lot for your efforts.
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Old 2020-09-19, 00:24   #28
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I've just added the discoveries posted to this thread to https://www.alpertron.com.ar/BRILLIANT.HTM and also fixed the errors detected at https://www.alpertron.com.ar/BRILLIANT2.HTM
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Old 2020-09-19, 21:21   #29
fivemack
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Thank you! May I also point you at https://mersenneforum.org/showthread.php?t=22626 ?
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Old 2020-09-25, 03:28   #30
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I've just updated the page https://www.alpertron.com.ar/BRILLIANT3.HTM with your results. Thanks a lot.
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Old 2020-09-30, 16:03   #31
fivemack
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The smallest 400-bit number with two 200-bit prime factors is

0x98B1A3CA31877A7140FEFFA30608FBAB17232646BEC3BAA167 * 0xD699697AC5B27CD0A75D35F9E19320D82A4F4101B550C65E97 = 2^399+198081

(about 15 curves at b1=1e6 for 2^399+{1..10^6} and then SNFS on about 800 400-bit numbers taking a median of 15740 seconds on one thread of i9/7940X)

I've got an evidence file with a prime factor of less than 200 bits for every 2^399+N which is composite and coprime to (2^23)! but am not quite sure where's best to put it

Last fiddled with by fivemack on 2020-09-30 at 16:03
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Old 2020-10-14, 08:54   #32
Alfred
 
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Quote:
Originally Posted by Dr Sardonicus View Post
It is intuitively obvious that, if k is "sufficiently large", the smallest "2-brilliant" number n > 22k is n = p1*p2, where p1 = nextprime(2k) and p2 = nextprime(p1 + 1). Numerical evidence suggests that "sufficiently large" is k > 3. This notion "obviously" applies to any base.
Dr Sardonicus,

does this statement apply to largest 2-brilliant numbers in base 10?

If yes, please give an example.
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Old 2020-10-14, 12:52   #33
swishzzz
 
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2^293 - 33769 is the product of two 147-bit primes:

Quote:
P45 = 126510626365064224002933822140885711272659161
P45 = 125794520368511128755464888278721782242924143
Factor file for 2^293 - c with c > 0 attached.
Attached Files
File Type: txt 293b_brilliant.txt (33.8 KB, 10 views)

Last fiddled with by swishzzz on 2020-10-14 at 12:55
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