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Old 2018-12-01, 23:18   #12
Batalov
 
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Phi(3,3^1118781+1)/3

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Quote:
Originally Posted by Batalov View Post
looks like a Chernick-like recipe for 3-prime factor Carmichael numbers: "if 40*q + 3, 200*q + 11 and 320*q + 17 are all prime, then their product is a Carmichael number".
This can be proven, easily, too, using Korselt's criterion.
Code:
? m=20*q+1
? ((2*m+1)*(10*m+1)*(16*m+1)-1)/(2*m)
64000*q^2 + 8520*q + 280
? ((2*m+1)*(10*m+1)*(16*m+1)-1)/(10*m)
12800*q^2 + 1704*q + 56
? ((2*m+1)*(10*m+1)*(16*m+1)-1)/(16*m)
8000*q^2 + 1065*q + 35

\\ --> all three divide
But there is probably a thousand forms similar to this one known in the literature since 1885 (Václav Šimerka). Note: before Korselt and before Carmichael.
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Old 2018-12-02, 02:51   #13
CRGreathouse
 
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What does Václav Šimerka prove there, Serge?
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Old 2018-12-02, 05:17   #14
Batalov
 
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Phi(3,3^1118781+1)/3

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I don't read Czech, sadly, only numbers... but this fragment on page 224 seems interesting...
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Old 2018-12-02, 09:27   #15
science_man_88
 
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Quote:
Originally Posted by Batalov View Post
I don't read Czech, sadly, only numbers... but this fragment on page 224 seems interesting...
google translate gives :

Quote:
Similar we find at p = 193.
the offer, according to the inventor, is the one of the most important in vague analysis by Fermatov; but it does not give a characteristic mark of truncated numbers (which would differ in all of them), similar to some divisible numbers. so we can find at p. We also find the same number at any time b with the module
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