Carmichael numbers and Šimerka numbers
 

As you are aware Carmichael numbers pertain to the property of composite numbers
behaving like prime numbers with regard to Fermat's theorem. They are Devaraj numbers
I.e. if N = p_1*p_2....p_r ( where p_i is prime) then

(P_1-1)*(N-1)/(p_2-1)......... (p_r-1) is an integer.
See A104016 and A104017.
a) conjecture: the least value of k, the degree to which atleast two of a Devaraj number's prime factors are
Inverses, is 2 (example 561 = 3*11*17 -here 3 and 17 are inverses (mod 5^2).

b) 5 and 11 are impossible cofactors of Devaraj numbers (including Carmichael numbers).
(to be continued)

Carmichael numbers and Devaraj numbers
 

[QUOTE=devarajkandadai;499098]As you are aware Carmichael numbers pertain to the property of composite numbers
behaving like prime numbers with regard to Fermat's theorem. They are Devaraj numbers
I.e. if N = p_1*p_2....p_r ( where p_i is prime) then

(P_1-1)*(N-1)^(r-2)*(p_2-1)......... (p_r-1) is an integer.
See A104016 and A104017.
a) conjecture: the least value of k, the degree to which atleast two of a Devaraj number's prime factors are
Inverses, is 2 (example 561 = 3*11*17 -here 3 and 17 are inverses (mod 5^2).

b) 5 and 11 are impossible cofactors of Devaraj numbers (including Carmichael numbers).
(to be continued)[/QUOTE]

C) 7 and 31 are inverses of 3rd degree

Carmichael numbers and Devaraj numbers
 

[QUOTE=devarajkandadai;499098]As you are aware Carmichael numbers pertain to the property of composite numbers
behaving like prime numbers with regard to Fermat's theorem. They are Devaraj numbers
I.e. if N = p_1*p_2....p_r ( where p_i is prime) then

(P_1-1)*(N-1)^(r-2)*(p_2-1)......... (p_r-1) is an integer.
See A104016 and A104017.
a) conjecture: the least value of k, the degree to which atleast two of a Devaraj number's prime factors are
Inverses, is 2 (example 561 = 3*11*17 -here 3 and 17 are inverses (mod 5^2).

b) 5 and 11 are impossible cofactors of Devaraj numbers (including Carmichael numbers).
(to be continued)[/QUOTE]

C) 7 and 31 are inverses of 3rd degree since 7 and 31 are inverses (mod 3^3).

Carmichael numbers and Devaraj numbers
 

[QUOTE=devarajkandadai;499105]C) 7 and 31 are inverses of 3rd degree since 7 and 31 are inverses (mod 3^3).[/QUOTE]

Carmichael numbers are subset of Devaraj numbers
Devaraj numbers subset of tortionfree numbers of degree k.

Carmichael numbers and Devaraj numbers
 

[QUOTE=devarajkandadai;499321]Carmichael numbers are subset of Devaraj numbers
Devaraj numbers subset of tortionfree numbers of degree k.[/QUOTE]

41and 61 are inverses of degree 4 (mod 5^4).
17 and 6947 are inverses of degree 10 (mod 3^10).

[QUOTE=devarajkandadai;499520]41and 61 are inverses of degree 4 (mod 5^4).
17 and 6947 are inverses of degree 10 (mod 3^10).[/QUOTE]

175129 and 3403470857219 are inverses of degree 25 (mod 5^25)

Well, 5 and [COLOR=Black][URL="http://factordb.com/index.php?id=1100000000033117248"]7469128023...77[/URL][/COLOR][SUB]<181>[/SUB] are goddamn inverses of degree 600.
131 and [COLOR=Black][URL="http://factordb.com/index.php?id=1100000001186922137"]1289338297...07[/URL][/COLOR][SUB]<1808>[/SUB] are inverses of degree 6002.
3 and (4025*2^66666+1)/3 are inverses of degree 66666.
7 and (3*2^320008+1)/7 are inverses of degree 320008.
There are thousands of similar anecdotal cases.

Do you have a point to make other than torture random semiprime numbers?

[QUOTE=devarajkandadai;499098]As you are aware Carmichael numbers pertain to the property of composite numbers
behaving like prime numbers with regard to Fermat's theorem. They are Devaraj numbers
I.e. if N = p_1*p_2....p_r ( where p_i is prime) then

(P_1-1)*(N-1)/(p_2-1)......... (p_r-1) is an integer.
See A104016 and A104017.
a) conjecture: the least value of k, the degree to which atleast two of a Devaraj number's prime factors are
Inverses, is 2 (example 561 = 3*11*17 -here 3 and 17 are inverses (mod 5^2).

b) 5 and 11 are impossible cofactors of Devaraj numbers (including Carmichael numbers).
(to be continued)[/QUOTE]

C) let N = (2*m+1)*(10*m+1)*(16*m+1)- here m is a natural nnumber. Then N is a Carmichael number if a) for a given value of m, 2*m+1, 10*m+1 and 16*m+1 are prime and b) 80*m^2 + 53*m + 7 is exactly divisible by 20.

[QUOTE=devarajkandadai;501409]... and b) 80*m^2 + 53*m + 7 is exactly divisible by 20.[/QUOTE]
This simply means that m=20*q+1. And therefore what you are trying to say 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".

With a difference that Chernick proved his and you are "just saying". To what limit did you even test it?

[QUOTE=Batalov;501447]And therefore what you are trying to say 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".[/QUOTE]

which only works if q is 1 mod 3, because the first defeats 0 mod 3 and the others fail for 2 mod 3.

[QUOTE=science_man_88;501450]which only works if q is 1 mod 3, because the first defeats 0 mod 3 and the others fail for 2 mod 3.[/QUOTE]
...except q=0 :rolleyes:

(because 3 is allowed to be divisible by 3 and still be prime)

[QUOTE=Batalov;501447]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".[/QUOTE]
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[/CODE] But there is probably a thousand forms similar to this one known in the literature since 1885 ([URL="https://dml.cz/handle/10338.dmlcz/122245"]Václav Šimerka[/URL]). Note: before Korselt and before Carmichael.

What does Václav Šimerka prove there, Serge?

I don't read Czech, sadly, only numbers... but this fragment on page 224 seems interesting...

[QUOTE=Batalov;501495]I don't read Czech, sadly, only numbers... but this fragment on page 224 seems interesting...[/QUOTE]

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[/QUOTE]