Carmichael numbers and Šimerka numbers

[devarajkandadai]

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)

[devarajkandadai]

Originally Posted by devarajkandadai

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)

C) 7 and 31 are inverses of 3rd degree

[devarajkandadai]

Originally Posted by devarajkandadai

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)

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

[devarajkandadai]

Originally Posted by devarajkandadai

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

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

[devarajkandadai]

Originally Posted by devarajkandadai

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

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

[devarajkandadai]

Originally Posted by devarajkandadai

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

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

[Batalov]

Well, 5 and 7469128023...77_<181> are goddamn inverses of degree 600.
131 and 1289338297...07_<1808> 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?

[devarajkandadai]

Originally Posted by devarajkandadai

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)

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.

[Batalov]

Originally Posted by devarajkandadai

... and b) 80*m^2 + 53*m + 7 is exactly divisible by 20.

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?

[science_man_88]

Originally Posted by Batalov

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".

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

[Batalov]

Originally Posted by science_man_88

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

...except q=0

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