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Old 2018-10-30, 12:08   #1
devarajkandadai
 
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Default 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)
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Old 2018-10-30, 14:18   #2
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Default Carmichael numbers and Devaraj numbers

Quote:
Originally Posted by devarajkandadai View Post
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

Last fiddled with by devarajkandadai on 2018-10-30 at 14:19 Reason: Corrected a slip
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Old 2018-10-30, 14:24   #3
devarajkandadai
 
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Default Carmichael numbers and Devaraj numbers

Quote:
Originally Posted by devarajkandadai View Post
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).
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Old 2018-11-02, 05:54   #4
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Default Carmichael numbers and Devaraj numbers

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Originally Posted by devarajkandadai View Post
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.
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Old 2018-11-04, 05:30   #5
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Default Carmichael numbers and Devaraj numbers

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Originally Posted by devarajkandadai View Post
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).
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Old 2018-11-05, 04:16   #6
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Originally Posted by devarajkandadai View Post
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)
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Old 2018-11-05, 06:05   #7
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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?
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Old 2018-12-01, 06:44   #8
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Quote:
Originally Posted by devarajkandadai View Post
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.
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Old 2018-12-01, 16:43   #9
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Quote:
Originally Posted by devarajkandadai View Post
... 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?
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Old 2018-12-01, 17:25   #10
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Quote:
Originally Posted by Batalov View Post
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.
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Old 2018-12-01, 20:59   #11
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Quote:
Originally Posted by science_man_88 View Post
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)
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