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Old 2018-10-20, 05:43   #1
devarajkandadai
 
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Default A tentative definition

Let N be a squarefree composite number with r factors, p_1,...p_r.
Then we can define N as a tortionfree number if atleast two of its
factors are inverses mod(P),where P is a prime number less than the largest prime factor of N.
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Old 2018-10-20, 15:15   #2
science_man_88
 
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Quote:
Originally Posted by devarajkandadai View Post
Let N be a squarefree composite number with r factors, p_1,...p_r.
Then we can define N as a tortionfree number if atleast two of its
factors are inverses mod(P),where P is a prime number less than the largest prime factor of N.
6p with p prime and 1 mod 3 are tortion free
10p with p>3; p 3 mod 5
at least using additive inverses. multiplicayive inverses are different.
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Old 2018-10-20, 19:37   #3
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What is tortion? Did you mean torsion?
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Old 2018-10-20, 20:53   #4
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No, he meant torture. Seriously.
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Old 2018-10-21, 01:00   #5
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Default Inverse (mod P)? With resect to what operation?

If "inverse" means "multiplicative inverse" we have the following:

If N is odd and the product of at least two prime factors, any two of its factors are multiplicative inverses (mod 2), and P = 2 is less than the largest prime factor of N.

If N has at least three prime factors, then at least two of them are multiplicative inverses (mod 2), and P = 2 is less than the largest prime factor of N.

(In the above two cases, of course, any odd factor is self-inverse (mod 2).)

This leaves N = 2*p, where p is prime. If N - 1 = 2*p - 1 is prime, it satisfies the definition. Otherwise, it does not.

In short, the numbers fitting the definition are of the form N = 2*p, where p > 2 is prime, and 2*p - 1 is also prime.

If you don't like the choice P = 2, you should have said so. However, even if you exclude P = 2 by fiat, your options are still limited:

If N has at least four prime factors if 3 divides N, or at least three prime factors if 3 does not divide N, then N has at least one pair of factors which are multiplicative inverses (mod 3), and P = 3 is less than the largest prime factor of N.
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Old 2018-10-21, 05:56   #6
devarajkandadai
 
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Quote:
Originally Posted by science_man_88 View Post
6p with p prime and 1 mod 3 are tortion free
10p with p>3; p 3 mod 5
at least using additive inverses. multiplicayive inverses are different.
Rightm now refering only to multplicative inverses
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Old 2018-10-21, 06:06   #7
devarajkandadai
 
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Quote:
Originally Posted by devarajkandadai View Post
Let N be a squarefree composite number with r factors, p_1,...p_r.
Then we can define N as a tortionfree number if atleast two of its
factors are inverses mod(P),where P is a prime number less than the largest prime factor of N.
Inspiration comes from group theory where normal sub-groups are tortion free.
Am refering only to multiplicative inverses.
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Old 2018-10-21, 23:48   #8
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Up to 100, I find: 10, 15, 21, 22, 26, 30, 33, 34, 35, 39, 42, 46, 51, 55, 57, 58, 65, 66, 69, 70, 77, 78, 82, 85, 86, 87, 91, 93, 94, 95.

Up to a million there are 607926 squarefree numbers, of which 525128 meet your definition.

Asymptotically the fraction is 1.

Edit: I forgot to post, so Sardonicus beat me to it.
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Old 2018-10-22, 00:03   #9
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Quote:
Originally Posted by Dr Sardonicus View Post
In short, the numbers fitting the definition are of the form N = 2*p, where p > 2 is prime, and 2*p - 1 is also prime.
AKA prime p such that p-1 is not of form 2ij+i+j via the sieve of sundaram.
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Old 2018-10-22, 04:52   #10
devarajkandadai
 
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Quote:
Originally Posted by CRGreathouse View Post
Up to 100, I find: 10, 15, 21, 22, 26, 30, 33, 34, 35, 39, 42, 46, 51, 55, 57, 58, 65, 66, 69, 70, 77, 78, 82, 85, 86, 87, 91, 93, 94, 95.

Up to a million there are 607926 squarefree numbers, of which 525128 meet your definition.

Asymptotically the fraction is 1.

Edit: I forgot to post, so Sardonicus beat me to it.
Asymtotically the fraction may be 1. Now I am able to conjecture
that a necessary condition for a squarefee composite ( with minimum 3 prime factors) to be a Devaraj number (which include Carmichael numbers) is that it should be tortion free.
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Old 2018-10-22, 14:43   #11
CRGreathouse
 
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Quote:
Originally Posted by devarajkandadai View Post
Asymtotically the fraction may be 1. Now I am able to conjecture
that a necessary condition for a squarefee composite ( with minimum 3 prime factors) to be a Devaraj number (which include Carmichael numbers) is that it should be tortion free.
A squarefree number with at least three prime factors has at least two distinct prime factors, which are multiplicative inverses mod 2.
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