20181020, 05:43  #1 
May 2004
2^{2}·79 Posts 
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. 
20181020, 15:15  #2  
"Forget I exist"
Jul 2009
Dumbassville
10000011000000_{2} Posts 
Quote:
10p with p>3; p 3 mod 5 at least using additive inverses. multiplicayive inverses are different. 

20181020, 19:37  #3 
"Jane Sullivan"
Jan 2011
Beckenham, UK
3^{2}·31 Posts 
What is tortion? Did you mean torsion?

20181020, 20:53  #4 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{3}×1,201 Posts 
No, he meant torture. Seriously.

20181021, 01:00  #5 
Feb 2017
Nowhere
1001111110000_{2} Posts 
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 selfinverse (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. 
20181021, 05:56  #6 
May 2004
13C_{16} Posts 

20181021, 06:06  #7  
May 2004
2^{2}×79 Posts 
Quote:
Am refering only to multiplicative inverses. 

20181021, 23:48  #8 
Aug 2006
5979_{10} Posts 
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. 
20181022, 00:03  #9 
"Forget I exist"
Jul 2009
Dumbassville
20300_{8} Posts 

20181022, 04:52  #10  
May 2004
2^{2}×79 Posts 
Quote:
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. 

20181022, 14:43  #11 
Aug 2006
1011101011011_{2} Posts 
A squarefree number with at least three prime factors has at least two distinct prime factors, which are multiplicative inverses mod 2.

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Tentative conjecture  devarajkandadai  Number Theory Discussion Group  10  20180722 05:38 
benchmarks overclock definition?  lfm  PrimeNet  4  20091115 00:43 
Mersenne Numbers: Definition  R.D. Silverman  Math  47  20090924 05:23 
Project Definition  Greenbank  Octoproth Search  4  20071207 18:41 
Mathematics definition  Damian  Lounge  1  20070527 13:30 