 mersenneforum.org > Math A tentative definition
 Register FAQ Search Today's Posts Mark Forums Read  2018-10-20, 05:43 #1 devarajkandadai   May 2004 22×79 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.   2018-10-20, 15:15   #2
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts Quote:
 Originally Posted by devarajkandadai 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.   2018-10-20, 19:37 #3 BudgieJane   "Jane Sullivan" Jan 2011 Beckenham, UK 27710 Posts What is tortion? Did you mean torsion?   2018-10-20, 20:53 #4 Batalov   "Serge" Mar 2008 Phi(4,2^7658614+1)/2 17·563 Posts No, he meant torture. Seriously.   2018-10-21, 01:00 #5 Dr Sardonicus   Feb 2017 Nowhere 33×5×37 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 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.   2018-10-21, 05:56   #6

May 2004

22·79 Posts Quote:
 Originally Posted by science_man_88 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   2018-10-21, 06:06   #7

May 2004

22·79 Posts Quote:
 Originally Posted by devarajkandadai 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.   2018-10-21, 23:48 #8 CRGreathouse   Aug 2006 3×1,993 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.   2018-10-22, 00:03   #9
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts Quote:
 Originally Posted by Dr Sardonicus 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.   2018-10-22, 04:52   #10

May 2004

22·79 Posts Quote:
 Originally Posted by CRGreathouse 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.   2018-10-22, 14:43   #11
CRGreathouse

Aug 2006

3×1,993 Posts Quote:
 Originally Posted by devarajkandadai 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.   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post devarajkandadai Number Theory Discussion Group 10 2018-07-22 05:38 lfm PrimeNet 4 2009-11-15 00:43 R.D. Silverman Math 47 2009-09-24 05:23 Greenbank Octoproth Search 4 2007-12-07 18:41 Damian Lounge 1 2007-05-27 13:30

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