20140529, 11:50  #1 
May 2013
1001_{2} Posts 
Can 1227133513 be the only composite number matching the conditions?
Conditions:
n such that O  n  1 and O  1 = 2^x , n ∈ 2N + 1, O = Ord_n(2), x ∈ Z≥0. Can 1227133513 be the only composite number matching the conditions? Erick Wong has check up to 10^20, see: Are there composite numbers matching the conditions? More info of 1227133513: Ord_{1227133513}(2) = 33 and 33  2^33  1. 
20140529, 12:46  #2 
Aug 2006
3·1,993 Posts 
I suspect he checked only to 2^64 using Jan's list.
Probably there are others but searching seems futile. 
20140529, 15:23  #3 
Nov 2003
2^{2}×5×373 Posts 

20140529, 17:48  #4 
Aug 2006
3·1,993 Posts 
Abusively assuming the bigO constant to be 1, the expected number of new examples up to 10^25 is 1/4. To get a 95% chance of finding and example you'd need to go above 10^385.

20140529, 20:05  #5 
Nov 2003
2^{2}×5×373 Posts 

20140812, 00:41  #6 
Apr 2011
2×5 Posts 
Sorry, I must be confused because n = 601 * 1801 * 6151 = 6657848551 seems to work. This is the same order of magnitude as the other solution quoted so it's unlikely to have been missed. It's less than 10^{10} and certainly less than 10^{20} that Erick Wong is supposed to have checked to.
I claim that O = 1025 = 2^{10} + 1 = 5^{2} * 41 2^{1025} = 1 mod n 2^{1025 / 5} = 5533233944 mod n and 2^{1025 / 41} = 33554432 (which seem to have more than their fair share of digit pairs 33, 44 and 55) This proves that the order of 2 is 1025 as required. Finally n = 1025 * 6495462 + 1 so O  n1 as required. The next solution seems to be 13857901601 = 6151 * 2252951 = 1025 * 13519904 + 1 2^{1025 / 5} = 12903300634 mod 13857901601 (hmm.. more digit pairs) while 2^{1025} = 1 mod 13857901601 This proves the order of 2 is 1025 again while n1 is of the required form. I believe there are a total of 4079 solutions with O=1025 of which an impressive one (but by no means the largest) is n = 19858924506932274923192217121413840253555986701099656443876494287833804013531009915252291156116144835199447717419022262305584364955410801 Indeed shouldn't any product of the primes where 2 has the required order qualify as a composite of the required form? As the orders get larger, the number of primes with the required order increase as well. Contrary to what Bob claims, I believe that when we start talking about numbers n where log(log(n)) is not small then these sorts of numbers are actually quite common. Last fiddled with by Rich on 20140812 at 00:52 Reason: Adding material 
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