20150531, 15:57  #12 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 

20150531, 16:29  #13 
Jul 2014
1BC_{16} Posts 
Thanks very much for all the replies.
I'm grateful for silverman pointing out the ambiguity of what I was saying. 
20150531, 21:28  #14 
∂^{2}ω=0
Sep 2002
República de California
2CFE_{16} Posts 
No  in this particular context it is widely established that 2^p1 means 'p a prime' and 2^n1 means 'n natural'. Claiming otherwise is the hallmark of a crank.

20150531, 23:31  #15  
May 2013
East. Always East.
11×157 Posts 
Quote:
So any set of numbers whose general form can't be factorized algebraically is guaranteed to contain a prime? (While I did understand the meaning of "mostly prime" I appreciate the subtleties so I apologize for butchering the wording in my question) 

20150601, 00:10  #16  
Nov 2003
2^{2}·5·373 Posts 
Quote:
Consider: f(x) = 1st composite odd integer greater than 2^2^x + 1. This is a well defined infinite sequence,with no algebraic factorization, yet it contains no primes.... Indeed. There is a known Fibonacci (i.e. an additive recursive sequence) that has no algebraic factorization (see Guy's book on unsolved problems) yet contains no primes. But in absence of other information, sequences of integers generated by algebraic expressions are expected to contain primes just on probability grounds. I am also curious as to how you leaped from what I said before to the use of the words "any set" and "guaranteed". Last fiddled with by R.D. Silverman on 20150601 at 00:19 

20150601, 09:03  #17  
Jul 2014
2^{2}·3·37 Posts 
misplaced modifier
Quote:
Is a "misplaced modifier" an error in grammar or an error in mathematical language. It sounds like a GCC error message :) 

20150601, 09:46  #18 
Nov 2003
2^{2}×5×373 Posts 

20150601, 10:38  #19 
Jul 2014
2^{2}·3·37 Posts 
Cool. Thanks.
BTW I came up with a proof a long time ago and I don't know if it's known or not. I guess I'm being naïve like I often am but I'd like to know for sure. The proof is a proof that proves that : (where sigma(n) is the sum of divisors function) for all real k > 0 there exist n such that sigma(n) > kn The reason I was interested in knowing why some 2^p 1 must be prime is because I wanted to find a proof that "perfect numbers", for which sigma(n) = 2n must exist. The reason I wanted to do that was because I wanted to know if could be proved that kmultiperfect numbers must exist for larger natural k e.g 100 
20150601, 11:58  #20  
Nov 2003
1110100100100_{2} Posts 
Quote:


20150601, 12:17  #21 
Jul 2014
2^{2}·3·37 Posts 
Great. I showed mine to a Lecturer at a university I used to go to for maths and he didn't believe
it. I aught to ask : is it known whether 100perfect numbers exist (as an example) ? They must be big. Last fiddled with by wildrabbitt on 20150601 at 12:18 
20150601, 12:46  #22  
Nov 2003
7460_{10} Posts 
Quote:
sigma(n)/n is clearly unbounded. I can't answer your second question as I have not looked at the literature for multiperfect numbers. My guess (based on very very limited knowledge) would be yes. But if one were to ask: Let g(n) be the smallest kperfect integer. How does k grow with n? I would expect that the result would be an iterated log function of n... My intuition (whatever that is worth) says that k grows very slowly with n. Note: I could be very wrong here. Last fiddled with by R.D. Silverman on 20150601 at 12:46 

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