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2015-05-31, 15:57   #12
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts

Quote:
 Originally Posted by R.D. Silverman Typical gibberish from you.
no it's not, numbers could include naturals, whole numbers, integers, reals, complex numbers and in theory even Quaternion

 2015-05-31, 16:29 #13 wildrabbitt   Jul 2014 1BC16 Posts Thanks very much for all the replies. I'm grateful for silverman pointing out the ambiguity of what I was saying.
2015-05-31, 21:28   #14
ewmayer
2ω=0

Sep 2002
República de California

2CFE16 Posts

Quote:
 Originally Posted by science_man_88 no it's not, numbers could include naturals, whole numbers, integers, reals, complex numbers and in theory even Quaternion
No - in this particular context it is widely established that 2^p-1 means 'p a prime' and 2^n-1 means 'n natural'. Claiming otherwise is the hallmark of a crank.

2015-05-31, 23:31   #15
TheMawn

May 2013
East. Always East.

11×157 Posts

Quote:
 Originally Posted by R.D. Silverman (1) 2^p-1 does not have any algebraic factors that force it to be composite. Consider instead, x^p-1, for p prime. Since p = 2k+1 for some k, we have the algebraic factor (x-1) of (x^(2k+1) - 1). Now, when x equals 2, this algebraic factor has the value 1 whereas for all other x > 1, the algebraic factor is non-trivial.
Interesting.

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)

2015-06-01, 00:10   #16
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by TheMawn Interesting. 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)
Guaranteed? No. Clearly one can construct infinite sets of numbers that contain no primes.
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 2015-06-01 at 00:19

2015-06-01, 09:03   #17
wildrabbitt

Jul 2014

22·3·37 Posts
misplaced modifier

Quote:
 I will offer a hint: In English, your wording would consist of a "misplaced modifier".
I would like to learn from what you have said in case I can benefit :

Is a "misplaced modifier" an error in grammar or an error in mathematical language.

It sounds like a GCC error message :)

2015-06-01, 09:46   #18
R.D. Silverman

Nov 2003

22×5×373 Posts

Quote:
 Originally Posted by wildrabbitt I would like to learn from what you have said in case I can benefit : Is a "misplaced modifier" an error in grammar or an error in mathematical language. It sounds like a GCC error message :)
It is both.

 2015-06-01, 10:38 #19 wildrabbitt   Jul 2014 22·3·37 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 k-multiperfect numbers must exist for larger natural k e.g 100
2015-06-01, 11:58   #20
R.D. Silverman

Nov 2003

11101001001002 Posts

Quote:
 Originally Posted by wildrabbitt 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
Yes, it is known. The proof might be given as a homework exercise in a number theory class.

 2015-06-01, 12:17 #21 wildrabbitt   Jul 2014 22·3·37 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 100-perfect numbers exist (as an example) ? They must be big. Last fiddled with by wildrabbitt on 2015-06-01 at 12:18
2015-06-01, 12:46   #22
R.D. Silverman

Nov 2003

746010 Posts

Quote:
 Originally Posted by wildrabbitt 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 100-perfect numbers exist (as an example) ? They must be big.
Huh? The proof is easy. Just look at sigma(2*3*5*7* ........n), then let n--> oo.
sigma(n)/n is clearly unbounded.

I can't answer your second question as I have not looked at the literature for multi-perfect numbers.

My guess (based on very very limited knowledge) would be yes. But if one were to ask:

Let g(n) be the smallest k-perfect 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 2015-06-01 at 12:46

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