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February 15th, 2011, 02:51 PM   #1
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n!*(n+1)! - 1 composite?

Hello everybody,

n>5

I have found that for n=6 to 30 :
n!*(n+1)! - 1 is always composite.

I want to know if that is true or not with n>30
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February 15th, 2011, 05:11 PM   #2
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Re: n!*(n+1)! - 1 composite?

It's prime for 76, 166, 344, 394 and so forth. There should be infinitely many, though for precise asymptotics I'd need to do some calculations.
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February 15th, 2011, 05:56 PM   #3
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Re: n!*(n+1)! - 1 composite?

For a random number near to n!*(n+1)!-1, the probability that it is a prime is around
For any random number near n, the probability that it is a prime is around
Given n!*(n+1)!-1 has no prime factor less than n, this should be a similar probability as a number less than n to be prime, we could estimate that the probability that n!(n+1)!-1 to be prime is around . Although the probability is low, it supports CRG's claim that there're an infinity of numbers.
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February 16th, 2011, 02:27 AM   #4
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Re: n!*(n+1)! - 1 composite?

Right. Even without the Mertens adjustment, though, we expect infinitely many -- just not as dense.

I don't find any more terms to 3265.
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February 16th, 2011, 03:29 AM   #5
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Re: n!*(n+1)! - 1 composite?

Hi bogauss, for instance, this polynomial gives prime numbers from (x=1,1000) at the following 189 values:
{for(x=1,1000,if(isprime(x^2+x+1),print(x)))}

Code:
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and this one , switching the signal of 1, gives prime numbers from (x=1,1000) at the following 312 values:
{for(x=1,1000,if(isprime(x^2+x-1),print(x)))}

Code:
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now the curious thing, could someone explain why the difference of ~ 64%? something to do with polynomial cyclotomicness I guess (do this word exists???)

ps: for x = 1.000.000 is ~ 63,2%
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February 16th, 2011, 05:15 AM   #6
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Re: n!*(n+1)! - 1 composite?

x^2 + x + 1 is divisible by 3 1/3 of the time, while x^2 + x - 1 is never divisible by 3. That accounts for 33% of the difference. If you find residues over all the primes you'd presumably find the required number.
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February 16th, 2011, 06:43 AM   #7
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Re: n!*(n+1)! - 1 composite?

Quote:
Originally Posted by CRGreathouse
x^2 + x + 1 is divisible by 3 1/3 of the time, while x^2 + x - 1 is never divisible by 3. That accounts for 33% of the difference. If you find residues over all the primes you'd presumably find the required number.
I understand, you're correct.

I was curious about cyclotomic polynomials which are clearly divisible by p 1/p of the time and why, lets say, , or any p such that p+1 is not a prime number, and how this behave changing the signals arbitrarily. That is, some cyclotomic polynomials yields composite numbers all the time, and I suspect it is because not all polynomials of the form are irreductible by eiseinstein's criterium... am I correct?
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February 16th, 2011, 10:30 AM   #8
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Re: n!*(n+1)! - 1 composite?

Quote:
Originally Posted by al-mahed
I was curious about cyclotomic polynomials which are clearly divisible by p 1/p of the time
Not a big deal; that's just one prime, right?

Quote:
Originally Posted by al-mahed
signals
I don't know what you mean when you write this.

Quote:
Originally Posted by al-mahed
That is, some cyclotomic polynomials yields composite numbers all the time
What, you mean like cyclic polynomials at composite values like ? Sure, those are reducible polynomials.
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February 16th, 2011, 02:34 PM   #9
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Re: n!*(n+1)! - 1 composite?

Sorry, my message was very confusing, let me try to be concise and to ask the correct question: is there polynomials that are not reductible but even so yields only composite numbers? I know there is no polynomial that yields only primes, so I was curious about it, just a mere curiosity. I suspect there is something trivial about it, but I'm not "getting there". So I was thinking about cyclotomic polynomials because they seems to be directly related to what I'm wondering.
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February 16th, 2011, 02:57 PM   #10
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Re: n!*(n+1)! - 1 composite?

Quote:
Originally Posted by al-mahed
Sorry, my message was very confusing, let me try to be concise and to ask the correct question: is there polynomials that are not reductible but even so yields only composite numbers?
Good question!

The answer is yes. One example is x^2 + x + 2 = x(x +1) + 2. Both summands are even, so its value is even, and for positive x, x^2 + x + 2 > 2. (If you want to get the result for all integers, use x^4 + 6*x^3 + 11*x^2 + 6*x + 4 instead, but the real interest is in the infinite part of the behavior, where you can ignore finite numbers of primes.)
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