[seqfan] Re: Proof
Andrew Weimholt
andrew.weimholt at gmail.com
Tue Dec 1 10:08:38 CET 2009
Spotted an error!
for some n, gcd( n, (n-1)! ) = (n-1)!
This invalidates the (attempted) proof.
Got caught doing my thinking on the keyboard instead of on paper again :-)
Andrew
On 11/30/09, Andrew Weimholt <andrew.weimholt at gmail.com> wrote:
> Franklin T. Adams-Watters recently submitted A167234, and poses a
> question in the comments:
>
> "What can we say about the asymptotic behavior of this sequence? Does
> it contain every integer > 2 infinitely often?"
>
> The answer is yes, and here is the proof...
>
> By Dirichlet's Theorem, there are an infinite number of primes of the
> form dk + b, for any positive coprimes d & b
> Furthermore, Dirichlet showed that the primes are evenly distributed
> over the phi(d) arithmetic progressions with difference d coprime to
> the first term.
> Thus the for a given d & b, the primes of the form dk + b account for
> 1 / phi(d) of the primes.
>
> If we let d = (n-1)!, for n > 2
> and b = 1,
>
> it follows that there are an infinite number of primes of the form
> k*(n-1)! + 1,
> and they account for 1 / phi( (n-1)! ) of the primes.
>
> In order to prove that the answer to Franklin's question is "yes", we will need
> to show that the number of primes of the form k*(n-1)! + 1 is still
> infinite when
> we add the restriction that n does not divide k*(n-1)!
>
> For each k, where n | ( k*(n-1)! )
>
> there exists some t > 0, such that k = t * n / gcd(n, (n-1)! )
>
> Substituting into k(n-1)! + 1, we get
>
> t * n! / gcd ( n, (n-1)! ) + 1
>
> which is also an arithmetic progression for a given n, and t = 1, 2, 3, ...
>
> If there are not an infinite number of primes of the form k(n-1)! + 1,
> where n does not divide k(n-1)!,
>
> then all of the primes of the form, k(n-1)! + 1 are also of the form t
> * n! / gcd( n, (n-1)! )
>
> implying that
>
> phi( (n-1)! ) = phi( n! / gcd( n, (n-1)! ) )
>
> which we will now show is false
>
> case 1: n is composite:
>
> then n / gcd( n, (n-1)! ) is at least 2, meaning, n! / gcd( n, (n-1)!
> ) is at least 2(n-1)!
>
> For x>1, there is always a prime in the interval (x, 2x).
>
> Therefore we have coprimes of n! / gcd( n, (n-1)! ) less than n! /
> gcd( n, (n-1)! ) and greater than (n-1)!
>
> Also, since n is not prime, there are no prime factors of n, not
> already in (n-1)!,
> so all numbers less than (n-1)! which are coprime to (n-1)! are also
> coprime to n! / gcd( n, (n-1)! )
>
> Therefore, phi( n! / gcd( n, (n-1) ) ) > phi( (n-1)! )
>
> case 2: n is prime:
>
> then phi( n! / gcd( n, (n-1)! ) ) = phi( n! ) > phi( (n-1)! )
>
> (Cf. A048855 : a(n) = phi ( n! ). Formula: a(n) = a(n-1)*n for
> composite n, and a(n) = a(n-1)*(n-1) for prime n.)
>
> Therefore, there are an infinite number of primes of the form k(n-1)!
> + 1 which are not congruent to 1 mod n.
>
> These primes are congruent to 1 mod x for all x in {1,2,...,n-1},
> therefore, for these primes, n is the smallest modulus,
> for which their two divisors are not congruent.
>
> Therefore, all numbers > 2 appear infinitely often in A167234.
>
>
> Andrew
>
More information about the SeqFan
mailing list