[seqfan] Proof

[Andrew Weimholt]

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