[seqfan] Re: Phil Scovis's problem

[Charles Greathouse]

I'm pretty sure you can get equality just by throwing all kinds of primes
in the remaining terms.

Let's see. Take the divisors of A061799(n-1) as the starting point for the
remaining terms, and find the product of the n-1 primes larger than the
greatest prime dividing A061799(n-1). Multiply each of the remaining terms
by the product divided by one of the primes. Then the gcd of the first term
with any of the other terms is distinct by construction, and the gcd of any
other two is the large product divided by two primes times some divisor of
A061799(n-1), but these must be distinct because the two primes are
distinct for any pair.

So that proves it, yes? The only 'deep' question is b(n); a(n) is simple.

Charles Greathouse
Analyst/Programmer
Case Western Reserve University


On Tue, Mar 5, 2013 at 2:48 PM, Maximilian Hasler <
maximilian.hasler at gmail.com> wrote:

> On Tue, Mar 5, 2013 at 3:22 PM, Neil Sloane <njasloane at gmail.com> wrote:
> > Charles, I agree a(n) >= A061799(n-1). Can you prove equality?
>
> Prior to recieving more recent messages, I proposed a comment in A061799:
>
> Also: Smallest possible member of a set of n+1 numbers with pairwise
> distinct GCD's. [Following an observation by Charles R Greathouse IV]
> (Proof: If the smallest number m(S) of the set (with card(S)=n+1) has
> a distinct GCD with each of the n other numbers, then it must have at
> least n distinct divisors (because any GCD is a divisor). It is then
> easy to choose larger members of the set as to have all elements with
> pairwise different GCD's, e.g., by successively multiplying with
> distinct and large enough primes.) - M. F. Hasler, Mar 05 2013
>
> Comments, improvements etc are welcome @ https://oeis.org/draft/A061799
>
> Maximilian
>