Max GCD Permutation Product

[David Wasserman]

The two sequences first disagree at m = 14.  To get the product up to
m!/LCM(1,2,3,...,m), it is necessary and sufficient to arrange the numbers
so that for any k, all the multiples of k are consecutive.
Here's an example for 13:
	1 3 9 6 12 4 8 2 10 5 7 11 13
But for 14, we would have to have the seven even numbers together, with
10 at one end next to 5, and 14 at the other end next to 7.  That would
leave no way to put 3, 6, 9, and 12 together.

  - David

>Leroy Quet:
> > If we take a permutation [a(1),a(2),...a(m)] of
> > [1,2,3,...m],
> >
> > we can form the product:
> >
> > P = product{k=1 to m} GCD(a(k),a(k-1)),
> >
> > where a(0) = a(m).
> >
> > But what would the maximum P be,taken over all permutations of the first
> > m positive integers?
> >
> > By hand, I get:
> >
> > 1, 1, 1, 2, 2, 12, 12, 48, 144,...
> >
> > which corresponds with sequence A025527,
> > m!/LCM(1,2,3,...,m).