Question on A057960

[Max]

On 4/25/06, Mitch Harris <Harris.Mitchell at mgh.harvard.edu> wrote:
> Max [mailto:maxale at gmail.com] wrote:
> >
> > On 4/24/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
> > >
> > > All published terms of A057960 satisfy a(n) =
> > > floor(a(n-1)*(a(n-1)+1/2)/a(n-2).  (In fact, I have
> > > verified the next 7 terms as well.)  Note that for A057960, a(n) =
> > > floor(a(n-1)*(sqrt(3)+1)).  (This is the formula from DELEHAM Philippe -
> > > writing sqrt(3) as "3^0, 5" - ugh.)
> > >
> > > Can anyone prove that this recurrence does, in fact, produce this sequence?
> >
> > It's enough to show that
> > 0 <= a(n-1)*(a(n-1)+1/2)/a(n-2) - a(n) < 1
> > or
> > 0 <= a(n-1)*(a(n-1)+1/2) - a(n-2)*a(n) < a(n-2)
> >
> > Using the explicit formula
> > a(n) = (4 + (1+sqrt(3))^(n+2) + (1-sqrt(3))^(n+2)) / 12
>
>  From this, it would also be enough to notice that
>
>    1-sqrt(3) ~= -.73
>
> so that
>
>    |(1-sqrt(3))^n| < 1,
>
> where it also then follows that
>
>    1/4 < ((-.73)^n + 4)/12 < 5/12
>
> which is removed by the floor function.

First off, are you proving the formula from DELEHAM Philippe or what?

And what do you mean saying "removed by the floor function" ?
In the formula
a(n) = (4 + (1+sqrt(3))^(n+2) + (1-sqrt(3))^(n+2)) / 12
~= (1+sqrt(3))^(n+2) / 12 + ((-.73)^(n+2) + 4)/12
the second summand represents a number about 4/12=1/3 that makes a(n) integer.
In other words, (1+sqrt(3))^(n+2) / 12 is close to an integer minus
1/3, and adding ((-.73)^n + 4)/12 makes it integer.
But I do not see any floor function here and why it should remove
((-.73)^(n+2) + 4)/12.
Conversely, we can state that
ceiling( (1+sqrt(3))^(n+2)/12 ) = (1+sqrt(3))^(n+2)/12 +
((1-sqrt(3))^(n+2)+4)/12

Max