Question on A057960
Mitch Harris
Harris.Mitchell at mgh.harvard.edu
Tue Apr 25 19:47:36 CEST 2006
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.
Mitch
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