Question on A057960
Max
maxale at gmail.com
Tue Apr 25 09:32:39 CEST 2006
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
one can show that
a(n-1)*(a(n-1)+1/2) - a(n-2)*a(n) = a(n-3) - (2+(-2)^n) / 12
Therefore, we need to show that
a(n-3) >= (2+(-2)^n) / 12
and
- (2+(-2)^n) / 12 < a(n-2) - a(n-3)
which easily follow from the same explicit formula.
Max
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