FW: A101368

[Max A.]

On 8/1/06, N. J. A. Sloane <njas at research.att.com> wrote:
> Presumably Floor's conjectures would follow from Ralf Stephan's conjectures
> that are already mentioned in A101368.

Oh, I did not pay attention to the fact that the listed g.f. and
formula are conjectures.
But any of mentioned recurrence formulae can proved by induction (see
below) while the g.f. easily follows from them.

As an example, I will prove that for n>2,
a(n) = 5a(n-1)-a(n-2)-1.     (*)

I will use the following definition of A101368:
a(1)=a(2)=1, a(n+1)=(1+a(n)+a(n)^2)/a(n-1) for n>2.

It is easy to verify formula (*) numerically for n=3 and n=4.

Let m>=5. Suppose that formula (*) holds for all n<m. Then
a(m) = (1+a(m-1)+a(m-1)^2)/a(m-2) =
(1+(5a(m-2)-a(m-3)-1)+(5a(m-2)-a(m-3)-1)^2)/a(m-2)
= 25a(m-2) - 10a(m-3) - 5 + (a(m-3)^2+a(m-3)+1)/a(m-2)
= 25a(m-2) - 10a(m-3) - 5 + a(m-4) = 5(5a(m-2) - a(m-3) - 1) -
(5a(m-3) - a(m-4) - 1) - 1
= 5a(m-1) - a(m-2) - 1.
Now it follows by induction that for any n>2, a(n)=5a(n-1)-a(n-2)-1.

Max