Gathering for Gardner

[Bob Hearn]

So will I, but the main purpose of this message is
to submit a sequence, with apologies for my
laziness/ineptness at not using the aproved form.
I used to rely on members of Sloane's Dream Team
to take the load off Neil himself, but I believe
that this address now gets bumped (tho I'm trying it
again).

The sequence is

0, 0, 0, 1, 1, 2, 4, 6, 11, 19, 32, 56, 96, 165, 285,
490, 844, 1454, 2503, 4311, 7424, 12784, 22016, 37913,
65289, 112434, 193620,

by which time my hand calculations are suspect (I'm
at home, away from PARI, and have no hand calculator).

It wasn't arrived at in the heat of battle, but just
from the recurrence
       a(n) = a(n-1) + a(n-2) + a(n-3) - a(n-4)
so that its generating function has
              x^4 - x^3 - x^2 - x + 1
in its denominator  (as does the sequence
0,1,1,1,3,4,7,13,21,37,64,109,..., also not in OEIS?)

So it's an example of a `symmetric' quartic recurrence
and has some expected divisibility properties (tho not
as spectacular as some I've seen).

It's close to  A000786 (& A048239), A115992, A115993,
but there's unlikely to be any connexion).  Best,   R.

On Sun, 23 Mar 2008, N. J. A. Sloane wrote:

> Tanya,  I will be at G4G8!
>
> Neil