Sequence

[Richard Guy]

Thankyou!  Some other odds & ends:  I've never
understood ``offset''.  For this sequence the
``right'' initial values are a(0) = a(1) = 0,
a(2) = a(3) = 1, since then a(-n) = -a(n).
Another mystery that eludes me is what to do
with two-way infinite sequences, though this
one is easy and can be dismissed with the
remark I've just made.

As to divisibility properties, it's easy to see
that any n-th order recurring sequence will be
periodic modulo  m  with period bounded by m^n.
If there are zeroes in the sequence, then  m
divides the corresponding terms.  Here
   2|a(n)  just if  n = 0,1,-1 mod 5 = 2^2 + 1
   3|a(n)    ,,    n = 0,1,-1,6 mod 12 = 3(3+1)
   5|a(n)    ,,    n = 0,1,-1 mod 13 = (5^2 + 1)/2
and so on, with corresponding results for higher
powers of the primes.  For  13  the period is
13(13 + 1)/2  --  compare 3.  I suspect that 3
and 13 are divisors of the discriminant of
     [(2x^2 - x + 2)^2 - 13x^2]/4
which, so far, I seem to have miscalculated.  R.

On Sun, 23 Mar 2008, Max Alekseyev wrote:

> Richard,
>
> Your sequence is well done.
>
> This is its o.g.f.:
>
> x^3/(x^4 - x^3 - x^2 - x + 1)
>
> and first 50 terms:
>
> 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, 333430,
> 574195, 988811, 1702816, 2932392, 5049824, 8696221, 14975621,
> 25789274, 44411292, 76479966, 131704911, 226806895, 390580480,
> 672612320, 1158294784, 1994680689, 3435007313, 5915370466,
> 10186763684, 17542460774, 30209587611, 52023441603
>
> btw, what particular divisibility properties of this sequence you have in mind?
>
> Regards,
> Max
>
> On Sun, Mar 23, 2008 at 10:48 AM, Richard Guy <rkg at cpsc.ucalgary.ca> wrote:
>> 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
>>
>>
>
>
>