[seqfan] Re: linear recurrence question

israel at math.ubc.ca israel at math.ubc.ca
Mon Apr 16 23:05:41 CEST 2012

To make it correspond to an Ehrhart polynomial, you'd first make a change
of variables X=x-1,Y=y-1,Z=z-1, and then you need s=p+q+r to get the
homogeneous form p X + q Y + r Z = 0.

Robert Israel
University of British Columbia

On Apr 16 2012, Max Alekseyev wrote:

>Constrains p*x + q*y + r*z = s, 1 <= x,y,z <= n define a 2-d polygon
>in 3-d space.
>So a(n) may be even a polynomial in n -- see
>http://en.wikipedia.org/wiki/Ehrhart_polynomial
>Max
>
> On Sun, Apr 15, 2012 at 2:57 PM, Kimberling, Clark <ck6 at evansville.edu>
> wrote:
>> Hello SeqFans,
>>
>>
>>
>> Suppose that p,q,r,s are integers, and let a(n) be the number of
>> triples (x,y,z) such that
>>
>>
>>
>> p*x + q*y + r*z = s,
>>
>>
>>
>> where x,y,z are all in {1,...,n}.  It appears that (a(n)) is a linear
>> recurrence sequence -- and that x,y,z  can be replaced by
>> x(1),x(2),...,x(k), etc.
>>
>>
>>
>> Can someone provide a reference?
>>
>>
>>
>> Thanks!
>>
>>
>>
>> Clark Kimberling
>>
>>
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>>
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