What's going on with this thing?
mike3
I was wondering about this. Let
a_0 = q_0
a_n = q_n sum_{k=0...n-1} a_k a_(n-k).
Note that if q_0 = 1, this yields the Catalan numbers. For general
q_n, the a_n appear to be a sum of various products composed of
combinations of n of the q_n with n not including 0, multiplied by
q_0^(n+1). But is there any way to describe this more explicitly?
Gottfried Helms
What is q_n ?
Gottfried
Robert Israel
I think you mean a_n = q_n sum_{k=0..n-1} a_k a_{n-1-k}. The Catalan
numbers C(n) satisfy this if all q_i = 1. If all q_i are equal to q,
you get q^(2n+1) C(n).
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
mike3
q_n is just a sequence, though I'm especially curious about
q_n = 1/(1 - u^n), q_0 = 1.
mike3
<isr...@math.MyUniversitysInitials.ca> wrote:
> mike3 <mike4...@yahoo.com> writes:
> > Hi.
>
> > I was wondering about this. Let
>
> > a_0 = q_0
> > a_n = q_n sum_{k=0...n-1} a_k a_(n-k).
>
> > Note that if q_0 = 1, this yields the Catalan numbers. For general
> > q_n, the a_n appear to be a sum of various products composed of
> > combinations of n of the q_n with n not including 0, multiplied by
> > q_0^(n+1). But is there any way to describe this more explicitly?
>
> I think you mean a_n = q_n sum_{k=0..n-1} a_k a_{n-1-k}. The Catalan
> numbers C(n) satisfy this if all q_i = 1. If all q_i are equal to q,
> you get q^(2n+1) C(n).
Yes, sorry, my bad. But what about unequal q_n, (i.e. a general
sequence
q_n) or the specific one q_n = 1/(1 - u^n), q_0 = 1?
In the case of general q_n, the first few terms are
a_0 = q_0
a_1 = q_1 q_0^2
a_2 = 2 q_2 q_1 q_0^3
a_3 = (q_3 q_1 q_1 + 4 q_3 q_2 q_1) q_0^4
a_4 = (4 q_4 q_2 q_1 q_1 + 2 q_4 q_3 q_1 q_1 + 8 q_4 q_3 q_2 q_1)
q_0^5
a_5 = <6 terms> q_0^6
...
so it appears as though
a_n = (sum_{(j_1, j_2, ..., j_n) e S_n} prod_{k=1...n} q_(j_k)) q_0^n
and S_n is some subset of all the combinations of the natural numbers
from
1 to n. What is the rule governing the subsets for a given n, or how
can one
approach the problem of trying to derive it?
mike3
Any answers? What kind of methods could be used to try and tackle
something
like this?
mike3
> On Jan 4, 6:34 pm, Gottfried Helms <he...@uni-kassel.de> wrote:
>
>
>
> > Am 05.01.2011 01:42 schriebmike3:
>
> > > Hi.
>
> > > I was wondering about this. Let
>
> > > a_0 = q_0
> > > a_n = q_n sum_{k=0...n-1} a_k a_(n-k).
>
> > > q_n, the a_n appear to be a sum of various products composed of
> > > combinations of n of the q_n with n not including 0, multiplied by
> > > q_0^(n+1). But is there any way to describe this more explicitly?
>
> > What is q_n ?
>
> > Gottfried
>
> q_n is just a sequence, though I'm especially curious about
> q_n = 1/(1 - u^n), q_0 = 1.
"just a sequence" = "an arbitrary sequence, unspecified".
Ilmari Karonen
> On Jan 4, 8:59 pm, mike3 <mike4...@yahoo.com> wrote:
>> On Jan 4, 6:34 pm, Gottfried Helms <he...@uni-kassel.de> wrote:
>> > Am 05.01.2011 01:42 schriebmike3:
>>
>> > > a_n = q_n sum_{k=0...n-1} a_k a_(n-k).
>>
>> > > Note that if q_0 = 1, this yields theCatalannumbers. For general
>> > > q_n, the a_n appear to be a sum of various products composed of
>> > > combinations of n of the q_n with n not including 0, multiplied by
>> > > q_0^(n+1). But is there any way to describe this more explicitly?
>>
>> > What is q_n ?
>>
>> q_n = 1/(1 - u^n), q_0 = 1.
>
> "just a sequence" = "an arbitrary sequence, unspecified".
If q_n can be anything, then a_n can also be (almost) anything. To
see this, take an arbitrary sequence a_n and define q_0 = a_0, q_n =
a_n / sum_{k=0...n-1} a_k a_(n-k) for n > 0.
The only problem that might happen is is sum_{k=0...n-1} a_k a_(n-k) =
0 (and a_n /= 0) for some n. A sufficient condition to avoid that is
e.g. a_n > 0 for all n. Even if one does not wish to impose such
restrictions, it's still a rather unlikely occurrence unless the a_n
are specifically chosen to make it happen.
--
Ilmari Karonen
To reply by e-mail, please replace ".invalid" with ".net" in address.
mike3
> On 2011-01-09,mike3<mike4...@yahoo.com> wrote:
>
>
>
> > On Jan 4, 8:59 pm,mike3<mike4...@yahoo.com> wrote:
> >> On Jan 4, 6:34 pm, Gottfried Helms <he...@uni-kassel.de> wrote:
> >> > Am 05.01.2011 01:42 schriebmike3:
>
> >> > > a_0 = q_0
>
> >> > > Note that if q_0 = 1, this yields theCatalannumbers. For general
> >> > >q_n, the a_n appear to be a sum of various products composed of
> >> > > q_0^(n+1). But is there any way to describe this more explicitly?
>
> >> > What isq_n?
>
> >>q_n= 1/(1 - u^n), q_0 = 1.
>
> > "just a sequence" = "an arbitrary sequence, unspecified".
>
> If q_n can be anything, then a_n can also be (almost) anything. To
> see this, take an arbitrary sequence a_n and define q_0 = a_0,q_n=
> a_n / sum_{k=0...n-1} a_k a_(n-k) for n > 0.
>
> The only problem that might happen is is sum_{k=0...n-1} a_k a_(n-k) =
> 0 (and a_n /= 0) for some n. A sufficient condition to avoid that is
> e.g. a_n > 0 for all n. Even if one does not wish to impose such
> restrictions, it's still a rather unlikely occurrence unless the a_n
> are specifically chosen to make it happen.
>
What I was curious about, though, was how to generate a_n from the
q_n,
in a "non-recursive" manner (like as a sum or something). In another
post
here, I mention some patterns that appear in the indexes of the q_n in
the terms, but I'm at a loss as to how to describe them with a formula
or method.
But I'm especially interested in the specific case q_n = 1/(1 - u^n).
What happens then? Is there a non-recurrent formula for that
particular
case? How would one go about approaching problems like this?