[seqfan] Re: Help with posting comment for partition numbers

[Olivier Gerard]

On Fri, Aug 22, 2014 at 1:49 AM, Bob Selcoe <rselcoe at entouchonline.net>
wrote:

> Hi,
>
> Hoping for a little assistance before offering a comment on A000041
> (partition numbers).  My purpose is to show (IMHO) a simple process to
> generate multiple recurrence equations which solve A000041(n).
>
> Here's what I intend to post. I think this is clear enough (if not, all
> constructive criticism is welcome):
>
> (Start)
> Referring to the equations f(k) in sequence A026794: a(n) = 1 +
> sum{k=1..k}  f(k), when k+1<=n<=2k+1.
>
>
Bob,

At least two things here:

1 - f(k) is your own notation, introduced recently by you in A026794.
When you are commenting a core sequence, try to use something already
used and clear. And do not just assume what you found is different of
classical
results, try to prove it is different (or equivalent).
For instance p(n,k) = number of integer partitions of n with k parts.


2 - There are really two variables here, because you are using the
decomposition
of partitions by least part (k) of a number (n).  But there are clearly
some confusion in your
explanations :    sum(k=1..k)   is a kind of self-referring limit notation
I cannot interpret.

Other members have already quoted the pentagonal theorem and Euler works
on partitions.  These results are a lot more powerful than you think in
that context,
with deep connections to number theory and analysis.
Also there is the important duality between restricting on sizes of parts
and restricting on
numbers of parts.  You should really study these.
Lookup their name and Ferrers diagram, Partition generating function, etc.
in Wikipedia,
Mathworld or equivalent resources, and let's talk again.



> For example, k=3: f(1) = a(n-1); f(2) = a(n-2) - a(n-3); f(3) = a(n-3) -
> a(n-4) - a(n-5) + a(n-6).  Therefore, when 4<=n<=7: a(n) = 1 + a(n-1) +
> a(n-2) - a(n-4) - a(n-5) + a(n-6) .
>
>
What you have here is a potential set of triangles of coefficients and
sequences.

I suggest you construct them with enough terms and look if they are
already present in the OEIS.



These equations solve a(n) when floor(n/2)<=k<=n-1; therefore, the number
> of equations which solve a(n) using this method is ceiling(n/2).  For
> example, a(7)=15 is solved by four equations: 1 + sum_{k=1..k}  f(k), when
> 3<=k<=6. (End)
>
> Does this look OK?  Appropriate to post in A000041 rather than  A026794?
>
>
No, this is not ready nor OK to post in A000041.

With my best regards,


Olivier