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

[Bob Selcoe]

Thank you very much, Olivier, for your helpful critique.

You've given me much to consider.  You may have guessed that I am not a 
trained mathematician, so I'm trying to learn standard notation, terms and 
conventions as I go.  This can be daunting; but I will look into what you've 
suggested and try to absorb as much as possible.  Using my own approach and 
notation is the best I can do at the moment; if there is a way to approach 
these equations in a more conventional and meaningful way, I would like to 
learn.

In the meantime, there are a few of your comments  I can address now. 
First, I am somewhat familiar with the pentagonal theorem and understand 
some of how it applies here; clearly, the initial terms of the equations 
involve pentagonals (i.e., MacMahon's recurrence), with variations after. 
As I see it, while the 1+ sum_f(k) equations (for lack of a better term) are 
less elegant than MacMahon's; they offer more variety.  So for P(250), 
MacMahon offers one beautiful equation; the 1+sum_f(k)'s offer 125 different 
equations (albeit generally not as beautiful), all related to each other and 
generated by one simple process.    Perhaps that is per se interesting, or 
even potentially useful, and (eventually) worthy of an oeis comment?

I had considered that it would be better to create a triangle, based on the 
numeric values in the terms with their attendant coefficients  - I've 
checked and there don't appear to be any corresponding sequences in the 
oeis.  But I don't know the format for including coefficients in triangles, 
so I avoided doing it.  Is there a way to format such a triangle?

re:   >> Referring to the equations f(k) in sequence A026794: a(n) = 1 +
>> sum{k=1..k}  f(k), when k+1<=n<=2k+1.

Yes, that is not clear.  What I mean is: sum the terms in the equations 
f(1), f(2), f(3)... f(k).  So perhaps sum_{i=1..k} f(i) when  k+1<=n<=2k+1 
makes sense?

Cheers,
Bob


--------------------------------------------------
From: "Olivier Gerard" <olivier.gerard at gmail.com>
Sent: Friday, August 22, 2014 12:04 AM
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: Help with posting comment for partition numbers

> 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
>
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