[seqfan] Re: A000041 partition sequence

[Bob Selcoe]

Hi William,

Thanks for the input.  First, yes, the observation about subtracting the 
pentagonal recurrence to find 1 makes sense -  in fact, I also included it 
in my initial posting, which the editors reverted because (among other 
things) it contained too much info.  I intended to make a second posting 
(after this one gets published) with this and one or two additional 
observations.  But first things first.

Second, I disagree that this is an "obfuscated" version of the Pentagonal 
sequence.  As I mentioned in our earlier correspondence:

>>>>Specifically, the initial terms of F(G) when G > the G-th pentagonal 
>>>>number (P) are: a(n-1) + a(n-2) - a(n-5) - a(n-7) + a(n-12), - 
>>>>a(n-15)...  +/- a(n-P). (following the "+ + - -" pattern).

As the sequence gets larger there are many variations - if you prefer to 
call them "baroque", that's fine - but they are defined and easily 
calculable;  and I think interesting in and of themselves.   The point 
wasn't to show the most efficient calculations of a(n) but a novel way of 
approaching them.  But I also expect there are subpatterns which can be seen 
more easily with this approach than using a more "efficient" calculation 
method (like pentagonals) - so in that regard they may have additional 
merit.

I also think it would be useful to add two triangles on the site containing 
j-values:
1) the rows of f(g) (pertains to A026794 - Triangular array T read by rows: 
T(n,k) = number of partitions of n in which least part is k, 1<=k<=n);
and 2) the rows of F(G).  My only issue is how to show multipliers in the 
sequences. (For example, row 5 of f(g) would be   {5, -6, -7, 
 2*10, -13, -14, 15}.  does "2*10" make sense?).  But again, that's a 
question for another time.

Anyway, the reason I asked for input on the Seqfan list was at the urging of 
an editor; I was hoping for clarification on the proper way to format the 
ideas (notation, terms, phrasing, etc.) and to see if the comment or formula 
is preferred as a post on A000041 (partition sequence).   I intend to offer 
a comment or formula one way or the other and let the editors decide, 
ultimately, if it should be posted.  I just want to use the best format and 
most understandable language.

Cheers,
Bob
PS - as I mentioned in our off-list correspondence I made a couple typos in 
the posted comment.  For others reading this email: Most importantly, the 
j<=2g+1 constraint applies only for F(G), not f(g); so f(4) =  a(n-4) - 
a(n-5) - a(n-6) + a(n-7) - a(n-7) + a(n-8) + a(n-9) - a(n-10) (final term 
added);    My apologies for the error.



--------------------------------------------------
From: "William Keith" <william.keith at gmail.com>
Sent: Sunday, August 03, 2014 1:54 AM
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: A000041 partition sequence

> Bob:
>
>     Okay.  It seems to me that what you have is simply an obfuscated
> version of the pentagonal recurrence.  For example, for 6 <= n <= 11, the
> PNT says
>
> a(n) = a(n-1) + a(n-2) - a(n-5) - a(n-7)
>
> and your recurrence says
>
> a(n) = 1 + a(n-1) + a(n-2) - a(n-5) - a(n-6) - a(n-7) + a(n-8) + a(n-9) +
> a(n-10) - a(n-13) - a(n-14) + a(n-15).
>
> But of course for 6 <= n <= 11, a(n-13) and lower terms are zero.  So drop
> them.  Then if we subtract, we get
>
> 0 = 1 - a(n-6) + a(n-8) + a(n-9) + a(n-10)
>
> or
>
> a(n-6) - a(n-8) - a(n-9) + a(n-10) = 1
>
> for this segment of the partition numbers, which is true.
>
> There might be a nugget of combinatorial interest here involving sets of
> partition numbers which sum to 1 under suitable signs, but I don't really
> feel inclined to hunt for it.  Sorry.  One can come up with all kinds of
> increasingly baroque constructions involving known sequences.  Since the
> expression for each F(G) implies that some moving frame of coefficients
> must sum to 1, perhaps the triangle of terms which sum to 1 would make a
> suitable sequence.
>
> Cordially,
> William Keith
>
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