[seqfan] Re: A000041 partition sequence

[Bob Selcoe]

Hello William,

I describe what F(G) is so this is an equation.  As I said, suggestions are appreciated if there are better ways to notate/phrase things, so if you have any I'd like to hear them.

re: pentagonal numbers - yes ,they are related and I had discussed this in the initial version of my comment, but later deleted it due to length.  Anyway, the sequences are related to pentagonal numbers/Euler's theorem but different - or at least the equations are different. 
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). 

Also, I realize there are a couple of typos in the comment and formula as they were presented. I have highlighted in red where they have been made.  Hopefully this clarifies things:  


(Start)
Let f(1) = a(n-j), j=1.  Find f(g) g>1 by summing the terms of f(g-1) substituting j+1 for j, and subtracting the terms substituting j+g for j. (deleted " j<=2*g+1" here.)  So f(2) = a(n-2) - a(n-3); f(3) = a(n-3) - a(n-4) - a(n-5) + a(n-6); 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); etc.

Let F(G) denote sum_{1..g} (f(g)), j<=2g+1.  Then A000041(n) = a(n) = 1 + F(G), G+1<=n<=2*G+1, j<=n,  j<=2*g+1.  For example, when 5<=n<=9, j<=n: a(n) = 1 + F(4) = 1 + a(n-1) + a(n-2) - 2*a(n-5) + a(n-8) + a(n-9). (ex. a(8) = 22 = 1 + 15 + 11 - 2*3 + 1).  

1 + F(G) solves for a(n) when floor(n/2)<=G<=n-1.  So for example, when n=17, a(17) can be solved when G={8..16} (nine different ways).  (End)



(Start)
For f(g), let f(1) = a(n-j) = 1 when j=1; and f(g) = f(g-1) (substituting j+1 for j) - f(g-1) (substituting j+g for j). Then a(n) = 1 + F(G): F(G) = sum_{G=1..g} (f(g)), j<=n,  G+1<=n<=2*G+1,  j<=2*g+1.

So f(2) = a(n-2) - a(n-3); f(3) = a(n-3) - a(n-4) - a(n-5) + a(n-6); 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); etc. Then when 5<=n<=9, j<=n, j<=2*g+1:  a(n) = 1 + F(4) = 1 + a(n-1) + a(n-2) - 2*a(n-5) + a(n-8) + a(n-9).  (ex. a(8) = 22 = 1 + 15 + 11 - 2*3 + 1).

1 + F(G) solves for a(n) when floor(n/2)<=G<=n-1.  So for example, when n=17, a(17) can be solved when G={8..16} (nine different ways).  (End)  

@William, I will send a separate email off list which hopefully will clarify what I'm trying to say.  

Cheers, Bob



 


--------------------------------------------------
From: "William Keith" <william.keith at gmail.com>
Sent: Saturday, August 02, 2014 7:29 PM
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: A000041 partition sequence

> Bob:
> 
>     I'm afraid I don't understand what you're trying to do.  Several of
> your statements don't make sense to me.  For instance, "1 + F(G) solves for
> a(n)"?  One typically solves equations, and this is not an equation.  Is
> this some way of recovering the sequence of partition numbers after a
> series of operations on itself?
> 
>     Could you give some explicit examples of the calculations you are
> describing?
> 
>     It is known that any partition number p(n) can be written as
> 
> p(n) =  p(n-1) + p(n-2) - p(n-5) - p(n-7) + p(n-12) + p(n-15) - ...
> 
> where the numbers being subtracted are known as the pentagonal numbers, and
> eventually the terms become zero.  Perhaps this is related to the ideas you
> are having.
> 
> William Keith
> 
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