[seqfan] Re: A000041 partition sequence

[Bob Selcoe]

Hello

First time on the list - posting here because an editor suggested I do so.

I have proposed the following comment for A000041 (partition sequence): 


(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; 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); 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.  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)





The editor suggested I make into a formula, hence:



(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.



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); etc. Then 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)





I have not yet proposed the formula.  Does anyone think the formula is necessary? Also, I'm pretty new to standard terms and notation, so any suggestions on how to improve terms, phrasing or notation is greatly appreciated.



Thanks in advance,

Bob Selcoe