Re: partition function
- From: se16@xxxxxxxxxxxxxx
- Date: Tue, 7 Oct 2008 14:14:05 -0700 (PDT)
On 7 Oct, 09:53, nikl...@xxxxxxxxx wrote:
Any thoughts on the combinatorial ways for the same question
lets take a+2b+3c +4d = n , with a,b,c,d non-negative integers , i
have a feeling that it can be done using inclusion exclusion
principle. If not for 4 variables how about for 3 , a+2b+3c=n , can we
do some transformations so that it ends up like y1 + y2 + y3 = z and
the number of solutions of that can be found out using the principles
for y1+y2+....yn=r as C(n+r-1,r) .
a + (y2 -y3) + 3c = n , where 0<=y2<=n but 0<=y3<=n/2 but again y2-
y3 takes odd values but in the original equation it takes only even
values , this is what stops me from proceeding further.
It is not as easy as that.
Look at the formulae in http://www.research.att.com/~njas/sequences/A001400
http://www.research.att.com/~njas/sequences/A008284 is also related to
the general case
.
- References:
- partition function
- From: niklaus
- Re: partition function
- From: victor_meldrew_666
- Re: partition function
- From: Mariano Suárez-Alvarez
- Re: partition function
- From: niklaus
- Re: partition function
- From: se16
- Re: partition function
- From: niklaus
- partition function
- Prev by Date: Re: lie algebra
- Next by Date: Re: Out-of-print math books: An Update
- Previous by thread: Re: partition function
- Next by thread: -- boundary of an open set
- Index(es):
Relevant Pages
|
