A lot more goodies on this at

http://www.research.att.com/~njas/sequences/A001791

It is possible this formula could be another also for that sequence.

I reduced the LHS to sum(k=1,n,C(n,k)^2*k/(n-k+1))

However, I like

a(n)=sum(i=1,n,c1(i+n-1, n))

found in the Link.


Cino

--- In mathforfun@yahoogroups.com, "video_ranger" <video_ranger@...>
wrote:
>
> Not about this problem per se, but just as a philosophical point:
Yes
> it's useful to define C(n,k) for all integers k with the
> understanding that C(n,k) = 0 for k < 0 or k > n, because as you
say
> there are zero ways of choosing less than 0 (or more than n)
objects
> from n. More generally 1/k! = 0 for all negative integers k because
> Gamma(k+1) = "infinity" for negative integers k.
>
> IMO, what makes this nice is that when there's a (finite or
infinite)
> sum of terms involving factorials in the denominators, more often
> than not the index can be imagined to extend to all negative and
> positive integers, for example exp(x) = Sum_k=-oo^+oo: x^k/k! with
> the understanding that reciprocals of negative factorials are
always
> zero. Then it's not necessary to keep track of the summation limits
> of index variables and index translation substitution (like k-->k'
=
> k + c) can be made easily without changing the summation limits (-
oo
> to +oo). It's quite a relief to be able to forget about summation
> limits when doing these types of manipulations.
>
>
> --- In mathforfun@yahoogroups.com, "ramsey2879" <ramsey2879@>
> wrote:
> >
> > --- In mathforfun@yahoogroups.com, "slim_the_dude" wrote
> > > Here is such an argument.
> > >
> > > Take a set of A of 2n objects, and arbitraily divide into two
> sets
> > B
> > > and C such that each of B and C contain n objects.
> > >
> > > C(2n,n+1) = # ways to choose n+1 objects from A.
> > > = sum(#ways to choose k objects from A, and the remaining
n+1-
> k
> > > objects from B) k = 0 to n
> > > = sum(C(n,k)*C(n,n+1-k)
> > > = sum(C(n,k)*C(n,k-1)
> > >
> > > (The last line follows from noting that for any a,b C(a,b) = C
> (a,a-
> > > b) )
> > >
> > Regarding my last post, first I rewrite my identity using Slim's
> > variables.
> >
> > C(2n,n+a) = Sum(k = a to n)(C(n,k)*C(n,k-a))
> >
> > As one can see except for the fact that I wrote k = a to n
> whereas
> > Slim wrote k = 0 to n, all I did was substitute a for 1 in Slim's
> > argument. There is no distinction in the argument; however, for
if
> k
> > is less than "a" then the product being summed is zero for k-a
> would
> > be less than Zero and there are Zero ways to chose less than 0
> items
> > from n items. Thus Slim's argument does prove my identity.
> > Of course there is also the other argument that C(2n,n) = Sum(k
=
> 0
> > to n)(C(n,k))^2 and that every binomial coefficient for the
> expansion
> > of (a + b)^2n can be writen as a sum of products of terms from
the
> > binomial coeficients of C(a+b)^n simply because that is how one
> would
> > arrive at the terms of (a+b)^2n from the terms of (a+b)^n.
> >
>