Error in A115866

[Dan Dima]

Also f(a(1),a(2),...,a(k)) can be computed as the coefficient of
x(1)^a(1)...x(k)^a(k) in the expansion:
1/2 * 1/(1 - (1+x(1))*...*(1+x(k))/2)



Dan



On 3/4/07, Dan Dima <dimad72 at gmail.com> wrote:
> Hi all,
>
>
>
> I found this more than a year ago when I tried to solve the following puzzle:
> http://faculty.missouristate.edu/l/lesreid/POW08_0506.html
>
>
> However I found a very simple (although infinite sum) formula for the
> number of paths from (0,0,...,0) to (a(1),a(2),...,a(k)) using
> "nonzero" (2^k-1) steps of the form (x(1),x(2),...,x(k)) where x(i) is
> in {0,1} for 1<=i<=k, k-dimensions.
>
> I have looked carefully on the web and I found many articles related
> to this issue - Multi-Dimensional Lattice Paths with Diagonal Steps
> (or various kind of steps) - but none of them matches my simple
> infinite sum:
>
>
> f(a(1),a(2),...,a(k))  =
> Sum( (C(n;a(1)) * C(n;a(2)) * ... C(n;a(k))) / 2^{n+1} , {n,
> max(a(1),a(2),...,a(k)), infinity}),
> Sum( (C(n;a(1)) * C(n;a(2)) * ... C(n;a(k))) / 2^{n+1} , {n, 0, infinity}),
> C(n;a)=n!/a!(n-a)! & we assumed C(n;a)=0 if n<a
>
>
> Please can someone correct me if I am wrong!
>
>
> Nick: If you want to compute larger terms for those sequences please
> avoid recursivity - a lot of redundant work will be done ;) ... just
> use straightforward "for loops" instead...
>
> Neil: what things happen at about 8 or 9 dimensions?
>
>
>
> Best regards,
> Dan
>
>
>
>
>
> On 3/3/07, N. J. A. Sloane <njas at research.att.com> wrote:
> > David,  strange things happen at about 8 or 9 dimensions,
> > so I suggest you go up to dim 10
> >
> > of course the arrays and diagonal sequence(s) should
> > also be submitted!
> >
> > Best
> >
> > Neil
> >
>