[seqfan] Re: A001924

[Ed Jeffery]

Alois,

After I posted that message on Seqfan, I edited Paolo's formula in A001924
to make it work properly. Since you are editing today, maybe you could go
ahead and make the correction, then delete my comment in the extension
field.

Thanks,

Ed

On Wed, Dec 28, 2011 at 12:53 AM, Ed Jeffery <lejeffery7 at gmail.com> wrote:

> Seqfans,
>
> A001924={0,1,3,7,14,26,...}. I am working on a formula for the nonzero
> terms of this sequence based on binomial coefficients only. But Paolo Lava
> has a formula already which doesn't seem to work properly: it is
>
> a(n)=Sum_{k=1..n} {C(n-k+2,k+1)}, with n>=0,
>
> as written by Paolo Lava. I assume that C(m,j) is a binomial coefficient.
> Is there something wrong with this formula and could someone else please
> check it, or am I having problems with Mathcad again (as usual)?
>
> At any rate, this sequence also arises in the following way. Interlace two
> copies of the binomial coefficients, like this:
>
> 1;
> 1;
> 1,1;
> 1,1;
> 1,2,1;
> 1,2,1;
> etc.,
>
> indexing the rows and columns by 0,1,2,... Next, for each k in {0,1,...},
> shift column k up k rows to get the new triangle
>
> T=
> 1;
> 1,1;
> 1,1,1;
> 1,2,1,1;
> 1,2,3,1,1;
> 1,3,3,4,1,1;
> 1,3,6,4,5,1,1;
> etc.
>
> Finally, omitting the first zero term, A001924 is given by the row sums of
> the fourth "sub-triangle" extracted from the above triangle and marked by
> braces in the following,
>
> 1;
> 1,1;
> 1,1,1;
> {1},2,1,1;
> {1,2},3,1,1;
> {1,3,3},4,1,1;
> {1,3,6,4},5,1,1;
> etc.
>
> in which the entries are obviously all still binomial coefficients,
> although a bit scrambled. (However there is a pattern.) Since the row sums
> of the original triangle T give the Fibonacci sequence (starting with
> a(0)=1, a(1)=2), any of the row sums of sub-triangles defined as above can
> also be expressed in terms of Fibonacci numbers, but I think their
> expression in terms of binomial coefficients is equally interesting.
>
> Regards,
>
> Ed Jeffery
>