# [seqfan] A001924

Ed Jeffery lejeffery7 at gmail.com
Wed Dec 28 07:53:48 CET 2011

```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

```