# [seqfan] Re: A019575 challenge

Ron Hardin rhhardin at att.net
Mon Aug 23 12:29:40 CEST 2010

```
> > -----  Original Message ----
> > > From: Robert Gerbicz  <robert.gerbicz at gmail.com>
> >
> > >
> > > That suggests that there can be a  polynom for  each k value, and  in fact,
>
> my
> > > conjecture is that:   T(n,k)=n*binomial(2*n-k-2,n-2) if  2*k>n, so for
>your
> >  >  a(n)=T(n,n-k)=n*binomial(n+k-2,k)=n^(k+1)/k!+O(n^k) (it is true,  if
> >n>2*k),
> > > that gives for the (k+1)-th difference  sequence: k+1, what  we  needed.
> > >
> The formula minus  the empirical result:
>
> 1  0
> 2  1 0
> 3  8 0  0
> 4  39 6 0 0
> 5  174 50 0 0 0
> 6  755 280 15 0 0  0
> 7  3233 1372 147 0 0 0 0
> 8  13727 6286 980 28 0 0 0  0
> 9  57914 27750 5532 324 0 0 0 0 0
> 10  243099 119748 28500 2430  45 0 0 0 0 0
> 11  1016157 509168 138545 15125 605 0 0 0 0 0 0
> 12   4232591 2143284 647361 84722 5082 66 0 0 0 0 0 0
> 13  17577013 8957676  2940093 443456 34476 1014 0 0 0 0 0 0 0

>
> The last  nonzero entry on even lines is  k*(n-1).

The last nonzero entry on odd lines n>3 is k*n^2

```