[seqfan] Re: A019575 challenge

[Ron Hardin]

----- Original Message ----
> From: Ron Hardin <rhhardin at att.net>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Sent: Sun, August 22, 2010 10:03:10 PM
> Subject: [seqfan] Re: A019575 challenge
> 
> 
> 
> 
> 
> 
> ----- 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
14  72804199 37241958 13073151 2214212 207025 9464 91 0 0 0 0 0 0 0
15  300874499 154221394 57194347 10679515 1149085 69825 1575 0 0 0 0 0 0 0 0
16  1240940159 636668974 247062052 50166220 6037600 448800 16200 120 0 0 0 0 0 0 
0 0
17  5109183314 2621862910 1056469556 230810428 30469848 2635680 129472 2312 0 0 
0 0 0 0 0 0 0
18  21002455979 10775294212 4480692006 1044351378 149106660 14528370 889542 
26010 153 0 0 0 0 0 0 0 0 0
19  86213785349 44209521328 18876138753 4661134530 712177551 76427310 5526549 
224181 3249 0 0 0 0 0 0 0 0 0 0
20  353452637999 181125326632 79079145738 20567649036 3335677335 387925850 
31966550 1642550 39710 190 0 0 0 0 0 0
21  1447388552609 741145189072 329751874848 89887445536 15374310315 1914310650 
175327670 10785390 367500 4410 0 0

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