[seqfan] Re: Guess the Recurrence
Maximilian Hasler
maximilian.hasler at gmail.com
Sat Dec 4 04:33:29 CET 2010
look it up on OEIS ! ;-)
the second one is A000914
the penultime is -A081048
The whole seems to be A125714.
(lines read from right to left).
Maximilian
On Fri, Dec 3, 2010 at 11:19 PM, Ron Hardin <rhhardin at att.net> wrote:
> Computing T(n,k; z), the number of nXk 0..z arrays with every 2X2 subblock
> summing to 2z,
> it comes out that there's a single recurrence for every n,k row or column for a
> given z
>
> z=0 Empirical: a(n)=1*a(n-1)
> z=1 Empirical: a(n)=3*a(n-1)-2*a(n-2)
> z=2 Empirical: a(n)=6*a(n-1)-11*a(n-2)+6*a(n-3)
> z=3 Empirical: a(n)=10*a(n-1)-35*a(n-2)+50*a(n-3)-24*a(n-4)
> z=4 Empirical: a(n)=15*a(n-1)-85*a(n-2)+225*a(n-3)-274*a(n-4)+120*a(n-5)
> z=5 Empirical:
> a(n)=21*a(n-1)-175*a(n-2)+735*a(n-3)-1624*a(n-4)+1764*a(n-5)-720*a(n-6)
>
> (so the initial (z+1)X(z+1) of the T(n,k) array determines the whole thing)
>
> The question is what's the system of the coefficients of the recurrences vs z?
>
> The first coef is (z+1)(z+2)/2 and the last is (z+1)!
>
> Can we guess the middle ones?
>
>
>
> rhhardin at mindspring.com
> rhhardin at att.net (either)
>
>
>
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