[math-fun] some exact formulas for integer sequences,
Emeric Deutsch
deutsch at duke.poly.edu
Wed Mar 24 02:35:51 CET 2004
Hello,
In A001006 there is the recursion
(n+2)a(n)=(2n+1)a(n-1)+(3n-3)a(n-2)
Using it with Maple, I got the first 1000 terms in a fraction of
a second.
It occurs, with a hint for the proof, in
E. Barcucci, R. Pinzani, R. Sprugnoli, The Motzkin family,
P.U.M.A. Ser. A, Vol. 2, 1991, No. 3-4, pp. 249-279.
which will be added to A001006.
I will ask Neil to add also
lim(a(n)/a(n-1), n->infinity) = 3
(from the Aigner paper) and
a(n) ~ 3^(n+1)sqrt(3)[1+1/(16n)]/[(2n+3)sqrt((n+2)Pi)]
from the Barcucci et al. paper
Emeric Deutsch
On Tue, 23 Mar 2004, Simon Plouffe wrote:
> hello,
>
> (this is old material from 1993),
>
> I collected some of my notes about exact formulas for
> integer sequences. They are somewhat usefull since it can be
> used to compute the n'th term of a sequence.
>
> it uses [ ] , { } and some constants like exp(1).
>
> http://www.lacim.uqam.ca/%7Eplouffe/exact.htm
>
> note : most of those formulas do not appear in the current On-line
> Encyclopedia of integer sequences.
>
>
> By the way, can anybody do this : Can we find an EXACT formula for
> the n'th term of A001006 (Motzkin numbers), the only useful formula
> we have is a recurrence but with the index that grows with n which
> makes it difficult to compute for large n.
> http://www.research.att.com/projects/OEIS?Anum=A001006 is
> 1,1,2,4,9,21,51,127,323,835,2188,5798, ... and it grows roughly like
> 3^n.
>
>
> Simon Plouffe
>
>
>
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