[seqfan] Re: Integer Sequence Analysis in Mathematica 7
zak seidov
zakseidov at yahoo.com
Sat Nov 22 08:36:42 CET 2008
Also, see
N A001609 a (n) = a (n - 1) + a (n - 3).
% C A001609 This comment covers a family of sequences which satisfy a recurrence of the form
a (n) = a (n - 1) + a (n - m), with a (n) = 1 for n = 1 ... m - 1, a (m) = m + 1.
The generating function is (x + m*x^m)/(1 - x - x^m).
a (n) = 1 + n*sum (binomial (n - 1 - (m - 1)*i, i - 1)/i, i = 1 .. n/m).
a(n) is in a close form as finite sum!
In Nmca:
a[1] = a[2] = 1; a[3] = 4; m = 3; a[n_] := 1 + n*Sum [Binomial [n - 1 - (m - 1)*i, i - 1]/i, {i, n/m}]
A001609 = Table[a[n], {n, 8}]
{1, 1, 4, 5, 6, 10, 15, 21}
zak
--- On Fri, 11/21/08, Maximilian Hasler <maximilian.hasler at gmail.com> wrote:
> From: Maximilian Hasler <maximilian.hasler at gmail.com>
> Subject: [seqfan] Re: Integer Sequence Analysis in Mathematica 7
> To: "Alexander Povolotsky" <apovolot at gmail.com>
> Cc: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
> Date: Friday, November 21, 2008, 8:59 PM
> [sorry, please ignore the beginning of my last reply, I
> didn't notice
> that PURRS input were not the 30 values but the recurrent
> equation...]
> M.
>
> On Fri, Nov 21, 2008 at 9:54 PM, Maximilian Hasler
> <maximilian.hasler at gmail.com> wrote:
> > Well, for 30 given values there's always a
> polynomial of degree 29
> > that fits *exactly* through these points, but in
> general not through
> > the 31st point...
> (...)
> > Well, actually A001609(n) is the most concise and most
> informative way
> > to code that function.
> (...)
> >> http://www.cs.unipr.it/purrs/ yields the
> following:
> >>
> >> Exact solution for x(n) = x(-1+n)+x(-3+n)
> >> x(n) =
> 1/2*(I*(29/54+1/18*sqrt(93))^(1/3)*sqrt(3)-I*sqrt(3)*(29/54-1/18*sqrt(93))^
> >> ...*(29/54-1/18*sqrt(93))^(1/3))^n*x(0)
> >> for each n >= 0
>
>
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