A000426 recurrence

[Mitch Harris]

On Dec 27, 2007 2:46 PM, David W. Wilson <wilson.d at anseri.com> wrote:
>
> Can anyone make heads or tails of the A000426 recurrence?

Yes. The presentation is not the most conducive to computation, it is
given as a relation assumed to be equal to 0. To get a recurrence that
one can actually apply, equate to 0 and solve for the highest indexed
term, a[n+5]. My preference would then be to shift by 5 to get the
more easily computable recurrence:

a[n_] := 1/2 ((2 n - 5) a[n - 1] + (5 n - 11) a[n - 2] + (5 n - 14)
a[n - 3] + (2 n - 5) a[n - 4] + 2 a[n - 5])

(in mathematica syntax)

-- 
Mitch