Polynomial sequence?

[Jonathan Post]

a(n) = (n+1)^n - n! = A000169(n) - A000142(n).

Since A000169 is the number of labeled rooted trees with n nodes, you should
be able to come up with a pure graph enumeration description of your
sequence.  When you do, I agree that it is nice.

On 12/17/06, Nick Hobson <nickh at qbyte.org> wrote:
>
> Hi,
>
> What do people think of this sequence: 1, 7, 58, 601, 7656, 116929,
> 2092112, 43006401, 999637120, 25933795801, ... ?
>
> The formula is a(n) = (n+1)^n - n!.  Lest that seem completely artificial,
> I should comment on how I stumbled across the sequence!  Fit a polynomial
> f of degree n-1 to the first n nth powers of positive integers.  Then
> f(n+1) = a(n).
>
> For example, the quadratic that fits (1,1), (2,8), and (3,27) is f(n) =
> 6n^2-11n+6.  Then f(4) = a(3) = 58.  Of course, it's not necessary to
> actually determine the polynomial f; a(n) can be found by considering
> differences.
>
> Nick
>
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