Formula for A120014(n), Anyone?

[Paul D. Hanna]

Max, 
     Thanks for that formula!  I see now how you obtained it. 
When ready, please submit your formula to A120014. 
 
Restating your formula, and substituting C_k with C(2k,k)/(k+1):
a(n) = (n^2/(n-1))^(n+1)/n^4*(2*n-1) + 
1/n^5*[Sum_{k=0..n-1} (2*(k+1) -
n*(k+3))*((n-1)/n^2)^(k-n-1)*(2*k)!/k!/(k+1)!]
 
Although this is not any simpler, the presence of the factorials suggest 
that simplification may require some identity of LambertW. 
I will try to apply some LambertW identities to your formula to see if it
simplifies. 
 
If LambertW is involved, it may also give a g.f. for A120014 - 
which is what I am *really* after!  
 
I feel that there is a general (although complex) g.f. for sequences 
generated by a(n) = [x^n] n-th self-composition of G(x). 
My hope is that A120014 is the first step toward such a generalization. 
  
Thanks again, Max! 
     Paul
 
Thanks, too, to Graeme McRae and others who have given attention to my
query.
   
 
On Wed, 7 Jun 2006 18:18:51 -0700 Max <maxale at gmail.com> writes:
> On 6/7/06, Paul D. Hanna <pauldhanna at juno.com> wrote:
> > Seqfans,
> >      Perhaps someone can find a formula for a(n) of the following
> > sequence?
> > It is not yet visible in OEIS, so I copy the entry below.
> > The best I can do so far is:
> >
> > a(n) = [x^n] x*((1-n+n^2) - n^2*(n+1)*x - n*(1-(n+2)*x)*C(x)
> > )/(1-n+n^2*x)^2,
> > where C(x) = (1-sqrt(1-4*x))/(2*x) is the Catalan function 
> (A000108).
> >
> > I believe that there is a nice closed form for a(n), but it 
> escapes me
> > ...
> 
> It follows directly from your formula that
> 
> a(n) = (n^2/(n-1))^(n+1) / n^4 * ( 2*n-1  + \sum_{k=0}^{n-1}
> (2*(k+1)/n - k - 3) * ((n-1)/n^2)^k * C_k )
> 
> where C_k is the k-th Catalan number.
> 
> But this formula still leaves two questions:
> 1. How to see (directly from the formula) that a(n) is integer?
> 2. How to prove that a(n) is divisible by n.
> 
> Max