Formula for A120014(n), Anyone?

[Paul D. Hanna]

Max (and Seqfans),
      Interleaving my replies below ... 
On Wed, 7 Jun 2006 20:23:29 -0700 Max <maxale at gmail.com> writes:
>
> If you bring C_k as C(2k,k)/(k+1) it is natural to split the 
> formula
> into three parts:
> n^2/(n-1))^(n+1) / n^4 * (2*n-1)
> (n^2/(n-1))^(n+1) / n^4 * (2-n)/n * Sum_{k=0..n-1} ((n-1)/n^2)^k
C(2k,k)
> - (n^2/(n-1))^(n+1) / n^4 * 2 * Sum_{k=0..n-1} ((n-1)/n^2)^k
C(2k,k)/(k+1)
 
This is helpful.  Yet some identity is needed to simplify. 
 
> These sums can give also an approximation for a(n) as n tends to
infinity.
 
Good insight. 
  
> > Although this is not any simpler, the presence of the factorials 
> suggest
> > that simplification may require some identity of LambertW.
> 
> Could you please explain this point (maybe with a smaller example)?
> What is the relationship between factorials and Lambert W-function?
 
Yes, my wording was vague at best.  
 
What I was refering to here was the following lovely 
LambertW convolution identity: 
 
Sum_{k=0..n} p*(n-k+p)^(n-k-1)* q*(k+q)^(k-1)* C(n,k) 
 
  = (p+q)*(n+p+q)^(n-1) 
 
which is the coefficient of x^n/n! in: 
 
W(x)^(p+q) = W(x)^p * W(x)^q 
 
where W(x) = -LambertW(-x)/x. 
  
My objective was to restate your formula into a form 
where this identity may be applied. 
  
We may restate your 3 parts into a convolution form: 
a(n) = (n^2/(n-1))^(n+1)/n^4*(2*n-1) 
+(n^2/(n-1))^2/n^4*(2-n)/n * Sum_{k=0..n-1} (n^2/(n-1))^(n-1-k) *
C(2*k,k)) 
-(n^2/(n-1))^2/n^4*2 * Sum_{k=0..n-1} (n^2/(n-1))^(n-1-k) *
C(2*k,k)/(k+1)) 
 
and then the sums include factors like (n-1)^(n-1-k) ... 
which reminded me of LambertW. 
 
But LambertW may not be involved at all - 
I will have to check. 
   
Thanks,
     Paul
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