Formula for A120014(n), Anyone?

[Max]

On 6/7/06, Paul D. Hanna <pauldhanna at juno.com> wrote:

> 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)!]

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)

If the involved sums were up to infinity, we could computed them, and
the fact that 1-4*(n-1)/n^2 = ((n-2)/n)^2 makes the results very
simple, i.e.,

Sum_{k=0..oo} ((n-1)/n^2)^k C(2k,k) = 1 / sqrt(1-4*(n-1)/n^2) = n/(n-2)

Sum_{k=0..oo} ((n-1)/n^2)^k C(2k,k) / (k+1) = (1 -
sqrt(1-4*(n-1)/n^2)) / (2 * (n-1)/n^2) = n/(n-1)

These sums can give also an approximation for a(n) as n tends to infinity.

> 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?

Thanks,
Max