Formula for A120014(n) -- Simplified

[Paul D. Hanna]

Max, 
     Thank you for the alternate approach.  Starting with the same 
formula:  a(n) = [x^n] { xC/(1-nxC) - [xC/1-nxC]^2 } 
(where C is Catalan function), I see that you 
collected terms involving the same Catalan coefficients, 
while I collected terms involving the same powers of n. 
 
> > a(n) = n^(n-1) - Sum_{k=1..n-1} 
> > n^(k-1)*k*(k-1)*(n-k-1)*(2*n-k-2)!/(n-k)!/n! 
>  
The goal of my approach here was to separate n from k 
as much as possible, hoping it would help in isolating a g.f.
 
My approach leads to the formula for the 
table of self-compositions of A120009: 
T(n,k) = [x^k] { xC/(1-nxC) - [xC/1-nxC]^2 }  
to be 
T(n,k) = n^(k-1) - Sum_{j=2..k-2}
n^(j-1)*j*(j-1)*(k-j-1)*(2k-j-2)!/(k-j)!/k! 
 
where the main diagonal is A120014(n) = T(n,n). 
 
> I've slightly different and somewhat simpler formula:
> 
> a(n) = Sum_{k=1..n} n^(k-3)*(n-k+1)*k*C(2*n-k-1,n-1)
> 
I concede, your approach yields simpler formulas; it leads to 
the formula for the table of self-compositions of A120009 
to be:  
T(n,k) = Sum_{j=1..k} n^(j-2)*(n-j+1)*j*(2k-j-1)!/(k-j)!/k! 
  
which is preferred over the formula for T(n,k) above. 
  
> implying that a(n) is the coefficient of x^n in the expansion of
> 
> (x*(1-x+x*n)) / (n^2 * (1-n*x)^2 * (1-x)^(n+1))
> 
Max, this is a nice result - good work!  
 
> > Thus, the e.g.f. of A120014 does involve the LambertW function as
> > suspected
> 
> It still does not sound to me as a strong evidence.
> 
Well, consider e.g.f.:
Sum_{n>=1} a(n)*x^n/n! = Sum_{n>=1} n^(n-1)*x^n/n! 
- Sum_{n>=1} Sum_{k=1..n-1}
n^(k-1)*k*(k-1)*(n-k-1)*(2*n-k-2)!/(n-k)!*x^n/(n!)^2
 
Not very pretty - but LambertW clearly is found in the first sum on the
right. 
Yet the ugly double sum may be subtracting it back out ... 
So, no, I do not think there is much hope of finding e.g.f. using this
formula! 
 
> > (I am still working on obtaining that e.g.f.).
> Any luck so far?
> 
Not yet, and it may require someone more clever than I. 
But I will continue to try.  
 
Thanks for sharing your nice results! 
I will credit you in any new submissions using your formulas. 
    Paul