Formula for A120014(n), Anyone?

[Graeme McRae]

a(n) = (n+1)*A001870(n)

----- Original Message ----- 
From: "Paul D. Hanna" <pauldhanna at juno.com>
To: <seqfan at ext.jussieu.fr>
Sent: Wednesday, June 07, 2006 3:50 PM
Subject: Formula for A120014(n), Anyone?


> 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
> ... 
>  
> Thanks, 
>     Paul 
> 
> ID: A120014
> NAME: 
> Coefficients of x^n in the n-th self-composition of the g.f. of A120009
> so that: 
> a(n) = [x^n] { (x-x^2) o x/(1-n*x) o (1-sqrt(1-4*x))/2 } for n>=1. 
> 
> SEQUENCE: 
> 1,2,9,60,530,5892,79681,1276760,23729310,502780580,11974950746,
> 316917570312,9230453871756,293492484431720,10117826259791025,
> 375952605020796720,14980065429077943734,637215061582781559972,
> 28822846054807004190990,1381481231061589066092200,
> 
> EXAMPLE: 
> Successive self-compositions of F(x), the g.f. of A120009, begin: 
> F(x) = (1)x + x^2 + x^3 - 6x^5 - 33x^6 - 143x^7 - 572x^8 - 2210x^9 +...
> F(F(x)) = x + (2)x^2 + 4x^3 + 6x^4 - 4x^5 - 100x^6 - 664x^7 +... 
> F(F(F(x))) = x + 3x^2 + (9)x^3 + 24x^4 + 42x^5 - 87x^6 - 1575x^7 +... 
> F(F(F(F(x)))) = x + 4x^2 + 16x^3 + (60)x^4 + 192x^5 + 360x^6 +... 
> F(F(F(F(F(x))))) = x + 5x^2 + 25x^3 + 120x^4 + (530)x^5 +1955x^6 +... 
> F(F(F(F(F(F(x)))))) = x + 6x^2 + 36x^3 +210x^4 +1164x^5 + (5892)x^6 +... 
> 
> COMMENT:
> a(n) is divisible by n for n>=1; a(n)/n = A120016(n). 
>