A179498 E.g.f. satisfies: A(x) = A(x*A(x))^2 - x*A'(x).

1, 1, 6, 78, 1648, 49500, 1957968, 97097336, 5834581632, 414370221696, 34127635732800, 3211425586911168, 341164552018811904, 40517022329819203584, 5335290940894955228160, 773591071307555130451200

Offset: 0

Author

Paul D. Hanna, Jul 31 2010

Keywords

• nonn

Examples

			E.g.f.: A(x) = 1 + x + 6*x^2/2! + 78*x^3/3! + 1648*x^4/4! + 49500*x^5/5! +...
Related expansions:
. x*A(x) = x + 2*x^2/2! + 18*x^3/3! + 312*x^4/4! + 8240*x^5/5! +...
. x*A(x)^2 = x + 4*x^2/2! + 42*x^3/3! + 768*x^4/4! + 20680*x^5/5! +..
. x*A'(x) = x + 12*x^2/2! + 234*x^3/3! + 6592*x^4/4! + 247500*x^5/5! +...
. A(x*A(x)) = 1 + x + 8*x^2/2! + 132*x^3/3! + 3400*x^4/4! + 120940*x^5/5! +...
. A(x*A(x))^2 = 1 + 2*x + 18*x^2/2! + 312*x^3/3! + 8240*x^4/4! + 297000*x^5/5! +...
Illustrate the iterations G_n(x) of G(x) = x*A(x) by:
. [G_3(x)/x]^2 = A(x)^2 * G_2'(x);
. [G_4(x)/x]^2 = A(x)^2 * G_3'(x);
. [G_5(x)/x]^2 = A(x)^2 * G_4'(x); ...
which can be shown by the chain rule of differentiation.
...
The RIORDAN ARRAY (A(x), x*A(x)) begins:
. 1;
. 1, 1;
. 6/2!, 2, 1;
. 78/3!, 14/2!, 3, 1;
. 1648/4!, 192/3!, 24/2!, 4, 1;
. 49500/5!, 4136/4!, 348/3!, 36/2!, 5, 1;
. 1957968/6!, 124840/5!, 7680/4!, 552/3!, 50/2!, 6, 1;
. 97097336/7!, 4928256/6!, 233940/5!, 12520/4!, 810/3!, 66/2!, 7, 1; ...
where the g.f. of column k = A(x)^(k+1) for k>=0. ...
The MATRIX LOG of the above Riordan array (A(x), x*A(x)) begins:
. 0;
. 1, 0;
. 4/2!, 2, 0;
. 42/3!, 8/2!, 3, 0;
. 768/4!, 84/3!, 12/2!, 4, 0;
. 20680/5!, 1536/4!, 126/3!, 16/2!, 5, 0;
. 749040/6!, 41360/5!, 2304/4!, 168/3!, 20/2!, 6, 0;
. 34497792/7!, 1498080/6!, 62040/5!, 3072/4!, 210/3!, 24/2!, 7, 0; ...
where the g.f. of column k = (k+1)*x*A(x)^2 for k>=0.
		

Crossrefs

Cf. A179497, A179499, A179421.

Programs

PARI

{a(n)=local(A=1+x+sum(m=2,n-1,a(m)*x^m/m!)+x*O(x^(n+5)));if(n<2,n!*polcoeff(A,n),n!*polcoeff(subst(A,x,x*A)^2,n)/(n-1))}

Formula

E.g.f. satisfies: x*A(x)^2 equals the g.f. of column 0 in the matrix log of the Riordan array (A(x), x*A(x)).

E.g.f.: A(x) = G(x)/x where G(x) = e.g.f. of A179497.

Let G_n(x) denote the n-th iteration of x*A(x) with G_0(x)=x, then

. [G_{n+1}(x)/x]^2 = A(x)^2*G_n'(x) for all n,

and L=x*A(x)^2 satisfies the series:

. A(x) = 1 + L + L*Dx(L)/2! + L*Dx(L*Dx(L))/3! + L*Dx(L*Dx(L*Dx(L)))/4! +...

. G_{-1}(x)/x = 1 - L + L*Dx(L)/2! - L*Dx(L*Dx(L))/3! + L*Dx(L*Dx(L*Dx(L)))/4! -+...

. G_n(x)/x = 1 + n*L + n^2*L*Dx(L)/2! + n^3*L*Dx(L*Dx(L))/3! + n^4*L*Dx(L*Dx(L*Dx(L)))/4! +...

where Dx(F) = d/dx(x*F).