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G.f. A(x) satisfies: A(x) = B(x)*(A(x) - x*C(x)) where B(x) = A(x/B(x)) and C(x) = A(x*C(x)).
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%I #20 Dec 20 2014 17:24:38

%S 1,1,2,6,26,149,1024,7965,68192,632724,6294190,66579501,744194484,

%T 8747497833,107718981328,1385436413289,18563761993762,258579817821799,

%U 3737335096804136,55957031888334621,866632465992896412,13865193902724224273,228875892203793317404,3893773927147402337094

%N G.f. A(x) satisfies: A(x) = B(x)*(A(x) - x*C(x)) where B(x) = A(x/B(x)) and C(x) = A(x*C(x)).

%C Compare to: G(x) = Series_Reversion( x - Series_Reversion(x*G(x)) * x*G(x) )/x, which is satisfied by the g.f. G(x) = 1 + x*G(x) * G(x*G(x)) of A030266 with offset 0.

%H Vaclav Kotesovec, <a href="/A247224/b247224.txt">Table of n, a(n) for n = 0..300</a>

%F G.f. A(x) satisfies:

%F (1) A(x) = Series_Reversion( x - Series_Reversion(x/A(x)) * x/A(x) )/x.

%F (2) A(x) = x/Series_Reversion( (x - Series_Reversion(x*A(x))) * A(x)/x ).

%F Given B(x) = A(x/B(x)) and C(x) = A(x*C(x)), then:

%F (3.a) A(x) = B(x*A(x)) and A(x) = C(x/A(x)),

%F (3.b) B(x) = x/Series_Reversion(x*A(x)),

%F (3.c) C(x) = Series_Reversion(x/A(x))/x,

%F (3.d) B(x) = A(x)/(A(x) - x*C(x)),

%F (3.e) C(x) = A(x)*(1 - 1/B(x))/x.

%e G.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 26*x^4 + 149*x^5 + 1024*x^6 +...

%e Let B(x) = A(x/B(x)) and C(x) = A(x*C(x)), where B(x) and C(x) begin:

%e B(x) = 1 + x + x^2 + 2*x^3 + 9*x^4 + 53*x^5 + 354*x^6 + 2651*x^7 + 21951*x^8 + 197666*x^9 + 1911091*x^10 + 19665622*x^11 + 214060860*x^12 +...

%e C(x) = 1 + x + 3*x^2 + 13*x^3 + 71*x^4 + 460*x^5 + 3399*x^6 + 27867*x^7 + 248789*x^8 + 2388199*x^9 + 24432778*x^10 + 264682253*x^11 + 3021086129*x^12 +...

%e then A(x) = B(x) * (A(x) - x*C(x)).

%o (PARI) {a(n)=local(A=1);for(i=1,n,A=serreverse(x-x/A*serreverse(x/(A +x^2*O(x^n))))/x);polcoeff(A,n)}

%o for(n=0,30,print1(a(n),", "))

%K nonn

%O 0,3

%A _Paul D. Hanna_, Dec 19 2014