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G.f. A(x) satisfies: A(x) = 1 + x * (A(x) + 4*x*A'(x)) / (A(x) + x*A'(x)).
2

%I #22 Aug 31 2024 12:47:26

%S 1,1,3,12,57,318,2190,20298,265557,4683954,102511182,2622117720,

%T 75857421522,2436766744728,85885843375992,3292322691248310,

%U 136306472824555725,6059583761048830290,287832499902612461910,14546503423112218184280,779260996651844076919230,44104811071523533909346100,2629647430438623322437523380,164732335319839237173383723220

%N G.f. A(x) satisfies: A(x) = 1 + x * (A(x) + 4*x*A'(x)) / (A(x) + x*A'(x)).

%C Compare to: C(x) = 1 + x*C(x) * (C(x) + 2*x*C'(x)) / (C(x) + x*C'(x)) holds when C(x) = 1 + x*C(x)^2 is a g.f. of the Catalan numbers (A000108).

%C If G(x) = 1 + x * (G(x) + y*x*G'(x)) / (G(x) + x*G'(x)), then negative coefficients are present in the power series expansion of G(x) when y < t where constant t = 3.79622607101172329768286422639804088884917373644497484011451904053377050...

%H Robert Israel, <a href="/A302101/b302101.txt">Table of n, a(n) for n = 0..382</a>

%F G.f. A(x) satisfies: A'(x) = A(x) * (A(x) - 1 - x) / (x*(1 + 4*x - A(x))).

%F G.f.: A(x) = exp( Integral (A(x) - 1 - x) / (x*(1 + 4*x - A(x))) dx ).

%F a(n) ~ n! * c * 3^n / n^2, where c = 0.030801423287090143490227542092... - _Vaclav Kotesovec_, Aug 31 2024

%e G.f.: A(x) = 1 + x + 3*x^2 + 12*x^3 + 57*x^4 + 318*x^5 + 2190*x^6 + 20298*x^7 + 265557*x^8 + 4683954*x^9 + 102511182*x^10 + ...

%e such that A(x) = 1 + x * (A(x) + 4*x*A'(x)) / (A(x) + x*A'(x)).

%p G:= 1:

%p for n from 1 to 30 do

%p G:= convert(series(1+x*(G+4*x*diff(G,x))/(G +x*diff(G,x)),x,n+1),polynom);

%p od:

%p seq(coeff(G,x,n),n=0..30); # _Robert Israel_, Apr 08 2018

%t nmax = 23; sol = {a[0] -> 1};

%t Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x] - (1 + x*(A[x] + 4*x*A'[x])/(A[x] + x*A'[x])) + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];

%t sol /. Rule -> Set;

%t a /@ Range[0, nmax] (* _Jean-François Alcover_, Nov 01 2019 *)

%o (PARI) /* Differential Equation */

%o {a(n) = my(A=1); for(i=0,n, A = 1 + x*(A + 4*x*A')/(A + x*A' +x^2*O(x^n))); polcoeff(A,n)}

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

%Y Cf. A302102.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Apr 07 2018