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%I #2 Mar 30 2012 18:37:20
%S 1,1,-4,-4,-8220,-16910960,-220513689396,-19336259194582782,
%T -12353453824556774353132,-60817754630605750994570243653,
%U -2385117541132316928253481462547625452
%N G.f. A(x) satisfies: Sum_{n>=0} log(A(3^n*x))^n/n! = 1 + Sum_{n>=0} 3^(3^n-n)*x^(3^n).
%F The coefficient of x^(3^n) in A(x)^(3^(3^n)) equals 3^(3^n-n):
%F [x^(3^n)] A(x)^(3^(3^n)) = 3^(3^n-n); while
%F [x^n] A(x)^(3^n) = 0 when n>0 is not a power of 3, with A(0)=1.
%e G.f.: A(x) = 1 + x - 4*x^2 - 4*x^3 - 8220*x^4 - 16910960*x^5 +...
%e log(A(x)) = x - 9*x^2/2 + x^3/3 - 32913*x^4/4 - 84513699*x^5/5 +...
%e ILLUSTRATE THE SERIES DEFINITION:
%e 1 + log(A(3x)) + log(A(9x))^2/2! + log(A(27x))^3/3! + log(A(81x))^4/4! +...
%e = 1 + 3*x + 9*x^3 + 2187*x^9 + 282429536481*x^27 +...
%e = 1 + 3^(1-0)*x + 3^(3-1)*x^3 + 3^(9-2)*x^9 + 3^(27-3)*x^27 +...
%e ILLUSTRATE (3^n)-th POWERS OF G.F. A(x).
%e The coefficients in the expansion of A(x)^(3^n) for n>=0 begin:
%e n=0: [(1), 1, -4, -4, -8220, -16910960, -220513689396,...];
%e n=1: [1, (3), -9, -35, -24648, -50782068, -661642361248,...];
%e n=2: [1, 9, (0), -240, -74574, -152788194, -1985840486856,...];
%e n=3: [1, 27, 243, (9), -236682, -462449898, -5965789971726,...];
%e n=4: [1, 81, 2916, 59076, (0), -1420876404, -17973134801100,...];
%e n=5: [1, 243, 28431, 2125845, 111615732, (0), -54490964413644,...];
%e n=6: [1, 729, 262440, 62178840, 10895760846, 1491228760410, (0),...];
%e where the coefficients along the diagonal (in parenthesis) begin:
%e [1,3,0,9,0,0,0,0,0,2187,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 282429536481,...]
%e and equal 3^(3^m-m) at positions n=3^m for m>=0, with zeros elsewhere (except for the initial '1').
%o (PARI) {a(n)=local(A=[1,1],B=[1,3]);for(i=1,n,A=concat(A,0); B=Vec(sum(m=0,#A,log(subst(Ser(A),x,3^m*x))^m/m!)); A[ #A]=-floor(B[ #A]/3^(#A-1)));A[n+1]}
%Y Cf. A167004, A167005, variant: A167000.
%K sign
%O 0,3
%A _Paul D. Hanna_, Nov 14 2009