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A366226 O.g.f. A(x) satisfies: A(x) = 1 + x*Sum_{n>=0} 2^n * log( A(3^n*x) )^n / n!.

%I A366226 #12 Oct 17 2023 09:34:47
%S A366226 1,1,6,261,56070,56526498,334429044030,15777272891508021,
%T A366226 6500948711591606135796,22416650201723925643982814186,
%U A366226 632905244163070372226486183732882316,146120187946706698644410320973489902454862324,277121097159744219425840626808464318501357604841881466
%N A366226 O.g.f. A(x) satisfies: A(x) = 1 + x*Sum_{n>=0} 2^n * log( A(3^n*x) )^n / n!.
%C A366226 In general, we have the following identity:
%C A366226 given A(x) = Sum_{n>=0} a(n)*x^n satisfies
%C A366226 A(x) = 1 + x*Sum_{n>=0} p^n * log( A(q^n*x) )^n / n!,
%C A366226 then a(n+1) = [x^n] A(x)^(p*q^n) for n >= 0, with a(0)=1,
%C A366226 for arbitrary fixed parameters p and q.
%C A366226 Here, p = 2 and q = 3.
%H A366226 Paul D. Hanna, <a href="/A366226/b366226.txt">Table of n, a(n) for n = 0..50</a>
%F A366226 G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.
%F A366226 (1) A(x) = 1 + x*Sum_{n>=0} 2^n*log( A(3^n*x) )^n / n!.
%F A366226 (2) a(n+1) = [x^n] A(x)^(2*3^n) for n >= 0, with a(0)=1.
%e A366226 G.f.: A(x) = 1 + x + 6*x^2 + 261*x^3 + 56070*x^4 + 56526498*x^5 + 334429044030*x^6 + 15777272891508021*x^7 + 6500948711591606135796*x^8 + ...
%e A366226 where
%e A366226 A(x) = 1 + x*[1 + 2*log(A(3*x)) + 2^2*log(A(3^2*x))^2/2! + 2^3*log(A(3^2*x))^3/3! + ... + 2^n*log(A(3^n*x))^n/n! + ...].
%e A366226 RELATED SERIES.
%e A366226 log(A(x)) = x + 11*x^2/2 + 766*x^3/3 + 223187*x^4/4 + 282345766*x^5/5 + 2006233236098*x^6/6 + 110438567161208518*x^7/7 + ...
%e A366226 RELATED TABLE.
%e A366226 The table of coefficients of x^k in A(x)^(2*3^n) begins:
%e A366226 n=0: [1,   2,     13,      534,     112698,    113168268, ...];
%e A366226 n=1: [1,   6,     51,     1766,     345165,    340906254, ...];
%e A366226 n=2: [1,  18,    261,     7350,    1112382,   1035922644, ...];
%e A366226 n=3: [1,  54,   1755,    56070,    4589001,   3250238022, ...];
%e A366226 n=4: [1, 162,  14013,   894294,   56526498,  12817431900, ...];
%e A366226 n=5: [1, 486, 120771, 20555046, 2731197285, 334429044030, ...]; ...
%e A366226 in which the main diagonal equals this sequence shift left,
%e A366226 illustrating that a(n+1) = [x^n] A(x)^(2*3^n) for n >= 0.
%o A366226 (PARI) {a(n) = my(A=[1, 1]); for(i=1, n, A=concat(A, Vec(Ser(A)^(2*3^(#A-1)))[ #A])); A[n+1]}
%o A366226 for(n=0,15,print1(a(n),", "))
%o A366226 (PARI) {a(n) = my(A=1+x); for(i=1, n, A = 1 + x*sum(m=0,#A, 2^m*log( subst(Ser(A),x,3^m*x +x*O(x^n)))^m/m!) ); polcoeff(A,n)}
%o A366226 for(n=0,15,print1(a(n),", "))
%Y A366226 Cf. A132695, A156904, A156905, A181444.
%K A366226 nonn
%O A366226 0,3
%A A366226 _Paul D. Hanna_, Oct 16 2023