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A360336 G.f. A(x) satisfies: [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(3*n))^(n+1) for n >= 0.

%I A360336 #9 Feb 06 2023 12:09:16
%S A360336 1,1,6,99,2608,90800,3835458,187727106,10356030404,632391914502,
%T A360336 42217751766193,3053486035335835,237640678130730437,
%U A360336 19794116975373467259,1756875217029906875379,165552614838271944281933,16509692094523556884973416,1737510282985845400007263814
%N A360336 G.f. A(x) satisfies: [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(3*n))^(n+1) for n >= 0.
%H A360336 Paul D. Hanna, <a href="/A360336/b360336.txt">Table of n, a(n) for n = 0..300</a>
%F A360336 G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies:
%F A360336 (1) [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(3*n))^(n+1) for n>=0.
%F A360336 (2) A(x) = Sum_{n>=0} b(n) * x^n/A(x)^n, where b(n) = [x^n] (1 + x*A(x)^(3*n))^(n+1) / (n+1).
%F A360336 a(n) ~ c * d^n * n! * n^alpha, where d = 5.7189630165873859806..., alpha = 1.0541176773983..., c = 0.03951220887392... - _Vaclav Kotesovec_, Feb 06 2023
%e A360336 G.f.: A(x) = 1 + x + 6*x^2 + 99*x^3 + 2608*x^4 + 90800*x^5 + 3835458*x^6 + 187727106*x^7 + 10356030404*x^8 + 632391914502*x^9 + ...
%e A360336 RELATED SERIES.
%e A360336 G.f. A(x) = B(x/A(x)) where B(x) = B(x*A(x)) begins:
%e A360336 B(x) = 1 + x + 7*x^2 + 118*x^3 + 3113*x^4 + 108221*x^5 + 4564720*x^6 + 223208259*x^7 + 12307249017*x^8 + ... + b(n)*x^n + ...
%e A360336 such that b(n) = [x^n] (1 + x*A(x)^(3*n))^(n+1) / (n+1),
%e A360336 as well as b(n) = [x^n] A(x)^(n+1) / (n+1),
%e A360336 so that b(n) begin:
%e A360336 [1/1, 2/2, 21/3, 472/4, 15565/5, 649326/6, 31953040/7, 1785666072/8, ...].
%e A360336 ILLUSTRATION OF DEFINITION.
%e A360336 The table of coefficients of x^k in A(x)^(n+1) begins:
%e A360336 n=0: [1, 1,  6,   99,  2608,  90800,  3835458,  187727106, ...];
%e A360336 n=1: [1, 2, 13,  210,  5450, 188004,  7893613,  384731112, ...];
%e A360336 n=2: [1, 3, 21,  334,  8544, 292017, 12186069,  591418401, ...];
%e A360336 n=3: [1, 4, 30,  472, 11909, 403268, 16725042,  808213780, ...];
%e A360336 n=4: [1, 5, 40,  625, 15565, 522211, 21523390, 1035561335, ...];
%e A360336 n=5: [1, 6, 51,  794, 19533, 649326, 26594644, 1273925322, ...];
%e A360336 n=6: [1, 7, 63,  980, 23835, 785120, 31953040, 1523791095, ...];
%e A360336 n=7: [1, 8, 76, 1184, 28494, 930128, 37613552, 1785666072, ...]; ...
%e A360336 Compare to the table of coefficients in (1 + x*A(x)^(3*n))^(n+1):
%e A360336 n=0: [1, 1,   0,    0,     0,       0,        0,          0, ...];
%e A360336 n=1: [1, 2,   7,   48,   719,   17882,   603567,   25021464, ...];
%e A360336 n=2: [1, 3,  21,  190,  2814,   65460,  2105997,   84726534, ...];
%e A360336 n=3: [1, 4,  42,  472,  7303,  162828,  4982706,  193437168, ...];
%e A360336 n=4: [1, 5,  70,  940, 15565,  341796, 10002300,  373126910, ...];
%e A360336 n=5: [1, 6, 105, 1640, 29340,  649326, 18377374,  658075230, ...];
%e A360336 n=6: [1, 7, 147, 2618, 50729, 1150968, 31953040, 1101647800, ...];
%e A360336 n=7: [1, 8, 196, 3920, 82194, 1934296, 53433184, 1785666072, ...]; ...
%e A360336 to see that the main diagonals of the tables are the same.
%o A360336 (PARI) {a(n) = my(A=[1]); for(m=1, n, A=concat(A, 0); A[m+1] = (Vec((1+x*Ser(A)^(3*m))^(m+1))[m+1] - Vec(Ser(A)^(m+1))[m+1])/(m+1) ); A[n+1]}
%o A360336 for(n=0, 20, print1(a(n), ", "))
%Y A360336 Cf. A360337, A360338, A302702, A360344.
%K A360336 nonn
%O A360336 0,3
%A A360336 _Paul D. Hanna_, Feb 06 2023