%I #10 Nov 16 2023 12:54:20
%S 1,1,4,37,532,9994,226252,5910445,173581060,5634589906,199792389160,
%T 7671942375898,316936631324368,14011781050744984,660054967923455212,
%U 33008607551445324157,1746771084107236755604,97536010045722766992778,5731874036042145864368824
%N G.f. A(x) satisfies: A(x) = 1 + x*A(x)^2 / (A(x) - 3*x*A'(x)).
%H Vaclav Kotesovec, <a href="/A352237/b352237.txt">Table of n, a(n) for n = 0..378</a>
%F G.f. A(x) satisfies:
%F (1) [x^n] A(x)^(3*n+1) = [x^(n-1)] (3*n+1) * A(x)^(3*n+1) for n >= 1.
%F (2) A(x) = 1 + x*A(x)^2/(A(x) - 3*x*A'(x)).
%F (3) A'(x) = A(x) * (1 + x*A(x)/(1 - A(x))) / (3*x).
%F (4) A(x) = exp( Integral (1 + x*A(x)/(1 - A(x)))/(3*x) dx ).
%F a(n) ~ c * 3^n * n! * n^(4/3), where c = 0.0543186200722307001992331... - _Vaclav Kotesovec_, Nov 16 2023
%e G.f.: A(x) = 1 + x + 4*x^2 + 37*x^3 + 532*x^4 + 9994*x^5 + 226252*x^6 + 5910445*x^7 + 173581060*x^8 + ...
%e such that A(x) = 1 + x*A(x)^2/(A(x) - 3*x*A'(x)).
%e Related table.
%e The table of coefficients of x^k in A(x)^(3*n+1) begins:
%e n=0: [1, 1, 4, 37, 532, 9994, 226252, ...];
%e n=1: [1, 4, 22, 200, 2717, 48788, 1069122, ...];
%e n=2: [1, 7, 49, 462, 6069, 104664, 2219784, ...];
%e n=3: [1, 10, 85, 850, 11020, 183832, 3777355, ...];
%e n=4: [1, 13, 130, 1391, 18083, 294203, 5869734, ...];
%e n=5: [1, 16, 184, 2112, 27852, 445632, 8659920, ...];
%e n=6: [1, 19, 247, 3040, 41002, 650161, 12353059, ...]; ...
%e in which the following pattern holds:
%e [x^n] A(x)^(3*n+1) = [x^(n-1)] (3*n+1) * A(x)^(3*n+1), n >= 1,
%e as illustrated by
%e [x^1] A(x)^4 = 4 = [x^0] 4*A(x)^4 = 4*1;
%e [x^2] A(x)^7 = 49 = [x^1] 7*A(x)^7 = 7*7;
%e [x^3] A(x)^10 = 850 = [x^2] 10*A(x)^10 = 10*85;
%e [x^4] A(x)^13 = 18083 = [x^3] 13*A(x)^13 = 13*1391;
%e [x^5] A(x)^16 = 445632 = [x^4] 16*A(x)^16 = 16*27852;
%e [x^6] A(x)^19 = 12353059 = [x^5] 19*A(x)^19 = 19*650161; ...
%e Also, compare the above terms along the diagonal to the series
%e B(x) = A(x*B(x)^3) = 1 + x + 7*x^2 + 85*x^3 + 1391*x^4 + 27852*x^5 + 650161*x^6 + 17204220*x^7 + ...
%e where B(x)^3 = (1/x) * Series_Reversion( x/A(x)^3 ).
%o (PARI) /* Using A(x) = 1 + x*A(x)^2/(A(x) - 3*x*A'(x)) */
%o {a(n) = my(A=1); for(i=1,n, A = 1 + x*A^2/(A - 3*x*A' + x*O(x^n)) );
%o polcoeff(A,n)}
%o for(n=0,20, print1(a(n),", "))
%o (PARI) /* Using [x^n] A(x)^(3*n+1) = [x^(n-1)] (3*n+1)*A(x)^(3*n+1) */
%o {a(n) = my(A=[1]); for(i=1,n, A=concat(A,0);
%o A[#A] = polcoeff((x*Ser(A)^(3*(#A)-2) - Ser(A)^(3*(#A)-2)/(3*(#A)-2)),#A-1));A[n+1]}
%o for(n=0,20, print1(a(n),", "))
%Y Cf. A088715, A286797, A317352, A352235, A352236, A352238.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Mar 08 2022