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G.f. A(x) satisfies: [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(2*n+2))^(n+1) for n >= 0.
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%I #8 Feb 06 2023 04:19:12

%S 1,1,6,82,1724,47223,1555047,58892186,2496826094,116434989450,

%T 5900151126856,322048641354617,18810964989814291,1169843128503194025,

%U 77145176721564799777,5376524285402806746719,394887654026596322701724,30489608056346314234108286,2469347798211941105406473481

%N G.f. A(x) satisfies: [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(2*n+2))^(n+1) for n >= 0.

%H Paul D. Hanna, <a href="/A360346/b360346.txt">Table of n, a(n) for n = 0..300</a>

%F G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies:

%F (1) [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(2*n+2))^(n+1) for n>=0.

%F (2) A(x) = Sum_{n>=0} b(n) * x^n/A(x)^n, where b(n) = [x^n] (1 + x*A(x)^(2*n+2))^(n+1) / (n+1).

%F a(n) ~ c * d^n * n! * n^alpha, where d = 3.93464558322824528799..., alpha = 2.153902930660..., c = 0.01676305987174... - _Vaclav Kotesovec_, Feb 06 2023

%e G.f.: A(x) = 1 + x + 6*x^2 + 82*x^3 + 1724*x^4 + 47223*x^5 + 1555047*x^6 + 58892186*x^7 + 2496826094*x^8 + 116434989450*x^9 + ...

%e RELATED SERIES.

%e G.f. A(x) = B(x/A(x)) where B(x) = B(x*A(x)) begins:

%e B(x) = 1 + x + 7*x^2 + 101*x^3 + 2161*x^4 + 59544*x^5 + 1965132*x^6 + 74504861*x^7 + 3161424763*x^8 + ... + b(n)*x^n + ...

%e such that b(n) = [x^n] (1 + x*A(x)^(2*n+2))^(n+1) / (n+1),

%e as well as b(n) = [x^n] A(x)^(n+1) / (n+1),

%e so that b(n) begin:

%e [1/1, 2/2, 21/3, 404/4, 10805/5, 357264/6, 13755924/7, 596038888/8, ...].

%e ILLUSTRATION OF DEFINITION.

%e The table of coefficients of x^k in A(x)^(n+1) begins:

%e n=0: [1, 1, 6, 82, 1724, 47223, 1555047, 58892186, ...];

%e n=1: [1, 2, 13, 176, 3648, 98878, 3231952, 121743878, ...];

%e n=2: [1, 3, 21, 283, 5790, 155319, 5039055, 188787837, ...];

%e n=3: [1, 4, 30, 404, 8169, 216924, 6985240, 260270488, ...];

%e n=4: [1, 5, 40, 540, 10805, 284096, 9079965, 336452690, ...];

%e n=5: [1, 6, 51, 692, 13719, 357264, 11333293, 417610542, ...];

%e n=6: [1, 7, 63, 861, 16933, 436884, 13755924, 504036226, ...];

%e n=7: [1, 8, 76, 1048, 20470, 523440, 16359228, 596038888, ...]; ...

%e Compare to the table of coefficients in (1 + x*A(x)^(2*n+2))^(n+1):

%e n=0: [1, 1, 2, 13, 176, 3648, 98878, 3231952, ...];

%e n=1: [1, 2, 9, 68, 884, 17386, 454318, 14493920, ...];

%e n=2: [1, 3, 21, 190, 2508, 47406, 1190949, 36928479, ...];

%e n=3: [1, 4, 38, 404, 5585, 103464, 2504568, 75227160, ...];

%e n=4: [1, 5, 60, 735, 10805, 200001, 4698210, 136509465, ...];

%e n=5: [1, 6, 87, 1208, 19011, 357264, 8227591, 231595008, ...];

%e n=6: [1, 7, 119, 1848, 31199, 602427, 13755924, 376756199, ...];

%e n=7: [1, 8, 156, 2680, 48518, 970712, 22218108, 596038888, ...]; ...

%e to see that the main diagonals of the tables are the same.

%o (PARI) {a(n) = my(A=[1]); for(m=1, n, A=concat(A, 0); A[m+1] = (Vec((1+x*Ser(A)^(2*m+2))^(m+1))[m+1] - Vec(Ser(A)^(m+1))[m+1])/(m+1) ); A[n+1]}

%o for(n=0, 20, print1(a(n), ", "))

%Y Cf. A360342, A360343, A360344, A360345, A360347.

%Y Cf. A302703, A360234.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Feb 05 2023