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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 #7 Feb 06 2023 04:04:47

%S 1,1,2,20,316,6686,173379,5255624,180911070,6938866748,292678301988,

%T 13446616806957,668017569348751,35678261176871802,2038906890461704040,

%U 124171127134721710130,8030684434410398312840,549848454475826567644385,39744302449387229743134043

%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="/A360342/b360342.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 = 0.0798993252137..., c = 0.118957192149397... - _Vaclav Kotesovec_, Feb 06 2023

%e G.f.: A(x) = 1 + x + 2*x^2 + 20*x^3 + 316*x^4 + 6686*x^5 + 173379*x^6 + 5255624*x^7 + 180911070*x^8 + 6938866748*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 + 3*x^2 + 27*x^3 + 417*x^4 + 8727*x^5 + 225018*x^6 + 6800714*x^7 + 233778499*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, 9/3, 108/4, 2085/5, 52362/6, 1575126/7, 54405712/8, ...].

%e ILLUSTRATION OF DEFINITION.

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

%e n=0: [1, 1, 2, 20, 316, 6686, 173379, 5255624, ...];

%e n=1: [1, 2, 5, 44, 676, 14084, 361794, 10897390, ...];

%e n=2: [1, 3, 9, 73, 1086, 22266, 566441, 16950588, ...];

%e n=3: [1, 4, 14, 108, 1553, 31312, 788620, 23442284, ...];

%e n=4: [1, 5, 20, 150, 2085, 41311, 1029745, 30401460, ...];

%e n=5: [1, 6, 27, 200, 2691, 52362, 1291355, 37859166, ...];

%e n=6: [1, 7, 35, 259, 3381, 64575, 1575126, 45848685, ...];

%e n=7: [1, 8, 44, 328, 4166, 78072, 1882884, 54405712, ...]; ...

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

%e n=0: [1, 1, -2, -1, -32, -519, -11490, -305967, ...];

%e n=1: [1, 2, 1, 0, 0, 0, 0, 0, ...];

%e n=2: [1, 3, 9, 28, 180, 2379, 47111, 1182009, ...];

%e n=3: [1, 4, 22, 108, 745, 8556, 153292, 3658316, ...];

%e n=4: [1, 5, 40, 265, 2085, 22706, 366450, 8157230, ...];

%e n=5: [1, 6, 63, 524, 4743, 52362, 781973, 16041192, ...];

%e n=6: [1, 7, 91, 910, 9415, 109536, 1575126, 29886445, ...];

%e n=7: [1, 8, 124, 1448, 16950, 211840, 3042820, 54405712, ...]; ...

%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. A360343, A360344, A360345, A360346, A360347.

%Y Cf. A360231, A302702.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Feb 05 2023