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G.f. A(x) satisfies: [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(n+3))^(n+1) for n >= 0.
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%I #21 Feb 06 2023 11:28:54

%S 1,1,5,48,673,12057,256763,6232909,168035350,4945380012,157008686993,

%T 5331606427775,192417007138176,7344652874314128,295384546093569838,

%U 12478509340848604628,552330553975194126634,25560514938260757190962,1234444956694450007259989,62114842767595821207341042

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

%H Paul D. Hanna, <a href="/A360235/b360235.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)^(n+3))^(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)^(n+3))^(n+1) / (n+1).

%F a(n) ~ c * d^n * n! * n^alfa, where d = A360279 = 2.12460658362428979..., alfa = 4.09132043601400805425594207544980..., c = 0.00160512950354606176706886534963706... - _Vaclav Kotesovec_, Jan 31 2023

%e G.f.: A(x) = 1 + x + 5*x^2 + 48*x^3 + 673*x^4 + 12057*x^5 + 256763*x^6 + 6232909*x^7 + 168035350*x^8 + 4945380012*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 + 6*x^2 + 64*x^3 + 946*x^4 + 17403*x^5 + 375913*x^6 + 9203150*x^7 + 249561291*x^8 + ... + b(n)*x^n + ...

%e such that b(n) = [x^n] (1 + x*A(x)^(n+3))^(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, 18/3, 256/4, 4730/5, 104418/6, 2631391/7, 73625200/8, ...].

%e ILLUSTRATION OF DEFINITION.

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

%e n=0: [1, 1, 5, 48, 673, 12057, 256763, 6232909, ...];

%e n=1: [1, 2, 11, 106, 1467, 25940, 546674, 13164522, ...];

%e n=2: [1, 3, 18, 175, 2397, 41868, 873317, 20861712, ...];

%e n=3: [1, 4, 26, 256, 3479, 60080, 1240618, 29397424, ...];

%e n=4: [1, 5, 35, 350, 4730, 80836, 1652870, 38851165, ...];

%e n=5: [1, 6, 45, 458, 6168, 104418, 2114759, 49309524, ...];

%e n=6: [1, 7, 56, 581, 7812, 131131, 2631391, 60866723, ...];

%e n=7: [1, 8, 68, 720, 9682, 161304, 3208320, 73625200, ...]; ...

%e Compare to the table of coefficients in (1 + x*A(x)^(n+3))^(n+1):

%e n=0: [1, 1, 3, 18, 175, 2397, 41868, 873317, ...];

%e n=1: [1, 2, 9, 60, 580, 7678, 129842, 2642540, ...];

%e n=2: [1, 3, 18, 136, 1350, 17520, 287288, 5690016, ...];

%e n=3: [1, 4, 30, 256, 2661, 34404, 550050, 10593112, ...];

%e n=4: [1, 5, 45, 430, 4730, 61811, 971600, 18221525, ...];

%e n=5: [1, 6, 63, 668, 7815, 104418, 1629245, 29869968, ...];

%e n=6: [1, 7, 84, 980, 12215, 168294, 2631391, 47432554, ...];

%e n=7: [1, 8, 108, 1376, 18270, 261096, 4125864, 73625200, ...]; ...

%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)^(m+3))^(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. A360231, A302702, A302703, A360234, A360236, A360237.

%Y Cf. A360347, A360338.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Jan 30 2023