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A360583 Expansion of A(x) satisfying [x^n] A(x) / (1 + x*A(x)^(n+1)) = 0 for n > 0.

%I A360583 #9 Mar 13 2023 04:17:16
%S A360583 1,1,3,17,139,1455,18326,267700,4426686,81455357,1646941293,
%T A360583 36238989035,861298646217,21978627323651,599195351716464,
%U A360583 17379759869328515,534392854606942358,17363705283005593096,594513962872698955686,21395116871674867310280,807416275879430472577570
%N A360583 Expansion of A(x) satisfying [x^n] A(x) / (1 + x*A(x)^(n+1)) = 0 for n > 0.
%H A360583 Paul D. Hanna, <a href="/A360583/b360583.txt">Table of n, a(n) for n = 0..300</a>
%F A360583 a(n) ~ c * n! * n^(3*LambertW(1) - 1 + 1/(1 + LambertW(1))) / LambertW(1)^n, where c = 0.0685246538334926708088... - _Vaclav Kotesovec_, Mar 13 2023
%e A360583 G.f.: A(x) = 1 + x + 3*x^2 + 17*x^3 + 139*x^4 + 1455*x^5 + 18326*x^6 + 267700*x^7 + 4426686*x^8 + 81455357*x^9 + 1646941293*x^10 + ...
%e A360583 The table of coefficients in the successive powers of g.f. A(x) begins:
%e A360583 n = 1: [1, 1,  3,  17,  139,  1455,  18326, ...];
%e A360583 n = 2: [1, 2,  7,  40,  321,  3290,  40685, ...];
%e A360583 n = 3: [1, 3, 12,  70,  555,  5583,  67827, ...];
%e A360583 n = 4: [1, 4, 18, 108,  851,  8424, 100624, ...];
%e A360583 n = 5: [1, 5, 25, 155, 1220, 11916, 140085, ...];
%e A360583 n = 6: [1, 6, 33, 212, 1674, 16176, 187372, ...];
%e A360583 n = 7: [1, 7, 42, 280, 2226, 21336, 243817, ...];
%e A360583 ...
%e A360583 The table of coefficients in A(x)/(1 + x*A(x)^(n+1)) begins:
%e A360583 n = 1: [1, 0,  1,   9,   88, 1021, 13736, 209940, ...];
%e A360583 n = 2: [1, 0,  0,   5,   64,  821, 11670, 184622, ...];
%e A360583 n = 3: [1, 0, -1,   0,   35,  587,  9283, 155666, ...];
%e A360583 n = 4: [1, 0, -2,  -6,    0,  315,  6555, 122855, ...];
%e A360583 n = 5: [1, 0, -3, -13,  -42,    0,  3467,  86025, ...];
%e A360583 n = 6: [1, 0, -4, -21,  -92, -364,     0,  45079, ...];
%e A360583 n = 7: [1, 0, -5, -30, -151, -784, -3866,      0, ...];
%e A360583 ...
%e A360583 in which the diagonal of all zeros illustrates that
%e A360583 [x^n] A(x) / (1 + x*A(x)^(n+1)) = 0 for n > 0.
%o A360583 (PARI) {a(n) = my(A=[1]); for(i=1,n, A = concat(A,0);
%o A360583 A[#A] = -polcoeff( Ser(A)/(1 + x*Ser(A)^(#A)), #A-1) );A[n+1]}
%o A360583 for(n=0,30,print1(a(n),", "))
%Y A360583 Cf. A360582, A360584.
%K A360583 nonn
%O A360583 0,3
%A A360583 _Paul D. Hanna_, Mar 12 2023