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A303062 G.f. A(x) satisfies: [x^(n-1)] (1 + x*A(x)^(n-1))^n / A(x)^n = 0 for n>1.

%I A303062 #19 Aug 11 2021 06:41:56
%S A303062 1,1,2,8,60,643,8564,133890,2376261,46832442,1009739331,23564025488,
%T A303062 590503218735,15793704933899,448695132962248,13487808514722460,
%U A303062 427624923581550100,14260707306806609885,499071020445057149835,18290961984686434723480,700757535935308305865473,28017787701624063252219677
%N A303062 G.f. A(x) satisfies: [x^(n-1)] (1 + x*A(x)^(n-1))^n / A(x)^n = 0 for n>1.
%H A303062 Paul D. Hanna, <a href="/A303062/b303062.txt">Table of n, a(n) for n = 0..300</a>
%F A303062 a(n) ~ c * n! * n^(2*LambertW(1)) / LambertW(1)^n, where c = 0.153879081661359639962985708... - _Vaclav Kotesovec_, Aug 11 2021
%e A303062 G.f.: A(x) = 1 + x + 2*x^2 + 8*x^3 + 60*x^4 + 643*x^5 + 8564*x^6 + 133890*x^7 + 2376261*x^8 + 46832442*x^9 + 1009739331*x^10 + ...
%e A303062 ILLUSTRATION OF DEFINITION.
%e A303062 The table of coefficients in (1 + x*A(x)^(n-1))^n / A(x)^n begins:
%e A303062 n=1: [1, 0, -2, -6, -50, -565, -7731, -122983, ...];
%e A303062 n=2: [1, 0, -2, -10, -89, -1030, -14307, -230054, ...];
%e A303062 n=3: [1, 0, 0, -9, -111, -1380, -19677, -320958, ...];
%e A303062 n=4: [1, 0, 4, 0, -94, -1520, -23388, -392776, ...];
%e A303062 n=5: [1, 0, 10, 20, 0, -1210, -24030, -436250, ...];
%e A303062 n=6: [1, 0, 18, 54, 225, 0, -18345, -427944, ...];
%e A303062 n=7: [1, 0, 28, 105, 651, 2835, 0, -316344, ...];
%e A303062 n=8: [1, 0, 40, 176, 1364, 8360, 41976, 0, ...]; ...
%e A303062 in which the main diagonal equals all zeros after the initial term, illustrating that [x^(n-1)] (1 + x*A(x)^(n-1))^n / A(x)^n = 0 for n>1.
%o A303062 (PARI) {a(n) = my(A=[1]); for(m=1,n+1, A=concat(A,0); A[m] = Vec( (1 + x*Ser(A)^(m-1))^m/Ser(A)^m )[m]/m ); A[n+1]}
%o A303062 for(n=0,30, print1(a(n),", "))
%Y A303062 Cf. A303063, A302702.
%K A303062 nonn
%O A303062 0,3
%A A303062 _Paul D. Hanna_, Apr 17 2018