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A303063 G.f. A(x) satisfies: [x^(n-1)] (1 + x*A(x)^n)^n / A(x)^n = 0 for n>1.

%I A303063 #20 Sep 13 2023 10:48:08
%S A303063 1,1,3,17,151,1812,26766,461302,8978490,193200156,4529641423,
%T A303063 114510000515,3097375627215,89116723381943,2714808312021989,
%U A303063 87242980758842543,2948618278635037930,104544558380661516685,3880035778583841094470,150451784852703095162304,6084892588256393044757197,256294338370540915598727500
%N A303063 G.f. A(x) satisfies: [x^(n-1)] (1 + x*A(x)^n)^n / A(x)^n = 0 for n>1.
%H A303063 Paul D. Hanna, <a href="/A303063/b303063.txt">Table of n, a(n) for n = 0..300</a>
%F A303063 a(n) ~ c * n! * n^(3*LambertW(1) + 1/(1 + LambertW(1))) / LambertW(1)^n, where c = 0.03203091421745281863810572012... - _Vaclav Kotesovec_, Aug 11 2021
%e A303063 G.f.: A(x) = 1 + x + 3*x^2 + 17*x^3 + 151*x^4 + 1812*x^5 + 26766*x^6 + 461302*x^7 + 8978490*x^8 + 193200156*x^9 + 4529641423*x^10 + ...
%e A303063 ILLUSTRATION OF DEFINITION.
%e A303063 The table of coefficients in (1 + x*A(x)^n)^n / A(x)^n begins:
%e A303063 n=1: [1, 0, -2, -12, -116, -1475, -22625, -400078, ...];
%e A303063 n=2: [1, 0, -2, -18, -197, -2630, -41347, -742194, ...];
%e A303063 n=3: [1, 0, 0, -15, -228, -3390, -55716, -1022901, ...];
%e A303063 n=4: [1, 0, 4, 0, -178, -3536, -64144, -1228756, ...];
%e A303063 n=5: [1, 0, 10, 30, 0, -2640, -63025, -1327450, ...];
%e A303063 n=6: [1, 0, 18, 78, 369, 0, -45519, -1252758, ...];
%e A303063 n=7: [1, 0, 28, 147, 1008, 5425, 0, -881412, ...];
%e A303063 n=8: [1, 0, 40, 240, 2012, 15080, 91832, 0, ...]; ...
%e A303063 in which the main diagonal equals all zeros after the initial term, illustrating that [x^(n-1)] (1 + x*A(x)^n)^n / A(x)^n = 0 for n>1.
%o A303063 (PARI) {a(n) = my(A=[1]); for(m=1,n+1, A=concat(A,0); A[m] = Vec( (1 + x*Ser(A)^m)^m/Ser(A)^m )[m]/m ); A[n+1]}
%o A303063 for(n=0,30, print1(a(n),", "))
%Y A303063 Cf. A303062, A360578, A360581.
%K A303063 nonn
%O A303063 0,3
%A A303063 _Paul D. Hanna_, Apr 17 2018