%I #10 Mar 11 2018 20:09:56
%S 1,1,6,117,4108,212715,14836482,1325939874,147020077944,
%T 19756725674418,3159029077103310,592025067793406643,
%U 128397048202595732724,31876577262066747598778,8974199586989454813287292,2841516158252445942521650875,1004586033935983529849353475184,394020228153801869013062325976323,170474484645857230477080154611084378,80943600739603137192744457809054719388
%N G.f. A(x) satisfies: [x^n] A( x/A(x)^(3*n) ) = 0 for n>=1.
%C Conjecture: n divides a(n) for n>=1.
%H Paul D. Hanna, <a href="/A300733/b300733.txt">Table of n, a(n) for n = 0..300</a>
%e G.f.: A(x) = 1 + x + 6*x^2 + 117*x^3 + 4108*x^4 + 212715*x^5 + 14836482*x^6 + 1325939874*x^7 + 147020077944*x^8 + 19756725674418*x^9 +...
%e The table of coefficients in A( x/A(x)^(3*n) ) begins:
%e n=0: [1, 1, 6, 117, 4108, 212715, 14836482, 1325939874, ...];
%e n=1: [1, 1, 3, 69, 2676, 148461, 10887243, 1010333013, ...];
%e n=2: [1, 1, 0, 30, 1532, 96642, 7658442, 748806253, ...];
%e n=3: [1, 1, -3, 0, 649, 55638, 5045967, 533295288, ...];
%e n=4: [1, 1, -6, -21, 0, 23910, 2954778, 356620197, ...];
%e n=5: [1, 1, -9, -33, -442, 0, 1298664, 212433928, ...];
%e n=6: [1, 1, -12, -36, -704, -17469, 0, 95171511, ...];
%e n=7: [1, 1, -15, -30, -813, -29793, -1010496, 0, ...];
%e n=8: [1, 1, -18, -15, -796, -38187, -1794006, -77230856, 0, ...]; ...
%e such that the main diagonal consists of all zeros after the initial terms.
%e The terms a(n)/n for n>=1 begin:
%e [1, 3, 39, 1027, 42543, 2472747, 189419982, 18377509743, 2195191741602, ...].
%o (PARI) {a(n) = my(A=[1, 1]); for(i=2, n, A=concat(A, 0); A[#A] = -Vec(subst(Ser(A), x, x/Ser(A)^(3*(#A-1))))[#A]); A[n+1]}
%o for(n=0, 20, print1(a(n), ", "))
%Y Cf. A266489, A300732.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Mar 11 2018