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%I #20 Mar 31 2020 17:23:23
%S 1,2,9,78,1045,19320,458304,13306902,457649757,18202765482,
%T 822272600160,41592018711672,2329051560965532,143045976577538632,
%U 9561491720518777632,690994864767311671302,53688078414653072521485,4462898094035056790939070
%N a(n) equals the coefficient of x^(2n-1) in the n-th iteration of x+x^3 for n>=1.
%F Conjecture: a(m) = 0 (mod 3) everywhere except at m = (3^n+1)/2, n>=0.
%e The coefficients of x^(2k-1), k>=1, in the n-th iterations of x+x^3 begin:
%e n=1: [(1), 1, 0, 0, 0, 0, 0, 0, ...];
%e n=2: [1,(2), 3, 3, 1, 0, 0, 0, ...];
%e n=3: [1, 3,(9), 24, 54, 102, 156, 192, ...];
%e n=4: [1, 4, 18,(78), 315, 1182, 4107, 13215, ...];
%e n=5: [1, 5, 30, 180,(1045), 5835, 31269, 160824, ...];
%e n=6: [1, 6, 45, 345, 2610,(19320), 139524, 982356, ...];
%e n=7: [1, 7, 63, 588, 5481, 50505,(458304), 4090128, ...];
%e n=8: [1, 8, 84, 924, 10234, 112812, 1232070,(13306902), ...]; ...;
%e coefficients in parenthesis form the initial terms of this sequence.
%e The nonzero terms (mod 3) begin:
%e a(1)=1, a(2)=2, a(5)=2, a(14)=1, a(41)=2, a(122)=1, ...
%o (PARI) {a(n)=local(A=x,G=x+x^3); for(i=1,n, A=subst(G, x, A+x*O(x^(2*n)))); polcoeff(A, 2*n-1)}
%Y Cf. A184900.
%K nonn
%O 1,2
%A _Paul D. Hanna_, Feb 01 2011
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