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A266704 Coefficient of x^2 in minimal polynomial of the continued fraction [1^n,2/3,1,1,1,...], where 1^n means n ones.

%I A266704 #12 Sep 29 2016 14:56:36
%S A266704 -3,-19,17,-75,-165,-463,-1181,-3123,-8145,-21355,-55877,-146319,
%T A266704 -383037,-1002835,-2625425,-6873483,-17994981,-47111503,-123339485,
%U A266704 -322906995,-845381457,-2213237419,-5794330757,-15169754895,-39714933885,-103975046803,-272210206481
%N A266704 Coefficient of x^2 in minimal polynomial of the continued fraction [1^n,2/3,1,1,1,...], where 1^n means n ones.
%C A266704 See A265762 for a guide to related sequences.
%H A266704 Colin Barker, <a href="/A266704/b266704.txt">Table of n, a(n) for n = 0..1000</a>
%H A266704 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-1).
%F A266704 a(n) = 2*a(n-1) - 2*a(n-2) + a(n-3).
%F A266704 G.f.:  (-3 - 13 x + 61 x^2 - 74 x^3 - 68 x^4 + 34 x^5)/(1 - 2 x - 2 x^2 + x^3).
%F A266704 a(n) = 2^(-n)*(43*(-2)^n+2*(3-sqrt(5))^n*(-7+sqrt(5))-2*(3+sqrt(5))^n*(7+sqrt(5)))/5 for n>2. - _Colin Barker_, Sep 29 2016
%e A266704 Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
%e A266704 [2/3,1,1,1,...] = (1+3*sqrt(5))/6 has p(0,x) = -11 - 3 x + 9 x^2, so a(0) = 9;
%e A266704 [1,2/3,1,1,...] = (19+9*sqrt(5))/22 has p(1,x) = -1 - 19 x + 11 x^2, so a(1) = 11;
%e A266704 [1,1,2/3,1,...] = (-17+9*sqrt(5))/2 has p(2,x) = -29 + 17 x + x^2, so a(2) = 1.
%t A266704 u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {2/3}, {{1}}];
%t A266704 f[n_] := FromContinuedFraction[t[n]];
%t A266704 t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
%t A266704 Coefficient[t, x, 0] (* A266703 *)
%t A266704 Coefficient[t, x, 1] (* A266704 *)
%t A266704 Coefficient[t, x, 2] (* A266703 *)
%o A266704 (PARI) Vec(-(3+13*x-61*x^2+74*x^3+68*x^4-34*x^5)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ _Colin Barker_, Sep 29 2016
%Y A266704 Cf. A265762, A266703.
%K A266704 sign,easy
%O A266704 0,1
%A A266704 _Clark Kimberling_, Jan 09 2016