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%I #11 Sep 30 2016 09:49:42
%S 9,11,1,29,45,149,359,971,2511,6605,17261,45221,118359,309899,811295,
%T 2124029,5560749,14558261,38113991,99783755,261237231,683927981,
%U 1790546669,4687712069,12272589495,32130056459,84117579839,220222683101,576550469421,1509428725205
%N 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 See A265762 for a guide to related sequences.
%H Colin Barker, <a href="/A266703/b266703.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-1).
%F a(n) = 2*a(n-1) - 2*a(n-2) + a(n-3).
%F G.f.: (9 - 7 x - 39 x^2 + 14 x^3 - 4 x^4 + 2 x^5)/(1 - 2 x - 2 x^2 + x^3).
%F a(n) = 2^(-n)*(-43*(-2)^n+(3+sqrt(5))^n*(-1+3*sqrt(5))-(3-sqrt(5))^n*(1+3*sqrt(5)))/5 for n>2. - _Colin Barker_, Sep 29 2016
%e Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
%e [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 [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 [1,1,2/3,1,...] = (-17+9*sqrt(5))/2 has p(2,x) = -29 + 17 x + x^2, so a(2) = 1.
%t u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {2/3}, {{1}}];
%t f[n_] := FromContinuedFraction[t[n]];
%t t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
%t Coefficient[t, x, 0] (* A266703 *)
%t Coefficient[t, x, 1] (* A266704 *)
%t Coefficient[t, x, 2] (* A266703 *)
%o (PARI) Vec((9-7*x-39*x^2+14*x^3-4*x^4+2*x^5)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ _Colin Barker_, Sep 29 2016
%Y Cf. A265762, A266704.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, Jan 09 2016
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