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A266707 Coefficient of x^2 in minimal polynomial of the continued fraction [1^n,tau,1,1,1,...], where 1^n means n ones and tau = golden ratio = (1 + sqrt(5))/2.

%I A266707 #13 Sep 30 2016 04:15:15
%S A266707 1,5,4,19,41,116,295,781,2036,5339,13969,36580,95759,250709,656356,
%T A266707 1718371,4498745,11777876,30834871,80726749,211345364,553309355,
%U A266707 1448582689,3792438724,9928733471,25993761701,68052551620,178163893171,466439127881,1221153490484
%N A266707 Coefficient of x^2 in minimal polynomial of the continued fraction [1^n,tau,1,1,1,...], where 1^n means n ones and tau = golden ratio = (1 + sqrt(5))/2.
%C A266707 See A265762 for a guide to related sequences.
%H A266707 Colin Barker, <a href="/A266707/b266707.txt">Table of n, a(n) for n = 0..1000</a>
%H A266707 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-1).
%F A266707 a(n) = 2*a(n-1) - 2*a(n-2) + a(n-3).
%F A266707 G.f.:  (1 + 3 x - 8 x^2 + 2 x^3)/(1 - 2 x - 2 x^2 + x^3).
%F A266707 a(n) = (2^(-1-n)*(-3*(-1)^n*2^(3+n)-(3-sqrt(5))^n*(-7+sqrt(5))+(3+sqrt(5))^n*(7+sqrt(5))))/5 for n>0. - _Colin Barker_, Sep 29 2016
%e A266707 Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
%e A266707 [tau,1,1,1,1,...] = sqrt(5) has p(0,x) = -5 + x^2, so a(0) = 1;
%e A266707 [1,tau,1,1,1,...] = (5 + sqrt(5))/5 has p(1,x) = 4 - 10 x + 5 x^2, so a(1) = 5;
%e A266707 [1,1,tau,1,1,...] = (9 - sqrt(5))/4 has p(2,x) = 19 - 18 x + 4 x^2, so a(2) = 4.
%t A266707 u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {GoldenRatio}, {{1}}];
%t A266707 f[n_] := FromContinuedFraction[t[n]];
%t A266707 t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
%t A266707 Coefficient[t, x, 0] (* A266707 *)
%t A266707 Coefficient[t, x, 1] (* A266708 *)
%t A266707 Coefficient[t, x, 2] (* A266707 *)
%o A266707 (PARI) Vec((1+3*x-8*x^2+2*x^3)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ _Colin Barker_, Sep 29 2016
%Y A266707 Cf. A265762, A266708.
%K A266707 nonn,easy
%O A266707 0,2
%A A266707 _Clark Kimberling_, Jan 09 2016