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A266802 Coefficient of x^3 in the minimal polynomial of the continued fraction [1^n,sqrt(3),1,1,...], where 1^n means n ones.

%I A266802 #4 Jan 09 2016 19:59:11
%S A266802 2,-12,-16,-294,-1552,-11868,-78142,-543996,-3706624,-25463142,
%T A266802 -174376288,-1195587372,-8193644926,-56162781804,-384938354032,
%U A266802 -2638425262758,-18083987259952,-123949619666556,-849562999302334,-5822992294650972,-39911380656754528
%N A266802 Coefficient of x^3 in the minimal polynomial of the continued fraction [1^n,sqrt(3),1,1,...], where 1^n means n ones.
%C A266802 See A265762 for a guide to related sequences.
%H A266802 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,15,-15,-5,1).
%F A266802 a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5) .
%F A266802 G.f.:  (2 (-1 + 11 x - 7 x^2 + 2 x^3 + 6 x^4))/(-1 + 5 x + 15 x^2 - 15 x^3 - 5 x^4 + x^5).
%e A266802 Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
%e A266802 [sqrt(3),1,1,1,...] has p(0,x) = 1 - 8 x - 7 x^2 + 2 x^3 + x^4, so a(0) = 2;
%e A266802 [1,sqrt(3),1,1,1,...] has p(1,x) = 1 - 12 x + 23 x^2 - 12 x^3 + x^4, so a(1) = -12;
%e A266802 [1,1,sqrt(3),1,1,1...] has p(2,x) = 49 - 98 x + 65 x^2 - 16 x^3 + x^4, so a(2) = -16.
%t A266802 u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[3]}, {{1}}];
%t A266802 f[n_] := FromContinuedFraction[t[n]];
%t A266802 t = Table[MinimalPolynomial[f[n], x], {n, 0, 40}];
%t A266802 Coefficient[t, x, 0] ; (* A266799 *)
%t A266802 Coefficient[t, x, 1];  (* A266800 *)
%t A266802 Coefficient[t, x, 2];  (* A266801 *)
%t A266802 Coefficient[t, x, 3];  (* A266802 *)
%t A266802 Coefficient[t, x, 4];  (* A266799 *)
%Y A266802 Cf. A265762, A266299, A266800, A266801.
%K A266802 sign,easy
%O A266802 0,1
%A A266802 _Clark Kimberling_, Jan 09 2016