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%I A266807 #4 Jan 10 2016 18:13:37
%S A266807 2,-90,-166,-2166,-12010,-89598,-594910,-4127706,-28160326,-193357590,
%T A266807 -1324392298,-9079876830,-62228230846,-426534794586,-2923470679270,
%U A266807 -20037876860598,-137341361295850,-941352453457086,-6452123715212446,-44223519044857050
%N A266807 Coefficient of x^3 in the minimal polynomial of the continued fraction [1^n,sqrt(6),1,1,...], where 1^n means n ones.
%C A266807 See A265762 for a guide to related sequences.
%H A266807 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,15,-15,-5,1).
%F A266807 a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5) .
%F A266807 G.f.:  (2 (-1 + 50 x - 127 x^2 - 22 x^3 + 15 x^4))/(-1 + 5 x + 15 x^2 - 15 x^3 - 5 x^4 + x^5).
%e A266807 Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
%e A266807 [sqrt(6),1,1,1,...] has p(0,x)=19-14x-13x^2+2x^3+x^4, so a(0) = 2;
%e A266807 [1,sqrt(6),1,1,1,...] has p(1,x)=19-90x+143x^2-90x^3+19x^4, so a(1) = -90;
%e A266807 [1,1,sqrt(6),1,1,1...] has p(2,x)=361-722x+527x^2-166x^3+19x^4, so a(2) = -166. ~
%t A266807 u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[6]}, {{1}}];
%t A266807 f[n_] := FromContinuedFraction[t[n]];
%t A266807 t = Table[MinimalPolynomial[f[n], x], {n, 0, 40}];
%t A266807 Coefficient[t, x, 0] ; (* A266804 *)
%t A266807 Coefficient[t, x, 1];  (* A266805 *)
%t A266807 Coefficient[t, x, 2];  (* A266806 *)
%t A266807 Coefficient[t, x, 3];  (* A266807 *)
%t A266807 Coefficient[t, x, 4];  (* A266804 *)
%Y A266807 Cf. A265762, A266804, A266805, A266806.
%K A266807 sign,easy
%O A266807 0,1
%A A266807 _Clark Kimberling_, Jan 10 2016

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