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A266801 Coefficient of x^2 in the minimal polynomial of the continued fraction [1^n,sqrt(3),1,1,...], where 1^n means n ones.

%I A266801 #4 Jan 09 2016 19:59:05
%S A266801 -7,23,65,653,3935,28373,190793,1317335,9003953,61779965,423273503,
%T A266801 2901611813,19886759705,136308977303,934267517345,6403586065133,
%U A266801 43890776239583,300832001287925,2061932830446953,14132698865151575,96866956468010513,663936003630421853
%N A266801 Coefficient of x^2 in the minimal polynomial of the continued fraction [1^n,sqrt(3),1,1,...], where 1^n means n ones.
%C A266801 See A265762 for a guide to related sequences.
%H A266801 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,15,-15,-5,1).
%F A266801 a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5) .
%F A266801 G.f.:  (7 - 58 x - 55 x^2 + 122 x^3 - 5 x^4)/(-1 + 5 x + 15 x^2 - 15 x^3 - 5 x^4 + x^5).
%e A266801 Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
%e A266801 [sqrt(3),1,1,1,...] has p(0,x)=1-8x-7x^2+2x^3+x^4, so a(0) = -7;
%e A266801 [1,sqrt(3),1,1,1,...] has p(1,x)=1-12x+23x^2-12x^3+x^4, so a(1) = 23;
%e A266801 [1,1,sqrt(3),1,1,1...] has p(2,x)=49-98x+65x^2-16x^3+x^4, so a(2) = 65.
%t A266801 u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[3]}, {{1}}];
%t A266801 f[n_] := FromContinuedFraction[t[n]];
%t A266801 t = Table[MinimalPolynomial[f[n], x], {n, 0, 40}];
%t A266801 Coefficient[t, x, 0] ; (* A266799 *)
%t A266801 Coefficient[t, x, 1];  (* A266800 *)
%t A266801 Coefficient[t, x, 2];  (* A266801 *)
%t A266801 Coefficient[t, x, 3];  (* A266802 *)
%t A266801 Coefficient[t, x, 4];  (* A266799 *)
%Y A266801 Cf. A265762, A266799, A266800, A266802.
%K A266801 sign,easy
%O A266801 0,1
%A A266801 _Clark Kimberling_, Jan 09 2016