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A266800 Coefficient of x in the minimal polynomial of the continued fraction [1^n,sqrt(3),1,1,...], where 1^n means n ones.

%I A266800 #4 Jan 09 2016 19:58:57
%S A266800 -8,-12,-98,-636,-4424,-30138,-207032,-1417788,-9720866,-66619404,
%T A266800 -456638168,-3129787002,-21452029928,-147034005996,-1007787102434,
%U A266800 -6907472856348,-47344530365672,-324504220137018,-2224185061818776,-15244791078484764,-104489352838678178
%N A266800 Coefficient of x in the minimal polynomial of the continued fraction [1^n,sqrt(3),1,1,...], where 1^n means n ones.
%C A266800 See A265762 for a guide to related sequences.
%H A266800 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,15,-15,-5,1).
%F A266800 a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5) .
%F A266800 G.f.:  -((2 (-4 + 14 x + 41 x^2 - 43 x^3 + 3 x^4))/(-1 + 5 x + 15 x^2 - 15 x^3 - 5 x^4 + x^5)).
%e A266800 Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
%e A266800 [sqrt(3),1,1,1,...] has p(0,x)=1-8x-7x^2+2x^3+x^4, so a(0) = -8;
%e A266800 [1,sqrt(3),1,1,1,...] has p(1,x)=1-12x+23x^2-12x^3+x^4, so a(1) = -12;
%e A266800 [1,1,sqrt(3),1,1,1...] has p(2,x)=49-98x+65x^2-16x^3+x^4, so a(2) = -98.
%t A266800 u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[3]}, {{1}}];
%t A266800 f[n_] := FromContinuedFraction[t[n]];
%t A266800 t = Table[MinimalPolynomial[f[n], x], {n, 0, 40}];
%t A266800 Coefficient[t, x, 0] ; (* A266799 *)
%t A266800 Coefficient[t, x, 1];  (* A266800 *)
%t A266800 Coefficient[t, x, 2];  (* A266801 *)
%t A266800 Coefficient[t, x, 3];  (* A266802 *)
%t A266800 Coefficient[t, x, 4];  (* A266799 *)
%Y A266800 Cf. A265762, A266799, A266801, A266802.
%K A266800 sign,easy
%O A266800 0,1
%A A266800 _Clark Kimberling_, Jan 09 2016