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A266799 Coefficient of x^0 in the minimal polynomial of the continued fraction [1^n,sqrt(3),1,1,...], where 1^n means n ones.

%I A266799 #7 Oct 21 2019 15:22:59
%S A266799 1,1,49,229,1861,12001,84241,572209,3935569,26939221,184737301,
%T A266799 1265964481,8677687969,59476087009,407659540081,2794128600901,
%U A266799 19151272325221,131264694791329,899701808208049,6166647394567441,42266831441062801,289701168799073461
%N A266799 Coefficient of x^0 in the minimal polynomial of the continued fraction [1^n,sqrt(3),1,1,...], where 1^n means n ones.
%C A266799 See A265762 for a guide to related sequences.
%H A266799 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,15,-15,-5,1).
%F A266799 a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5) .
%F A266799 G.f.:  (-1 + 4 x - 29 x^2 + 16 x^3 - x^4)/(-1 + 5 x + 15 x^2 - 15 x^3 - 5 x^4 + x^5).
%e A266799 Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
%e A266799 [sqrt(3),1,1,1,...] has p(0,x)=1-8x-7x^2+2x^3+x^4, so a(0) = 1;
%e A266799 [1,sqrt(3),1,1,1,...] has p(1,x)=1-12x+23x^2-12x^3+x^4, so a(1) = 1;
%e A266799 [1,1,sqrt(3),1,1,1...] has p(2,x)=49-98x+65x^2-16x^3+x^4, so a(2) = 49.
%t A266799 u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[3]}, {{1}}];
%t A266799 f[n_] := FromContinuedFraction[t[n]];
%t A266799 t = Table[MinimalPolynomial[f[n], x], {n, 0, 40}];
%t A266799 Coefficient[t, x, 0] ; (* A266799 *)
%t A266799 Coefficient[t, x, 1];  (* A266800 *)
%t A266799 Coefficient[t, x, 2];  (* A266801 *)
%t A266799 Coefficient[t, x, 3];  (* A266802 *)
%t A266799 Coefficient[t, x, 4];  (* A266799 *)
%t A266799 LinearRecurrence[{5,15,-15,-5,1},{1,1,49,229,1861},30] (* _Harvey P. Dale_, Oct 21 2019 *)
%Y A266799 Cf. A265762, A266800, A266801, A266802.
%K A266799 nonn,easy
%O A266799 0,3
%A A266799 _Clark Kimberling_, Jan 09 2016