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A266806 Coefficient of x^2 in the minimal polynomial of the continued fraction [1^n,sqrt(6),1,1,...], where 1^n means n ones. S.

%I A266806 #8 Jul 23 2025 15:39:46
%S A266806 -13,143,527,4859,30119,214847,1450643,10000367,68393039,469166939,
%T A266806 3214686407,22036489343,151033273907,1035215971919,7095427362959,
%U A266806 48632909524667,333334588608743,2284710128883647,15659633909836499,107332733533045679,735669484346002127
%N A266806 Coefficient of x^2 in the minimal polynomial of the continued fraction [1^n,sqrt(6),1,1,...], where 1^n means n ones. S.
%C A266806 ee A265762 for a guide to related sequences.
%H A266806 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,15,-15,-5,1).
%F A266806 a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5) .
%F A266806 G.f.:  (13 - 208 x - 7 x^2 + 116 x^3 + x^4)/(-1 + 5 x + 15 x^2 - 15 x^3 - 5 x^4 + x^5).
%e A266806 Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
%e A266806 [sqrt(6),1,1,1,...] has p(0,x)=19-14x-13x^2+2x^3+x^4, so a(0) = -13;
%e A266806 [1,sqrt(6),1,1,1,...] has p(1,x)=19-90x+143x^2-90x^3+19x^4, so a(1) = 143;
%e A266806 [1,1,sqrt(6),1,1,1...] has p(2,x)=361-722x+527x^2-166x^3+19x^4, so a(2) = 527.
%t A266806 u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[6]}, {{1}}];
%t A266806 f[n_] := FromContinuedFraction[t[n]];
%t A266806 t = Table[MinimalPolynomial[f[n], x], {n, 0, 40}];
%t A266806 Coefficient[t, x, 0] ; (* A266804 *)
%t A266806 Coefficient[t, x, 1];  (* A266805 *)
%t A266806 Coefficient[t, x, 2];  (* A266806 *)
%t A266806 Coefficient[t, x, 3];  (* A266807 *)
%t A266806 Coefficient[t, x, 4];  (* A266804 *)
%o A266806 (PARI) Vec((13-208*x-7*x^2+116*x^3+x^4)/(-1+5*x+15*x^2-15*x^3-5*x^4+x^5) + O(x^200)) \\ _Altug Alkan_, Jan 10 2015
%Y A266806 Cf. A265762, A266804, A266805, A266807.
%K A266806 sign,easy
%O A266806 0,1
%A A266806 _Clark Kimberling_, Jan 10 2016