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A266804 Coefficient of x^0 in the minimal polynomial of the continued fraction [1^n,sqrt(6),1,1,...], where 1^n means n ones.

%I A266804 #4 Jan 09 2016 19:59:19
%S A266804 19,19,361,1795,14011,91489,638899,4348051,29883145,204609571,
%T A266804 1402971259,9614651329,65903614291,451700107795,3096024736681,
%U A266804 21220400800579,145446970016059,996907894114081,6832909585226995,46833455808339091,321001289959109449
%N A266804 Coefficient of x^0 in the minimal polynomial of the continued fraction [1^n,sqrt(6),1,1,...], where 1^n means n ones.
%C A266804 See A265762 for a guide to related sequences.
%H A266804 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,15,-15,-5,1).
%F A266804 a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5) .
%F A266804 G.f.:  (-19 + 76 x + 19 x^2 + 10 x^3 - x^4)/(-1 + 5 x + 15 x^2 - 15 x^3 - 5 x^4 + x^5).
%e A266804 Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
%e A266804 [sqrt(6),1,1,1,...] has p(0,x)=19-14x-13x^2+2x^3+x^4, so a(0) = 19;
%e A266804 [1,sqrt(6),1,1,1,...] has p(1,x)=19-90x+143x^2-90x^3+19x^4, so a(1) = 19;
%e A266804 [1,1,sqrt(6),1,1,1...] has p(2,x)=361-722x+527x^2-166x^3+19x^4, so a(2) = 361.
%t A266804 u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[6]}, {{1}}];
%t A266804 f[n_] := FromContinuedFraction[t[n]];
%t A266804 t = Table[MinimalPolynomial[f[n], x], {n, 0, 40}];
%t A266804 Coefficient[t, x, 0] ; (* A266804 *)
%t A266804 Coefficient[t, x, 1];  (* A266805 *)
%t A266804 Coefficient[t, x, 2];  (* A266806 *)
%t A266804 Coefficient[t, x, 3];  (* A266807 *)
%t A266804 Coefficient[t, x, 4];  (* A266804 *)
%Y A266804 Cf. A265762, A266805, A266806, A266807.
%K A266804 nonn,easy
%O A266804 0,1
%A A266804 _Clark Kimberling_, Jan 09 2016