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A266705 Coefficient of x^2 in minimal polynomial of the continued fraction [1^n,sqrt(5),1,1,1,...], where 1^n means n ones.

%I A266705 #18 Sep 29 2016 14:54:59
%S A266705 1,11,11,45,101,281,719,1899,4955,12989,33989,89001,232991,609995,
%T A266705 1596971,4180941,10945829,28656569,75023855,196415019,514221179,
%U A266705 1346248541,3524524421,9227324745,24157449791,63245024651,165577624139,433487847789,1134885919205
%N A266705 Coefficient of x^2 in minimal polynomial of the continued fraction [1^n,sqrt(5),1,1,1,...], where 1^n means n ones.
%C A266705 See A265762 for a guide to related sequences.
%H A266705 Colin Barker, <a href="/A266705/b266705.txt">Table of n, a(n) for n = 0..1000</a>
%H A266705 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-1).
%F A266705 a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3) for n > 3.
%F A266705 G.f.:  (1 + 9*x - 13*x^2 + 2*x^3)/(1 - 2*x - 2*x^2 + x^3).
%F A266705 a(n) = (2^(-n)*(-23*(-2)^n-(3-sqrt(5))^n*(-9+sqrt(5))+(3+sqrt(5))^n*(9+sqrt(5))))/5 for n>0. - _Colin Barker_, Sep 29 2016
%e A266705 Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
%e A266705 [sqrt(5),1,1,1,1,...] = (-1+3*sqrt(5))/2 has p(0,x)=-11+x+x^2, so a(0) = 1;
%e A266705 [1,sqrt(5),1,1,1,...] = (23+3*sqrt(5))/22 has p(1,x)=11-23x+11x^2, so a(1) = 11;
%e A266705 [1,1,sqrt(5),1,1,...] = (45-3*sqrt(5))/22 has p(2,x)=45-45x+11x^2, so a(2) = 11.
%t A266705 u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[5]}, {{1}}];
%t A266705 f[n_] := FromContinuedFraction[t[n]];
%t A266705 t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
%t A266705 Coefficient[t, x, 0] (* A266705 *)
%t A266705 Coefficient[t, x, 1] (* A266706 *)
%t A266705 Coefficient[t, x, 2] (* A266705 *)
%o A266705 (PARI) Vec((1+9*x-13*x^2+2*x^3)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ _Colin Barker_, Sep 29 2016
%Y A266705 Cf. A265762, A266706.
%K A266705 nonn,easy
%O A266705 0,2
%A A266705 _Clark Kimberling_, Jan 09 2016