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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. 3

%I #18 Sep 29 2016 14:54:59

%S 1,11,11,45,101,281,719,1899,4955,12989,33989,89001,232991,609995,

%T 1596971,4180941,10945829,28656569,75023855,196415019,514221179,

%U 1346248541,3524524421,9227324745,24157449791,63245024651,165577624139,433487847789,1134885919205

%N 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 See A265762 for a guide to related sequences.

%H Colin Barker, <a href="/A266705/b266705.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-1).

%F a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3) for n > 3.

%F G.f.: (1 + 9*x - 13*x^2 + 2*x^3)/(1 - 2*x - 2*x^2 + x^3).

%F 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 Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:

%e [sqrt(5),1,1,1,1,...] = (-1+3*sqrt(5))/2 has p(0,x)=-11+x+x^2, so a(0) = 1;

%e [1,sqrt(5),1,1,1,...] = (23+3*sqrt(5))/22 has p(1,x)=11-23x+11x^2, so a(1) = 11;

%e [1,1,sqrt(5),1,1,...] = (45-3*sqrt(5))/22 has p(2,x)=45-45x+11x^2, so a(2) = 11.

%t u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[5]}, {{1}}];

%t f[n_] := FromContinuedFraction[t[n]];

%t t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]

%t Coefficient[t, x, 0] (* A266705 *)

%t Coefficient[t, x, 1] (* A266706 *)

%t Coefficient[t, x, 2] (* A266705 *)

%o (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 Cf. A265762, A266706.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Jan 09 2016

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