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 A266704 Coefficient of x^2 in minimal polynomial of the continued fraction [1^n,2/3,1,1,1,...], where 1^n means n ones. 3

%I #12 Sep 29 2016 14:56:36

%S -3,-19,17,-75,-165,-463,-1181,-3123,-8145,-21355,-55877,-146319,

%T -383037,-1002835,-2625425,-6873483,-17994981,-47111503,-123339485,

%U -322906995,-845381457,-2213237419,-5794330757,-15169754895,-39714933885,-103975046803,-272210206481

%N Coefficient of x^2 in minimal polynomial of the continued fraction [1^n,2/3,1,1,1,...], where 1^n means n ones.

%C See A265762 for a guide to related sequences.

%H Colin Barker, <a href="/A266704/b266704.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).

%F G.f.: (-3 - 13 x + 61 x^2 - 74 x^3 - 68 x^4 + 34 x^5)/(1 - 2 x - 2 x^2 + x^3).

%F a(n) = 2^(-n)*(43*(-2)^n+2*(3-sqrt(5))^n*(-7+sqrt(5))-2*(3+sqrt(5))^n*(7+sqrt(5)))/5 for n>2. - _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 [2/3,1,1,1,...] = (1+3*sqrt(5))/6 has p(0,x) = -11 - 3 x + 9 x^2, so a(0) = 9;

%e [1,2/3,1,1,...] = (19+9*sqrt(5))/22 has p(1,x) = -1 - 19 x + 11 x^2, so a(1) = 11;

%e [1,1,2/3,1,...] = (-17+9*sqrt(5))/2 has p(2,x) = -29 + 17 x + x^2, so a(2) = 1.

%t u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {2/3}, {{1}}];

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

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

%t Coefficient[t, x, 0] (* A266703 *)

%t Coefficient[t, x, 1] (* A266704 *)

%t Coefficient[t, x, 2] (* A266703 *)

%o (PARI) Vec(-(3+13*x-61*x^2+74*x^3+68*x^4-34*x^5)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ _Colin Barker_, Sep 29 2016

%Y Cf. A265762, A266703.

%K sign,easy

%O 0,1

%A _Clark Kimberling_, Jan 09 2016

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Last modified July 12 09:33 EDT 2024. Contains 374239 sequences. (Running on oeis4.)