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%I #11 Oct 01 2016 04:40:19
%S -7,-25,-59,-161,-415,-1093,-2855,-7481,-19579,-51265,-134207,-351365,
%T -919879,-2408281,-6304955,-16506593,-43214815,-113137861,-296198759,
%U -775458425,-2030176507,-5315071105,-13915036799,-36430039301,-95375081095,-249695203993
%N Coefficient of x in minimal polynomial of the continued fraction [2,1^n,2,1,1,...], where 1^n means n ones.
%C See A265762 for a guide to related sequences.
%H Colin Barker, <a href="/A266709/b266709.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.: (1 + 3 x - x^2)/(1 - 2 x - 2 x^2 + x^3).
%F a(n) = (2^(-n)*(9*(-2)^n+2*(3-sqrt(5))^n*(-11+5*sqrt(5))-2*(3+sqrt(5))^n*(11+5*sqrt(5))))/5. - _Colin Barker_, Oct 01 2016
%e Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
%e [2,2,1,1,1,...] = (7-sqrt(5))/2 has p(0,x) = 11 - 7 x + x^2, so a(0) = -7;
%e [2,1,2,1,1,1,...] = (25+sqrt(5))/10 has p(1,x) = 31 - 25 x + 5 x^2, so a(1) = -25;
%e [2,1,1,2,1,...] = (59-sqrt(5))/22 has p(2,x) = 79 - 59 x + 11 x^2, so a(2) = -59.
%t u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[{2}, u[n], {2}, {{1}}];
%t f[n_] := FromContinuedFraction[t[n]];
%t t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
%t Coefficient[t, x, 0] (* A236428 *)
%t Coefficient[t, x, 1] (* A266709 *)
%t Coefficient[t, x, 2] (* A236428 *)
%o (PARI) a(n) = round((2^(-n)*(9*(-2)^n+2*(3-sqrt(5))^n*(-11+5*sqrt(5))-2*(3+sqrt(5))^n*(11+5*sqrt(5))))/5) \\ _Colin Barker_, Oct 01 2016
%o (PARI) Vec(-(7+11*x-5*x^2)/((1+x)*(1-3*x+x^2)) + O(x^40)) \\ _Colin Barker_, Oct 01 2016
%Y Cf. A265762, A236428.
%K sign,easy
%O 0,1
%A _Clark Kimberling_, Jan 09 2016
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