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%I #13 Sep 29 2016 16:47:37
%S 9,11,5,41,81,239,599,1595,4149,10889,28481,74591,195255,511211,
%T 1338341,3503849,9173169,24015695,62873879,164605979,430944021,
%U 1128226121,2953734305,7732976831,20245196151,53002611659,138762638789,363285304745,951093275409
%N Coefficient of x^2 in minimal polynomial of the continued fraction [1^n,1/3,1,1,1,...], where 1^n means n ones.
%C See A265762 for a guide to related sequences.
%H Colin Barker, <a href="/A266701/b266701.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.: (9 - 7 x - 35 x^2 + 18 x^3)/(1 - 2 x - 2 x^2 + x^3).
%F a(n) = (2^(-n)*(-37*(-2)^n-2*(3-sqrt(5))^n*(2+3*sqrt(5))+(3+sqrt(5))^n*(-4+6*sqrt(5))))/5. - _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 [1/3,1,1,1,...] = (-1 + 3 sqrt(5))/6 has p(0,x) = -11 + 3 x + 9 x^2, so a(0) = 9;
%e [1,1/3,1,1,...] = (25 + 9 sqrt(5))/22 has p(1,x) = 5 - 25 x + 11 x^2, so a(1) = 11;
%e [1,1,1/3,1,...] = (35 - 9 sqrt(5))/10 has p(2,x) = 41 - 35 x + 5 x^2, so a(2) = 5.
%t u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {1/3}, {{1}}];
%t f[n_] := FromContinuedFraction[t[n]];
%t t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
%t Coefficient[t, x, 0] (* A266701 *)
%t Coefficient[t, x, 1] (* A266702 *)
%t Coefficient[t, x, 2] (* A266701 *)
%o (PARI) a(n) = round((2^(-n)*(-37*(-2)^n-2*(3-sqrt(5))^n*(2+3*sqrt(5))+(3+sqrt(5))^n*(-4+6*sqrt(5))))/5) \\ _Colin Barker_, Sep 29 2016
%o (PARI) Vec((9-7*x-35*x^2+18*x^3)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ _Colin Barker_, Sep 29 2016
%Y Cf. A265762, A266702.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, Jan 09 2016
%E Three typos in data fixed by _Colin Barker_, Sep 29 2016
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