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A266702 Coefficient of x in minimal polynomial of the continued fraction [1^n,1/3,1,1,1,...], where 1^n means n ones.

%I A266702 #11 Sep 29 2016 14:29:40
%S A266702 3,-25,-35,-117,-279,-757,-1955,-5145,-13443,-35221,-92183,-241365,
%T A266702 -631875,-1654297,-4330979,-11338677,-29685015,-77716405,-203464163,
%U A266702 -532676121,-1394564163,-3651016405,-9558485015,-25024438677,-65514830979,-171520054297
%N A266702 Coefficient of x in minimal polynomial of the continued fraction [1^n,1/3,1,1,1,...], where 1^n means n ones.
%C A266702 See A265762 for a guide to related sequences.
%H A266702 Colin Barker, <a href="/A266702/b266702.txt">Table of n, a(n) for n = 0..1000</a>
%H A266702 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-1).
%F A266702 a(n) = 2*a(n-1) - 2*a(n-2) + a(n-3).
%F A266702 G.f.:  (3 - 31 x + 9 x^2 + 6 x^3)/(1 - 2 x - 2 x^2 + x^3).
%F A266702 a(n) = (2^(-n)*(37*(-2)^n+2*(3-sqrt(5))^n*(-13+sqrt(5))-2*(3+sqrt(5))^n*(13+sqrt(5))))/5 for n>0. - _Colin Barker_, Sep 29 2016
%e A266702 Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
%e A266702 [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 A266702 [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 A266702 [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 A266702 u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {1/3}, {{1}}];
%t A266702 f[n_] := FromContinuedFraction[t[n]];
%t A266702 t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
%t A266702 Coefficient[t, x, 0] (* A266701 *)
%t A266702 Coefficient[t, x, 1] (* A266702 *)
%t A266702 Coefficient[t, x, 2] (* A266701 *)
%o A266702 (PARI) Vec((3-31*x+9*x^2+6*x^3)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ _Colin Barker_, Sep 29 2016
%Y A266702 Cf. A265762, A266701.
%K A266702 sign,easy
%O A266702 0,1
%A A266702 _Clark Kimberling_, Jan 09 2016