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%I #16 Sep 08 2022 08:46:15
%S 0,-10,-12,-44,-102,-280,-720,-1898,-4956,-12988,-33990,-89000,
%T -232992,-609994,-1596972,-4180940,-10945830,-28656568,-75023856,
%U -196415018,-514221180,-1346248540,-3524524422,-9227324744,-24157449792,-63245024650,-165577624140
%N Coefficient of x in minimal polynomial of the continued fraction [1^n,1/2,1,1,1,...], where 1^n means n ones.
%C See A265762 for a guide to related sequences.
%H Colin Barker, <a href="/A266700/b266700.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.: (2 (-5 x + 4 x^2))/(1 - 2 x - 2 x^2 + x^3).
%F a(n) = (2^(-n)*(9*(-1)^n*2^(1+n) + (3-sqrt(5))^n*(-9+sqrt(5)) - (3+sqrt(5))^n*(9+sqrt(5))))/5. - _Colin Barker_, Oct 20 2016
%e Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
%e [1/2,1,1,1,1,...] = sqrt(5))/2 has p(0,x) = -5 + 4 x^2, so a(0) = 1;
%e [1,1/2,1,1,1,...] = (5 + 2 sqrt(5))/5 has p(1,x) = 1 - 10 x + 5 x^2, so a(1) = 19;
%e [1,1,1/2,1,1,...] = 6 - 2 sqrt(5) has p(2,x) = 16 - 12 x + x^2, so a(2) = 29.
%t u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {1/2}, {{1}}];
%t f[n_] := FromContinuedFraction[t[n]];
%t t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
%t Coefficient[t, x, 0] (* A266699 *)
%t Coefficient[t, x, 1] (* A266700 *)
%t Coefficient[t, x, 2] (* A266699 *)
%t LinearRecurrence[{2, 2, -1}, {0, -10, -12}, 30] (* _Vincenzo Librandi_, Jan 06 2016 *)
%o (Magma) I:=[0,-10,-12]; [n le 3 select I[n] else 2*Self(n-1)+2*Self(n-2)-Self(n-3): n in [1..30]]; // _Vincenzo Librandi_, Jan 06 2016
%o (PARI) concat(0, Vec((-10*x+8*x^2)/(1-2*x-2*x^2+x^3) + O(x^100))) \\ _Altug Alkan_, Jan 07 2016
%Y Cf. A265762, A266699.
%K sign,easy
%O 0,2
%A _Clark Kimberling_, Jan 05 2016
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