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A266802 Coefficient of x^3 in the minimal polynomial of the continued fraction [1^n,sqrt(3),1,1,...], where 1^n means n ones. 5

%I #4 Jan 09 2016 19:59:11

%S 2,-12,-16,-294,-1552,-11868,-78142,-543996,-3706624,-25463142,

%T -174376288,-1195587372,-8193644926,-56162781804,-384938354032,

%U -2638425262758,-18083987259952,-123949619666556,-849562999302334,-5822992294650972,-39911380656754528

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

%C See A265762 for a guide to related sequences.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,15,-15,-5,1).

%F a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5) .

%F G.f.: (2 (-1 + 11 x - 7 x^2 + 2 x^3 + 6 x^4))/(-1 + 5 x + 15 x^2 - 15 x^3 - 5 x^4 + x^5).

%e Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:

%e [sqrt(3),1,1,1,...] has p(0,x) = 1 - 8 x - 7 x^2 + 2 x^3 + x^4, so a(0) = 2;

%e [1,sqrt(3),1,1,1,...] has p(1,x) = 1 - 12 x + 23 x^2 - 12 x^3 + x^4, so a(1) = -12;

%e [1,1,sqrt(3),1,1,1...] has p(2,x) = 49 - 98 x + 65 x^2 - 16 x^3 + x^4, so a(2) = -16.

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

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

%t t = Table[MinimalPolynomial[f[n], x], {n, 0, 40}];

%t Coefficient[t, x, 0] ; (* A266799 *)

%t Coefficient[t, x, 1]; (* A266800 *)

%t Coefficient[t, x, 2]; (* A266801 *)

%t Coefficient[t, x, 3]; (* A266802 *)

%t Coefficient[t, x, 4]; (* A266799 *)

%Y Cf. A265762, A266299, A266800, A266801.

%K sign,easy

%O 0,1

%A _Clark Kimberling_, Jan 09 2016

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Last modified May 2 09:06 EDT 2024. Contains 372178 sequences. (Running on oeis4.)