|
%I #8 Sep 08 2022 08:46:15
%S -5,-7,-7,115,607,4615,30427,211687,1442695,9909907,67867135,
%T 465315847,3188935867,21858303175,149816390407,1026863749555,
%U 7038210692767,48240661271047,330646286854555,2266283690589607,15533338646986375,106467089195295187
%N Coefficient of x^2 in the minimal polynomial of the continued fraction [1^n,sqrt(2),1,1,...], where 1^n means n ones.
%C See A265762 for a guide to related sequences.
%H G. C. Greubel, <a href="/A266712/b266712.txt">Table of n, a(n) for n = 0..1000</a>
%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.: (5 -18*x -103*x^2 -180*x^3 -7*x^4 +280*x^5 +56*x^6 -14*x^7)/(-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(2),1,1,1,...] has p(0,x) = -1 - 6 x - 5 x^2 + 2 x^3 + x^4, so a(0) = -5;
%e [1,sqrt(2),1,1,1,...] has p(1,x) = 1 + 2 x - 7 x^2 + 2 x^3 + x^4, so a(1) = -7;
%e [1,1,sqrt(2),1,1,1...] has p(2,x) = -9 + 18 x - 7 x^2 - 2 x^3 + x^4, so a(2) = -7.
%t u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[2]}, {{1}}];
%t f[n_] := FromContinuedFraction[t[n]];
%t t = Table[MinimalPolynomial[f[n], x], {n, 0, 40}];
%t Coefficient[t, x, 0] ; (* A266710 *)
%t Coefficient[t, x, 1]; (* A266711 *)
%t Coefficient[t, x, 2]; (* A266712 *)
%t Coefficient[t, x, 3]; (* A266713 *)
%t Coefficient[t, x, 4]; (* A266710 *)
%t LinearRecurrence[{5,15,-15,-5,1}, {-5, -7, -7, 115, 607, 4615, 30427, 211687}, 30] (* _G. C. Greubel_, Jan 26 2018 *)
%o (PARI) x='x+O('x^30); Vec((5 -18*x -103*x^2 -180*x^3 -7*x^4 +280*x^5 +56*x^6 -14*x^7)/(-1 + 5*x +15*x^2 -15*x^3 -5*x^4 +x^5)) \\ _G. C. Greubel_, Jan 26 2018
%o (Magma) I:=[115, 607, 4615, 30427, 211687]; [-5, -7, -7] cat [n le 5 select I[n] else 5*Self(n-1) + 15*Self(n-2) - 15*Self(n-3) - 5*Self(n-4) + Self(n-5): n in [1..30]]; // _G. C. Greubel_, Jan 26 2018
%Y Cf. A265762, A266710, A266711, A266713.
%K sign,easy
%O 0,1
%A _Clark Kimberling_, Jan 09 2016
|
