%I #21 Sep 08 2022 08:46:15
%S -7,-29,-67,-185,-475,-1253,-3271,-8573,-22435,-58745,-153787,-402629,
%T -1054087,-2759645,-7224835,-18914873,-49519771,-129644453,-339413575,
%U -888596285,-2326375267,-6090529529,-15945213307,-41745110405,-109290117895,-286125243293
%N Coefficient of x in minimal polynomial of the continued fraction [1^n,4,1,1,1,...], where 1^n means n ones.
%C See A265762 for a guide to related sequences.
%H Colin Barker, <a href="/A265803/b265803.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.: (-1)*(7 + 15*x - 5*x^2)/(1 - 2*x - 2*x^2 + x^3).
%F a(n) = (2^(-n)*(13*(-2)^n + 12*(3-sqrt(5))^n*(-2+sqrt(5)) - 12*(2+sqrt(5))*(3+sqrt(5))^n))/5. - _Colin Barker_, Oct 20 2016
%F a(n) = (13*(-1)^n - 12*Lucas(2*n+3))/5 = 5*(-1)^n - 12*F(n+1)*F(n+2), F=Fibonacci. - _G. C. Greubel_, Dec 12 2019
%e Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
%e [4,1,1,1,1,...] = (7 + sqrt(5))/2 has p(0,x) = 11 - 7 x + x^2, so a(0) = 1;
%e [1,4,1,1,1,...] = (29 - sqrt(5))/22 has p(1,x) = 19 - 29 x + 11 x^2, so a(1) = 11;
%e [1,1,4,1,1,...] = (67 + sqrt(5))/38 has p(2,x) = 59 - 67 x + 19 x^2, so a(2) = 19.
%p with(combinat); f:=fibonacci; seq( 5*(-1)^n - 12*f(n+1)*f(n+2), n=0..30); # _G. C. Greubel_, Dec 12 2019
%t u[n_]:= Table[1, {k,n}]; t[n_]:= Join[u[n], {4}, {{1}}];
%t f[n_]:= FromContinuedFraction[t[n]];
%t t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
%t Coefficient[t, x, 0] (* A265802 *)
%t Coefficient[t, x, 1] (* A265803 *)
%t Coefficient[t, x, 2] (* A236802 *)
%t LinearRecurrence[{2,2,-1}, {-7,-29,-67}, 30] (* _Vincenzo Librandi_, Jan 06 2016 *)
%o (PARI) Vec((-7-15*x+5*x^2)/(1-2*x-2*x^2+x^3) + O(x^30)) \\ _Altug Alkan_, Jan 04 2016
%o (PARI) vector(31, n, f=fibonacci; -(5*(-1)^n + 12*f(n)*f(n+1)) ) \\ _G. C. Greubel_, Dec 12 2019
%o (Magma) I:=[-7,-29,-67]; [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 (Sage) [(13*(-1)^n -12*lucas_number2(2*n+3,1,-1))/5 for n in (0..30)] # _G. C. Greubel_, Dec 12 2019
%o (GAP) List([0..30], n-> (13*(-1)^n -12*Lucas(1,-1,2*n+3)[2])/5 ); # _G. C. Greubel_, Dec 12 2019
%Y Cf. A000032, A000045, A265762, A265802.
%K sign,easy
%O 0,1
%A _Clark Kimberling_, Jan 04 2016