|
%I #24 Aug 11 2022 07:25:45
%S 3,1,3,13,19,41,99,197,419,913,1923,4093,8755,18617,39651,84533,
%T 180035,383521,817155,1740781,3708499,7900745,16831587,35857829,
%U 76391651,162744241,346709379,738628573,1573570675,3352327385,7141783779
%N Sum of powers of roots of x^3 - x^2 - x - 3.
%C a(n) is the sum of the n-th powers of the three roots of x^3 - x^2 - x - 3. These roots are c1 = 2.130395..., c2 = -0.5651977... - i*1.0434274..., and c3 = -0.5651977... + i*1.0434274..., and so a(n) = c1^n + c2^n + c3^n. The real parts of c2 and c3 are A273065.
%C a(n) can also be determined by Vieta's formulas and Newton's identities. For example, a(3) by definition is c1^3 + c2^3 + c3^3, and from Newton's identities this equals e1^3 - 3*e1*e2 + 3*e3 for e1, e2, e3 the elementary symmetric polynomials of x^3 - x^2 - x - 3. From Vieta's formulas we have e1 = 1, e2 = -1, and e3 = 3, giving us e1^3 - 3*e1*e2 + 3*e3 = 1 + 3 + 9 = 13, as expected.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,3).
%F a(n) = a(n-1) + a(n-2) + 3*a(n-3) with a(0)=3, a(1)=1, a(2) = 3.
%F G.f.: (3 - 2*x - x^2)/(1 - x - x^2 - 3*x^3).
%e For n=3, a(3) = (2.130395...)^3 + (-0.5651977... - i*1.0434274...)^3 + (-0.5651977... + i*1.0434274...)^3 = 13.
%t LinearRecurrence[{1, 1, 3}, {3, 1, 3}, 40]
%o (PARI) polsym(x^3 - x^2 - x - 3, 35) \\ _Joerg Arndt_, Aug 11 2022
%Y Cf. A103143, A123102, A247594, A356463, A273065 (Re c2,c3).
%K nonn,easy
%O 0,1
%A _Greg Dresden_, Aug 05 2022
|
