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%I #32 Aug 10 2021 15:31:43
%S 1,1,-1,-3,-3,1,7,9,1,-15,-25,-11,29,65,47,-47,-159,-159,47,365,477,
%T 65,-777,-1319,-607,1489,3415,2533,-2371,-8319,-8481,2209,19009,25281,
%U 4063,-40227,-69571,-33407,76391,179369,136385
%N Expansion of (1+x)*(1-x)/(1 - x + x^2 + x^3).
%C Let r1 be the tribonacci constant A058265, and r2 = -0.41964... + 0.6062...*i, where i = sqrt(-1), and r3 the complex conjugate of r2, the other constants also defined in A058265.
%C A formula in terms of cubic roots is known for r1 (see A058265), and Re(r2) = Re(r3) = (1-r1)/2 and Im(r2) = -Im(r3) = sqrt( 1/r1-Re^2(r2)).
%C Then the denominator of the g.f. is (x+r1)*(x+r2)*(x+r3) = x^3 + x^2 + 1 - x,
%C and the Binet formula is a(n) = (r3^2-1)*(-r3)^(-n-1)/( (r2-r3)*(r1-r3) ) -(r2^2-1)*(-r2)^(-n-1)/( (r2-r3)*(r1-r2) ) +(r1^2-1)*(-r1)^(-n-1)/( (r1-r2)*(r1-r3) ). - _R. J. Mathar_, based on input from _Alexander R. Povolotsky_ and _T. D. Noe_
%H G. C. Greubel, <a href="/A180735/b180735.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 (1,-1,-1)
%F INVERT transform of (1, 0, -2, 0, 2, 0, -2, 0, 2, 0, ...) = INVERT transform of (1 - 2x^2 + 2x^4 - 2x^6 + 2x^8 - ...).
%F a(n) = a(n-1) - a(n-2) - a(n-3), n > 3.
%F a(n) = (-1)^n*(A057597(n+2) - A057597(n)). - _R. J. Mathar_, Jan 27 2011
%e a(6) = 7 = (1, 1, 1, -1, -3, -3, 1) dot (-2, 0, 2, 0, -2, 0, 1) = (-2, 0, 2, 0, 6, 0, 1) = 7.
%t CoefficientList[Series[(1 + x)*(1 - x)/(1 - x + x^2 + x^3), {x, 0, 50}], x] (* _G. C. Greubel_, Feb 22 2017 *)
%t LinearRecurrence[{1,-1,-1},{1,1,-1},50] (* _Harvey P. Dale_, Aug 10 2021 *)
%o (PARI) x='x+O('x^50); Vec((1 + x)*(1 - x)/(1 - x + x^2 + x^3)) \\ _G. C. Greubel_, Feb 22 2017
%K sign,easy
%O 0,4
%A _Gary W. Adamson_, Jan 22 2011
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