%I #32 Sep 08 2022 08:45:54
%S 5,5,5,2,2,2,8,8,8,2,2,2,5,5,5,2,2,2,8,8,8,2,2,2,5,5,5,2,2,2,8,8,8,2,
%T 2,2,5,5,5,2,2,2,8,8,8,2,2,2,5,5,5,2,2,2,8,8,8,2,2,2,5,5,5,2,2,2,8,8,
%U 8,2,2,2,5,5,5,2,2,2,8,8,8,2,2,2,5,5,5,2,2,2,8,8,8,2,2,2,5,5,5,2,2,2,8,8,8,2,2,2
%N Period 12: repeat [5,5,5,2,2,2,8,8,8,2,2,2].
%C Also represents the decimal expansion of 50020080020/9009009009.
%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (1, 0, -1, 1, 0, -1, 1, 0, -1, 1).
%F From _R. J. Mathar_, Feb 04 2011: (Start)
%F a(n) = a(n-1) - a(n-3) + a(n-4) - a(n-6) + a(n-7) - a(n-9) + a(n-10).
%F G.f.: x*(5 + 2*x^3 + 8*x^6 + 2*x^9) / ( (1-x)*(1+x)*(x^2+1)*(x^2 - x + 1)*(x^4 - x^2 + 1) ). (End)
%F Let b(n) = A061037(n+1) - 2*A061037(n) + A061037(n-1) denote the second differences of A061037. Then a(n+2) = b(n) mod 9. - _Paul Curtz_, Mar 02 2011
%F a(n) = (17 - 9*cos(t)^2 - 6*sin(t) + 9*sin(t)^2)/4, where t := Pi*(3*n + 3 - sqrt(3)*sin(2*Pi*(n + 1)/3) + sqrt(3)*sin(4*Pi*(n + 1)/3))/18. - _Wesley Ivan Hurt_, Sep 26 2018
%t LinearRecurrence[{1, 0, -1, 1, 0, -1, 1, 0, -1, 1},{5, 5, 5, 2, 2, 2, 8, 8, 8, 2},108] (* _Ray Chandler_, Aug 27 2015 *)
%t CoefficientList[Series[x*(5+2*x^3+8*x^6+2*x^9)/((1-x)*(1+x)*(x^2+1)*(x^2-x+1)*(x^4-x^2+1)), {x, 0, 50}], x] (* _G. C. Greubel_, Sep 20 2018 *)
%t PadRight[{},120,{5,5,5,2,2,2,8,8,8,2,2,2}] (* _Harvey P. Dale_, Jul 03 2021 *)
%o (PARI) x='x+O('x^50); Vec(x*(5+2*x^3+8*x^6+2*x^9)/((1-x)*(1+x)*(x^2+1)*(x^2-x+1)*(x^4-x^2+1))) \\ _G. C. Greubel_, Sep 20 2018
%o (Magma) m:=25; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!(x*(5+2*x^3+8*x^6+2*x^9)/((1-x)*(1+x)*(x^2+1)*(x^2-x+1)*(x^4-x^2+1)))); // _G. C. Greubel_, Sep 20 2018
%Y Cf. A061037.
%K nonn,easy
%O 1,1
%A _Paul Curtz_, Nov 18 2010
%E Extended by _Ray Chandler_, Aug 27 2015