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a(0)=1; for n>0, a(n) = 1 + 3*A117571(n-1).
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%I #19 Sep 08 2022 08:45:54

%S 1,4,4,10,13,13,19,22,22,28,31,31,37,40,40,46,49,49,55,58,58,64,67,67,

%T 73,76,76,82,85,85,91,94,94,100,103,103,109,112,112,118,121,121,127,

%U 130,130,136,139,139,145,148,148,154,157,157,163,166,166,172

%N a(0)=1; for n>0, a(n) = 1 + 3*A117571(n-1).

%H B. Berselli, <a href="/A180964/b180964.txt">Table of n, a(n) for n = 0..10000</a>.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,1,-1).

%F G.f.: (1 +3*x +5*x^3)/((1-x)^2*(1+x+x^2)).

%F a(n) = a(n-1) +a(n-3) -a(n-4) for n>3.

%F a(n) = (n-1)*(n mod 3) +(n+1)*(n+1 mod 3) +n*(n+2 mod 3).

%F a(n) = 3*n +sqrt(3)*cos((4*n-3)*Pi/6) -sin((4*n-3)*Pi/6).

%F a(n) - a(n-1) = 3*A080425(n+1) for n>0.

%F From _G. C. Greubel_, Apr 06 2021: (Start)

%F a(n) = 3*n - 2*cos(2*Pi*(n+1)/3) = 3*n + A061347(n+1).

%F a(n) = 3*n + ChebyshevU(n, -1/2) + 2*ChebyshevU(n-1, -1/2). (End)

%p m:=60; S:=series( (1+3*x+5*x^3)/((1-x)^2*(1+x+x^2)), x, m+1):

%p seq(coeff(S, x, j), j=1..m); # _G. C. Greubel_, Apr 06 2021

%t CoefficientList[Series[(1 +3x +5x^3)/((1-x)^2(1+x+x^2)), {x, 0, 60}], x] (* _Vincenzo Librandi_, Mar 26 2013 *)

%t LinearRecurrence[{1,0,1,-1},{1,4,4,10},60] (* _Harvey P. Dale_, Aug 05 2020 *)

%o (Magma) I:=[1, 4, 4, 10]; [n le 4 select I[n] else Self(n-1)+Self(n-3)-Self(n-4): n in [1..60]]; // _Vincenzo Librandi_, Mar 26 2013

%o (Sage) [3*n +chebyshev_U(n, -1/2) +2*chebyshev_U(n-1, -1/2) for n in (0..60)] # _G. C. Greubel_, Apr 06 2021

%Y Cf. A061347.

%K nonn,easy

%O 0,2

%A _Bruno Berselli_, Sep 28 2010 - Oct 01 2010