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a(n) = 3*n - a(n-1) - 1 for n>0, a(1)=1.
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%I #31 Sep 08 2022 08:45:48

%S 1,4,4,7,7,10,10,13,13,16,16,19,19,22,22,25,25,28,28,31,31,34,34,37,

%T 37,40,40,43,43,46,46,49,49,52,52,55,55,58,58,61,61,64,64,67,67,70,70,

%U 73,73,76,76,79,79,82,82,85,85,88,88,91,91,94,94,97,97,100,100,103,103,106

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

%H Vincenzo Librandi, <a href="/A168233/b168233.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).

%F From _Bruno Berselli_, Nov 15 2010: (Start)

%F a(n) = (6*n + 3*(-1)^n + 1)/4.

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

%F a(n) = a(n-1) + a(n-2) - a(n-3), for n>3.

%F a(n) + a(n-1) = A016789(n-1) for n>1.

%F a(n) - a(n-1-2*k) = A010674(n-1) + A008585(k) for n>2*k+1 and k in A001477.

%F a(n) - a(n-2*k) = A008585(k) for n>2*k and k in A001477. (End)

%F a(n+1) = A016777(floor((n+1)/2)). - _R. J. Mathar_, Jan 03 2011

%F E.g.f.: (1/4)*(3 - 4*exp(x) + (1 + 6*x)*exp(2*x))*exp(-x). - _G. C. Greubel_, Jul 16 2016

%p a:=n->3*floor(n/2)+1; seq(a(k), k = 1..70); # _Wesley Ivan Hurt_, Feb 01 2013

%t CoefficientList[Series[(1 + 3*x - x^2)/((1+x) * (1-x)^2), {x, 0, 100}], x] (* _Vincenzo Librandi_, Feb 02 2013 *)

%t LinearRecurrence[{1,1,-1},{1,4,4},80] (* _Harvey P. Dale_, Oct 13 2015 *)

%o (Magma) [(6*n + 3*(-1)^n + 1)/4: n in [1..70]]; // _Vincenzo Librandi_, Feb 02 2013

%Y Cf. A008585, A001477, A016777, A010674, A016789.

%K nonn,easy

%O 1,2

%A _Vincenzo Librandi_, Nov 21 2009