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A087508 Number of k such that mod(k*n,3) = 1 for 0 <= k <= n.

%I A087508 #40 Sep 08 2022 08:45:11
%S A087508 0,1,1,0,2,2,0,3,3,0,4,4,0,5,5,0,6,6,0,7,7,0,8,8,0,9,9,0,10,10,0,11,
%T A087508 11,0,12,12,0,13,13,0,14,14,0,15,15,0,16,16,0,17,17,0,18,18,0,19,19,0,
%U A087508 20,20,0,21,21,0,22,22,0,23,23,0,24,24,0,25,25,0,26,26,0,27,27,0,28,28,0
%N A087508 Number of k such that mod(k*n,3) = 1 for 0 <= k <= n.
%H A087508 G. C. Greubel, <a href="/A087508/b087508.txt">Table of n, a(n) for n = 0..10000</a>
%H A087508 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,2,0,0,-1).
%F A087508 a(n) = A000027(n) - A087509(n) - A087507(n).
%F A087508 a(n) = (2/3)*(floor(n/3)+1)*(1-cos(2*Pi*n/3)).
%F A087508 G.f.: x*(1 + x)/(1 - x^3)^2. - _Arkadiusz Wesolowski_, May 28 2013
%F A087508 a(n) = sin(n*Pi/3)*((4n+6)*sin(n*Pi/3)-sqrt(3)*cos(n*Pi))/9. - _Wesley Ivan Hurt_, Sep 24 2017
%e A087508 a(4) = 2 because k=1 and k=4 satisfy the equation.
%t A087508 LinearRecurrence[{0,0,2,0,0,-1}, {0,1,1,0,2,2}, 100] (* _Vincenzo Librandi_, Sep 22 2015 *)
%t A087508 Table[PadRight[{0},3,n],{n,30}]//Flatten (* _Harvey P. Dale_, Jan 27 2021 *)
%o A087508 (PARI) concat(0,Vec((1+x)/(1-x^3)^2 +O(x^99))) \\ _Charles R Greathouse IV_, Oct 24 2014
%o A087508 (PARI) a(n) = sum(k=0, n, Mod(k*n, 3)==1); \\ _Michel Marcus_, Sep 27 2017
%o A087508 (Magma) I:=[0,1,1,0,2,2]; [n le 6 select I[n] else 2*Self(n-3) - Self(n-6): n in [1..100]]; // _Vincenzo Librandi_, Sep 22 2015
%o A087508 (SageMath)
%o A087508 @CachedFunction
%o A087508 def A087508(n):
%o A087508     if (n<6): return (0,1,1,0,2,2)[n]
%o A087508     else: return 2*A087508(n-3) - A087508(n-6)
%o A087508 [A087508(n) for n in (0..100)] # _G. C. Greubel_, Sep 02 2022
%Y A087508 Cf. A000027, A087507, A087509.
%K A087508 easy,nonn
%O A087508 0,5
%A A087508 _Paul Barry_, Sep 11 2003