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A047226 Numbers that are congruent to {0, 1, 2, 3, 4} mod 6; a(n)=floor(6(n-1)/5).

%I A047226 #38 Sep 08 2022 08:44:56
%S A047226 0,1,2,3,4,6,7,8,9,10,12,13,14,15,16,18,19,20,21,22,24,25,26,27,28,30,
%T A047226 31,32,33,34,36,37,38,39,40,42,43,44,45,46,48,49,50,51,52,54,55,56,57,
%U A047226 58,60,61,62,63,64,66,67,68,69,70,72,73,74,75,76,78,79
%N A047226 Numbers that are congruent to {0, 1, 2, 3, 4} mod 6; a(n)=floor(6(n-1)/5).
%H A047226 Vincenzo Librandi, <a href="/A047226/b047226.txt">Table of n, a(n) for n = 1..1000</a>
%H A047226 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,1,-1).
%F A047226 G.f.: x^2*(1+x+x^2+x^3+2*x^4) / ( (x^4+x^3+x^2+x+1)*(x-1)^2 ). - _R. J. Mathar_, Oct 08 2011
%F A047226 From _Wesley Ivan Hurt_, Sep 17 2015, Jul 16 2013: (Start)
%F A047226 a(n) = floor( 6*(n-1)/5 ).
%F A047226 a(n) = a(n-1) + a(n-5) - a(n-6) for n>6.
%F A047226 a(n) = n - 1 + floor((n-1)/5). (End)
%F A047226 Sum_{n>=2} (-1)^n/a(n) = (9-4*sqrt(3))*Pi/36 + log(2+sqrt(3))/(2*sqrt(3)) + log(2)/6. - _Amiram Eldar_, Dec 17 2021
%p A047226 A047226 := proc(n)
%p A047226     option remember;
%p A047226     if n <= 6 then
%p A047226         op(n,[0,1,2,3,4,6]) ;
%p A047226     else
%p A047226         procname(n-1)+procname(n-5)-procname(n-6) ;
%p A047226     end if;
%p A047226 end proc: # _R. J. Mathar_, Jul 25 2013
%t A047226 Select[Range[0, 100], MemberQ[{0, 1, 2, 3, 4}, Mod[#, 6]]&] (* _Vincenzo Librandi_, Jan 06 2013 *)
%o A047226 (Magma) [n: n in [0..100] | n mod 6 in [0..4]]; // _Vincenzo Librandi_, Jan 06 2013
%Y A047226 Cf. A032766, A004773, A001068, this, A047368, A004777, A248375.
%K A047226 nonn,easy
%O A047226 1,3
%A A047226 _N. J. A. Sloane_
%E A047226 Explicit formula added to definition by _M. F. Hasler_, Oct 05 2014