%I #39 Sep 08 2022 08:44:56
%S 0,1,2,4,5,6,7,8,10,11,12,13,14,16,17,18,19,20,22,23,24,25,26,28,29,
%T 30,31,32,34,35,36,37,38,40,41,42,43,44,46,47,48,49,50,52,53,54,55,56,
%U 58,59,60,61,62,64,65,66,67,68,70,71,72,73,74,76,77,78,79,80,82,83,84,85
%N Numbers that are congruent to {0, 1, 2, 4, 5} mod 6.
%C Complement of A016945. - _R. J. Mathar_, Feb 25 2008
%C Nonnegative integers m such that floor(2*m^2/12) = 2*floor(m^2/12). See the Crossrefs field of A265187 for similar sequences. - _Bruno Berselli_, Dec 08 2015
%C Also, numbers k such that Fibonacci(k) mod 4 = 0, 1 or 3. - _Bruno Berselli_, Oct 17 2017
%H David A. Corneth, <a href="/A047263/b047263.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,1,-1).
%F O.g.f.: x*(x^2+1)*(x^2+x+1)/((x-1)^2*(x^4+x^3+x^2+x+1)). - _R. J. Mathar_, Feb 25 2008
%F a(n) = a(n-5) + 6 for n > 5. - _R. J. Mathar_, Feb 25 2008
%F a(n) = a(n-1) + a(n-5) - a(n-6) for n > 6. - _Harvey P. Dale_, Oct 05 2014
%F From _Wesley Ivan Hurt_, Aug 16 2016: (Start)
%F a(n) = n + floor((n-4)/5).
%F a(n) = (6*n - 4 - ((n+1) mod 5))/5.
%F a(5k) = 6k-1, a(5k-1) = 6k-2, a(5k-2) = 6k-4, a(5k-3) = 6k-5, a(5k-4) = 6k-6. (End)
%F Sum_{n>=2} (-1)^n/a(n) = log(2+sqrt(3))/sqrt(3) - log(2)/6. - _Amiram Eldar_, Dec 17 2021
%p for n from 0 to 200 do if n mod 6 <> 3 then printf(`%d,`,n) fi: od:
%p A047263:=n->6*floor(n/5)+[0, 1, 2, 4, 5][(n mod 5)+1]: seq(A047263(n), n=0..100); # _Wesley Ivan Hurt_, Aug 16 2016
%t Select[Range[0,100], Mod[#,6]!=3&] (* _Harvey P. Dale_, May 17 2011 *)
%t LinearRecurrence[{1,0,0,0,1,-1},{0,1,2,4,5,6},90] (* _Harvey P. Dale_, Oct 05 2014 *)
%o (Magma) [n : n in [0..100] | n mod 6 in [0, 1, 2, 4, 5]]; // _Wesley Ivan Hurt_, Aug 16 2016
%o (PARI) first(n) = {select(x->(x%6!=3), vector(6*n\5, i, i-1))} \\ _David A. Corneth_, Oct 17 2017
%Y Cf. A016945, A265187.
%K nonn,easy
%O 1,3
%A _N. J. A. Sloane_, Dec 11 1999
%E More terms from _James A. Sellers_, Feb 19 2001