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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A047303 Numbers that are congruent to {0, 1, 2, 3, 4, 6} mod 7.

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%I A047303 #29 Sep 08 2022 08:44:56
%S A047303 0,1,2,3,4,6,7,8,9,10,11,13,14,15,16,17,18,20,21,22,23,24,25,27,28,29,
%T A047303 30,31,32,34,35,36,37,38,39,41,42,43,44,45,46,48,49,50,51,52,53,55,56,
%U A047303 57,58,59,60,62,63,64,65,66,67,69,70,71,72,73,74,76
%N A047303 Numbers that are congruent to {0, 1, 2, 3, 4, 6} mod 7.
%C A047303 Complement of A017041. - _Michel Marcus_, Sep 08 2015
%H A047303 <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,0,1,-1).
%F A047303 G.f.: x^2*(1+x+x^2+x^3+2*x^4+x^5) / ( (1+x)*(1+x+x^2)*(x^2-x+1)*(x-1)^2 ). - _R. J. Mathar_, Oct 25 2011
%F A047303 From _Wesley Ivan Hurt_, Sep 07 2015: (Start)
%F A047303 a(n) = a(n-1) + a(n-6) - a(n-7) for n>7.
%F A047303 a(n) = n + 1 + floor((n-2)/6) - ceiling((n-1)/6) + floor((n-1)/6) - ceiling(n/6) + floor(n/6). (End)
%F A047303 From _Wesley Ivan Hurt_, Jun 15 2016: (Start)
%F A047303 a(n) = (42*n - 51 + 3*cos(n*Pi) + 4*sqrt(3)*cos((1-4*n)*Pi/6) + 12*sin((1+2*n)*Pi/6))/36.
%F A047303 a(6k) = 7k-1, a(6k-1) = 7k-3, a(6k-2) = 7k-4, a(6k-3) = 7k-5, a(6k-4) = 7k-6, a(6k-5) = 7k-7. (End)
%p A047303 A047303:=n->n+1+floor((n-2)/6)-ceil((n-1)/6)+floor((n-1)/6)-ceil(n/6)+floor(n/6): seq(A047303(n), n=1..100); # _Wesley Ivan Hurt_, Sep 07 2015
%t A047303 Select[Range[0, 100], MemberQ[{0, 1, 2, 3, 4, 6}, Mod[#, 7]] &] (* _Vincenzo Librandi_, Sep 08 2015 *)
%t A047303 LinearRecurrence[{1,0,0,0,0,1,-1},{0,1,2,3,4,6,7},80] (* _Harvey P. Dale_, Sep 24 2016 *)
%o A047303 (Magma) [n: n in [0..100] | n mod 7 in [0..4] cat [6]]; // _Vincenzo Librandi_, Sep 08 2015
%Y A047303 Cf. A017041 (7n+5).
%K A047303 nonn,easy
%O A047303 1,3
%A A047303 _N. J. A. Sloane_
%E A047303 More terms from _Vincenzo Librandi_, Sep 08 2015