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Numbers that are congruent to {0, 1, 2} mod 7.
2

%I #45 Sep 08 2022 08:44:57

%S 0,1,2,7,8,9,14,15,16,21,22,23,28,29,30,35,36,37,42,43,44,49,50,51,56,

%T 57,58,63,64,65,70,71,72,77,78,79,84,85,86,91,92,93,98,99,100,105,106,

%U 107,112,113,114,119,120,121,126,127,128,133,134,135,140,141

%N Numbers that are congruent to {0, 1, 2} mod 7.

%H Vincenzo Librandi, <a href="/A047354/b047354.txt">Table of n, a(n) for n = 1..3000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1, 0, 1, -1).

%F a(n) = 7*floor(n/3)+(n mod 3), with offset 0 and a(0)=0. - _Gary Detlefs_, Mar 09 2010

%F From _R. J. Mathar_, Mar 29 2010: (Start)

%F G.f.: x^2*(1+x+5*x^2)/((1+x+x^2) * (x-1)^2).

%F a(n) = a(n-1) + a(n-3) - a(n-4) for n>4. (End)

%F a(n+1) = Sum_{k>=0} A030341(n,k)*b(k) with b(0)=1 and b(k)=7*3^(k-1) for k>0. - _Philippe Deléham_, Oct 24 2011

%F From _Wesley Ivan Hurt_, Jun 08 2016: (Start)

%F a(n) = (21*n-33-12*cos(2*n*Pi/3)+4*sqrt(3)*sin(2*n*Pi/3))/9.

%F a(3k) = 7k-5, a(3k-1) = 7k-6, a(3k-2) = 7k-7. (End)

%F a(n) = n + 4*floor((n-1)/3) - 1. - _Bruno Berselli_, Feb 06 2017

%p seq(7*floor(n/3)+(n mod 3), n=0..60); # _Gary Detlefs_, Mar 09 2010

%t Flatten[{#,#+1,#+2}&/@(7Range[0,20])] (* _Harvey P. Dale_, Mar 05 2011 *)

%o (Magma) [n : n in [0..150] | n mod 7 in [0..2]]; // _Wesley Ivan Hurt_, Jun 08 2016

%Y Cf. A030341.

%Y Cf. similar sequences with formula n+i*floor(n/3) listed in A281899.

%K nonn,easy

%O 1,3

%A _N. J. A. Sloane_