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Numbers that are congruent to {0, 2, 7} mod 8.
1

%I #25 Dec 31 2022 08:15:18

%S 0,2,7,8,10,15,16,18,23,24,26,31,32,34,39,40,42,47,48,50,55,56,58,63,

%T 64,66,71,72,74,79,80,82,87,88,90,95,96,98,103,104,106,111,112,114,

%U 119,120,122,127,128,130,135,136,138,143,144,146,151,152

%N Numbers that are congruent to {0, 2, 7} mod 8.

%H Vincenzo Librandi, <a href="/A047525/b047525.txt">Table of n, a(n) for n = 1..1000</a>

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

%F From _Chai Wah Wu_, May 28 2016: (Start)

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

%F G.f.: x^2*(x^2 + 5*x + 2)/(x^4 - x^3 - x + 1). (End)

%F a(n) = (24*n + 2*sqrt(3)*sin(2*Pi*n/3) + 12*cos(2*Pi*n/3) - 21)/9. - _Ilya Gutkovskiy_, May 29 2016

%F a(3k) = 8k-1, a(3k-1) = 8k-6, a(3k-2) = 8k-8. - _Wesley Ivan Hurt_, Jun 10 2016

%F E.g.f.: 1 + (3*exp(x)*(8*x - 7) + 2*exp(-x/2)*(6*cos(sqrt(3)*x/2) + sqrt(3)*sin(sqrt(3)*x/2)))/9. - _Stefano Spezia_, Dec 31 2022

%p A047525:=n->(24*n+2*sqrt(3)*sin(2*Pi*n/3)+12*cos(2*Pi*n/3)-21)/9: seq(A047525(n), n=1..100); # _Wesley Ivan Hurt_, Jun 10 2016

%t Flatten[# + {0, 2, 7} &/@(8 Range[0, 30])] (* _Vincenzo Librandi_, May 29 2016 *)

%o (Magma) [n: n in [0..200] | n mod 8 in [0, 2, 7]]; // _Vincenzo Librandi_, May 29 2016

%K nonn,easy

%O 1,2

%A _N. J. A. Sloane_