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A033118 Base 8 digits are, in order, the first n terms of the periodic sequence with initial period 1,0.

%I A033118 #33 Sep 08 2022 08:44:51
%S A033118 1,8,65,520,4161,33288,266305,2130440,17043521,136348168,1090785345,
%T A033118 8726282760,69810262081,558482096648,4467856773185,35742854185480,
%U A033118 285942833483841,2287542667870728,18300341342965825,146402730743726600
%N A033118 Base 8 digits are, in order, the first n terms of the periodic sequence with initial period 1,0.
%C A033118 Partial sums of A015565. - _Mircea Merca_, Dec 28 2010
%H A033118 Vincenzo Librandi, <a href="/A033118/b033118.txt">Table of n, a(n) for n = 1..1000</a>
%H A033118 <a href="/index/Rec">Index entries for linear recurrences with constant coefficients</a>, signature (8,1,-8).
%F A033118 a(n) = 8*a(n-1) + a(n-2) - 8*a(n-3).
%F A033118 a(n) = 2^(3*n+3)/63 - 1/14 - (-1)^n/18. - _R. J. Mathar_, Jan 08 2011
%F A033118 From _Paul Barry_, Apr 04 2008: (Start)
%F A033118 G.f. x/((1-x^2)*(1-8*x));
%F A033118 a(n) = (1/3)*Sum_{k=0..n} A001045(3k). (End)
%F A033118 a(n) = floor(8^(n+1)/9)/7 = floor((8*8^n-1)/63) = round((8*8^n-8)/63) = round((16*8^n-9)/63) = ceiling((8*8^n-8)/63). a(n) = a(n-2) + 8^(n-1), n > 2. - _Mircea Merca_, Dec 28 2010
%p A033118 seq(1/7*floor(8^(n+1)/9),n=1..22); # _Mircea Merca_, Dec 27 2010
%t A033118 Table[FromDigits[PadRight[{},n,{1,0}],8],{n,20}] (* or *) LinearRecurrence[ {8,1,-8},{1,8,65},20] (* _Harvey P. Dale_, Jan 20 2021 *)
%o A033118 (Magma) [Round((8*8^n-8)/63): n in [1..30]]; // _Vincenzo Librandi_, Jun 25 2011
%Y A033118 Pairwise sums are (8^n - 1)/7 (A023001).
%K A033118 nonn,base,easy
%O A033118 1,2
%A A033118 _Clark Kimberling_