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A037577 Base 5 digits are, in order, the first n terms of the periodic sequence with initial period 1,3.

%I A037577 #25 Jun 13 2015 00:49:16
%S A037577 1,8,41,208,1041,5208,26041,130208,651041,3255208,16276041,81380208,
%T A037577 406901041,2034505208,10172526041,50862630208,254313151041,
%U A037577 1271565755208,6357828776041,31789143880208,158945719401041,794728597005208
%N A037577 Base 5 digits are, in order, the first n terms of the periodic sequence with initial period 1,3.
%C A037577 This is a particular case of the generalized sequence a(n)=((A^n) - B)/(A-B). Sometimes the primes of this form are of interest, see A001348, A014224, A028491. - _Ctibor O. Zizka_, Apr 15 2008
%H A037577 Vincenzo Librandi, <a href="/A037577/b037577.txt">Table of n, a(n) for n = 1..1000</a>
%H A037577 <a href="/index/Rec">Index entries for linear recurrences with constant coefficients</a>, signature (5,1,-5).
%F A037577 a(n) = (5^n - 2)/3 for n odd ; a(n) = (5^n - 1)/3 for n even. - _Ctibor O. Zizka_, Apr 15 2008
%F A037577 a(n) = floor(5^n/3). - _Gary Detlefs_, Sep 06 2010
%F A037577 a(n) = 5*a(n-1) + a(n-2) - 5*a(n-3). - _Charles R Greathouse IV_, Jan 15 2011
%F A037577 G.f.: x*(3*x+1) / ((x-1)*(x+1)*(5*x-1)). - _Colin Barker_, Dec 27 2012
%e A037577 a(1) = (5-1)/3 = 1, a(2) = (5^2-1)/3 = 8. - _Philippe Deléham_, Nov 15 2013
%t A037577 CoefficientList[Series[(3 x + 1)/((x - 1) (x + 1) (5 x - 1)), {x, 0, 30}], x] (* _Vincenzo Librandi_, Oct 21 2013 *)
%K A037577 nonn,base,easy
%O A037577 1,2
%A A037577 _Clark Kimberling_
%E A037577 First formula corrected by _Philippe Deléham_, Nov 14 2013