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%I #25 Aug 08 2021 09:36:37
%S 3,13,55,221,887,3549,14199,56797,227191,908765,3635063,14540253,
%T 58161015,232644061,930576247,3722304989,14889219959,59556879837,
%U 238227519351,952910077405,3811640309623,15246561238493,60986244953975
%N Base 4 digits are, in order, the first n terms of the periodic sequence with initial period 3,1.
%H Harvey P. Dale, <a href="/A037583/b037583.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (4,1,-4).
%F G.f.: x*(3+x) / ( (x-1)*(4*x-1)*(1+x) ). - _R. J. Mathar_, Apr 26 2015
%F a(n) = floor(13*4^n/15). - _Karl V. Keller, Jr._, Aug 04 2021
%e In base 4, we get: 3, 31, 313, 3131, 31313, 313131, 3131313, 31313131, ... = A037589.
%t Table[FromDigits[PadRight[{},n,{3,1}],4],{n,30}] (* or *) LinearRecurrence[ {4,1,-4},{3,13,55},30] (* _Harvey P. Dale_, Sep 20 2020 *)
%o (Python) print([13*4**n//15 for n in range(1,30)]) # _Karl V. Keller, Jr._, Aug 04 2021
%Y Cf. A007090, A037584, A037585, A037586, A037587, A037588, A037589.
%K nonn,base,easy
%O 1,1
%A _Clark Kimberling_