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Base 6 digits are, in order, the first n terms of the periodic sequence with initial period 1,3.
1

%I #24 Sep 08 2022 08:44:52

%S 1,9,55,333,1999,11997,71983,431901,2591407,15548445,93290671,

%T 559744029,3358464175,20150785053,120904710319,725428261917,

%U 4352569571503,26115417429021,156692504574127,940155027444765,5640930164668591

%N Base 6 digits are, in order, the first n terms of the periodic sequence with initial period 1,3.

%H Vincenzo Librandi, <a href="/A037578/b037578.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec">Index entries for linear recurrences with constant coefficients</a>, signature (6,1,-6).

%F a(n):=5*a(n-1)+6*a(n-2)+4, a(0)=0, a(1)=1. [_Zerinvary Lajos_, Dec 14 2008]

%F a(n)= 6*a(n-1) +a(n-2) -6*a(n-3). a(n) = 9*6^n/35 - 2/5 + (-1)^n/7. [_R. J. Mathar_, Oct 05 2009]

%F G.f.: x*(3*x+1) / ((x-1)*(x+1)*(6*x-1)). [_Colin Barker_, Dec 27 2012]

%p a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=5*a[n-1]+6*a[n-2]+4 od: seq(a[n], n=1..33);# _Zerinvary Lajos_, Dec 14 2008

%t CoefficientList[Series[(3 x + 1)/((x - 1) (x + 1) (6 x - 1)), {x, 0, 30}], x] (* _Vincenzo Librandi_, Oct 21 2013 *)

%o (Magma) [9*6^n/35-2/5+(-1)^n/7: n in [1..30]]; // _Vincenzo Librandi_, Oct 21 2013

%K nonn,base,easy

%O 1,2

%A _Clark Kimberling_