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A033116 Base-6 digits are, in order, the first n terms of the periodic sequence with initial period 1,0.

%I A033116 #31 Sep 08 2022 08:44:51
%S A033116 1,6,37,222,1333,7998,47989,287934,1727605,10365630,62193781,
%T A033116 373162686,2238976117,13433856702,80603140213,483618841278,
%U A033116 2901713047669,17410278286014,104461669716085,626770018296510,3760620109779061
%N A033116 Base-6 digits are, in order, the first n terms of the periodic sequence with initial period 1,0.
%C A033116 Partial sums of A015540. - _Mircea Merca_, Dec 28 2010
%H A033116 Vincenzo Librandi, <a href="/A033116/b033116.txt">Table of n, a(n) for n = 1..1000</a>
%H A033116 <a href="/index/Rec">Index entries for linear recurrences with constant coefficients</a>, signature (6,1,-6).
%F A033116 From _R. J. Mathar_, Jan 08 2011: (Start)
%F A033116 G.f.: x / ( (1-x)*(1-6*x)*(1+x) ).
%F A033116 a(n) = 6^(n+1)/35 -1/10 -(-1)^n/14. (End)
%F A033116 a(n)=floor(6^(n+1)/35). a(n+1)=sum{k=0..floor(n/2)} 6^(n-2*k). a(n+1)=sum{k=0..n} sum{j=0..k} (-1)^(j+k)*6^j. - _Paul Barry_, Nov 12 2003, index corrected _R. J. Mathar_, Jan 08 2011
%F A033116 a(n) = 5*a(n-1) +6*a(n-2)+1. - _Zerinvary Lajos_, Dec 14 2008
%F A033116 a(n) = floor(6^(n+1)/7)/5 = floor((6*6^n-1)/35) = round((12*6^n-7)/70) = round((6*6^n-6)/35) = ceiling((6*6^n-6)/35). a(n)=a(n-2)+6^(n-1), n>2. - _Mircea Merca_, Dec 28 2010
%p A033116 a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=5*a[n-1]+6*a[n-2]+1 od: seq(a[n], n=1..33);# _Zerinvary Lajos_, Dec 14 2008
%p A033116 A033116 := proc(n) 6^(n+1)/35 -1/10 -(-1)^n/14 ; end proc: # _R. J. Mathar_, Jan 08 2011
%t A033116 Join[{a=1,b=6},Table[c=5*b+6*a+1;a=b;b=c,{n,60}]] (* _Vladimir Joseph Stephan Orlovsky_, Feb 06 2011 *)
%o A033116 (Magma) [Round((12*6^n-7)/70): n in [1..30]]; // _Vincenzo Librandi_, Jun 25 2011
%Y A033116 Cf. A015540
%K A033116 nonn,easy,base
%O A033116 1,2
%A A033116 _Clark Kimberling_