A033116 Base-6 digits are, in order, the first n terms of the periodic sequence with initial period 1,0.
1, 6, 37, 222, 1333, 7998, 47989, 287934, 1727605, 10365630, 62193781, 373162686, 2238976117, 13433856702, 80603140213, 483618841278, 2901713047669, 17410278286014, 104461669716085, 626770018296510, 3760620109779061
Offset: 1
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (6,1,-6).
Crossrefs
Cf. A015540
Programs
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Magma
[Round((12*6^n-7)/70): n in [1..30]]; // Vincenzo Librandi, Jun 25 2011
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Maple
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 A033116 := proc(n) 6^(n+1)/35 -1/10 -(-1)^n/14 ; end proc: # R. J. Mathar, Jan 08 2011
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Mathematica
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 *)
Formula
From R. J. Mathar, Jan 08 2011: (Start)
G.f.: x / ( (1-x)*(1-6*x)*(1+x) ).
a(n) = 6^(n+1)/35 -1/10 -(-1)^n/14. (End)
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
a(n) = 5*a(n-1) +6*a(n-2)+1. - Zerinvary Lajos, Dec 14 2008
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
Comments