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%I #14 Jun 05 2015 10:20:00
%S 1,4,16,64,13,52,208,832,3328,13312,53248,212992,851968,3407872,
%T 13631488,11479231,45916924,183667696,734670784,2938683136,1294379341,
%U 5177517364,20710069456,82840277824,331361111296,1325444445184
%N Let x(1)=1, x(n+1) = (4/3)*x(n) - floor((4/3)*x(n)); then a(n)=x(n)*3^n.
%C It seems that the sequence x(n) = a(n)/3^n which satisfies 0<x(n)<1 is not equidistributed in (0,1) and perhaps lim n -> infinity sum(k=1,n,x(k))/n = C < 0.38 < 1/2
%C It appears that a(n) = 13*4^(n-5) for n > 4. - _Creighton Dement_, Nov 04 2004
%C This is not true, as a(16) mod 13 = 10. - _Reinhard Zumkeller_, Jun 05 2015
%H Reinhard Zumkeller, <a href="/A073533/b073533.txt">Table of n, a(n) for n = 1..1000</a>
%o (Haskell)
%o import Data.Ratio (numerator, (%))
%o a073533 n = a073533_list !! (n-1)
%o a073533_list = f 1 3 1 where
%o f n p3 x = numerator(y * fromIntegral p3) : f (n + 1) (p3 * 3) y
%o where y = z - fromIntegral (floor z); z = 4%3 * x
%o -- _Reinhard Zumkeller_, Jun 05 2015
%Y Cf. A058842.
%K easy,nonn
%O 1,2
%A _Benoit Cloitre_, Aug 27 2002
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