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%I #9 Apr 25 2017 18:57:37
%S 0,1,2,3,4,5,6,7,8,1,2,3,4,5,6,7,8,9,2,3,4,5,6,7,8,9,10,11,12,13,14,
%T 15,16,1,2,3,4,5,6,7,8,9,2,3,4,5,6,7,8,9,10,3,4,5,6,7,8,1,2,3,4,5,6,7,
%U 8,9,2,3,4,5,6,7,8,9,10,3,4,5,6,7,8,9,10,11,4,5,6,7,8,9,10,11,12,5,6,7,8,9,10
%N a(0) = 0; a(n) = a(n/3)/3 if n = a(n/3) = 0 (mod 3); a(n) = a(n-1)+1 otherwise.
%C Bounded, conjectured maximum is a(257) = 17.
%H Robert Israel, <a href="/A025482/b025482.txt">Table of n, a(n) for n = 0..10000</a>
%p A[0]:= 0;
%p for n from 1 to 1000 do
%p if n mod 3 = 0 and A[n/3] mod 3 = 0 then A[n]:= A[n/3]/3
%p else A[n]:= A[n-1]+1
%p fi od:
%p seq(A[i],i=0..1000); # _Robert Israel_, Apr 25 2017
%K nonn
%O 0,3
%A _David W. Wilson_
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