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f(n)=f(n/3)+1                          f(1)=1

f(3)=f(3/3)+1 => f(1)+1 => 1+1=2

f(9)=f(3)+1 => 2+1=3

f(27)=f(9)+1 => 3+1=4

f(81)=f(27)+1=>5

f(243)=f(81)+1 => 6

f(729)=f(243)+1=>7

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