Iteration of Cof_p(Sigma(m))

[y.kohmoto]

    Hello, seqfans.
    I defined an idea "Sigma p height" as follows.

    It is related with a calculation of Amicable number.

    a(n) : 0, 2, 1, 2, 2, 2, 1, 3, 3, 4, 2, 2, 2, 2, 2, 2, 4, 3, 3, 2

    Let f(m) denote Sigma(m)/p^r. Where p^r is the highest power of p
dividing Sigma(m).
    a(n) gives the minimal number of f(m)s that f(f(.....(m)))=1 for p=2.
    example :
    f(10)= Sigma(10)/2^r=3^2, f(3^2)=13 , f(13 )=7, f(7)=1 , four f(m)s are
needed. So, a(10)=4

    I conjectured  " If p=2 then for any number n, the a(n) is finite ".  Is
it correct?

    But,  it is not correct for p=5 .
    Because, f(28)=Sigma(28)/5^r=2^3*7, f(2^3*7)=2^3*3, f(2^3*3)=2^2*3,
f(2^2*3)=2^2*7=28 .

    Yasutoshi