Iteration of Cof_p(Sigma(m))
y.kohmoto
zbi74583 at boat.zero.ad.jp
Mon Mar 22 08:36:53 CET 2004
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
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