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Iterating the sum-of-divisors function. (English) Zbl 0866.11003
Let $$\sigma$$ be the sum-of-divisors function and define $$\sigma^m(n)= \sigma(\sigma^{m-1}(n))$$ for $$m>1$$. A number $$n$$ is said to be $$(m,k)$$-perfect if $$\sigma^m(n)=kn$$, so that the classical perfect numbers are $$(1,2)$$-perfect. The authors give a table of all $$(2,k)$$-perfect numbers up to $$10^9$$; for example $$2^5\cdot 3^2\cdot 5\cdot 7^2\cdot 11\cdot 13\cdot 31$$ is $$(2,19)$$-perfect.
Besides the notorious problem of whether there are infinitely many perfect numbers, one may now ask whether every number is $$(m,k)$$-perfect for some suitable values of $$m$$ and $$k$$. The authors have verified that this is so for $$n\leq 1000$$, and some results for $$n\leq 400$$ are given. For example, $$n=389$$ is $$(m,k)$$-perfect with the least value $$m=296$$ and $$k= 2^{93}\cdot 3^{10}\cdots\approx 5\cdot 10^{232}$$. There are 14 values for $$n\leq 400$$ whose corresponding least values for $$m$$ exceed those for $$n$$, namely $$n=3,11,29,53, 58,59,67, 101,109,131, 149,173, 202,239$$. Naturally, the computations for such cases are relatively more complicated.
It had been suggested that, corresponding to any $$n_1,n_2$$, there are $$m_1,m_2$$ such that $$\sigma^{m_1}(n_1)= \sigma^{m_2}(n_2)$$. However, from their computational evidence, the authors believe that this may be false.
##### MSC:
 11A25 Arithmetic functions; related numbers; inversion formulas 11Y70 Values of arithmetic functions; tables
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##### References:
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