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Gerhard Paseman
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I've decided to collect some basic observations and references for the benefit of future readers.

A more challenging problem is to ask for integers $m$ and $p$ such that for all integers $k$, $p_0 = p, p_{k+1}=\sigma(p_k),$ and $p_k = 0 \bmod m$. The current problem adds the restriction that $p=m$, which implies $m$ is a multiperfect number. Since multiperfect numbers are rare, it should be hard to find a metaperfect number, a number $m$ that satisfies $\sigma^k(m) = 0 \bmod m$ for all iterations of $\sigma$.

Indeed, $\sigma(m) \lt \omega(m)$$\sigma(m) \lt m\omega(m)$ for most values of $m$, so for a potentially metaperfect number to exist, we can't depend on $\sigma(p_k)/m$ to be coprime to $m$ for very many $k$. More likely, $\sigma(p_k)/m$, if integral, will share a small factor with $m$ and further iterations of $\sigma$ will avoid certain large prime factors of $m$. This is what was observed, and what I hoped to prove and did not in the other answer.

It is an interesting side question to determine $\min_k g_k$ where $g_0=p_0$ and $g_{k+1} = \gcd(p_{k+1},g_k)$. In particular, do the iterates of $\sigma$ encounter a square or twice a square, regardless of the starting point? If so, then the minimum is odd and likely 1. Otherwise $p$ is a seed for $m$, and $p$ might be useful in looking for multiperfect numbers which are multiples of $m$.

Cohen and te Riele investigated a weaker question: Given $n$ is there a $k$ for which $\sigma^k(n) = 0 \mod n$? They did this in a 1996 paper and asserted through computation that the answer was yes for $n \leq 400$. Their data suggest to me both that the weaker question has an affirmative answer, and that there are no metaperfect numbers or even seeds for a number.

Regarding Gerry Myerson's example in a comment, I think one can construct arbitrarily long finite sequences $p_k$ which satisfy the congruence conditions, but to find such $p$ with $p=m$ will result in very large values even for $k$ as small as $3$. Toward this end, there is a 2009 paper of Katai that I have not found but think will be useful in this study.

I recommend as a starting point for a reference search

Cohen, Graeme L., and Herman JJ te Riele. "Iterating the sum-of-divisors function."

Experimental Mathematics 5.2 (1996): 91-100.

and (if you can get it)

Kátai, I. "On the prime power divisors of the iterates of \phi(n) and \sigma(n), Šiauliai

Math." Semin 4.12 (2009): 125-143.

Gerhard "Not Quite A Research Announcement" Paseman, 2015.07.21

I've decided to collect some basic observations and references for the benefit of future readers.

A more challenging problem is to ask for integers $m$ and $p$ such that for all integers $k$, $p_0 = p, p_{k+1}=\sigma(p_k),$ and $p_k = 0 \bmod m$. The current problem adds the restriction that $p=m$, which implies $m$ is a multiperfect number. Since multiperfect numbers are rare, it should be hard to find a metaperfect number, a number $m$ that satisfies $\sigma^k(m) = 0 \bmod m$ for all iterations of $\sigma$.

Indeed, $\sigma(m) \lt \omega(m)$ for most values of $m$, so for a potentially metaperfect number to exist, we can't depend on $\sigma(p_k)/m$ to be coprime to $m$ for very many $k$. More likely, $\sigma(p_k)/m$, if integral, will share a small factor with $m$ and further iterations of $\sigma$ will avoid certain large prime factors of $m$. This is what was observed, and what I hoped to prove and did not in the other answer.

It is an interesting side question to determine $\min_k g_k$ where $g_0=p_0$ and $g_{k+1} = \gcd(p_{k+1},g_k)$. In particular, do the iterates of $\sigma$ encounter a square or twice a square, regardless of the starting point? If so, then the minimum is odd and likely 1. Otherwise $p$ is a seed for $m$, and $p$ might be useful in looking for multiperfect numbers which are multiples of $m$.

Cohen and te Riele investigated a weaker question: Given $n$ is there a $k$ for which $\sigma^k(n) = 0 \mod n$? They did this in a 1996 paper and asserted through computation that the answer was yes for $n \leq 400$. Their data suggest to me both that the weaker question has an affirmative answer, and that there are no metaperfect numbers or even seeds for a number.

Regarding Gerry Myerson's example in a comment, I think one can construct arbitrarily long finite sequences $p_k$ which satisfy the congruence conditions, but to find such $p$ with $p=m$ will result in very large values even for $k$ as small as $3$. Toward this end, there is a 2009 paper of Katai that I have not found but think will be useful in this study.

I recommend as a starting point for a reference search

Cohen, Graeme L., and Herman JJ te Riele. "Iterating the sum-of-divisors function."

Experimental Mathematics 5.2 (1996): 91-100.

and (if you can get it)

Kátai, I. "On the prime power divisors of the iterates of \phi(n) and \sigma(n), Šiauliai

Math." Semin 4.12 (2009): 125-143.

Gerhard "Not Quite A Research Announcement" Paseman, 2015.07.21

I've decided to collect some basic observations and references for the benefit of future readers.

A more challenging problem is to ask for integers $m$ and $p$ such that for all integers $k$, $p_0 = p, p_{k+1}=\sigma(p_k),$ and $p_k = 0 \bmod m$. The current problem adds the restriction that $p=m$, which implies $m$ is a multiperfect number. Since multiperfect numbers are rare, it should be hard to find a metaperfect number, a number $m$ that satisfies $\sigma^k(m) = 0 \bmod m$ for all iterations of $\sigma$.

Indeed, $\sigma(m) \lt m\omega(m)$ for most values of $m$, so for a potentially metaperfect number to exist, we can't depend on $\sigma(p_k)/m$ to be coprime to $m$ for very many $k$. More likely, $\sigma(p_k)/m$, if integral, will share a small factor with $m$ and further iterations of $\sigma$ will avoid certain large prime factors of $m$. This is what was observed, and what I hoped to prove and did not in the other answer.

It is an interesting side question to determine $\min_k g_k$ where $g_0=p_0$ and $g_{k+1} = \gcd(p_{k+1},g_k)$. In particular, do the iterates of $\sigma$ encounter a square or twice a square, regardless of the starting point? If so, then the minimum is odd and likely 1. Otherwise $p$ is a seed for $m$, and $p$ might be useful in looking for multiperfect numbers which are multiples of $m$.

Cohen and te Riele investigated a weaker question: Given $n$ is there a $k$ for which $\sigma^k(n) = 0 \mod n$? They did this in a 1996 paper and asserted through computation that the answer was yes for $n \leq 400$. Their data suggest to me both that the weaker question has an affirmative answer, and that there are no metaperfect numbers or even seeds for a number.

Regarding Gerry Myerson's example in a comment, I think one can construct arbitrarily long finite sequences $p_k$ which satisfy the congruence conditions, but to find such $p$ with $p=m$ will result in very large values even for $k$ as small as $3$. Toward this end, there is a 2009 paper of Katai that I have not found but think will be useful in this study.

I recommend as a starting point for a reference search

Cohen, Graeme L., and Herman JJ te Riele. "Iterating the sum-of-divisors function."

Experimental Mathematics 5.2 (1996): 91-100.

and (if you can get it)

Kátai, I. "On the prime power divisors of the iterates of \phi(n) and \sigma(n), Šiauliai

Math." Semin 4.12 (2009): 125-143.

Gerhard "Not Quite A Research Announcement" Paseman, 2015.07.21

Source Link
Gerhard Paseman
  • 3.2k
  • 1
  • 18
  • 29

I've decided to collect some basic observations and references for the benefit of future readers.

A more challenging problem is to ask for integers $m$ and $p$ such that for all integers $k$, $p_0 = p, p_{k+1}=\sigma(p_k),$ and $p_k = 0 \bmod m$. The current problem adds the restriction that $p=m$, which implies $m$ is a multiperfect number. Since multiperfect numbers are rare, it should be hard to find a metaperfect number, a number $m$ that satisfies $\sigma^k(m) = 0 \bmod m$ for all iterations of $\sigma$.

Indeed, $\sigma(m) \lt \omega(m)$ for most values of $m$, so for a potentially metaperfect number to exist, we can't depend on $\sigma(p_k)/m$ to be coprime to $m$ for very many $k$. More likely, $\sigma(p_k)/m$, if integral, will share a small factor with $m$ and further iterations of $\sigma$ will avoid certain large prime factors of $m$. This is what was observed, and what I hoped to prove and did not in the other answer.

It is an interesting side question to determine $\min_k g_k$ where $g_0=p_0$ and $g_{k+1} = \gcd(p_{k+1},g_k)$. In particular, do the iterates of $\sigma$ encounter a square or twice a square, regardless of the starting point? If so, then the minimum is odd and likely 1. Otherwise $p$ is a seed for $m$, and $p$ might be useful in looking for multiperfect numbers which are multiples of $m$.

Cohen and te Riele investigated a weaker question: Given $n$ is there a $k$ for which $\sigma^k(n) = 0 \mod n$? They did this in a 1996 paper and asserted through computation that the answer was yes for $n \leq 400$. Their data suggest to me both that the weaker question has an affirmative answer, and that there are no metaperfect numbers or even seeds for a number.

Regarding Gerry Myerson's example in a comment, I think one can construct arbitrarily long finite sequences $p_k$ which satisfy the congruence conditions, but to find such $p$ with $p=m$ will result in very large values even for $k$ as small as $3$. Toward this end, there is a 2009 paper of Katai that I have not found but think will be useful in this study.

I recommend as a starting point for a reference search

Cohen, Graeme L., and Herman JJ te Riele. "Iterating the sum-of-divisors function."

Experimental Mathematics 5.2 (1996): 91-100.

and (if you can get it)

Kátai, I. "On the prime power divisors of the iterates of \phi(n) and \sigma(n), Šiauliai

Math." Semin 4.12 (2009): 125-143.

Gerhard "Not Quite A Research Announcement" Paseman, 2015.07.21