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A number $n \in \mathbb{N}$ is called quasi-perfect if $\sigma(n) = 2n+1$, where $\sigma$ is the sum of divisors function. It is known that if $n$ is quasi-perfect, then it must be the square of an odd integer (an oddly beautiful Putnam problem from the 1970s). To date no number is known to be quasi-perfect.

My question concerns the existence of 'generalized' quasi-perfect numbers, or rather, let $a,b \in \mathbb{N} \cup \{0\}$ be fixed integers, and call an integer $n$ $(a,b)$ quasi-perfect if it satisfies $\sigma(n) = an+b$.

Are there any known values of $a,b$ for which the number of $(a,b)$ quasi-perfect numbers are known to be infinite?

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I wondered about this a while ago and dblues had some nice preliminary observations in the very last post of this AoPS thread: – Daniel m3 Sep 6 '13 at 16:42
Primes. You might consider negative b, otherwise the answer may turn out to be no. – The Masked Avenger Sep 6 '13 at 16:44

Clearly, if $n=p$ whith $p$ being prime answers the question. $\sigma(p)=p+1$ which proves the case for $a=b=1$

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Or, more generally, if you take any $m$, then $n=pm$ with $p$ coprime to $m$ gives $\sigma(pm)=(p+1)\sigma(m)$, which means $a=b=\sigma(m)$ works. – James Cranch Sep 7 '13 at 2:38
Sorry James, but that doesn't work. (Try m=5 and various p.) – The Masked Avenger Sep 7 '13 at 3:13
@JamesCranch Can you give a numeric example? You need $\sigma(pm)=a p m + b$ and your example doesn't appear of this form. – joro Sep 7 '13 at 12:06
Sorry: you're right, I'm wrong! – James Cranch Sep 7 '13 at 13:06

Unfortunately, I can only give some non-answers to this question.

In the case $b=0$, I believe the answer is still unknown. There are some references in Sandor, et al, The Handbook of Number Theory starting on page 105. In particular, the best that appears to be known is the following upper bound: $$ \#\{n \le x | \sigma(n) = a n\} \le c x^{c' \log \log \log x/ \log \log x} $$ for all $a \in \mathbb{Q}$. ($c,c'$ are independent of $a$.)

I believe Konstantinos's example could be extended slightly: if $a=1$ and $b\neq 1$, then the number of solutions should be finite.

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For similar questions searching OEIS might help.

Trying $\sigma(n) = 2 n + 2$ returns A088831 Numbers n whose abundance sigma(n)-2n=+2

A comment in the sequence:

If $2^n-3$ is prime (n is a term of A050414) then $2^{n-1}(2^n-3)$ is in the sequence.

I supposes it plausible that there infinitely many primes of the form $2^n-3$.


There are provable solutions for $a=2$ and for $a=k$ when k-multiperfect numbers $\sigma(m) = k m$ exist.

Let $m$ be perfect number ($\sigma(m)=2m$) and $p$ a prime coprime to $m$.

$\sigma( m p ) = 2 m ( p + 1 ) = 2 m p + 2 m$

$a=2, b=2 m$.

For k-multiperfect numbers $m$, $a=k, b= k m$.

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That seems to differ from what I was trying to do, insofar as it's actually correct. – James Cranch Sep 7 '13 at 13:57
@JamesCranch maybe your method will work for prime $p$ and $a$ rational (depending on sigma), yet the question is about integer $a$. – joro Sep 7 '13 at 14:08

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