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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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See the abstract of my PhD thesis, "Generalised quasiperfect numbers", Bulletin Australian Math. Soc., 27 (1983), 153-156, where I consider numbers $n$ with $\sigma(n) = 2n + k^2$, $k$ odd, $(n,k)=1$.

Graeme Cohen

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  • $\begingroup$ Welcome to MathOverflow, Graeme! $\endgroup$ Aug 6, 2017 at 12:14
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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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    $\begingroup$ 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. $\endgroup$ Sep 7, 2013 at 2:38
  • $\begingroup$ Sorry James, but that doesn't work. (Try m=5 and various p.) $\endgroup$ Sep 7, 2013 at 3:13
  • $\begingroup$ @JamesCranch Can you give a numeric example? You need $\sigma(pm)=a p m + b$ and your example doesn't appear of this form. $\endgroup$
    – joro
    Sep 7, 2013 at 12:06
  • $\begingroup$ Sorry: you're right, I'm wrong! $\endgroup$ Sep 7, 2013 at 13:06
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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 $2^n-3$.

Added

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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  • $\begingroup$ That seems to differ from what I was trying to do, insofar as it's actually correct. $\endgroup$ Sep 7, 2013 at 13:57
  • $\begingroup$ @JamesCranch maybe your method will work for prime $p$ and $a$ rational (depending on sigma), yet the question is about integer $a$. $\endgroup$
    – joro
    Sep 7, 2013 at 14:08
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The related equation $$\sigma(n) = An + B(n)$$ where $B(n)$ "is a function that may depend on properties of $n$" is considered in the paper Variations on Euclid’s Formula for Perfect Numbers by Farideh Firoozbakht and Maximilian F. Hasler, published in the Journal of Integer Sequences.

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