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In what follows, we let $n$ be a positive integer. The classical sum of divisors of $n$ is given by $\sigma_1(n)$.

Denote the deficiency of $n$ by $D(n)=2n-\sigma_1(n)$, and denote the sum of aliquot divisors of $n$ by $s(n)=\sigma_1(n)-n$.

(Hereinafter, I shall abbreviate the notation for the divisor sum $\sigma_1$ simply as $\sigma$.)

Here is my question:

Is the following inequality true in general, where $\gcd(a,b)=1$? $$D(ab) < D(a)s(b)$$

MY ATTEMPT

Since $\gcd(a,b)=1$, we can rewrite $$D(ab) = 2ab - \sigma(ab) = 2ab - \sigma(a)\sigma(b)$$ $$D(a)s(b) = (2a - \sigma(a))(\sigma(b) - b) = -2ab - \sigma(a)\sigma(b) + 2a\sigma(b) + b\sigma(a).$$

Therefore, we get $$D(ab) - D(a)s(b) = \bigg(2ab - \sigma(a)\sigma(b)\bigg) - \bigg(-2ab - \sigma(a)\sigma(b) + 2a\sigma(b) + b\sigma(a)\bigg),$$ from which we obtain $$D(ab) - D(a)s(b) = 4ab - 2a\sigma(b) - b\sigma(a) = ab + (2ab - 2a\sigma(b)) + (ab - b\sigma(a)) = ab + 2a(b - \sigma(b)) + b(a - \sigma(a)) = ab - 2as(b) - bs(a).$$ Alas, this is where I get stuck. I currently do not see an easy way to get an upper bound of $0$ for $$ab - 2as(b) - bs(a)$$ when $\gcd(a,b)=1$.

Jose Arnaldo Bebita Dris
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  • I am currently running a Pari-GP script in Sage Cell Server to check for counterexamples to the inequality. – Jose Arnaldo Bebita Dris Sep 09 '20 at 23:17
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    There are infinitely many counterexamples. If $a,b$ are distinct odd prime numbers, then $D(ab)-D(a)s(b)=(a-1)(b-2)-2\ge 0$. – mathlove Sep 10 '20 at 05:52
  • Thank you for your comment, @mathlove! This answer to a closely related question encapsulates what I intended to ask, and proves in particular that, for odd perfect numbers $p^k m^2$ with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$, the inequality $$\dfrac{D(p^k m)}{D(p^k)s(m)} < 2$$ holds. – Jose Arnaldo Bebita Dris Sep 10 '20 at 06:02

2 Answers2

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No, it isn't.

There are infinitely many counterexamples.

If $a,b$ are distinct odd prime numbers, then we have $\gcd(a,b)=1$ and $$\begin{align}D(ab)-D(a)s(b)&=(2ab-(a+1)(b+1))-(2a-a-1)(b+1-b) \\\\&=ab-2a-b \\\\&=(a-1)(b-2)-2 \\\\&\gt 0\end{align}$$

mathlove
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I tried to check for counterexamples to the inequality $$D(xy) < D(x)s(y)$$ where $\gcd(x,y)=1$, using the following Pari-GP script in Sage Cell Server:

for(x=1, 1000, for(y=1, 1000, if(2*x*y-sigma(x*y) >= (2*x-sigma(x))*(sigma(y)-y) && gcd(x,y)==1,print(x,"   ",y))))

There were several lines returned. This shows that the inequality is not true in general.

Jose Arnaldo Bebita Dris
  • 8,471
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    Don't worry this has happened to me many times before. Computer algebra systems update function definitions without informing the user of every single change made, just ensure that for every investigation you conduct, you algebraically define every function used yourself. – Adam Ledger Sep 16 '20 at 13:14