Define the D-Ratio as the ratio of a natural number $n$ as: the sum of $n$'s Divisors, excluding 1 and $n$ divided by $n$ itself. [Thus the D-Ratio of $24$ is $$\frac{2 + 3 + 4 + 6 + 8 + 12}{24} = \frac{35}{24}.]$$

Define the cumulative D-Ratio Average to $n$ as the average D-Ratio of the natural numbers up to and including $n$. [Thus the D-Ratio Average to $10$ is $$\frac{0 + 0 + 0 + \frac{1}{2} + 0 + \frac{5}{6} + 0 + \frac{3}{4} + \frac{1}{3} + \frac{7}{10}}{10} = \frac{187}{600}.]$$

As $n$ increases, the D-Ratio Average to $n$ also increases (not with each step, but generally speaking), but it converges because -- as far as I can tell -- D-Ratios don't rise above $5.113$.

To what value does the D-Ratio Average converge?

  • 2
    $\begingroup$ You can make the $D$-ratio as large as you like, you just need enough prime factors. $\endgroup$ – Daniel Fischer Aug 27 '15 at 19:55

These notes derive (equation ($4$) on p. $4$)

$$ \sum_{k\le n}\frac{\sigma(k)}k=\frac{\pi^2}6n+O(\log n)\;. $$

Thus for your ratio $D(n)$ we have

$$ \sum_{k\le n}\frac{\sigma(k)-k-1}k=\left(\frac{\pi^2}6-1\right)n+O(\log n)\;, $$

and dividing by $n$ shows that the average converges to

$$ \frac{\pi^2}6-1\approx0.645\;. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.