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someone not specifically mathematically trained just related this they heard:

Take all the fractions of the form $\frac{1}{d}$ with integers d. Select the ones that have periodic decimal expansion. Take the periodic bits and for each calculate the sum of digits.

They claim the distribution of these sums is somewhat unusual or interesting

Having no idea myself but good reason to be procrastinating (I swear!) I made an experiment in Python (rough code here) and behold, the distribution does look odd:

Histogram plot of the distribution of the digit sums of the periodic bits of the fraction 1/n with n<10^4

Can someone explain what is going on here?

Alessio K
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  • When you say "digit sum", do you mean repeatedly digit summing until you get to a single digit number? – Arthur Aug 13 '20 at 12:54
  • What do you mean "Select the ones with periodic decimal expansion" ? All the $1/d$ have a periodic decimal expansion. – TheSilverDoe Aug 13 '20 at 12:55
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    @TheSilverDoe No, that's not true. They're all eventually periodic, but not necessarily periodic. $\frac 12$ and $\frac 15$ do not have periodic decimal expansions. – Matt Samuel Aug 13 '20 at 13:00
  • @MattSamuel Yes, of course. Thanks for the precision. – TheSilverDoe Aug 13 '20 at 13:07
  • Very interesting. I dabbled in a related sequence about two decades ago, submitting it to the OEIS: https://oeis.org/A048997, the main differences being that 1) I included terminating decimals, 2) for periodic expansions, I took the entire string till first repeat and 3) I only considered the sequence (and then only a few terms) and not so much the distribution or density of the sums. – Deepak Aug 13 '20 at 13:42

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If $\frac 1d=.\overline S$ where $S$ has length $n$ then $$\frac {10^n}d=S+\frac 1d\implies S=\frac {10^n-1}d$$

And since $9\,|\,(10^n-1)$ we see that $9\,|\,S$ unless $3\,|\,d$...so at least $\frac 23$ of your numbers will have iterated digit sum $9$.

To make it even more dramatic, stick with primes $p>5$. Then in all cases $9$ divides the iterated digit sum.

lulu
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