59,281

The target audience of this blog (especially the mathematics) is usually professional mathematicians in the Langlands program. I do sometimes, however, have posts suitable for a broader mathematical audience. Very rarely though do I have anything (possibly) interesting to say to a popular audience. In my recent talk in the Number Theory Web Seminar, I gave a talk about some math that I’ve discussed with Soundararajan (and which will possibly be written up at some day) about the “average” digit of \(1/p\) in its decimal expansion, in particular, discussing the distribution of primes for which the average digit of \(1/p\) is less than, equal to, or greater than \(4.5\) respectively. An easy argument using Cebotarev shows that the density of primes for which the average is exactly \(4.5\) is \(2/3\). More subtle, however, is that there are more primes for which the average is less than \(4.5\) than greater than \(4.5\), but still the (upper and lower) density of primes for which the average is greater than \(4.5\) is still positive, assuming GRH (the actual percentages of primes with digit average less than, equal to, and greater than \(4.5\) are approximately 28%, 67%, and 5% respectively).

I think the talk went well, and one reason I suspect is that it was self-contained. Moreover, quite a lot of the setup was completely elementary, although certainly it did move towards deeper topics (Kummer’s Conjecture and work of Patterson and Heath-Brown on equidistribution of Gauss sums, and work of Granville–Soundararajan on the distribution of L-values), it was a result that could more or less be appreciated by an undergraduate.

I decided that this was the time — if ever — that I should make a video post. I decided to make a “numberphile” style video — complete with brown paper and a title consisting of a single number — by taking my talk and significantly scaling back the mathematical content. My first attempt was, to put it mildly, a bit of a disaster. First of all, the aspects of making a video that I know nothing about (lighting, audio, glare, video, editing) were unsurprisingly a complete mess and a distraction from the actual mathematics. Second, my resident expert felt that it was still a bit too long, a bit too much like a recording of some lecture, and lacking a hook. So I cut down the script and made a second even more elementary version. This version (unfortunately) no longer has me writing on physical brown paper, but it might at least reach a bare minimum audio/video quality.

Just in case you want to skip the video and skip straight to the challenge problem, here it is:

Guess/Conjecture: Let \(p \ne 2,5\) be prime, and let \(C(p)\) denote the average of the digits of \(1/p\) in its decimal expansion. (Since the digits repeat this makes sense.) Then the maximum of \(C(p)\) for all primes is achieved by \(p = 59281\), with:

\(\displaystyle{C(59281) = \frac{486}{95} = 5.11 \ldots} \)

\(\displaystyle{
\begin{aligned}
\frac{1}{59281} = & \ 0.\overline{000016868811254870869249843963495892444459438943337662994} \\
& \ \overline{88875018977412661729727906074458932879} \end{aligned}}\)

A rough heuristic why this should be true: if the period of \(p\) is sufficiently large, then, if the digits are sufficiently random, the probability that the average deviates that much from \(4.5\) becomes exponentially small. Since there are not that many primes with small period, this leads to the heuristic that all but finitely many primes should have \(C(p)\) very close to \(4.5\). Moreover, it suggests finding them as factors of \(10^n – 1\) for small values of \(n\). (\(59281\) is a factor of \(10^{95} – 1\).) Making the above idea more precise suggests that it is highly unlikely to find a counterexample with period more than \(400\) or so. Now pari/gp can’t factor most numbers of this form even for small \(n\), but there is a second competing heuristic. If \(p\) is too large and still has small period, then because \(1/p\) starts out with a bunch of zeros, this suppresses the digit average. So any big prime factors of \(10^{n} – 1\) that pari/gp doesn’t find probably won’t be counterexamples anyway. Note this secondary effect also explains why \(C(p)\) can be significantly less than \(4.5\) — if \(p = (10^q – 1)/9\) is prime, for example, then \(C(p) = 9/q\). Since one expects infinitely many primes of this form (\(q = 2, 19, 23, 317, 1031, \ldots\)) one expects that \(C(p)\) can be arbitrarily small.

That said, I certainly have not done any significant computation on this question — possibly pari/gp is not finding \(10\) digit factors of \(10^n – 1\) for odd \(n < 400\) --- it was just an idle question I added to the end of my talk for fun. Hence:

  1. I offer a beer to the person who finds the first counterexample.
  2. I offer a bottle of fine Australian wine to the first person who proves the result. Proofs assuming GRH, for example, are certainly acceptable.

Probably the first thing to try (in order to look for a counter-example) would be to test all primes \(p < 10^{10}\) (say) which are factors of \(10^n - 1\) for some odd \(n < 1000\) or so.

Edit 08/25/21: Update from the youtube link: Matthew Bolan has carried out the computation I suggested above, using in addition information about the factorization of \(10^n – 1\) for small \(n\) given at the Cunningham Project (Jonathan Webster told me about this link). The current records for the primes \(p\) with the six highest values of \(A(1/p)\) are given in the following table. (I had already found the four smallest of these primes in my initial search.) After this computation, it looks like my beer is pretty safe!

Prime \(A(1/p)\) Period
\(59281\) \(5.115789474\) \(95\)
\(307627\) \(4.898734177\) \(79\)
\(9269802917\) \(4.866028708\) \(209\)
\(53 \) \(4.846153846\) \(13\)
\(173\) \(4.813953488\) \(43\)
\(561470969\) \(4.803108808\) \(193\)
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14 Responses to 59,281

  1. José Hdz says:

    Do you know if your talk was recorded?

    • Persiflage says:

      It was not! (Actually I requested this, because it was a talk about some math for which all the details are not completely written down…). Those caveats don’t apply to the math in the video above, however.

      (Also slightly modified your comment because honorifics are not required on blog comments! and I also maintain the illusion of pseudo-anonymity of this blog — purely for google indexing purposes at least)

  2. Persiflage says:

    In binary, the corresponding prime could well be \(p = 4721\) with period \(295\), where the string of length \(295\) has \(160\) ones and \(135\) zeros for a corresponding average

    \(C_2(4721) = \frac{160}{295} = \frac{32}{59} = 0.542\ldots \)

  3. Jon Awbrey says:

    I can’t imagine this will help with the problem, it’s just a thing I do with interesting numbers I encounter …

    See Riffs and Rotes for the basic idea.

    \( \begin{array}{lll}
    59281 & = & p_{5995} \\
    & = & p_{5 \cdot 11 \cdot 109} \\
    & = & p_{p_{3} p_{5} p_{29}} \\
    & = & p_{p_{p_{2}} p_{p_{3}} p_{p_{10}}} \\
    & = & p_{p_{p_{p_{1}}} p_{p_{p_{2}}} p_{p_{p_{1} p_{3}}}} \\
    & = & p_{p_{p_{p_{1}}} p_{p_{p_{p_{1}}}} p_{p_{p_{1} p_{p_{2}}}}} \\
    & = & p_{p_{p_{p_{1}}} p_{p_{p_{p_{1}}}} p_{p_{p_{1} p_{p_{p_{1}}}}}} \\
    & = & p_{p_{p_{p}} p_{p_{p_{p}}} p_{p_{p p_{p_{p}}}}}
    \end{array}\)

    Here is the Rote for \(59281\) …

    https://oeis.org/w/images/5/51/Rote_59281.png

  4. JC says:

    This is a very cool problem.

  5. Chen Wang says:

    The factors of small \(10^n-1\) has been documented in, for example, The Cunningham Project.

    • Persiflage says:

      Yes, this was mentioned on the youtube comments. There the user Matthew Bolan reports on carrying out much more extensive computations using the ideas suggested here (and taking advantage of the known factorizations). I guess I should update the post here to give the current records.

      • Jon says:

        I did a somewhat more extensive search. I took the known factors of 10^n-1 up to n=300000 from Matoko Kamada’s site (https://stdkmd.net/nrr/repunit/). I tested each of them, except 75 really large ones that I didn’t feel like modifying my script to deal with. Here is the top 10 list:
        [5.116,59281,95]
        [4.899,307627,79]
        [4.866,9269802917,209]
        [4.846,53,13]
        [4.846,53,13]
        [4.846,53,13]
        [4.814,173,43]
        [4.814,173,43]
        [4.803,561470969,193]
        [4.785,6163,79]

        I don’t think it’s worthwhile to continue, given how much evidence has accrued.

        • Jon Grantham says:

          OK, I see you’ve edited the original post to include these. The point is that extending the search doesn’t turn up anything interesting.

          The other point is that I forgot to pipe the output through uniq.

  6. Jonathan Webster says:

    I ran a “quick” check on every prime less than 3.5*10^8 with odd period. I found nothing to contribute to the table above of records. I did the check regardless the size of the period.

  7. Jon Awbrey says:

    I collected the above graphics and notes on the following page.

    Riffs and Rotes • 59281

  8. Pingback: What would a good ICM talk look like? | Persiflage

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