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Why do the first 100,000 zeroes of the Riemann Zeta function have double-digit sequence count discontinuities at 00,11,22,33,44,55,66,77,88,99?

I was investigating Benford Law type behaviors and was running some analyses on the web page "Andrew Odlyzko: Tables of zeros of the Riemann zeta function" ... more specifically the first 100,000 zeroes. The web page is First 100k zeroes of Riemann Zeta function and I simply performed a web browser search for each of the 100 two-digit sequences from 00 to 99.

What I found was as follows:

Fig. 1.

I understand why the "01" through "09" is in general lower, since leading zeroes do not appear. I also understand the discontinuity from 74 to 75 since the first 100,000 zeros list only goes up to 74921.

But why are multiples of eleven incrementally less frequent than their nearby digit sequences.
Noticed that C("01") is greater than C("00"), C("10") and C("12") are greater than C("11"), C("21") and C("23") are greater than C("22"), et cetera, continuing on to C("98") is greater than C("99").

Seems curious to me.

Any insight here would be appreciated.

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    $\begingroup$ So, considering this sample, we could think that the probability that $n+1$-th digit is $k$ provided that $n$-th digit is $k$ is less than $1/10$, right? $\endgroup$
    – ajotatxe
    Jul 14, 2017 at 20:28
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    $\begingroup$ You write *xx*; does that mean you count things like $136.982532\color{red}{33}67$ and not just $\color{red}{33}6.9825321367$? $\endgroup$
    – Hagen von Eitzen
    Jul 14, 2017 at 20:30
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    $\begingroup$ yes .... Hagen von Eitzen ... exactly $\endgroup$
    – phdmba7of12
    Jul 14, 2017 at 20:31
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    $\begingroup$ In other words, if you choose a zero and an arbitrary digit of it, there are, obviously, 10 possibilities for the next digit. But the probability that this next digit is the same as the previous one is less than the probability of any of the other nine digits. $\endgroup$
    – ajotatxe
    Jul 14, 2017 at 20:44
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    $\begingroup$ I was very confused by this question for a few minutes, until I realized that in the expression "xx" above and xx below the graph, you allow x$\neq $x. $\endgroup$
    – Servaes
    Jul 15, 2017 at 23:15

3 Answers 3

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This has nothing to do with the Riemann zeta function, but is rather a property of random sequences of digits. (As a rule there's no reason to assume that there's anything significant about the decimal expansion of any number, unless there's some concrete reason to believe otherwise.)

In particular, assume that we have a random sequence of digits where all of 0, 1, 2, ..., 9 are equally likely. Then you expect any particular two-digit sequence to occur 1 time in 100 as a pair of consecutive digits. Most of the numbers at the web site you've linked to are five digits, then a decimal point, then nine more digits. So in each of these numbers there are 12 possible "slots" where, say, "42" can occur, four before the decimal point and eight after. Across all the numbers there are 1.2 million such "slots" (actually a few less because not all the numbers have five digits before the decimal point) and so you expect (1.2 million)/(100) = 12000 occurrences of "42". This is what you see in your plot, except for the deviation because the sequence of zeros cuts off at 74931.

However, how should your browser behave if the sequence "111" appears? You'll find that it counts that as a single occurrence of "11". But the analysis I just gave would count it twice. So the browser-based counts for a sequence of digits that can overlap itself, such as 11, will be lower. In the literature of combinatorics on words this phenomenon is called "autocorrelation".

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  • $\begingroup$ realized this after posting ... ugh. thanks $\endgroup$
    – phdmba7of12
    Jul 14, 2017 at 22:24
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    $\begingroup$ I would call it a property of text search algorithms rather than a property of random digit sequences, but that's just semantics. $\endgroup$
    – David Z
    Jul 14, 2017 at 22:36
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    $\begingroup$ @DavidZ It's not just semantics. It's a question of how the question is formulated exactly. I would start this answer by explaining what the numbers in the graph mean, then tell that it's a property of random strings. $\endgroup$
    – JiK
    Jul 15, 2017 at 1:29
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When counting all occurrences of digit pairs, I get this graph instead: enter image description here

It does show the 75 cutoff and the no-leading-zero effect, but nothing about double digits appears special. If you count a digit pair at most once per line, Michael Lugo's explanation tells you why things get skewed.

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  • $\begingroup$ I realized this after posting. my bad ... $\endgroup$
    – phdmba7of12
    Jul 14, 2017 at 22:22
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Unlikely that this phenomenon is related to the Riemann zeroes per se. Rather it seems to be related to the choice of listing nine digits after the decimal. How much would things change if the analysis were done on the first 8 digits, or 10, or 20?

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