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Fix g and consider primes p such that g is a generator of the (Z/pZ)* (so that the base-g expansion of 1/p has full period length p - 1). Heuristically the base-g digits of the periodic chunk of 1/p for such p should become uniformly distributed as p ---> infinity.

Has this result been proven? If not, have any partial results in this direction been proven?

[EDIT: I forgot to specify that I'm assuming Artin's primitive root conjecture here so that we know that there are infinitely many such primes.]

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2 Answers 2

The reciprocals of prime powers are good models for normal numbers (they satisfy a weak form of normality). The property described in Aaron Meyerowitz's answer extends to fractions $1/p^n$ and their corresponding primitive root base (at least asymptotically) as is proved in "The reciprocals of integral powers of primes and normal numbers" by R.G. Stoneham.

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For each pair $g,p$ the digits are as equally distributed as they can be. In some sense all the digits other than $0$ and $g-1$ are roughly equally likely to be above or below average, however the digits $0$ and $g-1$ are never above average and usually below. Roughly speaking, If we look over $N$ primes $p_1,\cdots,p_N$ then we will see $D=\sum(p_i-1)$ digits and $0$ and $g-1$ will occur about $\frac Dg-\frac N2.$ times.

Let $p=gq+r$ be a prime with $g$ a generator of $\mathbb{Z}/p\mathbb{Z}$. Then the first $p-1$ base-$g$ digits of $1/p$ repeat to give the full expansion. If each appeared equally often that would be $q+\frac{r-1}p$ times each. In fact if $r=1$ then they do appear equally often. In general, $r-1$ of them appear $q+1$ times (call these abundant for $(p,r)$) and the other $g-r$ appear $q$ times (call these deficient for $(p,r)$). $0$ and $g-1$ are always deficient (except for $r=1$).

Example: For $g=21$, if $21$ is a primitive root $\mod p$ then $p \mod 21$ is $2,8,10,11,13$ or $19.$ The various possible digits are:

  • $r=2$ : $10$ abundant
  • $r=8$ : $2,5,7,10,13,15,18$ abundant
  • $r=10$ : $2,4,6,8,10,12,14,16,18$ abundant
  • $r=11$ : $0,2,4,6,8,10,12,14,16,18,20$ deficient
  • $r=13$ : $0,2,5,7,10,13,15,18,21$ deficient
  • $r=19$ : $0,10,20$ deficient.

The first $600$ primes (starting at $p=23$ and ending at $p=13627$) with $21$ as a generator are distributed as follows: $[[2, 127], [8, 110], [10, 83], [11, 121], [13, 76], [19, 83]]$. This seems far from uniform but I can't say what the limiting distribution is.

For the first $60$ primes the distribution is $[[2, 13], [8, 13], [10, 7], [11, 10], [13, 10], [19, 7]].$

The sum of $p-1$ over these $60$ is $25818$ giving an average of $1229 \frac{9}{21}$ for each of the $21$ digits.

The actual counts are:

$[[0, 1205], [1, 1232], [2, 1232], [3, 1232], [4, 1229], [5, 1235], [6, 1229],$ $[7, 1235], [8, 1229], [9, 1232], [10, 1238], [11, 1232], [12, 1229], [13, 1235],$ $ [14, 1229], [15, 1235], [17, 1232], [16, 1229], [19, 1232], [18, 1232], [20, 1205]]$

The fact that $0$ and $20$ come $24 \frac{9}{21}$ below average is because they are below average $60$ times: by $1/21$ and $7/21$ $13$ times each, by $9/21$ and $18/21$ $7$ times each, and by $10/21$ and $12/21$ $10$ times each.

There is nothing very special about $g=21$ except that it is not prime, not too small, has several possible $r$, but not too many. (For $13$ there are also $6$ $r$ values. For $20$ just $4$ and for $22$ there are $10$)

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