Uniform distribution of digits of 1/p

Question

Fix $g$ and consider primes $p$ such that $g$ is a generator of the $(\mathbb{Z}/p\mathbb{Z})^*$ (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 $p\to \infty$.

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.]

Answer 1

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.

Answer 2

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 $10$)

Answer 3

This question was studied by Korobov in the article On the distribution of digits in periodic fractions. He used exponential sums and his asyptotic formula is more precise than Stoneham's result mentioned by Gjergji Zaimi. Korobov also proved asyptotic formula for the number of given blocks in the part of the period. These results are included in his book Exponential Sums and their Applications.