MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.]

share|cite|improve this question

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.

share|cite|improve this answer

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 $10$)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.