Artin's Constant

My diurnal age today is 28229, a prime, with a reciprocal that has a period of 28228. It is thus one of those primes \(p\) with a period that is given by \( \dfrac{1}{p-1}\).

I got to wondering what proportion of primes have reciprocals that are one less than the prime itself. I asked Gemini. It's response follows and comprises the rest of this post.

Approximately 37.3956% of all prime numbers have reciprocals with a repeating decimal period of maximum length, \(p - 1\). In number theory, primes with this property are known as full reptend primes (or long primes) in base 10.

The Number Theory Behind Maximal Periods

The number of digits in the repeating period of the reciprocal \(1/p\) is determined by the multiplicative order of \(10\) modulo \(p\), denoted as \(\text{ord}_p(10)\).

By Fermat's Little Theorem, for any prime \(p \neq 2, 5\), the period length \(k\) is the smallest positive integer such that:

\[10^k \equiv 1 \pmod{p}\]

Since the multiplicative group \((\mathbb{Z}/p\mathbb{Z})^\times\) has order \(\phi(p) = p - 1\), Lagrange's theorem dictates that the period length \(k\) must always divide \(p - 1\).

For the period to achieve the absolute maximum length of \(p - 1\), the integer \(10\) must be a primitive root modulo \(p\). When \(10\) is a primitive root, the successive powers of \(10 \pmod{p}\) cycle through every possible non-zero remainder from \(1\) to \(p - 1\) before repeating, generating a cyclic period of length \(p - 1\).

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Artin's Constant and Asymptotic Density

The exact asymptotic density of primes for which a given integer is a primitive root is governed by Artin's conjecture on primitive roots.

Because the base \(a = 10\) is neither a perfect square nor \(-1\), Artin's conjecture predicts that the proportion of full reptend primes among all primes up to \(N\) converges to a universal mathematical constant as \(N \to \infty\). This value is known as Artin's constant (\(C_{\text{Artin}}\)), defined by the infinite product over all prime numbers \(q\):

\[C_{\text{Artin}} = \prod_{q \text{ prime}} \left(1 - \frac{1}{q(q-1)}\right) \approx 0.3739558136\dots\]

While Emil Artin's conjecture remains unconditionally unproven for specific bases like 10, Christopher Hooley proved in 1967 that the asymptotic density is exactly \(C_{\text{Artin}}\) under the assumption of the Generalized Riemann Hypothesis (GRH). Extensive computational verification confirms that the distribution stabilizes at this ~37.3956% threshold (cataloged as decimal expansion A005596 in the OEIS).

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Period Lengths of Small Primes

Comparing the first few primes highlights how the multiplicative order behaves when \(10\) is (and is not) a primitive root:

Prime (\(p\))	Reciprocal (\(1/p\))	Period Length (\(\text{ord}_p(10)\))	Full Reptend? (\(k = p - 1\))
3	\(0.\overline{3}\)	\(1\)	No
7	\(0.\overline{142857}\)	\(6\)	Yes
11	\(0.\overline{09}\)	\(2\)	No
13	\(0.\overline{076923}\)	\(6\)	No
17	\(0.\overline{0588235294117647}\)	\(16\)	Yes
19	\(0.\overline{052631578947368421}\)	\(18\)	Yes
23	\(0.\overline{0434782608695652173913}\)	\(22\)	Yes

The sequence of base-10 full reptend primes continues: \(7, 17, 19, 23, 29, 47, 59, 61, 97, 109, 113, 131, 149, \dots\) (OEIS sequence A001122). Noticeably, primes like \(13\) (where \(10^6 \equiv 1 \pmod{13}\)) and \(37\) (where \(10^3 \equiv 1 \pmod{37}\)) fall into sub-cyclic periods because their multiplicative orders are proper divisors of \(p - 1\).