Permutable Prime with more than 3 Digits is Probably Repunit
Conjecture
Let $p$ be a permutable prime with more than $3$ digits.
Then $p$ is very probably repunit.
Hence the next permutable prime after $991$ is the repunit prime $R_{19}$:
- $1 \, 111 \, 111 \, 111 \, 111 \, 111 \, 111$
Historical Note
It is trivially true, as shown in Digits of Permutable Prime, that a permutable prime with more than $1$ digit can contain only the digits from the set $\set {1, 3, 7, 9}$.
T.N. Bhargava and P.H. Doyle demonstrated in a $1974$ paper that a permutable prime cannot contain all of $1$, $3$, $7$ and $9$.
Subsequently, Allan W. Johnson showed that a non-repunit permutable prime is of the form $\sqbrk {aaa \cdots aab}$ and have more than $9$ billion digits.
In $1995$, Dmitry Mavlo determined further restrictions on the structure of permutable primes.
However, it still has not been proven that a permutable prime with more than $3$ digits is always a repunit prime.
Sources
- Sep. 1974: T.N. Bhargava and P.H. Doyle: On the Existence of Absolute Primes (Math. Mag. Vol. 47, no. 4: p. 233) www.jstor.org/stable/2689222
- Sept. 1975: Allan W. Johnson: Problems: $953$: Absolute Primes (Math. Mag. Vol. 48, no. 4: pp. 239 – 240) www.jstor.org/stable/2690361
- Mar. 1977: Allan W. Johnson: Solutions: $953$: Absolute Primes (Math. Mag. Vol. 50, no. 2: pp. 100 – 103) www.jstor.org/stable/2689738
- Jul. 1995: Dmitry Mavlo: Absolute Prime Numbers (Math. Gazette Vol. 79, no. 485: pp. 299 – 304) www.jstor.org/stable/3618302
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $113$