Editing Repdigit (section)

==Primes and repunits==
{{main|Repunit prime}}
For a repdigit to be [[prime number|prime]], it must be a [[repunit]] (i.e. the repeating digit is 1) and have a prime number of digits in its base (except trivial single-digit numbers), since, for example, the repdigit 77777 is divisible by 7, in any base > 7. In particular, as Brazilian repunits do not allow the number of digits to be exactly two, Brazilian primes must have an odd prime number of digits.{{sfnp|Schott|2010|loc=Theorem 2}} Having an odd prime number of digits is not enough to guarantee that a repunit is prime; for instance, 21 = 111<sub>4</sub> = 3 × 7 and 111 = 111<sub>10</sub> = 3 × 37 are not prime. In any given base ''b'', every repunit prime in that base with the exception of 11<sub>''b''</sub> (if it is prime) is a Brazilian prime. The smallest Brazilian primes are
:7 = 111<sub>2</sub>, 13 = 111<sub>3</sub>, 31 = 11111<sub>2</sub> = 111<sub>5</sub>, 43 = 111<sub>6</sub>, 73 = 111<sub>8</sub>, 127 = 1111111<sub>2</sub>, 157 = 111<sub>12</sub>, ... {{OEIS|id=A085104}}

While [[Divergence of the sum of the reciprocals of the primes|the sum of the reciprocals of the prime numbers]] is a divergent series, the sum of the reciprocals of the Brazilian prime numbers is a convergent series whose value, called the "Brazilian primes constant", is slightly larger than 0.33 {{OEIS|id=A306759}}.{{sfnp|Schott|2010|loc=Theorem 4}} This convergence implies that the Brazilian primes form a vanishingly small fraction of all prime numbers. For instance, among the 3.7&times;10<sup>10</sup> prime numbers smaller than 10<sup>12</sup>, only 8.8&times;10<sup>4</sup> are Brazilian.

The [[decimal]] repunit primes have the form <math>R_n=\tfrac{10^n-1}9\ \mbox{with } n\ge3</math> for the values of ''n'' listed in {{OEIS2C|id=A004023}}. It has been conjectured that there are infinitely many decimal repunit primes.<ref>Chris Caldwell, "[http://primes.utm.edu/glossary/page.php?sort=Repunit The Prime Glossary: repunit]" at The [[Prime Pages]]</ref> The [[binary number|binary]] repunits are the [[Mersenne number]]s and the binary repunit primes are the [[Mersenne prime]]s.

It is unknown whether there are infinitely many Brazilian primes. If the [[Bateman–Horn conjecture]] is true, then for every prime number of digits there would exist infinitely many repunit primes with that number of digits (and consequentially infinitely many Brazilian primes). Alternatively, if there are infinitely many decimal repunit primes, or infinitely many Mersenne primes, then there are infinitely many Brazilian primes.{{sfnp|Schott|2010|loc=Sections V.1 and V.2}} Because a vanishingly small fraction of primes are Brazilian, there are infinitely many non-Brazilian primes, forming the sequence
:2, 3, 5, 11, 17, 19, 23, 29, 37, 41, 47, 53, ... {{OEIS|id=A220627}}

If a [[Fermat number]] <math>F_n = 2^{2^n} + 1</math> is prime, it is not Brazilian, but if it is composite, it is Brazilian.{{sfnp|Schott|2010|loc=Proposition 3}}
Contradicting a previous conjecture,{{sfnp|Schott|2010|loc=Conjecture 1}} Resta, Marcus, Grantham, and Graves found examples of [[Sophie Germain prime]]s that are Brazilian, the first one is 28792661 = 11111<sub>73</sub>.<ref>{{cite arXiv|title=Brazilian primes which are also Sophie Germain primes|first1=Jon|last1=Grantham|first2=Hester|last2=Graves|year=2019|class=math.NT|eprint=1903.04577}}</ref>