In recreational mathematics, a repdigit or sometimes monodigit[1] is a natural number composed of repeated instances of the same digit in a positional number system (often implicitly decimal). The word is a portmanteau of "repeated" and "digit".Examples are 11, 666, 4444, and 999999. All repdigits are palindromic numbers and are multiples of repunits. Other well-known repdigits include the repunit primes and in particular the Mersenne primes (which are repdigits when represented in binary).
B
| x | By-1 |
| B-1 |
0<x<B
1<y
| 7 x | 105-1 |
| 10-1 |
A variation of repdigits called Brazilian numbers are numbers that can be written as a repdigit in some base, not allowing the repdigit 11, and not allowing the single-digit numbers (or all numbers will be Brazilian). For example, 27 is a Brazilian number because 27 is the repdigit 33 in base 8, while 9 is not a Brazilian number because its only repdigit representation is 118, not allowed in the definition of Brazilian numbers. The representations of the form 11 are considered trivial and are disallowed in the definition of Brazilian numbers, because all natural numbers n greater than two have the representation 11n - 1.[2] The first twenty Brazilian numbers are
7, 8, 10, 12, 13, 14, 15, 16, 18, 20, 21, 22, 24, 26, 27, 28, 30, 31, 32, 33, ... .
On some websites (including imageboards like 4chan), it is considered an auspicious event when the sequentially-assigned ID number of a post is a repdigit, such as 22,222,222, which is one type of "GET" (others including round numbers like 34,000,000, or sequential digits like 12,345,678).[3] [4]
The concept of a repdigit has been studied under that name since at least 1974,[5] and earlier called them "monodigit numbers".[1] The Brazilian numbers were introduced later, in 1994, in the 9th Iberoamerican Mathematical Olympiad that took place in Fortaleza, Brazil. The first problem in this competition, proposed by Mexico, was as follows:[6]
A number is called "Brazilian" if there exists an integer b such that for which the representation of n in base b is written with all equal digits. Prove that 1994 is Brazilian and that 1993 is not Brazilian.
For a repdigit to be 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. Having an odd prime number of digits is not enough to guarantee that a repunit is prime; for instance, 21 = 1114 = 3 × 7 and 111 = 11110 = 3 × 37 are not prime. In any given base b, every repunit prime in that base with the exception of 11b (if it is prime) is a Brazilian prime. The smallest Brazilian primes are
7 = 1112, 13 = 1113, 31 = 111112 = 1115, 43 = 1116, 73 = 1118, 127 = 11111112, 157 = 11112, ...
While 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 . This convergence implies that the Brazilian primes form a vanishingly small fraction of all prime numbers. For instance, among the 3.7×1010 prime numbers smaller than 1012, only 8.8×104 are Brazilian.
The decimal repunit primes have the form
| n-1}9 with | |
| R | |
| n=\tfrac{10 |
n\ge3
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. 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, ...
Fn=
| 2n | |
| 2 |
+1
p2 = 1 + b + b2 + ... + bq-1 with p, q ≥ 3 primes and b >= 2.
Norwegian mathematician Trygve Nagell has proved[9] that this equation has only one solution when p is prime corresponding to . Therefore, the only squared prime that is Brazilian is 112 = 121 = 111113.There is also one more nontrivial repunit square, the solution (p, b, q) = (20, 7, 4) corresponding to 202 = 400 = 11117, but it is not exceptional with respect to the classification of Brazilian numbers because 20 is not prime.
Perfect powers that are repunits with three digits or more in some base b are described by the Diophantine equation of Nagell and Ljunggren[10]
nt = 1 + b + b2 +...+ bq-1 with b, n, t > 1 and q > 2.
Yann Bugeaud and Maurice Mignotte conjecture that only three perfect powers are Brazilian repunits. They are 121, 343, and 400, the two squares listed above and the cube 343 = 73 = 11118.[11]
| p= | xm-1 | = |
| x-1 |
| yn-1 | |
| y-1 |
Some popular media publications have published articles suggesting that repunit numbers have numerological significance, describing them as "angel numbers".[13] [14] [15]