Inrecreational mathematics, arepdigit or sometimesmonodigit^[1] is anatural number composed of repeated instances of the samedigit in apositional number system (often implicitlydecimal). The word is aportmanteau of "repeated" and "digit".Examples are11,666,4444, and999999. All repdigits arepalindromic numbers and are multiples ofrepunits. Other well-known repdigits include therepunit primes and in particular theMersenne primes (which are repdigits when represented in binary).

Any such number can be represented as follows

${\displaystyle \underset{k}{\underbrace{nn\dots nn}}=\frac{(nn-n{)}^k-n^k}{(nn-2\cdot n)\cdot n^{(k-2)}}}$

Where nn is the concatenation of n with n. k the number of concatenated n.

nn can be represented mathematically as

${\displaystyle n\cdot \left({10}^{\lfloor {\log }_{10}(n)\rfloor +1}+1\right)}$

for n = 23 and k = 5, the formula will look like this

${\displaystyle \frac{(2323-23{)}^5-{23}^5}{(2323-2\cdot 23)\cdot {23}^{(5-2)}}=\frac{64363429993563657}{27704259}=\underset{5}{\underbrace{2323232323}}}$

However, 2323232323 is not a repdigit.

Also, any number can be decomposed into the sum and difference of the repdigit numbers.

For example 3453455634 = 3333333333 + (111111111 + (9999999 - (999999 - (11111 + (77 + (2))))))

Repdigits are the representation inbase${\displaystyle B}$ of the number${\displaystyle x\frac{B^y-1}{B-1}}$ where is the repeated digit and is the number of repetitions. For example, the repdigit 77777 in base 10 is${\displaystyle 7\times \frac{{10}^5-1}{10-1}}$.

A variation of repdigits calledBrazilian 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 numbersn greater than two have the representation 11_{n − 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, ... (sequenceA125134 in theOEIS).

On some websites (includingimageboards like4chan), 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"^{[clarification needed]} (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] earlierBeiler (1966) called them "monodigit numbers".^[1] The Brazilian numbers were introduced later, in 1994, in the 9th Iberoamerican Mathematical Olympiad that took place inFortaleza, Brazil. The first problem in this competition, proposed by Mexico, was as follows:^[6]

A numbern > 0 is called "Brazilian" if there exists an integerb such that1 <b <n – 1 for which the representation ofn in baseb is written with all equal digits. Prove that 1994 is Brazilian and that 1993 is not Brazilian.

For a repdigit to beprime, it must be arepunit (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.^[7] 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 baseb, every repunit prime in that base with the exception of 11_b (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, ... (sequenceA085104 in theOEIS)

Whilethe 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 (sequenceA306759 in theOEIS).^[8] 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.

Thedecimal repunit primes have the form${\displaystyle R_n=\frac{{10}^n-1}{9}\text{with }n\ge 3}$ for the values ofn listed inOEIS: A004023. It has been conjectured that there are infinitely many decimal repunit primes.^[9] Thebinary repunits are theMersenne numbers and the binary repunit primes are theMersenne primes.

It is unknown whether there are infinitely many Brazilian primes. If theBateman–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.^[10] 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, ... (sequenceA220627 in theOEIS)

If aFermat number${\displaystyle F_n=2^{2^n}+1}$ is prime, it is not Brazilian, but if it is composite, it is Brazilian.^[11]Contradicting a previous conjecture,^[12] Resta, Marcus, Grantham, and Graves found examples ofSophie Germain primes that are Brazilian, the first one is 28792661 = 11111₇₃.^[13]

The only positive integers that can be non-Brazilian are 1, 6, theprimes, and thesquares of the primes, for every other number is the product of two factorsx andy with 1 <x <y − 1, and can be written asxx in basey − 1.^[14] If a square of a primep² is Brazilian, then primep must satisfy theDiophantine equation

Norwegian mathematicianTrygve Nagell has proved^[15] that this equation has only one solution whenp is prime corresponding to(p,b,q) = (11, 3, 5). 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 baseb are described by theDiophantine equation of Nagell andLjunggren^[16]

Yann Bugeaud and Maurice Mignotte conjecture that only three perfect powers are Brazilian repunits. They are 121, 343, and 400 (sequenceA208242 in theOEIS), the two squares listed above and the cube 343 = 7³ = 111₁₈.^[17]

• The number of ways such that a numbern is Brazilian is inOEIS: A220136. Hence, there exist numbers that are non-Brazilian and others that are Brazilian; among these last integers, some are once Brazilian, others are twice Brazilian, or three times, or more. A number that isk times Brazilian is calledk-Brazilian number.
• Non-Brazilian numbers or 0-Brazilian numbers are constituted with 1 and 6, together with some primes and some squares of primes. The sequence of the non-Brazilian numbers begins with 1, 2, 3, 4, 5, 6, 9, 11, 17, 19, 23, 25, ... (sequenceA220570 in theOEIS).
• The sequence of 1-Brazilian numbers is composed of other primes, the only square of prime that is Brazilian, 121, and composite numbers≥ 8 that are the product of only two distinct factors such thatn =a ×b =aa_b–1 with1 <a <b – 1. (sequenceA288783 in theOEIS).
• The 2-Brazilian numbers (sequenceA290015 in theOEIS) consists of composites and only two primes: 31 and 8191. Indeed, according toGoormaghtigh conjecture, these two primes are the only known solutions of theDiophantine equation:
  ${\displaystyle p=\frac{x^m-1}{x-1}=\frac{y^n-1}{y-1}}$ withx,y > 1 andn,m > 2 :
  • (p, x, y, m, n) = (31, 5, 2, 3, 5) corresponding to 31 = 11111₂ = 111₅, and,
  • (p, x, y, m, n) = (8191, 90, 2, 3, 13) corresponding to 8191 = 1111111111111₂ = 111₉₀, with 11111111111 is therepunit with thirteen digits 1.
• For each sequence ofk-Brazilian numbers, there exists a smallest term. The sequence with these smallestk-Brazilian numbers begins with 1, 7, 15, 24, 40, 60, 144, 120, 180, 336, 420, 360, ... and are inOEIS: A284758. For instance, 40 is the smallest4-Brazilian number with 40 = 1111₃ = 55₇ = 44₉ = 22₁₉.
• In theDictionnaire de (presque) tous les nombres entiers,^[18] Daniel Lignon proposes that an integer ishighly Brazilian if it is a positive integer with more Brazilian representations than any smaller positive integer has. This definition comes from the definition ofhighly composite numbers created bySrinivasa Ramanujan in 1915. The first numbershighly Brazilian are 1, 7, 15, 24, 40, 60, 120, 180, 336, 360, 720, ... and are exactly inOEIS: A329383. From 360 to 321253732800 (maybe more), there are 80 successivehighly composite numbers that are also highly Brazilian numbers, seeOEIS: A279930.

Some popular media publications have published articles suggesting that repunit numbers havenumerological significance, describing them as "angel numbers".^[19][20][21]

• Six nines in pi

• Weisstein, Eric W."Repdigit".MathWorld.
• Problemas IX Olimpíada Iberoamericana de Matemática

Look uprepdigit in Wiktionary, the free dictionary.

• Base-dependent integer sequences

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