Brazilian numbers are so called as they were first formally presented at the 1994 math Olympiad Olimpiada Iberoamericana de Matematica in Fortaleza, Brazil.

Brazilian numbers are defined as:

The set of positive integer numbers where each number N has at least one natural number B where 1 < B < N-1 where the representation of N in base B has all equal digits.



E.G.

 1, 2 & 3 can not be Brazilian; there is no base B that satisfies the condition 1 < B < N-1.
 4 is not Brazilian; 4 in base 2 is 100. The digits are not all the same.
 5 is not Brazilian; 5 in base 2 is 101, in base 3 is 12. There is no representation where the digits are the same.
 6 is not Brazilian; 6 in base 2 is 110, in base 3 is 20, in base 4 is 12. There is no representation where the digits are the same.
 7 is Brazilian; 7 in base 2 is 111. There is at least one representation where the digits are all the same.
 8 is Brazilian; 8 in base 3 is 22. There is at least one representation where the digits are all the same.
 and so on...


All even integers 2P >= 8 are Brazilian because 2P = 2(P-1) + 2, which is 22 in base P-1 when P-1 > 2. That becomes true when P >= 4.

More common: for all all integers R and S, where R > 1 and also S-1 > R, then R*S is Brazilian because R*S = R(S-1) + R, which is RR in base S-1

The only problematic numbers are squares of primes, where R = S. Only 11^2 is brazilian to base 3.

All prime integers, that are brazilian, can only have the digit 1. Otherwise one could factor out the digit, therefore it cannot be a prime number. Mostly in form of 111 to base Integer(sqrt(prime number)). Must be an odd count of 1 to stay odd like primes > 2


Task

Write a routine (function, whatever) to determine if a number is Brazilian and use the routine to show here, on this page;

 the first 20 Brazilian numbers;
 the first 20 odd Brazilian numbers;
 the first 20 prime Brazilian numbers;


See also

 OEIS:A125134 - Brazilian numbers
 OEIS:A257521 - Odd Brazilian numbers
 OEIS:A085104 - Prime Brazilian numbers


 
fn same_digits(x: u64, base: u64) -> bool {
    let f = x % base;
    let mut n = x;
    while n > 0 {
        if n % base != f {
            return false;
        }
        n /= base;
    }
 
    true
}
fn is_brazilian(x: u64) -> bool {
    if x < 7 {
        return false;
    };
    if x % 2 == 0 {
        return true;
    };
 
    for base in 2..(x - 1) {
        if same_digits(x, base) {
            return true;
        }
    }
    false
}
 
fn main() {
    let mut counter = 0;
    let limit = 20;
    let big_limit = 100_000;
    let mut big_result: u64 = 0;
    let mut br: Vec<u64> = Vec::new();
    let mut o: Vec<u64> = Vec::new();
    let mut p: Vec<u64> = Vec::new();
 
    for x in 7.. {
        if is_brazilian(x) {
            counter += 1;
            if br.len() < limit {
                br.push(x);
            }
            if o.len() < limit && x % 2 == 1 {
                o.push(x);
            }
            if p.len() < limit && primes::is_prime(x) {
                p.push(x);
            }
            if counter == big_limit {
                big_result = x;
                break;
            }
        }
    }
    println!("First {} Brazilian numbers:", limit);
    println!("{:?}", br);
    println!("\nFirst {} odd Brazilian numbers:", limit);
    println!("{:?}", o);
    println!("\nFirst {} prime Brazilian numbers:", limit);
    println!("{:?}", p);
 
    println!("\nThe {}th Brazilian number: {}", big_limit, big_result);
}
 