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


#include <stdio.h>
 
typedef char bool;
 
#define TRUE 1
#define FALSE 0
 
bool same_digits(int n, int b) {
    int f = n % b;
    n /= b;
    while (n > 0) {
        if (n % b != f) return FALSE;
        n /= b;
    }
    return TRUE;
}
 
bool is_brazilian(int n) {
    int b;
    if (n < 7) return FALSE;
    if (!(n % 2) && n >= 8) return TRUE;
    for (b = 2; b < n - 1; ++b) {
        if (same_digits(n, b)) return TRUE;
    }
    return FALSE;
}
 
bool is_prime(int n) {
    int d = 5;
    if (n < 2) return FALSE;
    if (!(n % 2)) return n == 2;
    if (!(n % 3)) return n == 3;
    while (d * d <= n) {
        if (!(n % d)) return FALSE;
        d += 2;
        if (!(n % d)) return FALSE;
        d += 4;
    }
    return TRUE;
}
 
int main() {
    int i, c, n;
    const char *kinds[3] = {" ", " odd ", " prime "};    
    for (i = 0; i < 3; ++i) {
        printf("First 20%sBrazilian numbers:\n", kinds[i]);
        c = 0;
        n = 7;
        while (TRUE) {
            if (is_brazilian(n)) {
                printf("%d ", n);
                if (++c == 20) {
                    printf("\n\n");
                    break;
                }
            }
            switch (i) {
                case 0: n++; break;
                case 1: n += 2; break;
                case 2: 
                    do {
                        n += 2;
                    } while (!is_prime(n));
                    break;
            }
        }
    }
 
    for (n = 7, c = 0; c < 100000; ++n) {
        if (is_brazilian(n)) c++; 
    }
    printf("The 100,000th Brazilian number: %d\n", n - 1);
    return 0;
}