Problem Statement

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.

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

Let's start with the solution:

// Generate Brazilian sequence. Nigel Galloway: August 13th., 2019
let isBraz α=let mutable n,i,g=α,α+1,1 in (fun β->(while (i*g)<β do if g<α-1 then g<-g+1 else (n<-n*α; i<-n+i; g<-1)); β=i*g)

let Brazilian()=let rec fN n g=seq{if List.exists(fun α->α n) g then yield n
                                   yield! fN (n+1) ((isBraz (n-1))::g)}
                fN 4 [isBraz 2]


Brazilian() |> Seq.take 20 |> Seq.iter(printf "%d "); printfn ""


Brazilian() |> Seq.filter(fun n->n%2=1) |> Seq.take 20 |> Seq.iter(printf "%d "); printfn ""


Brazilian() |> Seq.filter isPrime |> Seq.take 20 |> Seq.iter(printf "%d "); printfn ""


printfn "%d" (Seq.item 99999 Brazilian)