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{{draft task}} 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: all integers, that factor decomposition is '''RS >= 8''', with '''S+1 > R''', are 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
;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|OEIS:A125134 - Brazilian numbers]]''' :* '''[[oeis:A257521|OEIS:A257521 - Odd Brazilian numbers]]''' :* '''[[oeis:A085104|OEIS:A085104 - Prime Brazilian numbers]]'''
AWK
# syntax: GAWK -f BRAZILIAN_NUMBERS.AWK
# converted from C
BEGIN {
split(",odd ,prime ",kinds,",")
for (i=1; i<=3; ++i) {
printf("first 20 %sBrazilian numbers:",kinds[i])
c = 0
n = 7
while (1) {
if (is_brazilian(n)) {
printf(" %d",n)
if (++c == 20) {
printf("\n")
break
}
}
switch (i) {
case 1:
n++
break
case 2:
n += 2
break
case 3:
do {
n += 2
} while (!is_prime(n))
break
}
}
}
exit(0)
}
function is_brazilian(n, b) {
if (n < 7) { return(0) }
if (!(n % 2) && n >= 8) { return(1) }
for (b=2; b<n-1; ++b) {
if (same_digits(n,b)) { return(1) }
}
return(0)
}
function is_prime(n, d) {
d = 5
if (n < 2) { return(0) }
if (!(n % 2)) { return(n == 2) }
if (!(n % 3)) { return(n == 3) }
while (d*d <= n) {
if (!(n % d)) { return(0) }
d += 2
if (!(n % d)) { return(0) }
d += 4
}
return(1)
}
function same_digits(n,b, f) {
f = n % b
n = int(n/b)
while (n > 0) {
if (n % b != f) { return(0) }
n = int(n/b)
}
return(1)
}
{{out}}
first 20 Brazilian numbers: 7 8 10 12 13 14 15 16 18 20 21 22 24 26 27 28 30 31 32 33
first 20 odd Brazilian numbers: 7 13 15 21 27 31 33 35 39 43 45 51 55 57 63 65 69 73 75 77
first 20 prime Brazilian numbers: 7 13 31 43 73 127 157 211 241 307 421 463 601 757 1093 1123 1483 1723 2551 2801
C
{{trans|Go}}
#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; }
{{out}}
First 20 Brazilian numbers:
7 8 10 12 13 14 15 16 18 20 21 22 24 26 27 28 30 31 32 33
First 20 odd Brazilian numbers:
7 13 15 21 27 31 33 35 39 43 45 51 55 57 63 65 69 73 75 77
First 20 prime Brazilian numbers:
7 13 31 43 73 127 157 211 241 307 421 463 601 757 1093 1123 1483 1723 2551 2801
The 100,000th Brazilian number: 110468
C++
{{trans|D}}
#include <iostream> bool sameDigits(int n, int b) { int f = n % b; while ((n /= b) > 0) { if (n % b != f) { return false; } } return true; } bool isBrazilian(int n) { if (n < 7) return false; if (n % 2 == 0)return true; for (int b = 2; b < n - 1; b++) { if (sameDigits(n, b)) { return true; } } return false; } bool isPrime(int n) { if (n < 2)return false; if (n % 2 == 0)return n == 2; if (n % 3 == 0)return n == 3; int d = 5; while (d * d <= n) { if (n % d == 0)return false; d += 2; if (n % d == 0)return false; d += 4; } return true; } int main() { for (auto kind : { "", "odd ", "prime " }) { bool quiet = false; int BigLim = 99999; int limit = 20; std::cout << "First " << limit << ' ' << kind << "Brazillian numbers:\n"; int c = 0; int n = 7; while (c < BigLim) { if (isBrazilian(n)) { if (!quiet)std::cout << n << ' '; if (++c == limit) { std::cout << "\n\n"; quiet = true; } } if (quiet && kind != "") continue; if (kind == "") { n++; } else if (kind == "odd ") { n += 2; } else if (kind == "prime ") { while (true) { n += 2; if (isPrime(n)) break; } } else { throw new std::runtime_error("Unexpected"); } } if (kind == "") { std::cout << "The " << BigLim + 1 << "th Brazillian number is: " << n << "\n\n"; } } return 0; }
{{out}}
First 20 Brazillian numbers:
7 8 10 12 13 14 15 16 18 20 21 22 24 26 27 28 30 31 32 33
The 100000th Brazillian number is: 110468
First 20 odd Brazillian numbers:
7 13 15 21 27 31 33 35 39 43 45 51 55 57 63 65 69 73 75 77
First 20 prime Brazillian numbers:
7 13 31 43 73 127 157 211 241 307 421 463 601 757 1093 1123 1483 1723 2551 2801
=={{header|C#|CSharp}}== {{trans|Go}}
using System; class Program { static bool sameDigits(int n, int b) { int f = n % b; while ((n /= b) > 0) if (n % b != f) return false; return true; } static bool isBrazilian(int n) { if (n < 7) return false; if (n % 2 == 0) return true; for (int b = 2; b < n - 1; b++) if (sameDigits(n, b)) return true; return false; } static bool isPrime(int n) { if (n < 2) return false; if (n % 2 == 0) return n == 2; if (n % 3 == 0) return n == 3; int d = 5; while (d * d <= n) { if (n % d == 0) return false; d += 2; if (n % d == 0) return false; d += 4; } return true; } static void Main(string[] args) { foreach (string kind in ",odd ,prime ".Split(',')) { bool quiet = false; int BigLim = 99999, limit = 20; Console.WriteLine("First {0} {1}Brazilian numbers:", limit, kind); int c = 0, n = 7; while (c < BigLim) { if (isBrazilian(n)) { if (!quiet) Console.Write("{0:n0} ", n); if (++c == limit) { Console.Write("\n\n"); quiet = true; } } if (quiet && kind != "") continue; switch (kind) { case "": n++; break; case "odd ": n += 2; break; case "prime ": while (true) { n += 2; if (isPrime(n)) break; } break; } } if (kind == "") Console.WriteLine("The {0:n0}th Brazilian number is: {1:n0}\n", BigLim + 1, n); } } }
{{out}}
First 20 Brazilian numbers:
7 8 10 12 13 14 15 16 18 20 21 22 24 26 27 28 30 31 32 33
The 100,000th Brazilian number is: 110,468
First 20 odd Brazilian numbers:
7 13 15 21 27 31 33 35 39 43 45 51 55 57 63 65 69 73 75 77
First 20 prime Brazilian numbers:
7 13 31 43 73 127 157 211 241 307 421 463 601 757 1,093 1,123 1,483 1,723 2,551 2,801
Regarding the 100,000th number, there is a wee discrepancy here with the '''F#''' version. The OEIS reference only goes to 4000, which is 4618. (4000 being the highest result published elsewhere)
Speedier Version
Based on the Pascal version, with some shortcuts. Can calculate to one billion in under 4 1/2 seconds (on a core i7). This is faster than the Pascal version because the sieve is an array of '''SByte''' (8 bits) and not a '''NativeUInt''' (32 bits). Also this code does not preserve the base of each Brazilain number in the array, so the Pascal version is more flexible if desiring to quickly verify a quantity of Brazilian numbers.
using System; class Program { const // flags: int PrMk = 0, // a number that is prime SqMk = 1, // a number that is the square of a prime number UpMk = 2, // a number that can be factored (aka un-prime) BrMk = -2, // a prime number that is also a Brazilian number Excp = 121; // exception square - the only square prime that is a Brazilian static int pow = 9, // power of 10 to count to max; // maximum sieve array length // An upper limit of the required array length can be calculated like this: // power of 10 fraction limit actual result // 1 2 / 1 * 10 = 20 20 // 2 4 / 3 * 100 = 133 132 // 3 6 / 5 * 1000 = 1200 1191 // 4 8 / 7 * 10000 = 11428 11364 // 5 10/ 9 * 100000 = 111111 110468 // 6 12/11 * 1000000 = 1090909 1084566 // 7 14/13 * 10000000 = 10769230 10708453 // 8 16/15 * 100000000 = 106666666 106091516 // 9 18/17 * 1000000000 = 1058823529 1053421821 // powers above 9 are impractical because of the maximum array length in C#, // which is around the UInt32.MaxValue, or 4294967295 static SByte[] PS; // the prime/Brazilian number sieve // once the sieve is populated, primes are <= 0, non-primes are > 0, // Brazilian numbers are (< 0) or (> 1) // 121 is a special case, in the sieve it is marked with the BrMk (-2) // typical sieve of Eratosthenes algorithm static void PrimeSieve(int top) { PS = new SByte[top]; int i, ii, j; i = 2; PS[j = 4] = SqMk; while (j < top - 2) PS[j += 2] = UpMk; i = 3; PS[j = 9] = SqMk; while (j < top - 6) PS[j += 6] = UpMk; i = 5; while ((ii = i * i) < top) { if (PS[i] == PrMk) { j = (top - i) / i; if ((j & 1) == 0) j--; do if (PS[j] == PrMk) PS[i * j] = UpMk; while ((j -= 2) > i); PS[ii] = SqMk; } do ; while (PS[i += 2] != PrMk); } } // consults the sieve and returns whether a number is Brazilian static bool IsBr(int number) { return Math.Abs(PS[number]) > SqMk; } // shows the first few Brazilian numbers of several kinds static void FirstFew(string kind, int amt) { Console.WriteLine("\nThe first {0} {1}Brazilian Numbers are:", amt, kind); int i = 7; while (amt > 0) { if (IsBr(i)) { amt--; Console.Write("{0} ", i); } switch (kind) { case "odd ": i += 2; break; case "prime ": do i += 2; while (PS[i] != BrMk || i == Excp); break; default: i++; break; } } Console.WriteLine(); } // expands a 111_X number into an integer static int Expand(int NumberOfOnes, int Base) { int res = 1; while (NumberOfOnes-- > 1) res = res * Base + 1; if (res > max || res < 0) res = 0; return res; } // displays an elapsed time stamp static string TS(string fmt, ref DateTime st, bool reset = false) { DateTime n = DateTime.Now; string res = string.Format(fmt, (n - st).TotalMilliseconds); if (reset) st = n; return res; } static void Main(string[] args) { int p2 = pow << 1; DateTime st = DateTime.Now, st0 = st; int p10 = (int)Math.Pow(10, pow), p = 10, cnt = 0; max = (int)(((long)(p10) * p2) / (p2 - 1)); PrimeSieve(max); Console.WriteLine(TS("Sieving took {0} ms", ref st, true)); int[] primes = new int[7]; // make short list of primes before Brazilians are added int n = 3; for (int i = 0; i < primes.Length; i++) { primes[i] = n; do ; while (PS[n += 2] != 0); } Console.WriteLine("\nChecking first few prime numbers of sequential ones:\nones checked found"); // now check the '111_X' style numbers. many are factorable, but some are prime, // then re-mark the primes found in the sieve as Brazilian. // curiously, only the numbers with a prime number of ones will turn out, so // restricting the search to those saves time. no need to wast time on even numbers of ones, // or 9 ones, 15 ones, etc... foreach(int i in primes) { Console.Write("{0,4}", i); cnt = 0; n = 2; do { if ((n - 1) % i != 0) { long br = Expand(i, n); if (br > 0) { if (PS[br] < UpMk) { PS[br] = BrMk; cnt++; } } else { Console.WriteLine("{0,8}{1,6}", n, cnt); break; } } n++; } while (true); } Console.WriteLine(TS("Adding Brazilian primes to the sieve took {0} ms", ref st, true)); foreach (string s in ",odd ,prime ".Split(',')) FirstFew(s, 20); Console.WriteLine(TS("\nRequired output took {0} ms", ref st, true)); Console.WriteLine("\nDecade count of Brazilian numbers:"); n = 6; cnt = 0; do { while (cnt < p) if (IsBr(++n)) cnt++; Console.WriteLine("{0,15:n0}th is {1,-15:n0} {2}", cnt, n, TS("time: {0} ms", ref st)); } while ((p *= 10) <= p10); PS = new sbyte[0]; Console.WriteLine("\nTotal elapsed was {0} ms", (DateTime.Now - st0).TotalMilliseconds); if (System.Diagnostics.Debugger.IsAttached) Console.ReadKey(); } }
{{out}}
Sieving took 3009.2927 ms
Checking first few prime numbers of sequential ones:
ones checked found
3 32540 3923
5 182 44
7 32 9
11 8 1
13 6 3
17 4 1
19 4 1
Adding Brazilian primes to the sieve took 8.3535 ms
The first 20 Brazilian Numbers are:
7 8 10 12 13 14 15 16 18 20 21 22 24 26 27 28 30 31 32 33
The first 20 odd Brazilian Numbers are:
7 13 15 21 27 31 33 35 39 43 45 51 55 57 63 65 69 73 75 77
The first 20 prime Brazilian Numbers are:
7 13 31 43 73 127 157 211 241 307 421 463 601 757 1093 1123 1483 1723 2551 2801
Required output took 2.4881 ms
Decade count of Brazilian numbers:
10th is 20 time: 0.057 ms
100th is 132 time: 0.1022 ms
1,000th is 1,191 time: 0.1351 ms
10,000th is 11,364 time: 0.1823 ms
100,000th is 110,468 time: 0.3758 ms
1,000,000th is 1,084,566 time: 1.8601 ms
10,000,000th is 10,708,453 time: 17.8373 ms
100,000,000th is 106,091,516 time: 155.2622 ms
1,000,000,000th is 1,053,421,821 time: 1448.9392 ms
Total elapsed was 4469.1985 ms
P.S. The best speed on Tio.run is under 5 seconds for the 100 millionth count ('''pow''' = 8). If you are very persistent, the 1 billionth count ('''pow''' = 9) can be made to work on Tio.run, it usually overruns the 60 second timeout limit, and cannot finish completely - the sieving by itself takes over 32 seconds (best case), which usually doesn't leave enough time for all the counting.
D
{{trans|C#}}
import std.stdio; bool sameDigits(int n, int b) { int f = n % b; while ((n /= b) > 0) { if (n % b != f) { return false; } } return true; } bool isBrazilian(int n) { if (n < 7) return false; if (n % 2 == 0) return true; for (int b = 2; b < n - 1; ++b) { if (sameDigits(n, b)) { return true; } } return false; } bool isPrime(int n) { if (n < 2) return false; if (n % 2 == 0) return n == 2; if (n % 3 == 0) return n == 3; int d = 5; while (d * d <= n) { if (n % d == 0) return false; d += 2; if (n % d == 0) return false; d += 4; } return true; } void main() { foreach (kind; ["", "odd ", "prime "]) { bool quiet = false; int BigLim = 99999; int limit = 20; writefln("First %s %sBrazillion numbers:", limit, kind); int c = 0; int n = 7; while (c < BigLim) { if (isBrazilian(n)) { if (!quiet) write(n, ' '); if (++c == limit) { writeln("\n"); quiet = true; } } if (quiet && kind != "") continue; switch (kind) { case "": n++; break; case "odd ": n += 2; break; case "prime ": while (true) { n += 2; if (isPrime(n)) break; } break; default: assert(false); } } if (kind == "") writefln("The %sth Brazillian number is: %s\n", BigLim + 1, n); } }
{{out}}
First 20 Brazillion numbers:
7 8 10 12 13 14 15 16 18 20 21 22 24 26 27 28 30 31 32 33
The 100000th Brazillian number is: 110468
First 20 odd Brazillion numbers:
7 13 15 21 27 31 33 35 39 43 45 51 55 57 63 65 69 73 75 77
First 20 prime Brazillion numbers:
7 13 31 43 73 127 157 211 241 307 421 463 601 757 1093 1123 1483 1723 2551 2801
=={{header|F_Sharp|F#}}==
The functions
// 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]
The Tasks
;the first 20 Brazilian numbers
Brazilian() |> Seq.take 20 |> Seq.iter(printf "%d "); printfn ""
{{out}}
7 8 10 12 13 14 15 16 18 20 21 22 24 26 27 28 30 31 32 33
;the first 20 odd Brazilian numbers
Brazilian() |> Seq.filter(fun n->n%2=1) |> Seq.take 20 |> Seq.iter(printf "%d "); printfn ""
{{out}}
7 13 15 21 27 31 33 35 39 43 45 51 55 57 63 65 69 73 75 77
;the first 20 prime Brazilian numbers Using [http://www.rosettacode.org/wiki/Extensible_prime_generator#The_function Extensible Prime Generator (F#)]
Brazilian() |> Seq.filter isPrime |> Seq.take 20 |> Seq.iter(printf "%d "); printfn ""
{{out}}
7 13 31 43 73 127 157 211 241 307 421 463 601 757 1093 1123 1483 1723 2551 2801
;finally that which the crowd really want to know: What is the 100,000th Brazilian number?
printfn "%d" (Seq.item 99999 Brazilian)
{{out}}
110468
So up to 100,000 ~10% of numbers are non-Brazilian. The millionth Brazilian is 1084566 so less than 10% are non-Brazilian. Large non-Brazilians seem to be rare.
Factor
{{works with|Factor|0.99 development release 2019-07-10}}
USING: combinators grouping io kernel lists lists.lazy math
math.parser math.primes.lists math.ranges namespaces prettyprint
prettyprint.config sequences ;
: (brazilian?) ( n -- ? )
2 over 2 - [a,b] [ >base all-equal? ] with find nip >boolean ;
: brazilian? ( n -- ? )
{
{ [ dup 7 < ] [ drop f ] }
{ [ dup even? ] [ drop t ] }
[ (brazilian?) ]
} cond ;
: .20-brazilians ( list -- )
[ 20 ] dip [ brazilian? ] lfilter ltake list>array . ;
100 margin set
1 lfrom "First 20 Brazilian numbers:"
1 [ 2 + ] lfrom-by "First 20 odd Brazilian numbers:"
lprimes "First 20 prime Brazilian numbers:"
[ print .20-brazilians nl ] 2tri@
{{out}}
First 20 Brazilian numbers:
{ 7 8 10 12 13 14 15 16 18 20 21 22 24 26 27 28 30 31 32 33 }
First 20 odd Brazilian numbers:
{ 7 13 15 21 27 31 33 35 39 43 45 51 55 57 63 65 69 73 75 77 }
First 20 prime Brazilian numbers:
{ 7 13 31 43 73 127 157 211 241 307 421 463 601 757 1093 1123 1483 1723 2551 2801 }
=={{header|Fōrmulæ}}==
In [https://wiki.formulae.org/Brazilian_numbers this] page you can see the solution of this task.
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text ([http://wiki.formulae.org/Editing_F%C5%8Drmul%C3%A6_expressions more info]). Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for transportation effects more than visualization and edition.
The option to show Fōrmulæ programs and their results is showing images. Unfortunately images cannot be uploaded in Rosetta Code.
Go
Version 1
package main import "fmt" func sameDigits(n, b int) bool { f := n % b n /= b for n > 0 { if n%b != f { return false } n /= b } return true } func isBrazilian(n int) bool { if n < 7 { return false } if n%2 == 0 && n >= 8 { return true } for b := 2; b < n-1; b++ { if sameDigits(n, b) { return true } } return false } func isPrime(n int) bool { switch { case n < 2: return false case n%2 == 0: return n == 2 case n%3 == 0: return n == 3 default: d := 5 for d*d <= n { if n%d == 0 { return false } d += 2 if n%d == 0 { return false } d += 4 } return true } } func main() { kinds := []string{" ", " odd ", " prime "} for _, kind := range kinds { fmt.Printf("First 20%sBrazilian numbers:\n", kind) c := 0 n := 7 for { if isBrazilian(n) { fmt.Printf("%d ", n) c++ if c == 20 { fmt.Println("\n") break } } switch kind { case " ": n++ case " odd ": n += 2 case " prime ": for { n += 2 if isPrime(n) { break } } } } } n := 7 for c := 0; c < 100000; n++ { if isBrazilian(n) { c++ } } fmt.Println("The 100,000th Brazilian number:", n-1) }
{{out}}
First 20 Brazilian numbers:
7 8 10 12 13 14 15 16 18 20 21 22 24 26 27 28 30 31 32 33
First 20 odd Brazilian numbers:
7 13 15 21 27 31 33 35 39 43 45 51 55 57 63 65 69 73 75 77
First 20 prime Brazilian numbers:
7 13 31 43 73 127 157 211 241 307 421 463 601 757 1093 1123 1483 1723 2551 2801
The 100,000th Brazilian number: 110468
Version 2
{{trans|C# (speedier version)}}
Some of the comments have been omitted in the interests of brevity.
Running a bit quicker than the .NET versions though not due to any further improvements on my part.
package main import ( "fmt" "math" "time" ) // flags const ( prMk int8 = 0 // prime sqMk = 1 // prime square upMk = 2 // non-prime brMk = -2 // Brazilian prime excp = 121 // the only Brazilian square prime ) var ( pow = 9 max = 0 ps []int8 ) // typical sieve of Eratosthenes func primeSieve(top int) { ps = make([]int8, top) i, j := 2, 4 ps[j] = sqMk for j < top-2 { j += 2 ps[j] = upMk } i, j = 3, 9 ps[j] = sqMk for j < top-6 { j += 6 ps[j] = upMk } i = 5 for i*i < top { if ps[i] == prMk { j = (top - i) / i if (j & 1) == 0 { j-- } for { if ps[j] == prMk { ps[i*j] = upMk } j -= 2 if j <= i { break } } ps[i*i] = sqMk } for { i += 2 if ps[i] == prMk { break } } } } // returns whether a number is Brazilian func isBr(number int) bool { temp := ps[number] if temp < 0 { temp = -temp } return temp > sqMk } // shows the first few Brazilian numbers of several kinds func firstFew(kind string, amt int) { fmt.Printf("\nThe first %d %sBrazilian numbers are:\n", amt, kind) i := 7 for amt > 0 { if isBr(i) { amt-- fmt.Printf("%d ", i) } switch kind { case "odd ": i += 2 case "prime ": for { i += 2 if ps[i] == brMk && i != excp { break } } default: i++ } } fmt.Println() } // expands a 111_X number into an integer func expand(numberOfOnes, base int) int { res := 1 for numberOfOnes > 1 { numberOfOnes-- res = res*base + 1 } if res > max || res < 0 { res = 0 } return res } func toMs(d time.Duration) float64 { return float64(d) / 1e6 } func commatize(n int) string { s := fmt.Sprintf("%d", n) le := len(s) for i := le - 3; i >= 1; i -= 3 { s = s[0:i] + "," + s[i:] } return s } func main() { start := time.Now() st0 := start p2 := pow << 1 p10 := int(math.Pow10(pow)) p, cnt := 10, 0 max = p10 * p2 / (p2 - 1) primeSieve(max) fmt.Printf("Sieving took %.4f ms\n", toMs(time.Since(start))) start = time.Now() primes := make([]int, 7) n := 3 for i := 0; i < len(primes); i++ { primes[i] = n for { n += 2 if ps[n] == 0 { break } } } fmt.Println("\nChecking first few prime numbers of sequential ones:") fmt.Println("ones checked found") for _, i := range primes { fmt.Printf("%4d", i) cnt, n = 0, 2 for { if (n-1)%i != 0 { br := expand(i, n) if br > 0 { if ps[br] < upMk { ps[br] = brMk cnt++ } } else { fmt.Printf("%8d%6d\n", n, cnt) break } } n++ } } ms := toMs(time.Since(start)) fmt.Printf("Adding Brazilian primes to the sieve took %.4f ms\n", ms) start = time.Now() for _, s := range []string{"", "odd ", "prime "} { firstFew(s, 20) } fmt.Printf("\nRequired output took %.4f ms\n", toMs(time.Since(start))) fmt.Println("\nDecade count of Brazilian numbers:") n, cnt = 6, 0 for { for cnt < p { n++ if isBr(n) { cnt++ } } ms = toMs(time.Since(start)) fmt.Printf("%15sth is %-15s time: %8.4f ms\n", commatize(cnt), commatize(n), ms) p *= 10 if p > p10 { break } } fmt.Printf("\nTotal elapsed was %.4f ms\n", toMs(time.Since(st0))) }
{{out}} Timings are for an Intel Core i7-8565U machine using Go 1.12.9 on Ubuntu 18.04.
Sieving took 2489.6647 ms
Checking first few prime numbers of sequential ones:
ones checked found
3 32540 3923
5 182 44
7 32 9
11 8 1
13 6 3
17 4 1
19 4 1
Adding Brazilian primes to the sieve took 1.2049 ms
The first 20 Brazilian numbers are:
7 8 10 12 13 14 15 16 18 20 21 22 24 26 27 28 30 31 32 33
The first 20 odd Brazilian numbers are:
7 13 15 21 27 31 33 35 39 43 45 51 55 57 63 65 69 73 75 77
The first 20 prime Brazilian numbers are:
7 13 31 43 73 127 157 211 241 307 421 463 601 757 1093 1123 1483 1723 2551 2801
Required output took 0.0912 ms
Decade count of Brazilian numbers:
10th is 20 time: 0.0951 ms
100th is 132 time: 0.0982 ms
1,000th is 1,191 time: 0.1015 ms
10,000th is 11,364 time: 0.1121 ms
100,000th is 110,468 time: 0.2201 ms
1,000,000th is 1,084,566 time: 0.9421 ms
10,000,000th is 10,708,453 time: 8.0068 ms
100,000,000th is 106,091,516 time: 78.0114 ms
1,000,000,000th is 1,053,421,821 time: 758.0320 ms
Total elapsed was 3249.0197 ms
Haskell
import Data.Numbers.Primes (primes) isBrazil :: Int -> Bool isBrazil n = 7 <= n && (even n || any (monoDigit n) [2 .. n - 2]) monoDigit :: Int -> Int -> Bool monoDigit n b = let (q, d) = quotRem n b in d == snd (until (uncurry (flip ((||) . (d /=)) . (0 ==))) ((`quotRem` b) . fst) (q, d)) main :: IO () main = mapM_ (\(s, xs) -> (putStrLn . concat) [ "First 20 " , s , " Brazilians:\n" , show . take 20 $ filter isBrazil xs , "\n" ]) [([], [1 ..]), ("odd", [1,3 ..]), ("prime", primes)]
{{Out}}
First 20 Brazilians:
[7,8,10,12,13,14,15,16,18,20,21,22,24,26,27,28,30,31,32,33]
First 20 odd Brazilians:
[7,13,15,21,27,31,33,35,39,43,45,51,55,57,63,65,69,73,75,77]
First 20 prime Brazilians:
[7,13,31,43,73,127,157,211,241,307,421,463,601,757,1093,1123,1483,1723,2551,2801]
Julia
{{trans|Go}}
using Primes, Lazy function samedigits(n, b) n, f = divrem(n, b) while n > 0 n, f2 = divrem(n, b) if f2 != f return false end end true end isbrazilian(n) = n >= 7 && (iseven(n) || any(b -> samedigits(n, b), 2:n-2)) brazilians = filter(isbrazilian, Lazy.range()) oddbrazilians = filter(n -> isodd(n) && isbrazilian(n), Lazy.range()) primebrazilians = filter(n -> isprime(n) && isbrazilian(n), Lazy.range()) println("The first 20 Brazilian numbers are: ", take(20, brazilians)) println("The first 20 odd Brazilian numbers are: ", take(20, oddbrazilians)) println("The first 20 prime Brazilian numbers are: ", take(20, primebrazilians))
{{out}}
The first 20 Brazilian numbers are: (7 8 10 12 13 14 15 16 18 20 21 22 24 26 27 28 30 31 32 33)
The first 20 odd Brazilian numbers are: (7 13 15 21 27 31 33 35 39 43 45 51 55 57 63 65 69 73 75 77)
The first 20 prime Brazilian numbers are: (7 13 31 43 73 127 157 211 241 307 421 463 601 757 1093 1123 1483 1723 2551 2801)
There has been some discussion of larger numbers in the sequence. See below:
function braziliandensities(N, interval) count, intervalcount, icount = 0, 0, 0 intervalcounts = Int[] for i in 7:typemax(Int) intervalcount += 1 if intervalcount > interval push!(intervalcounts, icount) intervalcount = 0 icount = 0 end if isbrazilian(i) icount += 1 count += 1 if count == N println("The $N th brazilian is $i.") return [n/interval for n in intervalcounts] end end end end braziliandensities(10000, 100) braziliandensities(100000, 1000) plot(1:1000:1000000, braziliandensities(1000000, 1000))
{{out}}
The 10000 th brazilian is 11364.
The 100000 th brazilian is 110468.
The 1000000 th brazilian is 1084566.
[http://alahonua.com/temp/newplot.png link plot png]
Kotlin
{{trans|C#}}
fun sameDigits(n: Int, b: Int): Boolean { var n2 = n val f = n % b while (true) { n2 /= b if (n2 > 0) { if (n2 % b != f) { return false } } else { break } } return true } fun isBrazilian(n: Int): Boolean { if (n < 7) return false if (n % 2 == 0) return true for (b in 2 until n - 1) { if (sameDigits(n, b)) { return true } } return false } fun isPrime(n: Int): Boolean { if (n < 2) return false if (n % 2 == 0) return n == 2 if (n % 3 == 0) return n == 3 var d = 5 while (d * d <= n) { if (n % d == 0) return false d += 2 if (n % d == 0) return false d += 4 } return true } fun main() { val bigLim = 99999 val limit = 20 for (kind in ",odd ,prime".split(',')) { var quiet = false println("First $limit ${kind}Brazilian numbers:") var c = 0 var n = 7 while (c < bigLim) { if (isBrazilian(n)) { if (!quiet) print("%,d ".format(n)) if (++c == limit) { print("\n\n") quiet = true } } if (quiet && kind != "") continue when (kind) { "" -> n++ "odd " -> n += 2 "prime" -> { while (true) { n += 2 if (isPrime(n)) break } } } } if (kind == "") println("The %,dth Brazilian number is: %,d".format(bigLim + 1, n)) } }
{{out}}
First 20 Brazilian numbers:
7 8 10 12 13 14 15 16 18 20 21 22 24 26 27 28 30 31 32 33
The 100,000th Brazilian number is: 110,468
First 20 odd Brazilian numbers:
7 13 15 21 27 31 33 35 39 43 45 51 55 57 63 65 69 73 75 77
First 20 primeBrazilian numbers:
7 13 31 43 73 127 157 211 241 307 421 463 601 757 1,093 1,123 1,483 1,723 2,551 2,801
Pascal
{{works with|Free Pascal}}
Using a sieve of Erathostenes to memorize the smallest factor of a composite number.
Checking primes first for '''111''' to base and if not then to '''11111''' ( Base^4+Base^3..+^1 = (Base^5 -1) / (Base-1) )
extreme reduced runtime time for space.
At the end only primes and square of primes need to be tested, all others are Brazilian.
program brazilianNumbers; {$IFDEF FPC} {$MODE DELPHI}{$OPTIMIZATION ON,All} {$CODEALIGN proc=32,loop=4} {$ELSE} {$APPTYPE CONSOLE} {$ENDIF} uses SysUtils; const //Must not be a prime PrimeMarker = 0; SquareMarker = PrimeMarker + 1; //MAX = 110468;// 1E5 brazilian //MAX = 1084566;// 1E6 brazilian //MAX = 10708453;// 1E7 brazilian //MAX = 106091516;// 1E8 brazilian MAX = 1053421821;// 1E9 brazilian var isprime: array of word; procedure MarkSmallestFactor; //sieve of erathotenes //but saving the smallest factor var i, j, lmt: NativeUint; begin lmt := High(isPrime); fillWord(isPrime[0], lmt + 1, PrimeMarker); //mark even numbers i := 2; j := i * i; isPrime[j] := SquareMarker; Inc(j, 2); while j <= lmt do begin isPrime[j] := 2; Inc(j, 2); end; //mark 3 but not 2 i := 3; j := i * i; isPrime[j] := SquareMarker; Inc(j, 6); while j <= lmt do begin isPrime[j] := 3; Inc(j, 6); end; i := 5; while i * i <= lmt do begin if isPrime[i] = 0 then begin j := lmt div i; if not (odd(j)) then Dec(j); while j > i do begin if isPrime[j] = 0 then isPrime[i * j] := i; Dec(j, 2); end; //mark square prime isPrime[i * i] := SquareMarker; end; Inc(i, 2); end; end; procedure OutFactors(n: NativeUint); var divisor, Next, rest: NativeUint; pot: NativeUint; begin divisor := 2; Next := 3; rest := n; Write(n: 10, ' = '); while (rest <> 1) do begin if (rest mod divisor = 0) then begin Write(divisor); pot := 0; repeat rest := rest div divisor; Inc(pot) until rest mod divisor <> 0; if pot > 1 then Write('^', pot); if rest > 1 then Write('*'); end; divisor := Next; Next := Next + 2; // cut condition: avoid many useless iterations if (rest <> 1) and (rest < divisor * divisor) then begin Write(rest); rest := 1; end; end; Write(' ', #9#9#9); end; procedure OutToBase(number, base: NativeUint); var BaseDgt: array[0..63] of NativeUint; i, rest: NativeINt; begin OutFactors(number); i := 0; while number <> 0 do begin rest := number div base; BaseDgt[i] := number - rest * base; number := rest; Inc(i); end; while i > 1 do begin Dec(i); Write(BaseDgt[i]); end; writeln(BaseDgt[0], ' to base ', base); end; function PrimeBase(number: NativeUint): NativeUint; var lnN: extended; i, exponent, n: NativeUint; begin // primes are only brazilian if 111...11 to base > 2 // the count of "1" must be odd , because brazilian primes are odd lnN := ln(number); exponent := 4; //result := exponent.th root of number Result := trunc(exp(lnN*0.25)); while result >2 do Begin // calc sum(i= 0 to exponent ) base^i; n := Result + 1; i := 2; repeat Inc(i); n := n*result + 1; until i > exponent; if n = number then EXIT; Inc(exponent,2); Result := trunc(exp(lnN/exponent)); end; //not brazilian Result := 0; end; function GetPrimeBrazilianBase(number: NativeUint): NativeUint; //result is base begin // prime of 2^n - 1 if (Number and (number + 1)) = 0 then Result := 2 else begin Result := trunc(sqrt(number)); //most of the brazilian primes are of this type base^2+base+1 IF (sqr(result)+result+1) <> number then result := PrimeBase(number); end; end; function GetBrazilianBase(number: NativeUInt): NativeUint; inline; begin Result := isPrime[number]; if Result > SquareMarker then Result := (number div Result) - 1 else begin if Result = SquareMarker then begin if number = 121 then Result := 3 else Result := 0; end else Result := GetPrimeBrazilianBase(number); end; end; procedure First20Brazilian; var i, n, cnt: NativeUInt; begin writeln('first 20 brazilian numbers'); i := 7; cnt := 0; while cnt < 20 do begin n := GetBrazilianBase(i); if n <> 0 then begin Inc(cnt); OutToBase(i, n); end; Inc(i); end; writeln; end; procedure First33OddBrazilian; var i, n, cnt: NativeUInt; begin writeln('first 33 odd brazilian numbers'); i := 7; cnt := 0; while cnt < 33 do begin n := GetBrazilianBase(i); if N <> 0 then begin Inc(cnt); OutToBase(i, n); end; Inc(i, 2); end; writeln; end; procedure First20BrazilianPrimes; var i, n, cnt: NativeUInt; begin writeln('first 20 brazilian prime numbers'); i := 7; cnt := 0; while cnt < 20 do begin IF isPrime[i] = PrimeMarker then Begin n := GetBrazilianBase(i); if n <> 0 then begin Inc(cnt); OutToBase(i, n); end; end; Inc(i); end; writeln; end; var T1, T0: TDateTime; i, n, cnt, lmt: NativeUInt; begin lmt := MAX; setlength(isPrime, lmt + 1); MarkSmallestFactor; First20Brazilian; First33OddBrazilian; First20BrazilianPrimes; Write('count brazilian numbers up to ', lmt, ' = '); T0 := now; i := 7; cnt := 0; n := 0; while (i <= lmt) do begin Inc(n, Ord(isPrime[i] = PrimeMarker)); if GetBrazilianBase(i) <> 0 then Inc(cnt); Inc(i); end; T1 := now; writeln(cnt); writeln('Count of primes ', n: 11+13); writeln((T1 - T0) * 86400 * 1000: 10: 0, ' ms'); setlength(isPrime, 0); end.
{{out}}
first 20 brazilian numbers
7 = 7 111 to base 2
8 = 2^3 22 to base 3
10 = 2*5 22 to base 4
12 = 2^2*3 22 to base 5
13 = 13 111 to base 3
14 = 2*7 22 to base 6
15 = 3*5 33 to base 4
16 = 2^4 22 to base 7
18 = 2*3^2 22 to base 8
20 = 2^2*5 22 to base 9
21 = 3*7 33 to base 6
22 = 2*11 22 to base 10
24 = 2^3*3 22 to base 11
26 = 2*13 22 to base 12
27 = 3^3 33 to base 8
28 = 2^2*7 22 to base 13
30 = 2*3*5 22 to base 14
31 = 31 11111 to base 2
32 = 2^5 22 to base 15
33 = 3*11 33 to base 10
first 33 odd brazilian numbers
7 = 7 111 to base 2
13 = 13 111 to base 3
15 = 3*5 33 to base 4
21 = 3*7 33 to base 6
27 = 3^3 33 to base 8
31 = 31 11111 to base 2
33 = 3*11 33 to base 10
35 = 5*7 55 to base 6
39 = 3*13 33 to base 12
43 = 43 111 to base 6
45 = 3^2*5 33 to base 14
51 = 3*17 33 to base 16
55 = 5*11 55 to base 10
57 = 3*19 33 to base 18
63 = 3^2*7 33 to base 20
65 = 5*13 55 to base 12
69 = 3*23 33 to base 22
73 = 73 111 to base 8
75 = 3*5^2 33 to base 24
77 = 7*11 77 to base 10
81 = 3^4 33 to base 26
85 = 5*17 55 to base 16
87 = 3*29 33 to base 28
91 = 7*13 77 to base 12
93 = 3*31 33 to base 30
95 = 5*19 55 to base 18
99 = 3^2*11 33 to base 32
105 = 3*5*7 33 to base 34
111 = 3*37 33 to base 36
115 = 5*23 55 to base 22
117 = 3^2*13 33 to base 38
119 = 7*17 77 to base 16
121 = 11^2 11111 to base 3
first 20 brazilian prime numbers
7 = 7 111 to base 2
13 = 13 111 to base 3
31 = 31 11111 to base 2
43 = 43 111 to base 6
73 = 73 111 to base 8
127 = 127 1111111 to base 2
157 = 157 111 to base 12
211 = 211 111 to base 14
241 = 241 111 to base 15
307 = 307 111 to base 17
421 = 421 111 to base 20
463 = 463 111 to base 21
601 = 601 111 to base 24
757 = 757 111 to base 27
1093 = 1093 1111111 to base 3
1123 = 1123 111 to base 33
1483 = 1483 111 to base 38
1723 = 1723 111 to base 41
2551 = 2551 111 to base 50
2801 = 2801 11111 to base 7
count brazilian numbers up to 1053421821 = 1000000000
Count of primes 53422305
21657 ms ( from 30971 ms )
real 0m26,411s -> marking small factors improved 7.8-> 3.8 seconds
user 0m26,239s
sys 0m0,157s
Perl
{{libheader|ntheory}} {{trans|Perl 6}}
use strict; use warnings; use feature 'say'; use ntheory qw<is_prime>; sub is_Brazilian { my($n) = @_; return 1 if $n > 6 && 0 == $n%2; LOOP: for (my $base = 2; $base < $n - 1; ++$base) { my $digit; my $nn = $n; while (1) { my $x = $nn % $base; $digit //= $x; next LOOP if $digit != $x; $nn = int $nn / $base; if ($nn < $base) { return 1 if $digit == $nn; next LOOP; } } } } my $upto = 20; my $Inf = 10e10; my $n = 1; my $s = "First $upto Brazilian numbers:\n"; $s .= do { $n <= $upto ? is_Brazilian($_) && $n++ && "$_ " : last } for 1..$Inf; say $s; $n = 1; $s = "\nFirst $upto odd Brazilian numbers:\n"; $s .= do { $n <= $upto ? 0 != $_%2 && is_Brazilian($_) && $n++ && "$_ " : last } for 1..$Inf; say $s; $n = 1; $s = "\nFirst $upto prime Brazilian numbers:\n"; $s .= do { $n <= $upto ? !!is_prime($_) && is_Brazilian($_) && $n++ && "$_ " : last } for 1..$Inf; say $s;
{{out}}
First 20 Brazilian numbers:
7 8 10 12 13 14 15 16 18 20 21 22 24 26 27 28 30 31 32 33
First 20 odd Brazilian numbers:
7 13 15 21 27 31 33 35 39 43 45 51 55 57 63 65 69 73 75 77
First 20 prime Brazilian numbers:
7 13 31 43 73 127 157 211 241 307 421 463 601 757 1093 1123 1483 1723 2551 2801
Perl 6
{{works with|Rakudo|2019.07.1}}
multi is-Brazilian (Int $n where $n %% 2 && $n > 6) { True }
multi is-Brazilian (Int $n) {
LOOP: loop (my int $base = 2; $base < $n - 1; ++$base) {
my $digit;
for $n.polymod( $base xx * ) {
$digit //= $_;
next LOOP if $digit != $_;
}
return True
}
False
}
my $upto = 20;
put "First $upto Brazilian numbers:\n", (^Inf).hyper.grep( *.&is-Brazilian )[^$upto];
put "\nFirst $upto odd Brazilian numbers:\n", (^Inf).hyper.map( * * 2 + 1 ).grep( *.&is-Brazilian )[^$upto];
put "\nFirst $upto prime Brazilian numbers:\n", (^Inf).hyper(:8degree).grep( { .is-prime && .&is-Brazilian } )[^$upto];
{{out}}
First 20 Brazilian numbers:
7 8 10 12 13 14 15 16 18 20 21 22 24 26 27 28 30 31 32 33
First 20 odd Brazilian numbers:
7 13 15 21 27 31 33 35 39 43 45 51 55 57 63 65 69 73 75 77
First 20 prime Brazilian numbers:
7 13 31 43 73 127 157 211 241 307 421 463 601 757 1093 1123 1483 1723 2551 2801
Phix
{{trans|C}}
function same_digits(integer n, b)
integer f = remainder(n,b)
n = floor(n/b)
while n>0 do
if remainder(n,b)!=f then return false end if
n = floor(n/b)
end while
return true
end function
function is_brazilian(integer n)
if n>=7 then
if remainder(n,2)=0 then return true end if
for b=2 to n-2 do
if same_digits(n,b) then return true end if
end for
end if
return false
end function
constant kinds = {" ", " odd ", " prime "}
for i=1 to length(kinds) do
printf(1,"First 20%sBrazilian numbers:\n", {kinds[i]})
integer c = 0, n = 7, p = 4
while true do
if is_brazilian(n) then
printf(1,"%d ",n)
c += 1
if c==20 then
printf(1,"\n\n")
exit
end if
end if
switch i
case 1: n += 1
case 2: n += 2
case 3: p += 1; n = get_prime(p)
end switch
end while
end for
integer n = 7, c = 0
atom t0 = time(), t1 = time()+1
while c<100000 do
if time()>t1 then
printf(1,"checking %d [count:%d]...\r",{n,c})
t1 = time()+1
end if
c += is_brazilian(n)
n += 1
end while
printf(1,"The %,dth Brazilian number: %d\n", {c,n-1})
?elapsed(time()-t0)
{{out}} (not very fast, takes about 4 times as long as Go)
First 20 Brazilian numbers:
7 8 10 12 13 14 15 16 18 20 21 22 24 26 27 28 30 31 32 33
First 20 odd Brazilian numbers:
7 13 15 21 27 31 33 35 39 43 45 51 55 57 63 65 69 73 75 77
First 20 prime Brazilian numbers:
7 13 31 43 73 127 157 211 241 307 421 463 601 757 1093 1123 1483 1723 2551 2801
The 100,000th Brazilian number: 110468
"52.8s"
Python
'''Brazilian numbers''' from itertools import count, islice # main :: IO () def main(): '''First 20 members each of: OEIS:A125134 OEIS:A257521 OEIS:A085104 ''' for kxs in ([ (' ', count(1)), (' odd ', count(1, 2)), (' prime ', primes()) ]): print( 'First 20' + kxs[0] + 'Brazilians:\n' + showList(take(20)(filter(isBrazil, kxs[1]))) + '\n' ) # isBrazil :: Int -> Bool def isBrazil(n): '''True if n is a Brazilian number, in the sense of OEIS:A125134. ''' return 7 <= n and ( 0 == n % 2 or any( map(monoDigit(n), range(2, n - 1)) ) ) # monoDigit :: Int -> Int -> Bool def monoDigit(n): '''True if all the digits of n, in the given base, are the same. ''' def go(b, n): (q, d) = divmod(n, b) return d == until( lambda qr: d != qr[1] or 0 == qr[0] )( lambda qr: divmod(qr[0], b) )((q, d))[1] return lambda base: go(base, n) # GENERIC ------------------------------------------------- # primes :: [Int] def primes(): ''' Non finite sequence of prime numbers. ''' n = 2 dct = {} while True: if n in dct: for p in dct[n]: dct.setdefault(n + p, []).append(p) del dct[n] else: yield n dct[n * n] = [n] n = 1 + n # showList :: [a] -> String def showList(xs): '''Stringification of a list.''' return '[' + ','.join(str(x) for x in xs) + ']' # take :: Int -> [a] -> [a] # take :: Int -> String -> String def take(n): '''The prefix of xs of length n, or xs itself if n > length xs. ''' return lambda xs: ( xs[0:n] if isinstance(xs, (list, tuple)) else list(islice(xs, n)) ) # until :: (a -> Bool) -> (a -> a) -> a -> a def until(p): '''The result of repeatedly applying f until p holds. The initial seed value is x. ''' def go(f, x): v = x while not p(v): v = f(v) return v return lambda f: lambda x: go(f, x) # MAIN --- if __name__ == '__main__': main()
{{Out}}
First 20 Brazilians:
[7,8,10,12,13,14,15,16,18,20,21,22,24,26,27,28,30,31,32,33]
First 20 odd Brazilians:
[7,13,15,21,27,31,33,35,39,43,45,51,55,57,63,65,69,73,75,77]
First 20 prime Brazilians:
[7,13,31,43,73,127,157,211,241,307,421,463,601,757,1093,1123,1483,1723,2551,2801]
REXX
{{trans|GO}}
{{out|output|text= when using the inputs of: <tt> , , , 100000 </tt>}}
```txt
══════════════════════ The first 20 Brazilian numbers ═══════════════════════
7 8 10 12 13 14 15 16 18 20 21 22 24 26 27 28 30 31 32 33
════════════════════ The first odd 20 Brazilian numbers ═════════════════════
7 13 15 21 27 31 33 35 39 43 45 51 55 57 63 65 69 73 75 77
═══════════════════ The first prime 20 Brazilian numbers ════════════════════
7 13 31 43 73 127 157 211 241 307 421 463 601 757 1093 1123 1483 1723 2551 2801
═════════════════════════ The 100000th Brazilian number ═════════════════════
110468
Sidef
func is_Brazilian_prime(q) { static L = Set() static M = 0 return true if L.has(q) return false if (q < M) var N = (q<1000 ? 1000 : 2*q) for K in (primes(3, ilog2(N+1))) { for n in (2 .. iroot(N-1, K-1)) { var p = (n**K - 1)/(n-1) L << p if (p<N && p.is_prime) } } M = (L.max \\ 0) return L.has(q) } func is_Brazilian(n) { if (!n.is_prime) { n.is_square || return (n>6) var m = n.isqrt return (m>3 && (!m.is_prime || m==11)) } is_Brazilian_prime(n) } with (20) {|n| say "First #{n} Brazilian numbers:" say (^Inf -> lazy.grep(is_Brazilian).first(n)) say "\nFirst #{n} odd Brazilian numbers:" say (^Inf -> lazy.grep(is_Brazilian).grep{.is_odd}.first(n)) say "\nFirst #{n} prime Brazilian numbers" say (^Inf -> lazy.grep(is_Brazilian).grep{.is_prime}.first(n)) }
{{out}}
First 20 Brazilian numbers:
[7, 8, 10, 12, 13, 14, 15, 16, 18, 20, 21, 22, 24, 26, 27, 28, 30, 31, 32, 33]
First 20 odd Brazilian numbers:
[7, 13, 15, 21, 27, 31, 33, 35, 39, 43, 45, 51, 55, 57, 63, 65, 69, 73, 75, 77]
First 20 prime Brazilian numbers
[7, 13, 31, 43, 73, 127, 157, 211, 241, 307, 421, 463, 601, 757, 1093, 1123, 1483, 1723, 2551, 2801]
Extra:
for n in (1..6) { say ("#{10**n->commify}th Brazilian number = ", is_Brazilian.nth(10**n)) }
{{out}}
10th Brazilian number = 20
100th Brazilian number = 132
1,000th Brazilian number = 1191
10,000th Brazilian number = 11364
100,000th Brazilian number = 110468
1,000,000th Brazilian number = 1084566
Visual Basic .NET
{{trans|C#}}
Module Module1
Function sameDigits(ByVal n As Integer, ByVal b As Integer) As Boolean
Dim f As Integer = n Mod b : n \= b : While n > 0
If n Mod b <> f Then Return False Else n \= b
End While : Return True
End Function
Function isBrazilian(ByVal n As Integer) As Boolean
If n < 7 Then Return False
If n Mod 2 = 0 Then Return True
For b As Integer = 2 To n - 2
If sameDigits(n, b) Then Return True
Next : Return False
End Function
Function isPrime(ByVal n As Integer) As Boolean
If n < 2 Then Return False
If n Mod 2 = 0 Then Return n = 2
If n Mod 3 = 0 Then Return n = 3
Dim d As Integer = 5
While d * d <= n
If n Mod d = 0 Then Return False Else d += 2
If n Mod d = 0 Then Return False Else d += 4
End While : Return True
End Function
Sub Main(args As String())
For Each kind As String In {" ", " odd ", " prime "}
Console.WriteLine("First 20{0}Brazilian numbers:", kind)
Dim Limit As Integer = 20, n As Integer = 7
Do
If isBrazilian(n) Then Console.Write("{0} ", n) : Limit -= 1
Select Case kind
Case " " : n += 1
Case " odd " : n += 2
Case " prime " : Do : n += 2 : Loop Until isPrime(n)
End Select
Loop While Limit > 0
Console.Write(vbLf & vbLf)
Next
End Sub
End Module
{{out}}
First 20 Brazilian numbers:
7 8 10 12 13 14 15 16 18 20 21 22 24 26 27 28 30 31 32 33
First 20 odd Brazilian numbers:
7 13 15 21 27 31 33 35 39 43 45 51 55 57 63 65 69 73 75 77
First 20 prime Brazilian numbers:
7 13 31 43 73 127 157 211 241 307 421 463 601 757 1093 1123 1483 1723 2551 2801
Speedier Version
Based on the C# speedier version, performance is just as good, one billion Brazilian numbers counted in 4 1/2 seconds (on a core i7).
Imports System
Module Module1
' flags:
Const _
PrMk As Integer = 0, ' a number that is prime
SqMk As Integer = 1, ' a number that is the square of a prime number
UpMk As Integer = 2, ' a number that can be factored (aka un-prime)
BrMk As Integer = -2, ' a prime number that is also a Brazilian number
Excp As Integer = 121 ' exception square - the only square prime that is a Brazilian
Dim pow As Integer = 9,
max As Integer ' maximum sieve array length
' An upper limit of the required array length can be calculated Like this:
' power of 10 fraction limit actual result
' 1 2 / 1 * 10 = 20 20
' 2 4 / 3 * 100 = 133 132
' 3 6 / 5 * 1000 = 1200 1191
' 4 8 / 7 * 10000 = 11428 11364
' 5 10/ 9 * 100000 = 111111 110468
' 6 12/11 * 1000000 = 1090909 1084566
' 7 14/13 * 10000000 = 10769230 10708453
' 8 16/15 * 100000000 = 106666666 106091516
' 9 18/17 * 1000000000 = 1058823529 1053421821
' powers above 9 are impractical because of the maximum array length in VB.NET,
' which is around the UInt32.MaxValue, Or 4294967295
Dim PS As SByte() ' the prime/Brazilian number sieve
' once the sieve is populated, primes are <= 0, non-primes are > 0,
' Brazilian numbers are (< 0) or (> 1)
' 121 is a special case, in the sieve it is marked with the BrMk (-2)
' typical sieve of Eratosthenes algorithm
Sub PrimeSieve(ByVal top As Integer)
PS = New SByte(top) {} : Dim i, ii, j As Integer
i = 2 : j = 4 : PS(j) = SqMk : While j < top - 2 : j += 2 : PS(j) = UpMk : End While
i = 3 : j = 9 : PS(j) = SqMk : While j < top - 6 : j += 6 : PS(j) = UpMk : End While
i = 5 : ii = 25 : While ii < top
If PS(i) = PrMk Then
j = (top - i) / i : If (j And 1) = 0 Then j -= 1
Do : If PS(j) = PrMk Then PS(i * j) = UpMk
j -= 2 : Loop While j > i : PS(ii) = SqMk
End If
Do : i += 2 : Loop While PS(i) <> PrMk : ii = i * i
End While
End Sub
' consults the sieve and returns whether a number is Brazilian
Function IsBr(ByVal number As Integer) As Boolean
Return Math.Abs(PS(number)) > SqMk
End Function
' shows the first few Brazilian numbers of several kinds
Sub FirstFew(ByVal kind As String, ByVal amt As Integer)
Console.WriteLine(vbLf & "The first {0} {1}Brazilian Numbers are:", amt, kind)
Dim i As Integer = 7 : While amt > 0
If IsBr(i) Then amt -= 1 : Console.Write("{0} ", i)
Select Case kind : Case "odd " : i += 2
Case "prime " : Do : i += 2 : Loop While PS(i) <> BrMk OrElse i = Excp
Case Else : i += 1 : End Select : End While : Console.WriteLine()
End Sub
' expands a 111_X number into an integer
Function Expand(ByVal NumberOfOnes As Integer, ByVal Base As Integer) As Integer
Dim res As Integer = 1
While NumberOfOnes > 1 AndAlso res < Integer.MaxValue \ Base
res = res * Base + 1 : NumberOfOnes -= 1 : End While
If res > max OrElse res < 0 Then res = 0
Return res
End Function
' returns an elapsed time string
Function TS(ByVal fmt As String, ByRef st As DateTime, ByVal Optional reset As Boolean = False) As String
Dim n As DateTime = DateTime.Now,
res As String = String.Format(fmt, (n - st).TotalMilliseconds)
If reset Then st = n
Return res
End Function
Sub Main(args As String())
Dim p2 As Integer = pow << 1, primes(6) As Integer, n As Integer,
st As DateTime = DateTime.Now, st0 As DateTime = st,
p10 As Integer = CInt(Math.Pow(10, pow)), p As Integer = 10, cnt As Integer = 0
max = CInt(((CLng((p10)) * p2) / (p2 - 1))) : PrimeSieve(max)
Console.WriteLine(TS("Sieving took {0} ms", st, True))
' make short list of primes before Brazilians are added
n = 3 : For i As Integer = 0 To primes.Length - 1
primes(i) = n : Do : n += 2 : Loop While PS(n) <> 0 : Next
Console.WriteLine(vbLf & "Checking first few prime numbers of sequential ones:" &
vbLf & "ones checked found")
' now check the '111_X' style numbers. many are factorable, but some are prime,
' then re-mark the primes found in the sieve as Brazilian.
' curiously, only the numbers with a prime number of ones will turn out, so
' restricting the search to those saves time. no need to wast time on even numbers of ones,
' or 9 ones, 15 ones, etc...
For Each i As Integer In primes
Console.Write("{0,4}", i) : cnt = 0 : n = 2 : Do
If (n - 1) Mod i <> 0 Then
Dim br As Long = Expand(i, n)
If br > 0 Then
If PS(br) < UpMk Then PS(br) = BrMk : cnt += 1
Else
Console.WriteLine("{0,8}{1,6}", n, cnt) : Exit Do
End If
End If : n += 1 : Loop While True
Next
Console.WriteLine(TS("Adding Brazilian primes to the sieve took {0} ms", st, True))
For Each s As String In ",odd ,prime ".Split(",") : FirstFew(s, 20) : Next
Console.WriteLine(TS(vbLf & "Required output took {0} ms", st, True))
Console.WriteLine(vbLf & "Decade count of Brazilian numbers:")
n = 6 : cnt = 0 : Do : While cnt < p : n += 1 : If IsBr(n) Then cnt += 1
End While
Console.WriteLine("{0,15:n0}th is {1,-15:n0} {2}", cnt, n, TS("time: {0} ms", st))
If p < p10 Then p *= 10 Else Exit Do
Loop While (True) : PS = New SByte(-1) {}
Console.WriteLine(vbLf & "Total elapsed was {0} ms", (DateTime.Now - st0).TotalMilliseconds)
If System.Diagnostics.Debugger.IsAttached Then Console.ReadKey()
End Sub
End Module
{{out}}
Sieving took 2967.834 ms
Checking first few prime numbers of sequential ones:
ones checked found
3 32540 3923
5 182 44
7 32 9
11 8 1
13 8 3
17 4 1
19 4 1
Adding Brazilian primes to the sieve took 8.6242 ms
The first 20 Brazilian Numbers are:
7 8 10 12 13 14 15 16 18 20 21 22 24 26 27 28 30 31 32 33
The first 20 odd Brazilian Numbers are:
7 13 15 21 27 31 33 35 39 43 45 51 55 57 63 65 69 73 75 77
The first 20 prime Brazilian Numbers are:
7 13 31 43 73 127 157 211 241 307 421 463 601 757 1093 1123 1483 1723 2551 2801
Required output took 2.8256 ms
Decade count of Brazilian numbers:
10th is 20 time: 0.0625 ms
100th is 132 time: 0.1156 ms
1,000th is 1,191 time: 0.1499 ms
10,000th is 11,364 time: 0.1986 ms
100,000th is 110,468 time: 0.4081 ms
1,000,000th is 1,084,566 time: 1.9035 ms
10,000,000th is 10,708,453 time: 15.9129 ms
100,000,000th is 106,091,516 time: 149.8814 ms
1,000,000,000th is 1,053,421,821 time: 1412.3526 ms
Total elapsed was 4391.7287 ms
The point of utilizing a sieve is that it caches or memoizes the results. Since we are going through a long sequence of possible Brazilian numbers, it pays off to check the prime factoring in an efficient way, rather than one at a time.
zkl
fcn isBrazilian(n){
foreach b in ([2..n-2]){
f,m := n%b, n/b;
while(m){
if((m % b)!=f) continue(2);
m/=b;
}
return(True);
}
False
}
fcn isBrazilianW(n){ isBrazilian(n) and n or Void.Skip }
println("First 20 Brazilian numbers:");
[1..].tweak(isBrazilianW).walk(20).println();
println("\nFirst 20 odd Brazilian numbers:");
[1..*,2].tweak(isBrazilianW).walk(20).println();
{{out}}
First 20 Brazilian numbers:
L(7,8,10,12,13,14,15,16,18,20,21,22,24,26,27,28,30,31,32,33)
First 20 odd Brazilian numbers:
L(7,13,15,21,27,31,33,35,39,43,45,51,55,57,63,65,69,73,75,77)
{{libheader|GMP}} GNU Multiple Precision Arithmetic Library Using GMP ( probabilistic primes), because it is easy and fast to generate primes.
[[Extensible prime generator#zkl]] could be used instead.
var [const] BI=Import("zklBigNum"); // libGMP
println("\nFirst 20 prime Brazilian numbers:");
p:=BI(1);
Walker.zero().tweak('wrap{ p.nextPrime().toInt() })
.tweak(isBrazilianW).walk(20).println();
{{out}}
First 20 prime Brazilian numbers:
L(7,13,31,43,73,127,157,211,241,307,421,463,601,757,1093,1123,1483,1723,2551,2801)
println("The 100,00th Brazilian number: ",
[1..].tweak(isBrazilianW).drop(100_000).value);
{{out}}
The 100,00th Brazilian number: 110468