Brazilian numbers: Difference between revisions
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Not very fast, takes about 4 times as long as Go, at least on desktop/Phix, however under pwa/p2js it is about the same speed. |
Not very fast, takes about 4 times as long as Go(v1), at least on desktop/Phix, however under pwa/p2js it is about the same speed. |
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<pre> |
<pre> |
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First 20 Brazilian numbers: |
First 20 Brazilian numbers: |
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Revision as of 21:51, 1 January 2022
You are encouraged to solve this task according to the task description, using any language you may know.
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
11l
<lang 11l>F isPrime(n)
I n % 2 == 0
R n == 2
I n % 3 == 0
R n == 3
V d = 5
L d * d <= n
I n % d == 0
R 0B
I n % (d + 2) == 0
R 0B
d += 6
R 1B
F sameDigits(=n, b)
V d = n % b
n I/= b
I d == 0
R 0B
L n > 0
I n % b != d
R 0B
n I/= b
R 1B
F isBrazilian(n)
I n < 7
R 0B
I (n [&] 1) == 0
R 1B
L(b) 2 .. n - 2
I sameDigits(n, b)
R 1B
R 0B
F printList(title, check)
print(title)
V n = 7
[Int] l
L
I check(n) & isBrazilian(n)
l.append(n)
I l.len == 20 {L.break}
n++
print(l.join(‘, ’))
print()
printList(‘First 20 Brazilian numbers:’, n -> 1B) printList(‘First 20 odd Brazilian numbers:’, n -> (n [&] 1) != 0) printList(‘First 20 prime Brazilian numbers:’, n -> isPrime(n))</lang>
- Output:
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
Action!
Calculations on a real Atari 8-bit computer take quite long time. It is recommended to use an emulator capable with increasing speed of Atari CPU.
<lang Action!>INCLUDE "H6:SIEVE.ACT"
BYTE FUNC SameDigits(INT x,b)
INT d
d=x MOD b
x==/b
WHILE x>0
DO
IF x MOD b#d THEN
RETURN (0)
FI
x==/b
OD
RETURN (1)
BYTE FUNC IsBrazilian(INT x)
INT b
IF x<7 THEN RETURN (0) FI
IF x MOD 2=0 THEN RETURN (1) FI
FOR b=2 TO x-2
DO
IF SameDigits(x,b) THEN
RETURN (1)
FI
OD
RETURN (0)
PROC Main()
DEFINE COUNT="20" DEFINE MAXNUM="3000" BYTE ARRAY primes(MAXNUM+1) INT i,x,c CHAR ARRAY s
Put(125) PutE() ;clear the screen Sieve(primes,MAXNUM+1)
FOR i=0 TO 2
DO
IF i=0 THEN
s=" "
ELSEIF i=1 THEN
s=" odd "
ELSE
s=" prime "
FI
PrintF("First %I%SBrazilian numbers:%E",COUNT,s)
c=0 x=7
DO
IF IsBrazilian(x) THEN
PrintI(x) Put(32)
c==+1
IF c=COUNT THEN EXIT FI
FI
IF i=0 THEN
x==+1
ELSEIF i=1 THEN
x==+2
ELSE
DO
x==+2
UNTIL primes(x)
OD
FI
OD
PutE() PutE()
OD
RETURN</lang>
- Output:
Screenshot from Atari 8-bit computer
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
ALGOL W
Constructs a sieve of Brazilian numbers from the definition. <lang algolw>begin % find some Brazilian numbers - numbers N whose representation in some %
% base B ( 1 < B < N-1 ) has all the same digits %
% set b( 1 :: n ) to a sieve of Brazilian numbers where b( i ) is true %
% if i is Brazilian and false otherwise - n must be at least 8 %
procedure BrazilianSieve ( logical array b ( * ) ; integer value n ) ;
begin
logical isEven;
% start with even numbers flagged as Brazilian and odd numbers as %
% non-Brazilian %
isEven := false;
for i := 1 until n do begin
b( i ) := isEven;
isEven := not isEven
end for_i ;
% numbers below 7 are not Brazilian (see task notes) %
for i := 1 until 6 do b( i ) := false;
% flag all 33, 55, etc. numbers in each base as Brazilian %
% No Brazilian number can have a representation of 11 in any base B %
% as that would mean B + 1 = N, which contradicts B < N - 1 %
% also, no need to consider even digits as we know even numbers > 6 %
% are all Brazilian %
for base := 2 until n div 2 do begin
integer b11, bnn;
b11 := base + 1;
bnn := b11;
for digit := 3 step 2 until base - 1 do begin
bnn := bnn + b11 + b11;
if bnn <= n
then b( bnn ) := true
else goto end_for_digits
end for_digits ;
end_for_digits:
end for_base ;
% handle 111, 1111, 11111, ..., 333, 3333, ..., etc. %
for base := 2 until truncate( sqrt( n ) ) do begin
integer powerMax;
powerMax := MAXINTEGER div base; % avoid 32 bit %
if powerMax > n then powerMax := n; % integer overflow %
for digit := 1 step 2 until base - 1 do begin
integer bPower, bN;
bPower := base * base;
bN := digit * ( bPower + base + 1 ); % ddd %
while bN <= n and bPower <= powerMax do begin
if bN <= n then begin
b( bN ) := true
end if_bN_le_n ;
bPower := bPower * base;
bN := bN + ( digit * bPower )
end while_bStart_le_n
end for_digit
end for_base ;
end BrazilianSieve ;
% sets p( 1 :: n ) to a sieve of primes up to n %
procedure Eratosthenes ( logical array p( * ) ; integer value n ) ;
begin
p( 1 ) := false; p( 2 ) := true;
for i := 3 step 2 until n do p( i ) := true;
for i := 4 step 2 until n do p( i ) := false;
for i := 2 until truncate( sqrt( n ) ) do begin
integer ii; ii := i + i;
if p( i ) then for pr := i * i step ii until n do p( pr ) := false
end for_i ;
end Eratosthenes ;
integer MAX_NUMBER;
MAX_NUMBER := 2000000;
begin
logical array b ( 1 :: MAX_NUMBER );
logical array p ( 1 :: MAX_NUMBER );
integer bCount;
BrazilianSieve( b, MAX_NUMBER );
write( "The first 20 Brazilian numbers:" );write();
bCount := 0;
for bPos := 1 until MAX_NUMBER do begin
if b( bPos ) then begin
bCount := bCount + 1;
writeon( i_w := 1, s_w := 0, " ", bPos );
if bCount >= 20 then goto end_first_20
end if_b_bPos
end for_bPos ;
end_first_20:
write();write( "The first 20 odd Brazilian numbers:" );write();
bCount := 0;
for bPos := 1 step 2 until MAX_NUMBER do begin
if b( bPos ) then begin
bCount := bCount + 1;
writeon( i_w := 1, s_w := 0, " ", bPos );
if bCount >= 20 then goto end_first_20_odd
end if_b_bPos
end for_bPos ;
end_first_20_odd:
write();write( "The first 20 prime Brazilian numbers:" );write();
Eratosthenes( p, MAX_NUMBER );
bCount := 0;
for bPos := 1 until MAX_NUMBER do begin
if b( bPos ) and p( bPos ) then begin
bCount := bCount + 1;
writeon( i_w := 1, s_w := 0, " ", bPos );
if bCount >= 20 then goto end_first_20_prime
end if_b_bPos
end for_bPos ;
end_first_20_prime:
write();write( "Various Brazilian numbers:" );
bCount := 0;
for bPos := 1 until MAX_NUMBER do begin
if b( bPos ) then begin
bCount := bCount + 1;
if bCount = 100
or bCount = 1000
or bCount = 10000
or bCount = 100000
or bCount = 1000000
then write( s_w := 0, bCount, "th Brazilian number: ", bPos );
if bCount >= 1000000 then goto end_1000000
end if_b_bPos
end for_bPos ;
end_1000000:
end
end.</lang>
- Output:
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 20 odd Brazilian numbers:
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:
7 13 31 43 73 127 157 211 241 307 421 463 601 757 1093 1123 1483 1723 2551 2801
Various Brazilian numbers:
100th Brazilian number: 132
1000th Brazilian number: 1191
10000th Brazilian number: 11364
100000th Brazilian number: 110468
1000000th Brazilian number: 1084566
AppleScript
<lang applescript>on isBrazilian(n)
repeat with b from 2 to n - 2
set d to n mod b
set temp to n div b
repeat while (temp mod b = d) -- ((temp > 0) and (temp mod b = d))
set temp to temp div b
end repeat
if (temp = 0) then return true
end repeat
return false
end isBrazilian
on isPrime(n)
if (n < 4) then return (n > 1)
if ((n mod 2 is 0) or (n mod 3 is 0)) then return false
repeat with i from 5 to (n ^ 0.5) div 1 by 6
if ((n mod i is 0) or (n mod (i + 2) is 0)) then return false
end repeat
return true
end isPrime
-- Task code: on runTask()
set output to {}
set astid to AppleScript's text item delimiters
set AppleScript's text item delimiters to " "
set collector to {}
set n to 1
repeat until ((count collector) is 20)
if (isBrazilian(n)) then set end of collector to n
set n to n + 1
end repeat
set end of output to "First 20 Brazilian numbers: " & linefeed & collector
set collector to {}
set n to 1
repeat until ((count collector) is 20)
if (isBrazilian(n)) then set end of collector to n
set n to n + 2
end repeat
set end of output to "First 20 odd Brazilian numbers: " & linefeed & collector
set collector to {}
if (isBrazilian(2)) then set end of collector to 2
set n to 3
repeat until ((count collector) is 20)
if (isPrime(n)) and (isBrazilian(n)) then set end of collector to n
set n to n + 2
end repeat
set end of output to "First 20 prime Brazilian numbers: " & linefeed & collector
set AppleScript's text item delimiters to linefeed
set output to output as text
set AppleScript's text item delimiters to astid
return output
end runTask
return runTask()</lang>
- Output:
<lang applescript>"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"</lang>
Arturo
<lang rebol>brazilian?: function [n][
if n < 7 -> return false
if zero? and n 1 -> return true
loop 2..n-2 'b [
if 1 = size unique digits.base:b n ->
return true
]
return false
]
printFirstByRule: function [rule,title][
print ~"First 20 |title|brazilian numbers:"
i: 7
found: new []
while [20 > size found][
if brazilian? i ->
'found ++ i
do.import rule
]
print found
print ""
]
printFirstByRule [i: i+1] "" printFirstByRule [i: i+2] "odd " printFirstByRule [i: i+2, while [not? prime? i] -> i: i+2] "prime "</lang>
- Output:
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
AWK
<lang 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)
} </lang>
- Output:
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
<lang c>#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;
}</lang>
- Output:
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#
<lang csharp>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);
}
}
}</lang>
- Output:
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. <lang csharp>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();
}
}</lang>
- Output:
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 msP.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.
C++
<lang cpp>#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;
}</lang>
- Output:
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
D
<lang d>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);
}
}</lang>
- Output:
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
Delphi
<lang Delphi> program Brazilian_numbers;
{$APPTYPE CONSOLE}
{$R *.res}
uses
System.SysUtils;
type
TBrazilianNumber = record private FValue: Integer; FIsBrazilian: Boolean; FIsPrime: Boolean; class function SameDigits(a, b: Integer): Boolean; static; class function CheckIsBrazilian(a: Integer): Boolean; static; class function CheckIsPrime(a: Integer): Boolean; static; constructor Create(const Number: Integer); procedure SetValue(const Value: Integer); public property Value: Integer read FValue write SetValue; property IsBrazilian: Boolean read FIsBrazilian; property IsPrime: Boolean read FIsPrime; end;
{ TBrazilianNumber }
class function TBrazilianNumber.CheckIsBrazilian(a: Integer): Boolean; var
b: Integer;
begin
if (a < 7) then Exit(false);
if (a mod 2 = 0) then Exit(true);
for b := 2 to a - 2 do
begin
if (sameDigits(a, b)) then
exit(True);
end;
Result := False;
end;
constructor TBrazilianNumber.Create(const Number: Integer); begin
SetValue(Number);
end;
class function TBrazilianNumber.CheckIsPrime(a: Integer): Boolean; var
d: Integer;
begin
if (a < 2) then exit(False);
if (a mod 2) = 0 then exit(a = 2);
if (a mod 3) = 0 then exit(a = 3);
d := 5;
while (d * d <= a) do
begin
if (a mod d = 0) then
Exit(false);
inc(d, 2);
if (a mod d = 0) then
Exit(false);
inc(d, 4);
end;
Result := True;
end;
class function TBrazilianNumber.SameDigits(a, b: Integer): Boolean; var
f: Integer;
begin
f := a mod b;
a := a div b;
while a > 0 do
begin
if (a mod b) <> f then
exit(False);
a := a div b;
end;
Result := True;
end;
procedure TBrazilianNumber.SetValue(const Value: Integer); begin
if Value < 0 then FValue := 0 else FValue := Value; FIsBrazilian := CheckIsBrazilian(FValue); FIsPrime := CheckIsPrime(FValue);
end;
const
TextLabel: array[0..2] of string = (, 'odd', 'prime');
var
Number: TBrazilianNumber; Count: array[0..2] of Integer; i, j, left, Num: Integer; data: array[0..2] of string;
begin
left := 3;
for i := 0 to 99999 do
begin
if Number.Create(i).IsBrazilian then
for j := 0 to 2 do
begin
if (Count[j] >= 20) and (j > 0) then
continue;
case j of
0:
begin
inc(Count[j]);
Num := i;
if (Count[j] <= 20) then
data[j] := data[j] + i.ToString + ' '
else
Continue;
end;
1:
begin
if Odd(i) then
begin
inc(Count[j]);
data[j] := data[j] + i.ToString + ' ';
end;
end;
2:
begin
if Number.IsPrime then
begin
inc(Count[j]);
data[j] := data[j] + i.ToString + ' ';
end;
end;
end;
if Count[j] = 20 then
dec(left);
end;
if left = 0 then
Break;
end;
while Count[0] < 100000 do
begin
inc(Num);
if Number.Create(Num).IsBrazilian then
inc(Count[0]);
end;
for i := 0 to 2 do begin Writeln(#10'First 20 ' + TextLabel[i] + ' Brazilian numbers:'); Writeln(data[i]); end;
Writeln('The 100,000th Brazilian number: ', Num);
readln;
end.
</lang>
- Output:
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
F#
The functions
<lang fsharp> // 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]
</lang>
The Tasks
- the first 20 Brazilian numbers
<lang fsharp> Brazilian() |> Seq.take 20 |> Seq.iter(printf "%d "); printfn "" </lang>
- Output:
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
<lang fsharp> Brazilian() |> Seq.filter(fun n->n%2=1) |> Seq.take 20 |> Seq.iter(printf "%d "); printfn "" </lang>
- Output:
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 Extensible Prime Generator (F#) <lang fsharp> Brazilian() |> Seq.filter isPrime |> Seq.take 20 |> Seq.iter(printf "%d "); printfn "" </lang>
- Output:
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?
<lang fsharp> printfn "%d" (Seq.item 99999 Brazilian) </lang>
- Output:
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
<lang factor>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@</lang>
- Output:
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 }
Fōrmulæ
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. 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 storage and transfer purposes more than visualization and edition.
Programs in Fōrmulæ are created/edited online in its website, However they run on execution servers. By default remote servers are used, but they are limited in memory and processing power, since they are intended for demonstration and casual use. A local server can be downloaded and installed, it has no limitations (it runs in your own computer). Because of that, example programs can be fully visualized and edited, but some of them will not run if they require a moderate or heavy computation/memory resources, and no local server is being used.
In this page you can see the program(s) related to this task and their results.
Forth
<lang forth>: prime? ( n -- flag )
dup 2 < if drop false exit then dup 2 mod 0= if 2 = exit then dup 3 mod 0= if 3 = exit then 5 begin 2dup dup * >= while 2dup mod 0= if 2drop false exit then 2 + 2dup mod 0= if 2drop false exit then 4 + repeat 2drop true ;
- same_digits? ( n b -- ? )
2dup mod >r
begin
tuck / swap
over 0 >
while
2dup mod r@ <> if
2drop rdrop false exit
then
repeat
2drop rdrop true ;
- brazilian? ( n -- ? )
dup 7 < if drop false exit then
dup 1 and 0= if drop true exit then
dup 1- 2 do
dup i same_digits? if
unloop drop true exit
then
loop
drop false ;
- next_prime ( n -- n )
begin 2 + dup prime? until ;
- print_brazilian ( n1 n2 -- )
>r 7
begin
r@ 0 >
while
dup brazilian? if
dup .
r> 1- >r
then
over 0= if
next_prime
else
over +
then
repeat
2drop rdrop cr ;
." First 20 Brazilian numbers:" cr 1 20 print_brazilian cr
." First 20 odd Brazilian numbers:" cr 2 20 print_brazilian cr
." First 20 prime Brazilian numbers:" cr 0 20 print_brazilian
bye</lang>
- Output:
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
Fortran
<lang Fortran> !Constructs a sieve of Brazilian numbers from the definition. !From the Algol W algorithm, somewhat "Fortranized"
PROGRAM BRAZILIAN
IMPLICIT NONE
! ! PARAMETER definitions !
INTEGER , PARAMETER :: MAX_NUMBER = 2000000 , NUMVARS = 20
! ! Local variables !
LOGICAL , DIMENSION(1:MAX_NUMBER) :: b
INTEGER :: bcount
INTEGER :: bpos
CHARACTER(15) :: holder
CHARACTER(100) :: outline
LOGICAL , DIMENSION(1:MAX_NUMBER) :: p
! ! find some Brazilian numbers - numbers N whose representation in some ! ! base B ( 1 < B < N-1 ) has all the same digits ! ! set b( 1 :: n ) to a sieve of Brazilian numbers where b( i ) is true ! ! if i is Brazilian and false otherwise - n must be at least 8 ! ! sets p( 1 :: n ) to a sieve of primes up to n
CALL BRAZILIANSIEVE(b , MAX_NUMBER)
WRITE(6 , 34)"The first 20 Brazilian numbers:"
bcount = 0
outline =
holder =
bpos = 1
DO WHILE ( bcount<NUMVARS )
IF( b(bpos) )THEN
bcount = bcount + 1
WRITE(holder , *)bpos
outline = TRIM(outline) // " " // ADJUSTL(holder)
END IF
bpos = bpos + 1
END DO
WRITE(6 , 34)outline
WRITE(6 , 34)"The first 20 odd Brazilian numbers:"
outline =
holder =
bcount = 0
bpos = 1
DO WHILE ( bcount<NUMVARS )
IF( b(bpos) )THEN
bcount = bcount + 1
WRITE(holder , *)bpos
outline = TRIM(outline) // " " // ADJUSTL(holder)
END IF
bpos = bpos + 2
END DO
WRITE(6 , 34)outline
WRITE(6 , 34)"The first 20 prime Brazilian numbers:"
CALL ERATOSTHENES(p , MAX_NUMBER)
bcount = 0
outline =
holder =
bpos = 1
DO WHILE ( bcount<NUMVARS )
IF( b(bpos) .AND. p(bpos) )THEN
bcount = bcount + 1
WRITE(holder , *)bpos
outline = TRIM(outline) // " " // ADJUSTL(holder)
END IF
bpos = bpos + 1
END DO
WRITE(6 , 34)outline
WRITE(6 , 34)"Various Brazilian numbers:"
bcount = 0
bpos = 1
DO WHILE ( bcount<1000000 )
IF( b(bpos) )THEN
bcount = bcount + 1
IF( (bcount==100) .OR. (bcount==1000) .OR. (bcount==10000) .OR. &
& (bcount==100000) .OR. (bcount==1000000) )WRITE(* , *)bcount , &
&"th Brazilian number: " , bpos
END IF
bpos = bpos + 1
END DO
STOP
34 FORMAT(/ , a)
END PROGRAM BRAZILIAN
!
SUBROUTINE BRAZILIANSIEVE(B , N)
IMPLICIT NONE
! ! Dummy arguments !
INTEGER :: N
LOGICAL , DIMENSION(*) :: B
INTENT (IN) N
INTENT (OUT) B
! ! Local variables !
INTEGER :: b11
INTEGER :: base
INTEGER :: bn
INTEGER :: bnn
INTEGER :: bpower
INTEGER :: digit
INTEGER :: i
LOGICAL :: iseven
INTEGER :: powermax
!
iseven = .FALSE.
B(1:6) = .FALSE. ! numbers below 7 are not Brazilian (see task notes)
DO i = 7 , N
B(i) = iseven
iseven = .NOT.iseven
END DO
DO base = 2 , (N/2)
b11 = base + 1
bnn = b11
DO digit = 3 , base - 1 , 2
bnn = bnn + b11 + b11
IF( bnn>N )EXIT
B(bnn) = .TRUE.
END DO
END DO
DO base = 2 , INT(SQRT(FLOAT(N)))
powermax = HUGE(powermax)/base ! avoid 32 bit !
IF( powermax>N )powermax = N ! integer overflow !
DO digit = 1 , base - 1 , 2
bpower = base*base
bn = digit*(bpower + base + 1)
DO WHILE ( (bn<=N) .AND. (bpower<=powermax) )
IF( bn<=N )B(bn) = .TRUE.
bpower = bpower*base
bn = bn + (digit*bpower)
END DO
END DO
END DO
RETURN
END SUBROUTINE BRAZILIANSIEVE
!
SUBROUTINE ERATOSTHENES(P , N)
IMPLICIT NONE
! ! Dummy arguments !
INTEGER :: N
LOGICAL , DIMENSION(*) :: P
INTENT (IN) N
INTENT (INOUT) P
! ! Local variables !
INTEGER :: i
INTEGER :: ii
LOGICAL :: oddeven
INTEGER :: pr
!
P(1) = .FALSE.
P(2) = .TRUE.
oddeven = .TRUE.
DO i = 3 , N
P(i) = oddeven
oddeven = .NOT.oddeven
END DO
DO i = 2 , INT(SQRT(FLOAT(N)))
ii = i + i
IF( P(i) )THEN
DO pr = i*i , N , ii
P(pr) = .FALSE.
END DO
END IF
END DO
RETURN
END SUBROUTINE ERATOSTHENES
</lang>
- Output:
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
Various Brazilian numbers:
100 th Brazilian number: 132
1000 th Brazilian number: 1191
10000 th Brazilian number: 11364
100000 th Brazilian number: 110468
1000000 th Brazilian number: 1084566
FreeBASIC
<lang freebasic>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
Wend
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 b
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
Wend
Return True
End Function
Dim kind(2) As String ={"", "odd", "prime"} For i As Integer = 0 To 2
Print Using "First 20 & Brazilian numbers: "; kind(i)
Dim Limit As Integer = 20, n As Integer = 7
Do
If isBrazilian(n) Then Print Using "& "; n; : Limit -= 1
Select Case kind(i)
Case "" : n += 1
Case "odd" : n += 2
Case "prime" : Do : n += 2 : Loop Until isPrime(n)
End Select
Loop While Limit > 0
Next i Sleep</lang>
Go
Version 1
<lang go>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)
}</lang>
- Output:
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
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. <lang go>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)))
}</lang>
- Output:
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
Groovy
<lang groovy>import org.codehaus.groovy.GroovyBugError
class Brazilian {
private static final List<Integer> primeList = new ArrayList<>(Arrays.asList(
2, 3, 5, 7, 9, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89,
97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 169, 173, 179, 181,
191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 247, 251, 257, 263, 269, 271, 277, 281,
283, 293, 299, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 377, 379, 383, 389,
397, 401, 403, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 481, 487, 491,
499, 503, 509, 521, 523, 533, 541, 547, 557, 559, 563, 569, 571, 577, 587, 593, 599, 601, 607,
611, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 689, 691, 701, 709, 719,
727, 733, 739, 743, 751, 757, 761, 767, 769, 773, 787, 793, 797, 809, 811, 821, 823, 827, 829,
839, 853, 857, 859, 863, 871, 877, 881, 883, 887, 907, 911, 919, 923, 929, 937, 941, 947, 949,
953, 967, 971, 977, 983, 991, 997
))
static boolean isPrime(int n) {
if (n < 2) {
return false
}
for (Integer prime : primeList) {
if (n == prime) {
return true
}
if (n % prime == 0) {
return false
}
if (prime * prime > n) {
return true
}
}
BigInteger bi = BigInteger.valueOf(n)
return bi.isProbablePrime(10)
}
private static boolean sameDigits(int n, int b) {
int f = n % b
n = n.intdiv(b)
while (n > 0) {
if (n % b != f) {
return false
}
n = n.intdiv(b)
}
return true
}
private static boolean 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 void main(String[] args) {
for (String kind : Arrays.asList("", "odd ", "prime ")) {
boolean quiet = false
int bigLim = 99_999
int limit = 20
System.out.printf("First %d %sBrazilian numbers:\n", limit, kind)
int c = 0
int n = 7
while (c < bigLim) {
if (isBrazilian(n)) {
if (!quiet) System.out.printf("%d ", n)
if (++c == limit) {
System.out.println("\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:
throw new GroovyBugError("Oops")
}
}
if ("" == kind) {
System.out.printf("The %dth Brazilian number is: %d\n\n", bigLim + 1, n)
}
}
}
}</lang>
- Output:
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 100000th Brazilian number is: 110468 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
Haskell
<lang 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)]</lang>
- Output:
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]
Isabelle
Not the most beautiful proofs and the theorem about "R*S >= 8, with S+1 > R, are Brazilian" is missing.
<lang Isabelle>theory Brazilian imports Main begin
function (sequential) base :: "nat ⇒ nat ⇒ nat list" where
"base n 0 = undefined"
| "base n (Suc 0) = replicate n 1" | "base n b = (if n < b then [n]
else (base (n div b) b) @ [n mod b]
)"
by pat_completeness auto
termination base
apply(relation "measure (λ(n,b). n div b)", simp) using div_greater_zero_iff by auto
lemma base_simps:
"b > 1 ⟹ base n b = (if n < b then [n] else base (n div b) b @ [n mod b])" by (metis One_nat_def base.elims nat_neq_iff not_less_zero)
lemma "base 123 10 = [1, 2, 3]"
and "base 65536 2 = [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]" and "base 65535 2 = [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]" and "base 123 100 = [1, 23]" and "base 69 100 = [69]" and "base 255 16 = [15, 15]" by(simp add: base_simps)+
lemma "base 5 1 = [1, 1, 1, 1, 1]"
by (simp add: eval_nat_numeral(3) numeral_Bit0)
lemma base_2_numbers: "a < b ⟹ c < b ⟹ a > 0 ⟹ base (a * b + c) b = [a, c]"
apply(simp add: base_simps) using mult_eq_if by auto
lemma base_half_minus_one: "even n ⟹ n ≥ 8 ⟹ base n (n div 2 - 1) = [2, 2]" proof -
assume "even n" and "n ≥ 8" have n: "(2 * (n div 2 - 1) + 2) = n" using ‹n ≥ 8› ‹even n› add.commute dvd_mult_div_cancel le_0_eq by auto from ‹n ≥ 8› base_2_numbers[where b="n div 2 - 1" and a=2 and c=2] have "base (2 * (n div 2 - 1) + 2) (n div 2 - 1) = [2, 2]" by simp with n show ?thesis by simp
qed
lemma base_rs_numbers: "r > 0 ⟹ s - 1 > r ⟹ base (r*s) (s - 1) = [r, r]" proof -
assume "r > 0" and "s - 1 > r" from ‹s - 1 > r› have "r*s = r*(s - 1) + r" by (metis add.commute gr_implies_not0 less_diff_conv mult.commute mult_eq_if) from base_2_numbers[where a=r and b="s - 1" and c=r] have "s - 1 > r ⟹ 0 < r ⟹ base (r * (s - 1) + r) (s - 1) = [r, r]" . with ‹s - 1 > r› ‹r > 0› have "base (r * (s - 1) + r) (s - 1) = [r, r]" by(simp) with ‹r * s = r * (s - 1) + r› show "base (r*s) (s - 1) = [r, r]" by (simp)
qed
definition all_equal :: "nat list ⇒ bool" where
"all_equal xs ≡ ∀x∈set xs. ∀y∈set xs. x = y"
lemma all_equal_alt:
"all_equal xs ⟷ replicate (length xs) (hd xs) = xs" unfolding all_equal_def apply(induction xs) apply(simp) apply(simp) by (metis in_set_replicate replicate_eqI)
definition brazilian :: "nat set" where
"brazilian ≡ {n. ∃b. 1 < b ∧ Suc b < n ∧ all_equal (base n b)}"
lemma "0 ∉ brazilian"
and "1 ∉ brazilian" and "2 ∉ brazilian" and "3 ∉ brazilian" by (simp add: brazilian_def)+
lemma "4 ∉ brazilian"
apply (simp add: brazilian_def all_equal_def) apply(intro allI impI) apply(case_tac "b = 1", simp) apply(case_tac "b = 2", simp add: base_simps, blast) by(simp)
lemma "5 ∉ brazilian"
apply (simp add: brazilian_def all_equal_def) apply(intro allI impI) apply(case_tac "b = 1", simp) apply(case_tac "b = 2", simp add: base_simps, blast) apply(case_tac "b = 3", simp add: base_simps, blast) by(simp)
lemma "6 ∉ brazilian"
apply (simp add: brazilian_def all_equal_def) apply(intro allI impI) apply(case_tac "b = 1", simp) apply(case_tac "b = 2", simp add: base_simps, blast) apply(case_tac "b = 3", simp add: base_simps, blast) apply(case_tac "b = 4", simp add: base_simps, blast) by(simp)
lemma "7 ∈ brazilian"
apply(simp add: brazilian_def) apply(rule exI[where x=2]) by(simp add: base_simps all_equal_def)
lemma "8 ∈ brazilian"
apply(simp add: brazilian_def) apply(rule exI[where x=3]) by(simp add: base_simps all_equal_def)
lemma "9 ∉ brazilian"
apply (simp add: brazilian_def all_equal_def) apply(intro allI impI) apply(case_tac "b = 1", simp) apply(case_tac "b = 2", simp add: base_simps, blast) apply(case_tac "b = 3", simp add: base_simps, blast) apply(case_tac "b = 4", simp add: base_simps, blast) apply(case_tac "b = 5", simp add: base_simps, blast) apply(case_tac "b = 6", simp add: base_simps, blast) apply(case_tac "b = 7", simp add: base_simps, blast) by(simp)
theorem "even n ⟹ n ≥ 8 ⟹ n ∈ brazilian"
apply(simp add: brazilian_def) apply(rule_tac x="n div 2 - 1" in exI) by(simp add: base_half_minus_one[simplified] all_equal_def)
(* The problem description on Rosettacode was broken. Fortunately, we found the error when proving it correct with Isabelle. Rosettacode claimed: "all integers, that factor decomposition is R*S >= 8, with S+1 > R,
are Brazilian because R*S = R(S-1) + R, which is RR in base S-1"
This is wrong. Here are some counterexamples:
r = 3 s = 3 r*s = 9 ≥ 8 s+1 = 4 > 3 = r But (s*r) = 9 ∉ brazilian
The correct precondition would be s-1>r instead of s+1>r.
But this is not enough.
Also, r > 1 is needed additionally, because r=1 s=9 r*s = 9 ≥ 8 s+1 = 10 > 1 = r or s-1 = 8 > 1 = r But (s*r) = 9 ∉ brazilian
Doing the proof, we also learn that the precondition r*s ≥ 8 is not required.
- )
theorem "r > 1 ⟹ s-1 > r ⟹ r*s ∈ brazilian"
apply(simp add: brazilian_def) apply(rule_tac x="s - 1" in exI) apply(subst base_rs_numbers[of r s]) using not_numeral_le_zero apply fastforce apply(simp; fail) by(simp add: all_equal_def)
fun is_brazilian_for_base :: "nat ⇒ nat ⇒ bool" where
"is_brazilian_for_base n 0 ⟷ False"
| "is_brazilian_for_base n (Suc 0) ⟷ False" | "is_brazilian_for_base n (Suc b) ⟷ all_equal (base n (Suc b)) ∨ is_brazilian_for_base n b"
lemma "is_brazilian_for_base 7 2" and "is_brazilian_for_base 8 3" by code_simp+
lemma is_brazilian_for_base_Suc_simps:
"is_brazilian_for_base n (Suc b) ⟷ b ≠ 0 ∧ (all_equal (base n (Suc b)) ∨ is_brazilian_for_base n b)" by(cases b)(simp)+
lemma is_brazilian_for_base:
"is_brazilian_for_base n b ⟷ (∃w ∈ {2..b}. all_equal (base n w))"
proof(induction b)
case 0
show "is_brazilian_for_base n 0 = (∃w∈{2..0}. all_equal (base n w))" by simp
next
case (Suc b)
assume IH: "is_brazilian_for_base n b = (∃w∈{2..b}. all_equal (base n w))"
show "is_brazilian_for_base n (Suc b) = (∃w∈{2..Suc b}. all_equal (base n w))"
apply(simp add: is_brazilian_for_base_Suc_simps IH)
using le_Suc_eq by fastforce
qed
lemma is_brazilian_for_base_is:
"Suc (Suc n) ∈ brazilian ⟷ is_brazilian_for_base (Suc (Suc n)) n" apply(simp add: brazilian_def is_brazilian_for_base) using less_Suc_eq_le by force
definition brazilian_executable :: "nat ⇒ bool" where
"brazilian_executable n ≡ n > 1 ∧ is_brazilian_for_base n (n - 2)"
lemma brazilian_executable[code_unfold]:
"n ∈ brazilian ⟷ brazilian_executable n" apply(simp add: brazilian_executable_def) apply(cases "n = 0 ∨ n = 1") apply(simp add: brazilian_def) apply(blast) apply(simp) apply(case_tac n, simp, simp, rename_tac n2) apply(case_tac n2, simp, simp, rename_tac n3) apply(subst is_brazilian_for_base_is[symmetric]) apply(simp) done
text‹ In Isabelle/HOl, functions must be total, i.e. they must terminate. Therefore, we cannot simply write a function which enumerates the infinite set of natural numbers and stops when we found 20 Brazilian numbers, since it is not guaranteed that 20 Brazilian numbers exist and that the function will terminate. We could prove that and then write that function, but here is the lazy solution: ›
lemma "[n ← upt 0 34. n ∈ brazilian] =
[7,8,10,12,13,14,15,16,18,20,21,22,24,26,27,28,30,31,32,33]" by(code_simp)
lemma "[n ← upt 0 80. odd n ∧ n ∈ brazilian] =
[7,13,15,21,27,31,33,35,39,43,45,51,55,57,63,65,69,73,75,77]" by code_simp
(*TODO: the first 20 prime Brazilian numbers*)
end</lang>
J
The brazilian verb checks if 1 is the tally of one of the sets of values in the possible base representations. <lang>
Doc=: conjunction def 'u :: (n"_)'
brazilian=: (1 e. (#@~.@(#.^:_1&>)~ (2 + [: (i.) _3&+)))&> Doc 'brazilian y NB. is 1 if y is brazilian, else 0'
Filter=: (#~`)(`:6)
B=: brazilian Filter 4 + i. 300 NB. gather Brazilion numbers less than 304
20 {. B NB. first 20 Brazilion numbers
7 8 10 12 13 14 15 16 18 20 21 22 24 26 27 28 30 31 32 33
odd =: 1 = 2&|
20 {. odd Filter B NB. first 20 odd Brazilion numbers
7 13 15 21 27 31 33 35 39 43 45 51 55 57 63 65 69 73 75 77
prime=: 1&p:
20 {. prime Filter B NB. uh oh need a new technique
7 13 31 43 73 127 157 211 241 0 0 0 0 0 0 0 0 0 0 0
NB. p: y is the yth prime, with 2 being prime 0
20 {. brazilian Filter p: 2 + i. 500
7 13 31 43 73 127 157 211 241 307 421 463 601 757 1093 1123 1483 1723 2551 2801 </lang>
Java
<lang java>import java.math.BigInteger; import java.util.List;
public class Brazilian {
private static final List<Integer> primeList = List.of(
2, 3, 5, 7, 9, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89,
97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 169, 173, 179, 181,
191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 247, 251, 257, 263, 269, 271, 277, 281,
283, 293, 299, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 377, 379, 383, 389,
397, 401, 403, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 481, 487, 491,
499, 503, 509, 521, 523, 533, 541, 547, 557, 559, 563, 569, 571, 577, 587, 593, 599, 601, 607,
611, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 689, 691, 701, 709, 719,
727, 733, 739, 743, 751, 757, 761, 767, 769, 773, 787, 793, 797, 809, 811, 821, 823, 827, 829,
839, 853, 857, 859, 863, 871, 877, 881, 883, 887, 907, 911, 919, 923, 929, 937, 941, 947, 949,
953, 967, 971, 977, 983, 991, 997
);
public static boolean isPrime(int n) {
if (n < 2) {
return false;
}
for (Integer prime : primeList) {
if (n == prime) {
return true;
}
if (n % prime == 0) {
return false;
}
if (prime * prime > n) {
return true;
}
}
BigInteger bi = BigInteger.valueOf(n);
return bi.isProbablePrime(10);
}
private static boolean sameDigits(int n, int b) {
int f = n % b;
while ((n /= b) > 0) {
if (n % b != f) {
return false;
}
}
return true;
}
private static boolean 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;
}
public static void main(String[] args) {
for (String kind : List.of("", "odd ", "prime ")) {
boolean quiet = false;
int bigLim = 99_999;
int limit = 20;
System.out.printf("First %d %sBrazilian numbers:\n", limit, kind);
int c = 0;
int n = 7;
while (c < bigLim) {
if (isBrazilian(n)) {
if (!quiet) System.out.printf("%d ", n);
if (++c == limit) {
System.out.println("\n");
quiet = true;
}
}
if (quiet && !"".equals(kind)) continue;
switch (kind) {
case "":
n++;
break;
case "odd ":
n += 2;
break;
case "prime ":
do {
n += 2;
} while (!isPrime(n));
break;
default:
throw new AssertionError("Oops");
}
}
if ("".equals(kind)) {
System.out.printf("The %dth Brazilian number is: %d\n\n", bigLim + 1, n);
}
}
}
}</lang>
- Output:
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 100000th Brazilian number is: 110468 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
jq
Works with gojq, the Go implementation of jq
See Erdős-primes#jq for a suitable definition of `is_prime` as used here.
<lang jq># Output: a stream of digits, least significant digit first def to_base($base):
def butlast(s):
label $out
| foreach (s,null) as $x ({};
if $x == null then break $out else .emit = .prev | .prev = $x end;
select(.emit).emit);
if . == 0 then 0 else butlast(recurse( if . == 0 then empty else ./$base | floor end ) % $base) end ;
def sameDigits($n; $b):
($n % $b) as $f | all( ($n | to_base($b)); . == $f) ;
def isBrazilian:
. as $n | if . < 7 then false elif (.%2 == 0 and . >= 8) then true else any(range(2; $n-1); sameDigits($n; .) ) end;
def brazilian_numbers($m; $n; $step):
range($m; $n; $step) | select(isBrazilian);
def task($n):
"First \($n) Brazilian numbers:", limit($n; brazilian_numbers(7; infinite; 1)), "First \($n) odd Brazilian numbers:", limit($n; brazilian_numbers(7; infinite; 2)), "First \($n) prime Brazilian numbers:", limit($n; brazilian_numbers(7; infinite; 2) | select(is_prime)) ;
task(20)
</lang>
- Output:
As elsewhere, e.g. #Python.
Julia
<lang julia>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))
</lang>
- Output:
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: <lang julia>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))
</lang>
- Output:
The 10000 th brazilian is 11364. The 100000 th brazilian is 110468. The 1000000 th brazilian is 1084566.
Kotlin
<lang scala>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))
}
}</lang>
- Output:
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
Lua
<lang lua>function sameDigits(n,b)
local f = n % b
n = math.floor(n / b)
while n > 0 do
if n % b ~= f then
return false
end
n = math.floor(n / b)
end
return true
end
function isBrazilian(n)
if n < 7 then
return false
end
if (n % 2 == 0) and (n >= 8) then
return true
end
for b=2,n-2 do
if sameDigits(n,b) then
return true
end
end
return false
end
function isPrime(n)
if n < 2 then
return false
end
if n % 2 == 0 then
return n == 2
end
if n % 3 == 0 then
return n == 3
end
local d = 5
while d * d <= n do
if n % d == 0 then
return false
end
d = d + 2
if n % d == 0 then
return false
end
d = d + 4
end
return true
end
function main()
local kinds = {" ", " odd ", " prime "}
for i=1,3 do
print("First 20" .. kinds[i] .. "Brazillion numbers:")
local c = 0
local n = 7
while true do
if isBrazilian(n) then
io.write(n .. " ")
c = c + 1
if c == 20 then
print()
print()
break
end
end
if i == 1 then
n = n + 1
elseif i == 2 then
n = n + 2
elseif i == 3 then
repeat
n = n + 2
until isPrime(n)
end
end
end
end
main()</lang>
- Output:
First 20 Brazillion numbers: 7 8 10 12 13 14 15 16 18 20 21 22 24 26 27 28 30 31 32 33 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
Mathematica / Wolfram Language
<lang wolfram>brazilianQ[n_Integer /; n>6 ] := AnyTrue[
Range[2, n-2],
MatchQ[IntegerDigits[n, #], {x_ ...}] &
] Select[Range[100], brazilianQ, 20] Select[Range[100], brazilianQ@# && OddQ@# &, 20] Select[Range[10000], brazilianQ@# && PrimeQ@# &, 20] </lang>
- Output:
{7, 8, 10, 12, 13, 14, 15, 16, 18, 20, 21, 22, 24, 26, 27, 28, 30, 31, 32, 33}
{7, 13, 15, 21, 27, 31, 33, 35, 39, 43, 45, 51, 55, 57, 63, 65, 69, 73, 75, 77}
{7, 13, 31, 43, 73, 127, 157, 211, 241, 307, 421, 463, 601, 757, 1093, 1123, 1483, 1723, 2551, 2801}
Nim
<lang Nim>proc isPrime(n: Positive): bool =
## Check if a number is prime.
if n mod 2 == 0:
return n == 2
if n mod 3 == 0:
return n == 3
var d = 5
while d * d <= n:
if n mod d == 0:
return false
if n mod (d + 2) == 0:
return false
inc d, 6
result = true
proc sameDigits(n, b: Positive): bool =
## Check if the digits of "n" in base "b" are all the same.
var d = n mod b
var n = n div b
if d == 0:
return false
while n > 0:
if n mod b != d:
return false
n = n div b
result = true
proc isBrazilian(n: Positive): bool =
## Check if a number is brazilian.
if n < 7:
return false
if (n and 1) == 0:
return true
for b in 2..(n - 2):
if sameDigits(n, b):
return true
- ———————————————————————————————————————————————————————————————————————————————————————————————————
when isMainModule:
import strutils
template printList(title: string; findNextToCheck: untyped) =
## Template to print a list of brazilians numbers.
## "findNextTocheck" is a list of instructions to find the
## next candidate starting for the current one "n".
block:
echo '\n' & title
var n {.inject.} = 7
var list: seq[int]
while true:
if n.isBrazilian():
list.add(n)
if list.len == 20: break
findNextToCheck
echo list.join(", ")
printList("First 20 Brazilian numbers:"):
inc n
printList("First 20 odd Brazilian numbers:"):
inc n, 2
printList("First 20 prime Brazilian numbers:"):
inc n, 2
while not n.isPrime():
inc n, 2</lang>
- Output:
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
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.
<lang pascal>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.</lang>
- Output:
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
<lang perl>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; use constant Inf => 1e10;
my $b = "First $upto Brazilian numbers:\n"; my $n = 0; $b .= do { $n < $upto ? (is_Brazilian($_) and ++$n and "$_ ") : last } for 1 .. Inf;
$b .= "\n\nFirst $upto odd Brazilian numbers:\n"; $n = 0; $b .= do { $n < $upto ? (!!($_%2) and is_Brazilian($_) and ++$n and "$_ ") : last } for 1 .. Inf;
$b .= "\n\nFirst $upto prime Brazilian numbers:\n"; $n = 0; $b .= do { $n < $upto ? (!!is_prime($_) and is_Brazilian($_) and ++$n and "$_ ") : last } for 1 .. Inf;
say $b;</lang>
- Output:
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
with javascript_semantics 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 platform()!=JS and 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)
- Output:
Not very fast, takes about 4 times as long as Go(v1), at least on desktop/Phix, however under pwa/p2js it is about the same speed.
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
<lang python>Brazilian numbers
from itertools import count, islice
- 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(base):
def g(b, n):
(q, d) = divmod(n, b)
def p(qr):
return d != qr[1] or 0 == qr[0]
def f(qr):
return divmod(qr[0], b)
return d == until(p)(f)(
(q, d)
)[1]
return g(base, n)
return go
- -------------------------- TEST --------------------------
- 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'
)
- ------------------- GENERIC FUNCTIONS --------------------
- 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.
def go(xs):
return (
xs[0:n]
if isinstance(xs, (list, tuple))
else list(islice(xs, n))
)
return go
- 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):
def g(x):
v = x
while not p(v):
v = f(v)
return v
return g
return go
- MAIN ---
if __name__ == '__main__':
main()</lang>
- Output:
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]
Racket
<lang racket>#lang racket
(require math/number-theory)
(define (repeat-digit? n base d-must-be-1?)
(call-with-values
(λ () (quotient/remainder n base))
(λ (q d) (and (or (not d-must-be-1?) (= d 1))
(let loop ((n q))
(if (zero? n)
d
(call-with-values
(λ () (quotient/remainder n base))
(λ (q r) (and (= d r) (loop q))))))))))
(define (brazilian? n (for-prime? #f))
(for/first ((b (in-range 2 (sub1 n))) #:when (repeat-digit? n b for-prime?)) b))
(define (prime-brazilian? n)
(and (prime? n) (brazilian? n #t)))
(module+ main
(displayln "First 20 Brazilian numbers:") (stream->list (stream-take (stream-filter brazilian? (in-naturals)) 20)) (displayln "First 20 odd Brazilian numbers:") (stream->list (stream-take (stream-filter brazilian? (stream-filter odd? (in-naturals))) 20)) (displayln "First 20 prime Brazilian numbers:") (stream->list (stream-take (stream-filter prime-brazilian? (stream-filter odd? (in-naturals))) 20)))</lang>
- Output:
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)
Raku
(formerly Perl 6)
<lang perl6>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];</lang>
- Output:
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
REXX
<lang>/*REXX pgm finds: 1st N Brazilian #s; odd Brazilian #s; prime Brazilian #s; ZZZth #.*/ parse arg t.1 t.2 t.3 t.4 . /*obtain optional arguments from the CL*/ if t.4== | t.4=="," then t.4= 0 /*special test case of Nth Brazilian #.*/ hdr.1= 'first'; hdr.2= "first odd"; hdr.3= 'first prime'; hdr.4= /*four headers.*/
#p= 0 /*#P: the number of primes (so far).*/
do c=1 for 4 /*process each of the four cases. */
if t.c== | t.c=="," then t.c= 20 /*check if a target is null or a comma.*/
step= 1 + (c==2) /*STEP is set to unity or two (for ODD)*/
if t.c==0 then iterate /*check to see if this case target ≡ 0.*/
$=; #= 0 /*initialize list to null; counter to 0*/
do j=1 by step until #>= t.c /*search integers for Brazilian # type.*/
prime= 0 /*signify if J may not be prime. */
if c==3 then do /*is this a "case 3" calculation? */
if \isPrime(j) then iterate /*(case 3) Not a prime? Then skip it.*/
prime= 1 /*signify if J is definately a prime.*/
end /* [↓] J≡prime will be used for speedup*/
if \isBraz(j, prime) then iterate /*Not Brazilian number? " " " */
#= # + 1 /*bump the counter of Brazilian numbers*/
if c\==4 then $= $ j /*for most cases, append J to ($) list.*/
end /*j*/ /* [↑] cases 1──►3, $ has leading blank*/
say /* [↓] use a special header for cases.*/
if c==4 then do; $= j; t.c= th(t.c); end /*for Nth Brazilian number, just use J.*/
say center(' 'hdr.c" " t.c " Brazilian number"left('s', c\==4)" ", 79, '═')
say strip($) /*display a case result to the terminal*/
end /*c*/ /* [↑] cases 1──►3 have a leading blank*/
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ isBraz: procedure; parse arg x,p; if x<7 then return 0 /*Is # < seven? Nope. */
if x//2==0 then return 1 /*Is # even? Yup. _*/
if p then mx= iSqrt(x) /*X prime? Use integer √X*/
else mx= x%3 -1 /*X not known if prime. */
do b=2 for mx /*scan for base 2 ──► max*/
if sameDig(x, b) then return 1 /*it's a Brazilian number*/
end /*b*/; return 0 /*not " " " */
/*──────────────────────────────────────────────────────────────────────────────────────*/ isPrime: procedure expose @. !. #p; parse arg x -1 _ /*get 1st arg & last decimal dig*/
if #p==0 then do; !.=0; y= 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67
do i=1 for words(y); #p= #p+1; z=word(y,i); @.#p= z; !.z=1; end
end /*#P: is the number of primes. */
if !.x then return 1; if x<61 then return 0; if x//2==0 then return 0
if x//3==0 then return 0; if _==5 then return 0; if x//7==0 then return 0
do j=5 until @.j**2>x; if x//@.j ==0 then return 0
if x//(@.j+2) ==0 then return 0
end /*j*/; #p= #p + 1; @.#p= x; !.x= 1; return 1 /*it's a prime.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/ iSqrt: procedure; parse arg x; q= 1; r= 0; do while q<=x; q= q*4; end
do while q>1; q=q%4; _=x-r-q; r=r%2; if _>=0 then do;x=_;r=r+q;end;end; return r
/*──────────────────────────────────────────────────────────────────────────────────────*/ sameDig: procedure; parse arg x, b; f= x // b /* // ◄── the remainder.*/
x= x % b /* % ◄── is integer ÷ */
do while x>0; if x//b \==f then return 0
x= x % b
end /*while*/; return 1 /*it has all the same dig*/
/*──────────────────────────────────────────────────────────────────────────────────────*/ th: parse arg th; return th || word('th st nd rd', 1+(th//10)*(th//100%10\==1)*(th//10<4))</lang>
- output when using the inputs of: , , , 100000
══════════════════════ 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
Ring
<lang ring> load "stdlib.ring"
decList = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15] baseList = ["0","1","2","3","4","5","6","7","8","9","A","B","C","D","E","F"] brazil = [] brazilOdd = [] brazilPrime = [] num1 = 0 num2 = 0 num3 = 0 limit = 20
see "working..." + nl for n = 1 to 2802
for m = 2 to 16
flag = 1
basem = decimaltobase(n,m)
for p = 1 to len(basem)-1
if basem[p] != basem[p+1]
flag = 0
exit
ok
next
if flag = 1 and m < n - 1
add(brazil,n)
delBrazil(brazil)
ok
if flag = 1 and m < n - 1 and n % 2 = 1
add(brazilOdd,n)
delBrazil(brazilOdd)
ok
if flag = 1 and m < n - 1 and isprime(n)
add(brazilPrime,n)
delBrazil(brazilPrime)
ok
next
next
see "2 <= base <= 16" + nl see "first 20 brazilian numbers:" + nl showarray(brazil) see "first 20 odd brazilian numbers:" + nl showarray(brazilOdd) see "first 11 brazilian prime numbers:" + nl showarray(brazilPrime)
see "done..." + nl
func delBrazil(brazil)
for z = len(brazil) to 2 step -1
if brazil[z] = brazil[z-1]
del(brazil,z)
ok
next
func decimaltobase(nr,base)
binList = []
binary = 0
remainder = 1
while(nr != 0)
remainder = nr % base
ind = find(decList,remainder)
rem = baseList[ind]
add(binList,rem)
nr = floor(nr/base)
end
binlist = reverse(binList)
binList = list2str(binList)
binList = substr(binList,nl,"")
return binList
func showArray(array)
txt = ""
if len(array) < limit
limit = len(array)
ok
for n = 1 to limit
txt = txt + array[n] + ","
next
txt = left(txt,len(txt)-1)
see txt + nl
</lang> Output:
working... 2 <= base <= 16 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: 7,13,15,21,27,31,33,35,39,43,45,51,55,57,63,65,73,75,77,85 first 11 brazilian prime numbers: 7,13,31,43,73,127,157,211,241,1093,2801 done...
Ruby
<lang ruby>def sameDigits(n,b)
f = n % b
while (n /= b) > 0 do
if n % b != f then
return false
end
end
return true
end
def isBrazilian(n)
if n < 7 then
return false
end
if n % 2 == 0 then
return true
end
for b in 2 .. n - 2 do
if sameDigits(n, b) then
return true
end
end
return false
end
def isPrime(n)
if n < 2 then
return false
end
if n % 2 == 0 then
return n == 2
end
if n % 3 == 0 then
return n == 3
end
d = 5
while d * d <= n do
if n % d == 0 then
return false
end
d = d + 2
if n % d == 0 then
return false
end
d = d + 4
end
return true
end
def main
for kind in ["", "odd ", "prime "] do
quiet = false
bigLim = 99999
limit = 20
puts "First %d %sBrazilian numbers:" % [limit, kind]
c = 0
n = 7
while c < bigLim do
if isBrazilian(n) then
if not quiet then
print "%d " % [n]
end
c = c + 1
if c == limit then
puts
puts
quiet = true
end
end
if quiet and kind != "" then
next
end
if kind == "" then
n = n + 1
elsif kind == "odd " then
n = n + 2
elsif kind == "prime " then
loop do
n = n + 2
if isPrime(n) then
break
end
end
else
raise "Unexpected"
end
end
if kind == "" then
puts "The %dth Brazillian number is: %d" % [bigLim + 1, n]
puts
end
end
end
main()</lang>
- Output:
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 100000th Brazillian number is: 110468 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
Rust
<lang rust> 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);
} </lang>
- Output:
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 100000th Brazilian number: 110468
Scala
<lang scala>object BrazilianNumbers {
private val PRIME_LIST = List( 2, 3, 5, 7, 9, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 169, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 247, 251, 257, 263, 269, 271, 277, 281, 283, 293, 299, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 377, 379, 383, 389, 397, 401, 403, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 481, 487, 491, 499, 503, 509, 521, 523, 533, 541, 547, 557, 559, 563, 569, 571, 577, 587, 593, 599, 601, 607, 611, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 689, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 767, 769, 773, 787, 793, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 871, 877, 881, 883, 887, 907, 911, 919, 923, 929, 937, 941, 947, 949, 953, 967, 971, 977, 983, 991, 997 )
def isPrime(n: Int): Boolean = {
if (n < 2) {
return false
}
for (prime <- PRIME_LIST) {
if (n == prime) {
return true
}
if (n % prime == 0) {
return false
}
if (prime * prime > n) {
return true
}
}
val bigDecimal = BigInt.int2bigInt(n) bigDecimal.isProbablePrime(10) }
def sameDigits(n: Int, b: Int): Boolean = {
var n2 = n
val f = n % b
var done = false
while (!done) {
n2 /= b
if (n2 > 0) {
if (n2 % b != f) {
return false
}
} else {
done = true
}
}
true
}
def isBrazilian(n: Int): Boolean = {
if (n < 7) {
return false
}
if (n % 2 == 0) {
return true
}
for (b <- 2 until n - 1) {
if (sameDigits(n, b)) {
return true
}
}
false
}
def main(args: Array[String]): Unit = {
for (kind <- List("", "odd ", "prime ")) {
var quiet = false
var bigLim = 99999
var limit = 20
println(s"First $limit ${kind}Brazilian numbers:")
var c = 0
var n = 7
while (c < bigLim) {
if (isBrazilian(n)) {
if (!quiet) {
print(s"$n ")
}
c = c + 1
if (c == limit) {
println()
println()
quiet = true
}
}
if (!quiet || kind == "") {
if (kind == "") {
n = n + 1
} else if (kind == "odd ") {
n = n + 2
} else if (kind == "prime ") {
do {
n = n + 2
} while (!isPrime(n))
} else {
throw new AssertionError("Oops")
}
}
}
if (kind == "") {
println(s"The ${bigLim + 1}th Brazilian number is: $n")
println()
}
}
}
}</lang>
- Output:
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 100000th Brazilian number is: 110468 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
Sidef
<lang ruby>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))
}</lang>
- Output:
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: <lang ruby>for n in (1..6) {
say ("#{10**n->commify}th Brazilian number = ", is_Brazilian.nth(10**n))
}</lang>
- Output:
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
<lang vbnet>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</lang>
- Output:
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). <lang vbnet>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 </lang>
- Output:
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.
Wren
<lang ecmascript>import "/math" for Int
var sameDigits = Fn.new { |n, b|
var f = n % b
n = (n/b).floor
while (n > 0) {
if (n%b != f) return false
n = (n/b).floor
}
return true
}
var isBrazilian = Fn.new { |n|
if (n < 7) return false
if (n%2 == 0 && n >= 8) return true
for (b in 2...n-1) {
if (sameDigits.call(n, b)) return true
}
return false
}
for (kind in [" ", " odd ", " prime "]) {
System.print("First 20%(kind)Brazilian numbers:")
var c = 0
var n = 7
while (true) {
if (isBrazilian.call(n)) {
System.write("%(n) ")
c = c + 1
if (c == 20) {
System.print("\n")
break
}
}
if (kind == " ") {
n = n + 1
} else if (kind == " odd ") {
n = n + 2
} else {
while (true) {
n = n + 2
if (Int.isPrime(n)) break
}
}
}
}
var c = 0 var n = 7 while (c < 1e5) {
if (isBrazilian.call(n)) c = c + 1 n = n + 1
} System.print("The 100,000th Brazilian number: %(n-1)")</lang>
- Output:
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
zkl
<lang 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 }</lang> <lang zkl>println("First 20 Brazilian numbers:"); [1..].tweak(isBrazilianW).walk(20).println();
println("\nFirst 20 odd Brazilian numbers:"); [1..*,2].tweak(isBrazilianW).walk(20).println();</lang>
- Output:
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)
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. <lang zkl>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();</lang>
- Output:
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)
<lang zkl>println("The 100,00th Brazilian number: ",
[1..].tweak(isBrazilianW).drop(100_000).value);</lang>
- Output:
The 100,00th Brazilian number: 110468
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