Brazilian numbers: Difference between revisions
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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'''.<BR>
More common: for all all integers
The only problematic numbers are squares of primes, where R = S. Only 11^2 is brazilian to base 3.<BR>
All prime integers, that are brazilian, can only have the digit '''1'''
;Task:
Line 38:
:* '''[[oeis:A085104|OEIS:A085104 - Prime Brazilian numbers]]'''
<br><br>
=={{header|11l}}==
{{trans|Nim}}
<syntaxhighlight 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))</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|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.
{{libheader|Action! Sieve of Eratosthenes}}
<syntaxhighlight 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</syntaxhighlight>
{{out}}
[https://gitlab.com/amarok8bit/action-rosetta-code/-/raw/master/images/Brazilian_numbers.png Screenshot from Atari 8-bit computer]
<pre>
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
</pre>
=={{header|ALGOL W}}==
Constructs a sieve of Brazilian numbers from the definition.
<syntaxhighlight 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.</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|AppleScript}}==
<syntaxhighlight 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()</syntaxhighlight>
{{output}}
<syntaxhighlight 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"</syntaxhighlight>
=={{header|Arturo}}==
<syntaxhighlight 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 "</syntaxhighlight>
{{out}}
<pre>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</pre>
=={{header|AWK}}==
<syntaxhighlight lang="awk">
# syntax: GAWK -f BRAZILIAN_NUMBERS.AWK
# converted from C
Line 104 ⟶ 505:
return(1)
}
</syntaxhighlight>
{{out}}
<pre>
Line 111 ⟶ 512:
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
</pre>
=={{header|BASIC}}==
==={{header|ANSI BASIC}}===
{{trans|Modula-2}}
{{works with|Decimal BASIC}}
<syntaxhighlight lang="basic">
1000 REM Brazilian numbers
1010 DECLARE EXTERNAL FUNCTION IsBrazilian
1020 DECLARE EXTERNAL FUNCTION SameDigits
1030 DECLARE EXTERNAL FUNCTION IsPrime
1040 PRINT "First 20 Brazilian numbers:"
1050 LET C = 0
1060 LET N = 7
1070 DO WHILE C < 20
1080 IF IsBrazilian(N) <> 0 THEN
1090 PRINT N;
1100 LET C = C + 1
1110 END IF
1120 LET N = N + 1
1130 LOOP
1140 PRINT
1150 PRINT
1160 PRINT "First 20 odd Brazilian numbers:"
1170 LET C = 0
1180 LET N = 7
1190 DO WHILE C < 20
1200 IF IsBrazilian(N) <> 0 THEN
1210 PRINT N;
1220 LET C = C + 1
1230 END IF
1240 LET N = N + 2
1250 LOOP
1260 PRINT
1270 PRINT
1280 PRINT "First 20 prime Brazilian numbers:"
1290 LET C = 0
1300 LET N = 7
1310 DO WHILE C < 20
1320 IF IsBrazilian(N) <> 0 THEN
1330 PRINT N;
1340 LET C = C + 1
1350 END IF
1360 DO
1370 LET N = N + 2
1380 LOOP WHILE IsPrime(N) = 0
1390 LOOP
1400 PRINT
1410 END
1420 REM
1430 EXTERNAL FUNCTION IsBrazilian(N)
1440 REM Result: 1 if N is Brazilian, 0 otherwise
1450 IF N < 7 THEN
1460 LET IsBrazilian = 0
1470 ELSEIF (MOD(N, 2) = 0) AND (N >= 8) THEN
1480 LET IsBrazilian = 1
1490 ELSE
1500 FOR B = 2 TO N - 2
1510 IF SameDigits(N, B) <> 0 THEN
1520 LET IsBrazilian = 1
1530 EXIT FUNCTION
1540 END IF
1550 NEXT B
1560 LET IsBrazilian = 0
1570 END IF
1580 END FUNCTION
1590 REM
1600 EXTERNAL FUNCTION SameDigits(N, B)
1610 REM Result: 1 if N has same digits in the base B, 0 otherwise
1620 LET NL = N ! Local N
1630 LET F = MOD(NL, B)
1640 LET NL = INT(NL / B)
1650 DO WHILE NL > 0
1660 IF MOD(NL, B) <> F THEN
1670 LET SameDigits = 0
1680 EXIT FUNCTION
1690 END IF
1700 LET NL = INT(NL / B)
1710 LOOP
1720 LET SameDigits = 1
1730 END FUNCTION
1740 REM
1750 EXTERNAL FUNCTION IsPrime(N)
1760 REM Result: non-zero if N is prime, 0 otherwise
1770 IF N < 2 THEN
1780 LET IsPrime = 0
1790 ELSEIF MOD(N, 2) = 0 THEN
1800 REM IsPrime = (N = 2)
1810 IF N = 2 THEN
1820 LET IsPrime = 1
1830 ELSE
1840 LET IsPrime = 0
1850 END IF
1860 ELSEIF MOD(N, 3) = 0 THEN
1870 REM IsPrime = (N = 3)
1880 IF N = 3 THEN
1890 LET IsPrime = 1
1900 ELSE
1910 LET IsPrime = 0
1920 END IF
1930 ELSE
1940 LET D = 5
1950 DO WHILE D * D <= N
1960 IF MOD(N, D) = 0 THEN
1970 LET IsPrime = 0
1980 EXIT FUNCTION
1990 ELSE
2000 LET D = D + 2
2010 IF MOD(N, D) = 0 THEN
2020 LET IsPrime = 0
2030 EXIT FUNCTION
2040 ELSE
2050 LET D = D + 4
2060 END IF
2070 END IF
2080 LOOP
2090 LET IsPrime = 1
2100 END IF
2110 END FUNCTION
</syntaxhighlight>
{{out}}
<pre>
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
</pre>
==={{header|FreeBASIC}}===
{{trans|Visual Basic .NET}}
<syntaxhighlight 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</syntaxhighlight>
==={{header|Nascom BASIC}}===
{{trans|Modula-2}}
{{works with|Nascom ROM BASIC|4.7}}
This version shows only the first 12 prime Brazilian numbers because of slow work.
<syntaxhighlight lang="basic">
10 REM Brazilian numbers
20 PRINT "First 20 Brazilian numbers:"
30 C=0:N=7
40 IF C>=20 THEN 90
50 GOSUB 320
60 IF BR THEN PRINT N;:C=C+1
70 N=N+1
80 GOTO 40
90 PRINT:PRINT
100 PRINT "First 20 odd Brazilian numbers:"
110 C=0:N=7
120 IF C>=20 THEN 170
130 GOSUB 320
140 IF BR THEN PRINT N;:C=C+1
150 N=N+2
160 GOTO 120
170 PRINT:PRINT
180 PRINT "First 12 prime Brazilian numbers:"
190 C=0:N=7
200 IF C>=12 THEN 270
210 GOSUB 320
220 IF BR THEN PRINT N;:C=C+1
230 N=N+2
240 GOSUB 530
250 IF NOT PRM THEN 230
260 GOTO 200
270 PRINT
280 END
290 REM ** Is Brazilian?
300 REM Result: BR=-1 if N is Brazilian
310 REM 0 otherwise
320 IF N<7 THEN BR=0:RETURN
330 IF INT(N/2)*2=N AND N>=8 THEN BR=-1:RETURN
340 FOR B=2 TO N-2
350 GOSUB 430
360 IF SMD THEN BR=-1:RETURN
370 NEXT B
380 BR=0
390 RETURN
400 REM ** Same digits?
410 REM Result: SMD=-1 if N has same digits in B,
420 REM 0 otherwise
430 NL=N:REM "Local" N
440 NDB=INT(NL/B)
450 F=NL-NDB*B
460 NL=NDB:NDB=INT(NL/B)
470 IF NL<=0 THEN SMD=-1:RETURN
480 IF NL-NDB*B<>F THEN SMD=0:RETURN
490 NL=INT(NL/B):NDB=INT(NL/B)
500 GOTO 470
510 REM ** Is prime?
520 REM Result: PRM=-1 if N prime, 0 otherwise
530 IF N<2 THEN PRM=0:RETURN
540 IF INT(N/2)*2=N THEN PRM=N=2:RETURN
550 IF INT(N/3)*3=N THEN PRM=N=3:RETURN
560 D=5
570 IF D*D>N THEN PRM=-1:RETURN
580 IF INT(N/D)*D=N THEN PRM=0:RETURN
590 D=D+2
600 IF INT(N/D)*D=N THEN PRM=0:RETURN
610 D=D+4
620 GOTO 570
</syntaxhighlight>
{{out}}
<pre>
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 12 prime Brazilian numbers:
7 13 31 43 73 127 157 211 241 307 421 463
</pre>
==={{header|Visual Basic .NET}}===
{{trans|C#}}
<syntaxhighlight 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</syntaxhighlight>
{{out}}
<pre>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
</pre>
====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).
<syntaxhighlight 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
</syntaxhighlight>
{{out}}
<pre>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
</pre>
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.
=={{header|C}}==
{{trans|Go}}
<
typedef char bool;
Line 186 ⟶ 1,064:
printf("The 100,000th Brazilian number: %d\n", n - 1);
return 0;
}</
{{out}}
Line 201 ⟶ 1,079:
The 100,000th Brazilian number: 110468
</pre>
=={{header|C sharp|C#}}==
{{trans|Go}}
<
class Program {
Line 346 ⟶ 1,133:
}
}
}</
{{out}}
<pre>First 20 Brazilian numbers:
Line 363 ⟶ 1,150:
===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.
<
class Program
Line 461 ⟶ 1,248:
if (System.Diagnostics.Debugger.IsAttached) Console.ReadKey();
}
}</
{{out}}
<pre>Sieving took 3009.2927 ms
Line 499 ⟶ 1,286:
Total elapsed was 4469.1985 ms</pre> P.S. The best speed on Tio.run is under 5 seconds for the 100 millionth count ('''pow''' = 8). If you are very persistent, the 1 billionth count ('''pow''' = 9) can be made to work on Tio.run, it usually overruns the 60 second timeout limit, and cannot finish completely - the sieving by itself takes over 32 seconds (best case), which usually doesn't leave enough time for all the counting.
=={{header|C++}}==
{{trans|D}}
<syntaxhighlight 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;
}</syntaxhighlight>
{{out}}
<pre>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</pre>
=={{header|D}}==
{{trans|C#}}
<
bool sameDigits(int n, int b) {
Line 570 ⟶ 1,445:
if (kind == "") writefln("The %sth Brazillian number is: %s\n", BigLim + 1, n);
}
}</
{{out}}
<pre>First 20 Brazillion numbers:
Line 582 ⟶ 1,457:
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</pre>
=={{header|Delphi}}==
{{Trans|Go}}
<syntaxhighlight 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.
</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|EasyLang}}==
{{trans|C}}
<syntaxhighlight lang="easylang">
func sameDigits n b .
f = n mod b
repeat
n = n div b
until n = 0
if n mod b <> f
return 0
.
.
return 1
.
func isBrazilian7 n .
# n >= 7
if n mod 2 = 0
return 1
.
for b = 2 to n - 2
if sameDigits n b = 1
return 1
.
.
return 0
.
func prime n .
if n mod 2 = 0 and n > 2
return 0
.
i = 3
sq = sqrt n
while i <= sq
if n mod i = 0
return 0
.
i += 2
.
return 1
.
for kind$ in [ "" "odd" "prime" ]
print "First 20 " & kind$ & " Brazilian numbers:"
n = 7
cnt = 1
while cnt <= 20
if isBrazilian7 n = 1
write n & " "
cnt += 1
.
if kind$ = ""
n += 1
elif kind$ = "odd"
n += 2
else
repeat
n += 2
until prime n = 1
.
.
.
print ""
.
</syntaxhighlight>
=={{header|F_Sharp|F#}}==
===The functions===
<
// Generate Brazilian sequence. Nigel Galloway: August 13th., 2019
let isBraz α=let mutable n,i,g=α,α+1,1 in (fun β->(while (i*g)<β do if g<α-1 then g<-g+1 else (n<-n*α; i<-n+i; g<-1)); β=i*g)
Line 592 ⟶ 1,724:
yield! fN (n+1) ((isBraz (n-1))::g)}
fN 4 [isBraz 2]
</syntaxhighlight>
===The Tasks===
;the first 20 Brazilian numbers
<
Brazilian() |> Seq.take 20 |> Seq.iter(printf "%d "); printfn ""
</syntaxhighlight>
{{out}}
<pre>
Line 603 ⟶ 1,735:
</pre>
;the first 20 odd Brazilian numbers
<
Brazilian() |> Seq.filter(fun n->n%2=1) |> Seq.take 20 |> Seq.iter(printf "%d "); printfn ""
</syntaxhighlight>
{{out}}
<pre>
Line 612 ⟶ 1,744:
;the first 20 prime Brazilian numbers
Using [http://www.rosettacode.org/wiki/Extensible_prime_generator#The_function Extensible Prime Generator (F#)]
<
Brazilian() |> Seq.filter isPrime |> Seq.take 20 |> Seq.iter(printf "%d "); printfn ""
</syntaxhighlight>
{{out}}
<pre>
Line 620 ⟶ 1,752:
</pre>
;finally that which the crowd really want to know: What is the 100,000<sup>th</sup> Brazilian number?
<
printfn "%d" (Seq.item 99999 Brazilian)
</syntaxhighlight>
{{out}}
<pre>
Line 631 ⟶ 1,763:
=={{header|Factor}}==
{{works with|Factor|0.99 development release 2019-07-10}}
<
math.parser math.primes.lists math.ranges namespaces prettyprint
prettyprint.config sequences ;
Line 652 ⟶ 1,784:
1 [ 2 + ] lfrom-by "First 20 odd Brazilian numbers:"
lprimes "First 20 prime Brazilian numbers:"
[ print .20-brazilians nl ] 2tri@</
{{out}}
<pre>
Line 663 ⟶ 1,795:
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 }
</pre>
=={{header|Forth}}==
<syntaxhighlight 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</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|Fortran}}==
<syntaxhighlight 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
</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|Fōrmulæ}}==
'''Solution'''
Function to determine if the given number is Brazilian:
[[File:Fōrmulæ - Brazilian numbers 01.png]]
The next function calculates the first Brazilian numbers that agree the given condition. Notice that the condition is provided as a lambda expression:
[[File:Fōrmulæ - Brazilian numbers 02.png]]
'''Test case 1.''' First 20 Brazilian numbers:
[[File:Fōrmulæ - Brazilian numbers 03.png]]
[[File:Fōrmulæ - Brazilian numbers 04.png]]
'''Test case 2.''' First 20 odd Brazilian numbers:
[[File:Fōrmulæ - Brazilian numbers 05.png]]
[[File:Fōrmulæ - Brazilian numbers 06.png]]
'''Test case 3.''' First 20 prime Brazilian numbers:
[[File:Fōrmulæ - Brazilian numbers 07.png]]
[[File:Fōrmulæ - Brazilian numbers 08.png]]
=={{header|Go}}==
===Version 1===
<
import "fmt"
Line 768 ⟶ 2,202:
}
fmt.Println("The 100,000th Brazilian number:", n-1)
}</
{{out}}
Line 790 ⟶ 2,224:
Running a bit quicker than the .NET versions though not due to any further improvements on my part.
<
import (
Line 982 ⟶ 2,416:
}
fmt.Printf("\nTotal elapsed was %.4f ms\n", toMs(time.Since(st0)))
}</
{{out}}
Line 1,024 ⟶ 2,458:
Total elapsed was 3249.0197 ms
</pre>
=={{header|Groovy}}==
{{trans|Java}}
<syntaxhighlight 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)
}
}
}
}</syntaxhighlight>
{{out}}
<pre>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 </pre>
=={{header|Haskell}}==
<
isBrazil :: Int -> Bool
Line 1,052 ⟶ 2,598:
, "\n"
])
[([], [1 ..]), ("odd", [1,3 ..]), ("prime", primes)]</
{{Out}}
<pre>First 20 Brazilians:
Line 1,062 ⟶ 2,608:
First 20 prime Brazilians:
[7,13,31,43,73,127,157,211,241,307,421,463,601,757,1093,1123,1483,1723,2551,2801]</pre>
=={{header|Isabelle}}==
{{works with|Isabelle|2020}}
Not the most beautiful proofs and the theorem about "R*S >= 8, with S+1 > R, are Brazilian" is missing.
<syntaxhighlight 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</syntaxhighlight>
=={{header|J}}==
The brazilian verb checks if 1 is the tally of one of the sets of values in the possible base representations.
<syntaxhighlight lang="text">
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
</syntaxhighlight>
=={{header|Java}}==
<
import java.util.List;
Line 1,161 ⟶ 2,968:
}
}
}</
{{out}}
<pre>First 20 Brazilian numbers:
Line 1,173 ⟶ 2,980:
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 </pre>
=={{header|jq}}==
{{works with|jq}}
'''Works with gojq, the Go implementation of jq'''
See [[Erdős-primes#jq]] for a suitable definition of `is_prime` as used here.
<syntaxhighlight 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)
</syntaxhighlight>
{{out}}
As elsewhere, e.g. [[#Python]].
=={{header|Julia}}==
{{trans|Go}}
<
function samedigits(n, b)
Line 1,199 ⟶ 3,052:
println("The first 20 prime Brazilian numbers are: ", take(20, primebrazilians))
</
<pre>
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)
Line 1,207 ⟶ 3,060:
There has been some discussion of larger numbers in the sequence. See below:
<
count, intervalcount, icount = 0, 0, 0
intervalcounts = Int[]
Line 1,231 ⟶ 3,084:
braziliandensities(100000, 1000)
plot(1:1000:1000000, braziliandensities(1000000, 1000))
</
<pre>
The 10000 th brazilian is 11364.
Line 1,238 ⟶ 3,091:
</pre>
[[File:Brazilian densities.png|thumb|center]]
=={{header|Kotlin}}==
{{trans|C#}}
<
var n2 = n
val f = n % b
Line 1,313 ⟶ 3,166:
if (kind == "") println("The %,dth Brazilian number is: %,d".format(bigLim + 1, n))
}
}</
{{out}}
<pre>First 20 Brazilian numbers:
Line 1,324 ⟶ 3,177:
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 </pre>
=={{header|Lua}}==
{{trans|C}}
<
local f = n % b
n = math.floor(n / b)
Line 1,411 ⟶ 3,263:
end
main()</
{{out}}
<pre>First 20 Brazillion numbers:
Line 1,421 ⟶ 3,273:
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</pre>
=={{header|Mathematica}} / {{header|Wolfram Language}}==
<syntaxhighlight lang="mathematica">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]
</syntaxhighlight>
{{out}}
<pre>{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}</pre>
=={{header|MATLAB}}==
{{trans|Mathematica_/_Wolfram_Language}}
<syntaxhighlight lang="MATLAB">
clear all;close all;clc;
% Find the first 20 Brazilian numbers in the range 7 to 100.
brazilian_numbers = [];
for num = 7:100
if brazilianQ(num)
brazilian_numbers = [brazilian_numbers, num];
if length(brazilian_numbers) == 20
break;
end
end
end
% For Brazilian numbers
fprintf('% 8d', brazilian_numbers);
fprintf("\n");
% Find the first 20 odd Brazilian numbers in the range 7 to 100.
odd_brazilian_numbers = [];
for num = 7:100
if brazilianQ(num) && mod(num, 2) ~= 0
odd_brazilian_numbers = [odd_brazilian_numbers, num];
if length(odd_brazilian_numbers) == 20
break;
end
end
end
% For odd Brazilian numbers
fprintf('% 8d', odd_brazilian_numbers);
fprintf("\n");
% Find the first 20 Brazilian prime numbers in the range 7 to 10000.
brazilian_prime_numbers = [];
for num = 7:10000
if brazilianQ(num) && isprime(num)
brazilian_prime_numbers = [brazilian_prime_numbers, num];
if length(brazilian_prime_numbers) == 20
break;
end
end
end
% For Brazilian prime numbers
fprintf('% 8d', brazilian_prime_numbers);
fprintf("\n");
function isBrazilian = brazilianQ(n)
% Function to check if a number is a Brazilian number.
if n <= 6
error('Input must be greater than 6.');
end
isBrazilian = false;
for b = 2:(n-2)
base_b_digits = custom_dec2base(n, b) - '0'; % Convert number to base b and then to digits
if all(base_b_digits == base_b_digits(1))
isBrazilian = true;
break;
end
end
end
function digits = custom_dec2base(num, base)
% Custom function to convert number to any base representation
if base < 2 || base > num
error('Base must be at least 2 and less than the number itself.');
end
digits = [];
while num > 0
remainder = mod(num, base);
digits = [remainder, digits];
num = floor(num / base);
end
end
</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|Modula-2}}==
{{trans|C|<code>CARDINAL</code> (unsigned integer) used instead of signed integer.<code>ELSIF</code> used instead of sequential <code>IF</code>s with <code>RETURN</code> s. In the main program, the loop <code>for (i = 0; i < 3; ++i) </code> is replaced by subsequent actions, because the actions vary depending on <code>i</code>.}}
{{works with|ADW Modula-2|any (Compile with the linker option ''Console Application'').}}
<syntaxhighlight lang="modula2">
MODULE BrazilianNumbers;
FROM STextIO IMPORT
WriteLn, WriteString;
FROM SWholeIO IMPORT
WriteInt;
VAR
C, N: CARDINAL;
PROCEDURE SameDigits(N, B: CARDINAL): BOOLEAN;
VAR
F: CARDINAL;
BEGIN
F := N MOD B;
N := N / B;
WHILE N > 0 DO
IF N MOD B <> F THEN
RETURN FALSE;
END;
N := N / B;
END;
RETURN TRUE;
END SameDigits;
PROCEDURE IsBrazilian(N: CARDINAL): BOOLEAN;
VAR
B: CARDINAL;
BEGIN
IF N < 7 THEN
RETURN FALSE
ELSIF (N MOD 2 = 0) AND (N >= 8) THEN
RETURN TRUE
ELSE
FOR B := 2 TO N - 2 DO
IF SameDigits(N, B) THEN
RETURN TRUE
END
END;
RETURN FALSE;
END
END IsBrazilian;
PROCEDURE IsPrime(N: CARDINAL): BOOLEAN;
VAR
D: CARDINAL;
BEGIN
IF N < 2 THEN
RETURN FALSE;
ELSIF N MOD 2 = 0 THEN
RETURN N = 2;
ELSIF N MOD 3 = 0 THEN
RETURN N = 3;
ELSE
D := 5;
WHILE D * D <= N DO
IF N MOD D = 0 THEN
RETURN FALSE
ELSE
D := D + 2;
IF N MOD D = 0 THEN
RETURN FALSE
ELSE
D := D + 4
END
END
END;
RETURN TRUE;
END
END IsPrime;
BEGIN
WriteString("First 20 Brazilian numbers:");
WriteLn;
C := 0; N := 7;
WHILE C < 20 DO
IF IsBrazilian(N) THEN
WriteInt(N, 1);
WriteString(" ");
C := C + 1;
END;
N := N + 1;
END;
WriteLn; WriteLn;
WriteString("First 20 odd Brazilian numbers:");
WriteLn;
C := 0; N := 7;
WHILE C < 20 DO
IF IsBrazilian(N) THEN
WriteInt(N, 1);
WriteString(" ");
C := C + 1;
END;
N := N + 2;
END;
WriteLn; WriteLn;
WriteString("First 20 prime Brazilian numbers:");
WriteLn;
C := 0; N := 7;
WHILE C < 20 DO
IF IsBrazilian(N) THEN
WriteInt(N, 1);
WriteString(" ");
C := C + 1;
END;
REPEAT
N := N + 2
UNTIL IsPrime(N);
END;
WriteLn;
END BrazilianNumbers.
</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|Nim}}==
<syntaxhighlight 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</syntaxhighlight>
{{out}}
<pre>
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</pre>
=={{header|Pascal}}==
Line 1,428 ⟶ 3,591:
extreme reduced runtime time for space.<BR>
At the end only primes and square of primes need to be tested, all others are Brazilian.
<
{$IFDEF FPC}
Line 1,719 ⟶ 3,882:
setlength(isPrime, 0);
end.</
{{out}}
<pre>first 20 brazilian numbers
Line 1,808 ⟶ 3,971:
sys 0m0,157s
</pre>
=={{header|Perl}}==
{{libheader|ntheory}}
{{trans|
<
use warnings;
use ntheory qw<is_prime>;
use constant Inf => 1e10;
sub is_Brazilian {
Line 1,838 ⟶ 3,999:
my $upto = 20;
my $n = 0;
$n = 0;
$n = 0;
{{out}}
<pre>First 20 Brazilian numbers:
Line 1,862 ⟶ 4,020:
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</pre>
=={{header|Phix}}==
{{trans|C}}
<!--<syntaxhighlight lang="phix">(phixonline)-->
<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>
<span style="color: #008080;">function</span> <span style="color: #000000;">same_digits</span><span style="color: #0000FF;">(</span><span style="color: #004080;">integer</span> <span style="color: #000000;">n</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">b</span><span style="color: #0000FF;">)</span>
<span style="color: #004080;">integer</span> <span style="color: #000000;">f</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">remainder</span><span style="color: #0000FF;">(</span><span style="color: #000000;">n</span><span style="color: #0000FF;">,</span><span style="color: #000000;">b</span><span style="color: #0000FF;">)</span>
<span style="color: #000000;">n</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">floor</span><span style="color: #0000FF;">(</span><span style="color: #000000;">n</span><span style="color: #0000FF;">/</span><span style="color: #000000;">b</span><span style="color: #0000FF;">)</span>
<span style="color: #008080;">while</span> <span style="color: #000000;">n</span><span style="color: #0000FF;">></span><span style="color: #000000;">0</span> <span style="color: #008080;">do</span>
<span style="color: #008080;">if</span> <span style="color: #7060A8;">remainder</span><span style="color: #0000FF;">(</span><span style="color: #000000;">n</span><span style="color: #0000FF;">,</span><span style="color: #000000;">b</span><span style="color: #0000FF;">)!=</span><span style="color: #000000;">f</span> <span style="color: #008080;">then</span> <span style="color: #008080;">return</span> <span style="color: #004600;">false</span> <span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #000000;">n</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">floor</span><span style="color: #0000FF;">(</span><span style="color: #000000;">n</span><span style="color: #0000FF;">/</span><span style="color: #000000;">b</span><span style="color: #0000FF;">)</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">while</span>
<span style="color: #008080;">return</span> <span style="color: #004600;">true</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">function</span>
<span style="color: #008080;">function</span> <span style="color: #000000;">is_brazilian</span><span style="color: #0000FF;">(</span><span style="color: #004080;">integer</span> <span style="color: #000000;">n</span><span style="color: #0000FF;">)</span>
<span style="color: #008080;">if</span> <span style="color: #000000;">n</span><span style="color: #0000FF;">>=</span><span style="color: #000000;">7</span> <span style="color: #008080;">then</span>
<span style="color: #008080;">if</span> <span style="color: #7060A8;">remainder</span><span style="color: #0000FF;">(</span><span style="color: #000000;">n</span><span style="color: #0000FF;">,</span><span style="color: #000000;">2</span><span style="color: #0000FF;">)=</span><span style="color: #000000;">0</span> <span style="color: #008080;">then</span> <span style="color: #008080;">return</span> <span style="color: #004600;">true</span> <span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">b</span><span style="color: #0000FF;">=</span><span style="color: #000000;">2</span> <span style="color: #008080;">to</span> <span style="color: #000000;">n</span><span style="color: #0000FF;">-</span><span style="color: #000000;">2</span> <span style="color: #008080;">do</span>
<span style="color: #008080;">if</span> <span style="color: #000000;">same_digits</span><span style="color: #0000FF;">(</span><span style="color: #000000;">n</span><span style="color: #0000FF;">,</span><span style="color: #000000;">b</span><span style="color: #0000FF;">)</span> <span style="color: #008080;">then</span> <span style="color: #008080;">return</span> <span style="color: #004600;">true</span> <span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #008080;">return</span> <span style="color: #004600;">false</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">function</span>
<span style="color: #008080;">constant</span> <span style="color: #000000;">kinds</span> <span style="color: #0000FF;">=</span> <span style="color: #0000FF;">{</span><span style="color: #008000;">" "</span><span style="color: #0000FF;">,</span> <span style="color: #008000;">" odd "</span><span style="color: #0000FF;">,</span> <span style="color: #008000;">" prime "</span><span style="color: #0000FF;">}</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">i</span><span style="color: #0000FF;">=</span><span style="color: #000000;">1</span> <span style="color: #008080;">to</span> <span style="color: #7060A8;">length</span><span style="color: #0000FF;">(</span><span style="color: #000000;">kinds</span><span style="color: #0000FF;">)</span> <span style="color: #008080;">do</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"First 20%sBrazilian numbers:\n"</span><span style="color: #0000FF;">,</span> <span style="color: #0000FF;">{</span><span style="color: #000000;">kinds</span><span style="color: #0000FF;">[</span><span style="color: #000000;">i</span><span style="color: #0000FF;">]})</span>
<span style="color: #004080;">integer</span> <span style="color: #000000;">c</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">0</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">n</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">7</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">p</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">4</span>
<span style="color: #008080;">while</span> <span style="color: #004600;">true</span> <span style="color: #008080;">do</span>
<span style="color: #008080;">if</span> <span style="color: #000000;">is_brazilian</span><span style="color: #0000FF;">(</span><span style="color: #000000;">n</span><span style="color: #0000FF;">)</span> <span style="color: #008080;">then</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"%d "</span><span style="color: #0000FF;">,</span><span style="color: #000000;">n</span><span style="color: #0000FF;">)</span>
<span style="color: #000000;">c</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">1</span>
<span style="color: #008080;">if</span> <span style="color: #000000;">c</span><span style="color: #0000FF;">==</span><span style="color: #000000;">20</span> <span style="color: #008080;">then</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"\n\n"</span><span style="color: #0000FF;">)</span>
<span style="color: #008080;">exit</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #008080;">switch</span> <span style="color: #000000;">i</span>
<span style="color: #008080;">case</span> <span style="color: #000000;">1</span><span style="color: #0000FF;">:</span> <span style="color: #000000;">n</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">1</span>
<span style="color: #008080;">case</span> <span style="color: #000000;">2</span><span style="color: #0000FF;">:</span> <span style="color: #000000;">n</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">2</span>
<span style="color: #008080;">case</span> <span style="color: #000000;">3</span><span style="color: #0000FF;">:</span> <span style="color: #000000;">p</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">1</span><span style="color: #0000FF;">;</span> <span style="color: #000000;">n</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">get_prime</span><span style="color: #0000FF;">(</span><span style="color: #000000;">p</span><span style="color: #0000FF;">)</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">switch</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">while</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #004080;">integer</span> <span style="color: #000000;">n</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">7</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">c</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">0</span>
<span style="color: #004080;">atom</span> <span style="color: #000000;">t0</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">time</span><span style="color: #0000FF;">(),</span> <span style="color: #000000;">t1</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">time</span><span style="color: #0000FF;">()+</span><span style="color: #000000;">1</span>
<span style="color: #008080;">while</span> <span style="color: #000000;">c</span><span style="color: #0000FF;"><</span><span style="color: #000000;">100000</span> <span style="color: #008080;">do</span>
<span style="color: #008080;">if</span> <span style="color: #7060A8;">platform</span><span style="color: #0000FF;">()!=</span><span style="color: #004600;">JS</span> <span style="color: #008080;">and</span> <span style="color: #7060A8;">time</span><span style="color: #0000FF;">()></span><span style="color: #000000;">t1</span> <span style="color: #008080;">then</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"checking %d [count:%d]...\r"</span><span style="color: #0000FF;">,{</span><span style="color: #000000;">n</span><span style="color: #0000FF;">,</span><span style="color: #000000;">c</span><span style="color: #0000FF;">})</span>
<span style="color: #000000;">t1</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">time</span><span style="color: #0000FF;">()+</span><span style="color: #000000;">1</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #000000;">c</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">is_brazilian</span><span style="color: #0000FF;">(</span><span style="color: #000000;">n</span><span style="color: #0000FF;">)</span>
<span style="color: #000000;">n</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">1</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">while</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"The %,dth Brazilian number: %d\n"</span><span style="color: #0000FF;">,</span> <span style="color: #0000FF;">{</span><span style="color: #000000;">c</span><span style="color: #0000FF;">,</span><span style="color: #000000;">n</span><span style="color: #0000FF;">-</span><span style="color: #000000;">1</span><span style="color: #0000FF;">})</span>
<span style="color: #0000FF;">?</span><span style="color: #7060A8;">elapsed</span><span style="color: #0000FF;">(</span><span style="color: #7060A8;">time</span><span style="color: #0000FF;">()-</span><span style="color: #000000;">t0</span><span style="color: #0000FF;">)</span>
<!--</syntaxhighlight>-->
{{out}}
<pre>
First 20 Brazilian numbers:
Line 1,967 ⟶ 4,093:
"52.8s"
</pre>
=={{header|Python}}==
<
from itertools import count, islice
Line 2,009 ⟶ 4,116:
in the given base, are the same.
'''
def go(
def g(
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]
Line 2,050 ⟶ 4,182:
or xs itself if n > length xs.
'''
else list(islice(xs, n))
)
return go
Line 2,062 ⟶ 4,196:
The initial seed value is x.
'''
def go(f
return g
return go
# MAIN ---
if __name__ == '__main__':
main()</
{{Out}}
<pre>First 20 Brazilians:
Line 2,083 ⟶ 4,219:
[7,13,31,43,73,127,157,211,241,307,421,463,601,757,1093,1123,1483,1723,2551,2801]</pre>
=={{header|Quackery}}==
<code>isprime</code> is defined at [[Primality by trial division#Quackery]].
<syntaxhighlight lang="Quackery"> [ dup base put
/mod temp put
true swap
[ dup 0 > while
base share /mod
temp share != iff
[ dip not ]
done
again ]
drop
temp release
base release ] is allsame ( n n --> b )
[ false swap
dup 3 - times
[ dup i 2 +
allsame iff
[ dip not
conclude ]
done ]
drop ] is brazilian ( n --> b )
say "First 20 Brazilian numbers:" cr
[] 0
[ dup brazilian if
[ dup dip join ]
1+
over size 20 = until ]
drop echo
cr
cr
say "First 20 odd Brazilian numbers:" cr
[] 1
[ dup brazilian if
[ dup dip join ]
2 +
over size 20 = until ]
drop echo
cr
cr
say "First 20 prime Brazilian numbers:" cr
[] 1
[ dup isprime not iff
[ 2 + ] again
dup brazilian if
[ dup dip join ]
2 +
over size 20 = until ]
drop echo</syntaxhighlight>
{{out}}
<pre>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 ]
</pre>
=={{header|Racket}}==
<syntaxhighlight 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)))</syntaxhighlight>
{{out}}
<pre>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)</pre>
=={{header|Raku}}==
(formerly Perl 6)
{{works with|Rakudo|2019.07.1}}
<syntaxhighlight lang="raku" line>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];</syntaxhighlight>
{{out}}
<pre>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</pre>
=={{header|REXX}}==
{{trans|GO}}
<syntaxhighlight lang="text">/*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 #.*/
Line 2,106 ⟶ 4,380:
end /*j*/ /* [↑] cases 1──►3, $ has leading blank*/
say /* [↓] use a special header for cases.*/
if c==4 then do; $= j;
say center(' 'hdr.c" "
say strip($) /*display a case result to the terminal*/
end /*c*/ /* [↑] cases 1──►3 have a leading blank*/
Line 2,114 ⟶ 4,388:
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
else mx= x%3 -1 /*X not known if prime. */
do b=2 for mx /*scan for base 2 ──► max*/
Line 2,130 ⟶ 4,404:
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>1; q=q%4; _=x-r-q; r=r%2; if _>=0 then do;x=_;r=r+q;
/*──────────────────────────────────────────────────────────────────────────────────────*/
sameDig: procedure; parse arg x, b; f= x // b /* // ◄── the remainder.*/
Line 2,139 ⟶ 4,413:
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))</
{{out|output|text= when using the inputs of: <tt> , , , 100000 </tt>}}
<pre>
Line 2,153 ⟶ 4,427:
═════════════════════════ The 100000th Brazilian number ═════════════════════
110468
</pre>
=={{header|Ring}}==
<syntaxhighlight 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
</syntaxhighlight>
Output:
<pre>
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...
</pre>
=={{header|RPL}}==
{{works with|HP|49}}
≪ SWAP OVER IDIV2 ROT → digit base
≪ 1 SF
'''WHILE''' DUP 1 FS? AND '''REPEAT'''
'''IF''' base IDIV2 digit ≠ '''THEN''' 1 CF '''END'''
'''END'''
DROP 1 FS?
≫ ≫ '<span style="color:blue">SAMEDIGITS?</span>' STO
≪ → n
≪ 0
2 n 2 - '''FOR''' b
n b <span style="color:blue">SAMEDIGITS?</span> OR
'''IF''' DUP '''THEN''' n 'b' STO '''END'''
'''NEXT'''
≫ '<span style="color:blue">BRAZILIAN?</span>' STO
≪ → max inc
≪ { } 5
'''DO'''
'''IF''' DUP <span style="color:blue">BRAZILIAN?</span>' '''THEN''' SWAP OVER + SWAP '''END'''
inc EVAL
'''UNTIL''' OVER SIZE max ≥ '''END'''
DROP
≫ '<span style="color:blue">TASK</span>' STO
20 ≪ 1 + ≫ <span style="color:blue">TASK</span>
20 ≪ 2 + ≫ <span style="color:blue">TASK</span>
20 ≪ NEXTPRIME ≫ <span style="color:blue">TASK</span>
{{out}}
<pre>
3: {7 8 10 12 13 14 15 16 18 20 21 22 24 26 27 28 30 31 32 33}
2: {7 13 15 21 27 31 33 35 39 43 45 51 55 57 63 65 69 73 75 77}
1: {7 13 31 43 73 127 157 211 241 307 421 463 601 757 1093 1123 1483 1723 2551 2801}
</pre>
=={{header|Ruby}}==
{{trans|C++}}
<
f = n % b
while (n /= b) > 0 do
Line 2,252 ⟶ 4,658:
end
main()</
{{out}}
<pre>First 20 Brazilian numbers:
Line 2,264 ⟶ 4,670:
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</pre>
=={{header|Rust}}==
<syntaxhighlight 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);
}
</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|Scala}}==
{{trans|Java}}
<syntaxhighlight 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: $n")
println()
}
}
}
}</syntaxhighlight>
{{out}}
<pre>First 20 Brazilian numbers:
7 8 10 12 13 14 15 16 18 20 21 22 24 26 27 28 30 31 32 33
The 100000th Brazilian number is: 110468
First 20 odd Brazilian numbers:
7 13 15 21 27 31 33 35 39 43 45 51 55 57 63 65 69 73 75 77
First 20 prime Brazilian numbers:
7 13 31 43 73 127 157 211 241 307 421 463 601 757 1093 1123 1483 1723 2551 2801 </pre>
===functional===
use Scala 3
<syntaxhighlight lang="scala">object Brazilian extends App {
def oddPrimes =
LazyList.from(3, 2).filter(p => (3 to math.sqrt(p).ceil.toInt by 2).forall(p % _ > 0))
val primes = 2 #:: oddPrimes
def sameDigits(num: Int, base: Int): Boolean = {
val first = num % base
@annotation.tailrec
def iter(num: Int): Boolean = {
if (num % base) == first then iter(num / base)
else num == 0
}
iter(num / base)
}
def isBrazilian(num: Int): Boolean = {
num match {
case x if x < 7 => false
case x if (x & 1) == 0 => true
case _ => (2 to num - 2).exists(sameDigits(num,_))
}
}
val (limit, bigLimit) = (20, 100_000)
val doList = Seq(("brazilian", LazyList.from(7)),
("odd", LazyList.from(7, 2)),
("prime", primes))
for((listStr, stream) <- doList)
println(s"$listStr: " + stream.filter(isBrazilian(_)).take(limit).toList)
println("be a little patient, it will take some time")
val bigElement = LazyList.from(7).filter(isBrazilian(_)).drop(bigLimit - 1).head
println(s"brazilian($bigLimit): $bigElement")
}</syntaxhighlight>
{{out}}
<pre>brazilian: List(7, 8, 10, 12, 13, 14, 15, 16, 18, 20, 21, 22, 24, 26, 27, 28, 30, 31, 32, 33)
odd: List(7, 13, 15, 21, 27, 31, 33, 35, 39, 43, 45, 51, 55, 57, 63, 65, 69, 73, 75, 77)
prime: List(7, 13, 31, 43, 73, 127, 157, 211, 241, 307, 421, 463, 601, 757, 1093, 1123, 1483, 1723, 2551, 2801)
be a little patient, it will take some time
brazilian(100000): 110468
</pre>
=={{header|Sidef}}==
<
static L = Set()
Line 2,308 ⟶ 4,961:
say "\nFirst #{n} prime Brazilian numbers"
say (^Inf -> lazy.grep(is_Brazilian).grep{.is_prime}.first(n))
}</
{{out}}
<pre>
Line 2,322 ⟶ 4,975:
Extra:
<
say ("#{10**n->commify}th Brazilian number = ", is_Brazilian.nth(10**n))
}</
{{out}}
<pre>
Line 2,335 ⟶ 4,988:
</pre>
=={{header|
{{trans|
<syntaxhighlight lang="v (vlang)">fn same_digits(nn int, b int) bool {
f := nn % b
mut n := nn/b
for n > 0 {
if n%b != f {
return false
}
n /= b
}
return true
}
fn is_brazilian(n int) bool {
if n < 7 {
return false
}
if n%2 == 0 && n >= 8 {
return true
}
for b in 2..n-1 {
if same_digits(n, b) {
return true
}
}
return false
}
fn is_prime(n int) bool {
match true {
n < 2 {
return false
}
n%2 == 0 {
return n == 2
}
n%3 == 0 {
return n == 3
}
else {
mut d := 5
for d*d <= n {
if n%d == 0 {
return false
}
d += 2
if n%d == 0 {
return false
}
d += 4
}
return true
}
}
}
fn main() {
kinds := [" ", " odd ", " prime "]
for kind in kinds {
println("First 20${kind}Brazilian numbers:")
mut c := 0
mut n := 7
for {
if is_brazilian(n) {
print("$n ")
c++
if c == 20 {
println("\n")
break
}
}
match kind {
" " {
n++
}
" odd " {
n += 2
}
" prime "{
for {
n += 2
if is_prime(n) {
break
}
}
}
else{}
}
}
}
mut n := 7
for c := 0; c < 100000; n++ {
if is_brazilian(n) {
c++
}
}
println("The 100,000th Brazilian number: ${n-1}")
}</syntaxhighlight>
{{out}}
<pre>
First 20 Brazilian numbers:
7 8 10 12 13 14 15 16 18 20 21 22 24 26 27 28 30 31 32 33
Line 2,390 ⟶ 5,099:
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
</pre>
=={{header|Wren}}==
{{trans|Go}}
{{libheader|Wren-math}}
<syntaxhighlight lang="wren">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) {
}
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)")</syntaxhighlight>
{{out}}
<pre>
First
7 8 10 12
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
</pre>
=={{header|XPL0}}==
{{trans|C}}
<syntaxhighlight lang "XPL0">func SameDigits(N, B);
int N, B, F;
[F:= rem(N/B);
N:= N/B;
while N > 0 do
[if rem(N/B) # F then return false;
N:= N/B;
];
return true;
];
func IsBrazilian(N);
int N, B;
[if N < 7 then return false;
if rem(N/2) = 0 and N >= 8 then return true;
for B:= 2 to N-2 do
if SameDigits(N, B) then
return false;
];
func IsPrime(N); \Return 'true' if N is prime
int N, D;
[if N < 2 then return false;
if (N&1) = 0 then return N = 2;
if rem(N/3) = 0 then return N = 3;
D:= 5;
while D*D <= N do
[if rem(N/D) = 0 then return false;
D:= D+2;
if rem(N/D) = 0 then return false;
D:= D+4;
];
return true;
];
int I, C, N, Kinds;
[Kinds:= [" ", " odd ", " prime "];
for I:= 0 to 3-1 do
[Text(0, "First 20"); Text(0, Kinds(I));
Text(0, "Brazilian numbers:^m^j");
C:= 0; N:= 7;
loop [if IsBrazilian(N) then
[IntOut(0, N); ChOut(0, ^ );
C:= C+1;
if C = 20 then
[CrLf(0); CrLf(0); quit];
];
case I of
0: N:= N+1;
1: N:= N+2;
2: repeat N:= N+2 until IsPrime(N)
other [];
];
];
N:= 7; C:= 0;
while C < 100_000 do
[if IsBrazilian(N) then C:= C+1;
N:= N+1;
];
Text(0, "The 100,000th Brazilian number: "); IntOut(0, N-1); CrLf(0);
]</syntaxhighlight>
{{out}}
<pre>
First 20 Brazilian numbers:
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
The
</pre>
=={{header|zkl}}==
<
foreach b in ([2..n-2]){
f,m := n%b, n/b;
Line 2,566 ⟶ 5,264:
False
}
fcn isBrazilianW(n){ isBrazilian(n) and n or Void.Skip }</
<
[1..].tweak(isBrazilianW).walk(20).println();
println("\nFirst 20 odd Brazilian numbers:");
[1..*,2].tweak(isBrazilianW).walk(20).println();</
{{out}}
<pre>
Line 2,585 ⟶ 5,283:
[[Extensible prime generator#zkl]] could be used instead.
<
println("\nFirst 20 prime Brazilian numbers:");
p:=BI(1);
Walker.zero().tweak('wrap{ p.nextPrime().toInt() })
.tweak(isBrazilianW).walk(20).println();</
{{out}}
<pre>
Line 2,596 ⟶ 5,294:
L(7,13,31,43,73,127,157,211,241,307,421,463,601,757,1093,1123,1483,1723,2551,2801)
</pre>
<
[1..].tweak(isBrazilianW).drop(100_000).value);</
{{out}}
<pre>
| |||