Eban numbers
You are encouraged to solve this task according to the task description, using any language you may know.
- Definition
An eban number is a number that has no letter e in it when the number is spelled in English.
Or more literally, spelled numbers that contain the letter e are banned.
The American version of spelling numbers will be used here (as opposed to the British).
2,000,000,000 is two billion, not two milliard.
Only numbers less than one sextillion (1021) will be considered in/for this task.
This will allow optimizations to be used.
- Task
-
- show all eban numbers ≤ 1,000 (in a horizontal format), and a count
- show all eban numbers between 1,000 and 4,000 (inclusive), and a count
- show a count of all eban numbers up and including 10,000
- show a count of all eban numbers up and including 100,000
- show a count of all eban numbers up and including 1,000,000
- show a count of all eban numbers up and including 10,000,000
- show all output here.
- See also
-
- The MathWorld entry: eban numbers.
- The OEIS entry: A6933, eban numbers.
AWK
<lang AWK>
- syntax: GAWK -f EBAN_NUMBERS.AWK
- converted from FreeBASIC
BEGIN {
main(2,1000,1) main(1000,4000,1) main(2,10000,0) main(2,100000,0) main(2,1000000,0) main(2,10000000,0) main(2,100000000,0) exit(0)
} function main(start,stop,printable, b,count,i,m,r,t) {
printf("%d-%d:",start,stop)
for (i=start; i<=stop; i+=2) {
b = int(i / 1000000000)
r = i % 1000000000
m = int(r / 1000000)
r = i % 1000000
t = int(r / 1000)
r = r % 1000
if (m >= 30 && m <= 66) { m %= 10 }
if (t >= 30 && t <= 66) { t %= 10 }
if (r >= 30 && r <= 66) { r %= 10 }
if (x(b) && x(m) && x(t) && x(r)) {
count++
if (printable) {
printf(" %d",i)
}
}
}
printf(" (count=%d)\n",count)
} function x(n) {
return(n == 0 || n == 2 || n == 4 || n == 6)
} </lang>
- Output:
2-1000: 2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66 (count=19) 1000-4000: 2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000 (count=21) 2-10000: (count=79) 2-100000: (count=399) 2-1000000: (count=399) 2-10000000: (count=1599) 2-100000000: (count=7999)
C
<lang c>#include "stdio.h"
- include "stdbool.h"
- define ARRAY_LEN(a,T) (sizeof(a) / sizeof(T))
struct Interval {
int start, end; bool print;
};
int main() {
struct Interval intervals[] = {
{2, 1000, true},
{1000, 4000, true},
{2, 10000, false},
{2, 100000, false},
{2, 1000000, false},
{2, 10000000, false},
{2, 100000000, false},
{2, 1000000000, false},
};
int idx;
for (idx = 0; idx < ARRAY_LEN(intervals, struct Interval); ++idx) {
struct Interval intv = intervals[idx];
int count = 0, i;
if (intv.start == 2) {
printf("eban numbers up to and including %d:\n", intv.end);
} else {
printf("eban numbers between %d and %d (inclusive:)", intv.start, intv.end);
}
for (i = intv.start; i <= intv.end; i += 2) {
int b = i / 1000000000;
int r = i % 1000000000;
int m = r / 1000000;
int t;
r = i % 1000000;
t = r / 1000;
r %= 1000;
if (m >= 30 && m <= 66) m %= 10;
if (t >= 30 && t <= 66) t %= 10;
if (r >= 30 && r <= 66) r %= 10;
if (b == 0 || b == 2 || b == 4 || b == 6) {
if (m == 0 || m == 2 || m == 4 || m == 6) {
if (t == 0 || t == 2 || t == 4 || t == 6) {
if (r == 0 || r == 2 || r == 4 || r == 6) {
if (intv.print) printf("%d ", i);
count++;
}
}
}
}
}
if (intv.print) {
printf("\n");
}
printf("count = %d\n\n", count);
}
return 0;
}</lang>
- Output:
eban numbers up to and including 1000: 2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66 count = 19 eban numbers between 1000 and 4000 (inclusive:)2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000 count = 21 eban numbers up to and including 10000: count = 79 eban numbers up to and including 100000: count = 399 eban numbers up to and including 1000000: count = 399 eban numbers up to and including 10000000: count = 1599 eban numbers up to and including 100000000: count = 7999 eban numbers up to and including 1000000000: count = 7999
C#
<lang csharp>using System;
namespace EbanNumbers {
struct Interval {
public int start, end;
public bool print;
public Interval(int start, int end, bool print) {
this.start = start;
this.end = end;
this.print = print;
}
}
class Program {
static void Main() {
Interval[] intervals = {
new Interval(2, 1_000, true),
new Interval(1_000, 4_000, true),
new Interval(2, 10_000, false),
new Interval(2, 100_000, false),
new Interval(2, 1_000_000, false),
new Interval(2, 10_000_000, false),
new Interval(2, 100_000_000, false),
new Interval(2, 1_000_000_000, false),
};
foreach (var intv in intervals) {
if (intv.start == 2) {
Console.WriteLine("eban numbers up to and including {0}:", intv.end);
} else {
Console.WriteLine("eban numbers between {0} and {1} (inclusive):", intv.start, intv.end);
}
int count = 0;
for (int i = intv.start; i <= intv.end; i += 2) {
int b = i / 1_000_000_000;
int r = i % 1_000_000_000;
int m = r / 1_000_000;
r = i % 1_000_000;
int t = r / 1_000;
r %= 1_000;
if (m >= 30 && m <= 66) m %= 10;
if (t >= 30 && t <= 66) t %= 10;
if (r >= 30 && r <= 66) r %= 10;
if (b == 0 || b == 2 || b == 4 || b == 6) {
if (m == 0 || m == 2 || m == 4 || m == 6) {
if (t == 0 || t == 2 || t == 4 || t == 6) {
if (r == 0 || r == 2 || r == 4 || r == 6) {
if (intv.print) Console.Write("{0} ", i);
count++;
}
}
}
}
}
if (intv.print) {
Console.WriteLine();
}
Console.WriteLine("count = {0}\n", count);
}
}
}
}</lang>
- Output:
eban numbers up to and including 1000: 2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66 count = 19 eban numbers between 1000 and 4000 (inclusive): 2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000 count = 21 eban numbers up to and including 10000: count = 79 eban numbers up to and including 100000: count = 399 eban numbers up to and including 1000000: count = 399 eban numbers up to and including 10000000: count = 1599 eban numbers up to and including 100000000: count = 7999 eban numbers up to and including 1000000000: count = 7999
C++
<lang cpp>#include <iostream>
struct Interval {
int start, end; bool print;
};
int main() {
Interval intervals[] = {
{2, 1000, true},
{1000, 4000, true},
{2, 10000, false},
{2, 100000, false},
{2, 1000000, false},
{2, 10000000, false},
{2, 100000000, false},
{2, 1000000000, false},
};
for (auto intv : intervals) {
if (intv.start == 2) {
std::cout << "eban numbers up to and including " << intv.end << ":\n";
} else {
std::cout << "eban numbers bwteen " << intv.start << " and " << intv.end << " (inclusive):\n";
}
int count = 0;
for (int i = intv.start; i <= intv.end; i += 2) {
int b = i / 1000000000;
int r = i % 1000000000;
int m = r / 1000000;
r = i % 1000000;
int t = r / 1000;
r %= 1000;
if (m >= 30 && m <= 66) m %= 10;
if (t >= 30 && t <= 66) t %= 10;
if (r >= 30 && r <= 66) r %= 10;
if (b == 0 || b == 2 || b == 4 || b == 6) {
if (m == 0 || m == 2 || m == 4 || m == 6) {
if (t == 0 || t == 2 || t == 4 || t == 6) {
if (r == 0 || r == 2 || r == 4 || r == 6) {
if (intv.print) std::cout << i << ' ';
count++;
}
}
}
}
}
if (intv.print) {
std::cout << '\n';
}
std::cout << "count = " << count << "\n\n";
}
return 0;
}</lang>
- Output:
eban numbers up to and including 1000: 2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66 count = 19 eban numbers bwteen 1000 and 4000 (inclusive): 2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000 count = 21 eban numbers up to and including 10000: count = 79 eban numbers up to and including 100000: count = 399 eban numbers up to and including 1000000: count = 399 eban numbers up to and including 10000000: count = 1599 eban numbers up to and including 100000000: count = 7999 eban numbers up to and including 1000000000: count = 7999
D
<lang d>import std.stdio;
struct Interval {
int start, end; bool print;
}
void main() {
Interval[] intervals = [
{2, 1_000, true},
{1_000, 4_000, true},
{2, 10_000, false},
{2, 100_000, false},
{2, 1_000_000, false},
{2, 10_000_000, false},
{2, 100_000_000, false},
{2, 1_000_000_000, false},
];
foreach (intv; intervals) {
if (intv.start == 2) {
writeln("eban numbers up to an including ", intv.end, ':');
} else {
writeln("eban numbers between ", intv.start ," and ", intv.end, " (inclusive):");
}
int count;
for (int i = intv.start; i <= intv.end; i = i + 2) {
int b = i / 1_000_000_000;
int r = i % 1_000_000_000;
int m = r / 1_000_000;
r = i % 1_000_000;
int t = r / 1_000;
r %= 1_000;
if (m >= 30 && m <= 66) m %= 10;
if (t >= 30 && t <= 66) t %= 10;
if (r >= 30 && r <= 66) r %= 10;
if (b == 0 || b == 2 || b == 4 || b == 6) {
if (m == 0 || m == 2 || m == 4 || m == 6) {
if (t == 0 || t == 2 || t == 4 || t == 6) {
if (r == 0 || r == 2 || r == 4 || r == 6) {
if (intv.print) write(i, ' ');
count++;
}
}
}
}
}
if (intv.print) {
writeln();
}
writeln("count = ", count);
writeln;
}
}</lang>
- Output:
eban numbers up to an including 1000: 2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66 count = 19 eban numbers between 1000 and 4000 (inclusive): 2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000 count = 21 eban numbers up to an including 10000: count = 79 eban numbers up to an including 100000: count = 399 eban numbers up to an including 1000000: count = 399 eban numbers up to an including 10000000: count = 1599 eban numbers up to an including 100000000: count = 7999 eban numbers up to an including 1000000000: count = 7999
Factor
<lang factor>USING: arrays formatting fry io kernel math math.functions math.order math.ranges prettyprint sequences ;
- eban? ( n -- ? )
1000000000 /mod 1000000 /mod 1000 /mod
[ dup 30 66 between? [ 10 mod ] when ] tri@ 4array
[ { 0 2 4 6 } member? ] all? ;
- .eban ( m n -- ) "eban numbers in [%d, %d]: " printf ;
- eban ( m n q -- o ) '[ 2dup .eban [a,b] [ eban? ] @ ] call ; inline
- .eban-range ( m n -- ) [ filter ] eban "%[%d, %]\n" printf ;
- .eban-count ( m n -- ) "count of " write [ count ] eban . ;
1 1000 1000 4000 [ .eban-range ] 2bi@ 4 9 [a,b] [ [ 1 10 ] dip ^ .eban-count ] each</lang>
- Output:
eban numbers in [1, 1000]: { 2, 4, 6, 30, 32, 34, 36, 40, 42, 44, 46, 50, 52, 54, 56, 60, 62, 64, 66 }
eban numbers in [1000, 4000]: { 2000, 2002, 2004, 2006, 2030, 2032, 2034, 2036, 2040, 2042, 2044, 2046, 2050, 2052, 2054, 2056, 2060, 2062, 2064, 2066, 4000 }
count of eban numbers in [1, 10000]: 79
count of eban numbers in [1, 100000]: 399
count of eban numbers in [1, 1000000]: 399
count of eban numbers in [1, 10000000]: 1599
count of eban numbers in [1, 100000000]: 7999
count of eban numbers in [1, 1000000000]: 7999
FreeBASIC
<lang freebasic> ' Eban_numbers ' Un número eban es un número que no tiene la letra e cuando el número está escrito en inglés. ' O más literalmente, los números escritos que contienen la letra e están prohibidos. ' ' Usaremos la versión americana de los números de ortografía (a diferencia de los británicos). ' 2000000000 son dos billones, no dos millardos (mil millones). '
Data 2, 1000, 1 Data 1000, 4000, 1 Data 2, 10000, 0 Data 2, 100000, 0 Data 2, 1000000, 0 Data 2, 10000000, 0 Data 2, 100000000, 0 Data 0, 0, 0
Dim As Double tiempo = Timer Dim As Integer start, ended, printable, count Dim As Long i, b, r, m, t Do
Read start, ended, printable
If start = 0 Then Exit Do
If start = 2 Then
Print "eban numbers up to and including"; ended; ":"
Else
Print "eban numbers between "; start; " and "; ended; " (inclusive):"
End If
count = 0
For i = start To ended Step 2
b = Int(i / 1000000000)
r = (i Mod 1000000000)
m = Int(r / 1000000)
r = (i Mod 1000000)
t = Int(r / 1000)
r = (r Mod 1000)
If m >= 30 And m <= 66 Then m = (m Mod 10)
If t >= 30 And t <= 66 Then t = (t Mod 10)
If r >= 30 And r <= 66 Then r = (r Mod 10)
If b = 0 Or b = 2 Or b = 4 Or b = 6 Then
If m = 0 Or m = 2 Or m = 4 Or m = 6 Then
If t = 0 Or t = 2 Or t = 4 Or t = 6 Then
If r = 0 Or r = 2 Or r = 4 Or r = 6 Then
If printable Then Print i;
count += 1
End If
End If
End If
End If
Next i
If printable Then Print
Print "count = "; count & Chr(10)
Loop tiempo = Timer - tiempo Print "Run time: " & (tiempo) & " seconds." End </lang>
- Output:
eban numbers up to and including 100: 2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66 count = 19 eban numbers between 1000 and 4000 (inclusive): 2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000 count = 21 eban numbers up to and including 10000: count = 79 eban numbers up to and including 100000: count = 399 eban numbers up to and including 1000000: count = 399 eban numbers up to and including 10000000: count = 1599 eban numbers up to and including 100000000: count = 7999 Run time: 1.848286400010693 seconds.
Fōrmulæ
In this page you can see the solution of this task.
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The option to show Fōrmulæ programs and their results is showing images. Unfortunately images cannot be uploaded in Rosetta Code.
Go
<lang go>package main
import "fmt"
type Range struct {
start, end uint64 print bool
}
func main() {
rgs := []Range{
{2, 1000, true},
{1000, 4000, true},
{2, 1e4, false},
{2, 1e5, false},
{2, 1e6, false},
{2, 1e7, false},
{2, 1e8, false},
{2, 1e9, false},
}
for _, rg := range rgs {
if rg.start == 2 {
fmt.Printf("eban numbers up to and including %d:\n", rg.end)
} else {
fmt.Printf("eban numbers between %d and %d (inclusive):\n", rg.start, rg.end)
}
count := 0
for i := rg.start; i <= rg.end; i += 2 {
b := i / 1000000000
r := i % 1000000000
m := r / 1000000
r = i % 1000000
t := r / 1000
r %= 1000
if m >= 30 && m <= 66 {
m %= 10
}
if t >= 30 && t <= 66 {
t %= 10
}
if r >= 30 && r <= 66 {
r %= 10
}
if b == 0 || b == 2 || b == 4 || b == 6 {
if m == 0 || m == 2 || m == 4 || m == 6 {
if t == 0 || t == 2 || t == 4 || t == 6 {
if r == 0 || r == 2 || r == 4 || r == 6 {
if rg.print {
fmt.Printf("%d ", i)
}
count++
}
}
}
}
}
if rg.print {
fmt.Println()
}
fmt.Println("count =", count, "\n")
}
}</lang>
- Output:
eban numbers up to and including 1000: 2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66 count = 19 eban numbers between 1000 and 4000 (inclusive): 2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000 count = 21 eban numbers up to and including 10000: count = 79 eban numbers up to and including 100000: count = 399 eban numbers up to and including 1000000: count = 399 eban numbers up to and including 10000000: count = 1599 eban numbers up to and including 100000000: count = 7999 eban numbers up to and including 1000000000: count = 7999
Haskell
<lang haskell>{-# LANGUAGE NumericUnderscores #-} import Data.List (intercalate) import Text.Printf (printf) import Data.List.Split (chunksOf)
isEban :: Int -> Bool isEban n = all (`elem` [0, 2, 4, 6]) z
where (b, r1) = n `quotRem` (10 ^ 9) (m, r2) = r1 `quotRem` (10 ^ 6) (t, r3) = r2 `quotRem` (10 ^ 3) z = b : map (\x -> if x >= 30 && x <= 66 then x `mod` 10 else x) [m, t, r3]
ebans = map f
where f x = (thousands x, thousands $ length $ filter isEban [1..x])
thousands:: Int -> String thousands = reverse . intercalate "," . chunksOf 3 . reverse . show
main :: IO () main = do
uncurry (printf "eban numbers up to and including 1000: %2s\n%s\n\n") $ r [1..1000]
uncurry (printf "eban numbers between 1000 and 4000: %2s\n%s\n\n") $ r [1000..4000]
mapM_ (uncurry (printf "eban numbers up and including %13s: %5s\n")) ebanCounts
where
ebanCounts = ebans [ 10_000
, 100_000
, 1_000_000
, 10_000_000
, 100_000_000
, 1_000_000_000 ]
r = ((,) <$> thousands . length <*> show) . filter isEban</lang>
- Output:
eban numbers up to and including 1000: 19 [2,4,6,30,32,34,36,40,42,44,46,50,52,54,56,60,62,64,66] eban numbers between 1000 and 4000: 21 [2000,2002,2004,2006,2030,2032,2034,2036,2040,2042,2044,2046,2050,2052,2054,2056,2060,2062,2064,2066,4000] eban numbers up and including 10,000: 79 eban numbers up and including 100,000: 399 eban numbers up and including 1,000,000: 399 eban numbers up and including 10,000,000: 1,599 eban numbers up and including 100,000,000: 7,999 eban numbers up and including 1,000,000,000: 7,999
J
<lang J> Filter =: (#~`)(`:6)
itemAmend =: (29&< *. <&67)`(,: 10&|)} iseban =: [: *./ 0 2 4 6 e.~ [: itemAmend [: |: (4#1000)&#:
(;~ #) iseban Filter >: i. 1000
┌──┬─────────────────────────────────────────────────────┐ │19│2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66│ └──┴─────────────────────────────────────────────────────┘
NB. INPUT are the correct integers, head and tail shown
({. , {:) INPUT =: 1000 + i. 3001
1000 4000
(;~ #) iseban Filter INPUT
┌──┬────────────────────────────────────────────────────────────────────────────────────────────────────────┐ │21│2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000│ └──┴────────────────────────────────────────────────────────────────────────────────────────────────────────┘
(, ([: +/ [: iseban [: >: i.))&> 10000 * 10 ^ i. +:2 10000 79
100000 399
1e6 399 1e7 1599
</lang>
Java
<lang java>import java.util.List;
public class Main {
private static class Range {
int start;
int end;
boolean print;
public Range(int s, int e, boolean p) {
start = s;
end = e;
print = p;
}
}
public static void main(String[] args) {
List<Range> rgs = List.of(
new Range(2, 1000, true),
new Range(1000, 4000, true),
new Range(2, 10_000, false),
new Range(2, 100_000, false),
new Range(2, 1_000_000, false),
new Range(2, 10_000_000, false),
new Range(2, 100_000_000, false),
new Range(2, 1_000_000_000, false)
);
for (Range rg : rgs) {
if (rg.start == 2) {
System.out.printf("eban numbers up to and including %d\n", rg.end);
} else {
System.out.printf("eban numbers between %d and %d\n", rg.start, rg.end);
}
int count = 0;
for (int i = rg.start; i <= rg.end; ++i) {
int b = i / 1_000_000_000;
int r = i % 1_000_000_000;
int m = r / 1_000_000;
r = i % 1_000_000;
int t = r / 1_000;
r %= 1_000;
if (m >= 30 && m <= 66) m %= 10;
if (t >= 30 && t <= 66) t %= 10;
if (r >= 30 && r <= 66) r %= 10;
if (b == 0 || b == 2 || b == 4 || b == 6) {
if (m == 0 || m == 2 || m == 4 || m == 6) {
if (t == 0 || t == 2 || t == 4 || t == 6) {
if (r == 0 || r == 2 || r == 4 || r == 6) {
if (rg.print) System.out.printf("%d ", i);
count++;
}
}
}
}
}
if (rg.print) {
System.out.println();
}
System.out.printf("count = %d\n\n", count);
}
}
}</lang>
- Output:
eban numbers up to and including 1000 2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66 count = 19 eban numbers between 1000 and 4000 2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000 count = 21 eban numbers up to and including 10000 count = 79 eban numbers up to and including 100000 count = 399 eban numbers up to and including 1000000 count = 399 eban numbers up to and including 10000000 count = 1599 eban numbers up to and including 100000000 count = 7999 eban numbers up to and including 1000000000 count = 7999
Julia
<lang Julia> function iseban(n::Integer)
b, r = divrem(n, oftype(n, 10 ^ 9)) m, r = divrem(r, oftype(n, 10 ^ 6)) t, r = divrem(r, oftype(n, 10 ^ 3)) m, t, r = (30 <= x <= 66 ? x % 10 : x for x in (m, t, r)) return all(in((0, 2, 4, 6)), (b, m, t, r))
end
println("eban numbers up to and including 1000:") println(join(filter(iseban, 1:100), ", "))
println("eban numbers between 1000 and 4000 (inclusive):") println(join(filter(iseban, 1000:4000), ", "))
println("eban numbers up to and including 10000: ", count(iseban, 1:10000)) println("eban numbers up to and including 100000: ", count(iseban, 1:100000)) println("eban numbers up to and including 1000000: ", count(iseban, 1:1000000)) println("eban numbers up to and including 10000000: ", count(iseban, 1:10000000)) println("eban numbers up to and including 100000000: ", count(iseban, 1:100000000)) println("eban numbers up to and including 1000000000: ", count(iseban, 1:1000000000)) </lang>
- Output:
eban numbers up to and including 1000: 2, 4, 6, 30, 32, 34, 36, 40, 42, 44, 46, 50, 52, 54, 56, 60, 62, 64, 66 eban numbers between 1000 and 4000 (inclusive): 2000, 2002, 2004, 2006, 2030, 2032, 2034, 2036, 2040, 2042, 2044, 2046, 2050, 2052, 2054, 2056, 2060, 2062, 2064, 2066, 4000 eban numbers up to and including 10000: 79 eban numbers up to and including 100000: 399 eban numbers up to and including 1000000: 399 eban numbers up to and including 10000000: 1599 eban numbers up to and including 100000000: 7999 eban numbers up to and including 1000000000: 7999
Kotlin
<lang scala>// Version 1.3.21
typealias Range = Triple<Int, Int, Boolean>
fun main() {
val rgs = listOf<Range>(
Range(2, 1000, true),
Range(1000, 4000, true),
Range(2, 10_000, false),
Range(2, 100_000, false),
Range(2, 1_000_000, false),
Range(2, 10_000_000, false),
Range(2, 100_000_000, false),
Range(2, 1_000_000_000, false)
)
for (rg in rgs) {
val (start, end, prnt) = rg
if (start == 2) {
println("eban numbers up to and including $end:")
} else {
println("eban numbers between $start and $end (inclusive):")
}
var count = 0
for (i in start..end step 2) {
val b = i / 1_000_000_000
var r = i % 1_000_000_000
var m = r / 1_000_000
r = i % 1_000_000
var t = r / 1_000
r %= 1_000
if (m >= 30 && m <= 66) m %= 10
if (t >= 30 && t <= 66) t %= 10
if (r >= 30 && r <= 66) r %= 10
if (b == 0 || b == 2 || b == 4 || b == 6) {
if (m == 0 || m == 2 || m == 4 || m == 6) {
if (t == 0 || t == 2 || t == 4 || t == 6) {
if (r == 0 || r == 2 || r == 4 || r == 6) {
if (prnt) print("$i ")
count++
}
}
}
}
}
if (prnt) println()
println("count = $count\n")
}
}</lang>
- Output:
Same as Go example.
Lua
<lang lua>function makeInterval(s,e,p)
return {start=s, end_=e, print_=p}
end
function main()
local intervals = {
makeInterval( 2, 1000, true),
makeInterval(1000, 4000, true),
makeInterval( 2, 10000, false),
makeInterval( 2, 1000000, false),
makeInterval( 2, 10000000, false),
makeInterval( 2, 100000000, false),
makeInterval( 2, 1000000000, false)
}
for _,intv in pairs(intervals) do
if intv.start == 2 then
print("eban numbers up to and including " .. intv.end_ .. ":")
else
print("eban numbers between " .. intv.start .. " and " .. intv.end_ .. " (inclusive)")
end
local count = 0
for i=intv.start,intv.end_,2 do
local b = math.floor(i / 1000000000)
local r = i % 1000000000
local m = math.floor(r / 1000000)
r = i % 1000000
local t = math.floor(r / 1000)
r = r % 1000
if m >= 30 and m <= 66 then m = m % 10 end
if t >= 30 and t <= 66 then t = t % 10 end
if r >= 30 and r <= 66 then r = r % 10 end
if b == 0 or b == 2 or b == 4 or b == 6 then
if m == 0 or m == 2 or m == 4 or m == 6 then
if t == 0 or t == 2 or t == 4 or t == 6 then
if r == 0 or r == 2 or r == 4 or r == 6 then
if intv.print_ then io.write(i .. " ") end
count = count + 1
end
end
end
end
end
if intv.print_ then
print()
end
print("count = " .. count)
print()
end
end
main()</lang>
- Output:
eban numbers up to and including 1000: 2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66 count = 19 eban numbers between 1000 and 4000 (inclusive) 2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000 count = 21 eban numbers up to and including 10000: count = 79 eban numbers up to and including 1000000: count = 399 eban numbers up to and including 10000000: count = 1599 eban numbers up to and including 100000000: count = 7999 eban numbers up to and including 1000000000: count = 7999
Nim
<lang nim>import strformat
proc iseban(n: int): bool =
if n == 0:
return false
var b = n div 1_000_000_000
var r = n mod 1_000_000_000
var m = r div 1_000_000
r = r mod 1_000_000
var t = r div 1_000
r = r mod 1_000
m = if m in 30..66: m mod 10 else: m
t = if t in 30..66: t mod 10 else: t
r = if r in 30..66: r mod 10 else: r
return {b, m, t, r} <= {0, 2, 4, 6}
echo "eban numbers up to and including 1000:" for i in 0..100:
if iseban(i):
stdout.write(&"{i} ")
echo "\n\neban numbers between 1000 and 4000 (inclusive):" for i in 1_000..4_000:
if iseban(i):
stdout.write(&"{i} ")
var count = 0 for i in 0..10_000:
if iseban(i): inc count
echo &"\n\nNumber of eban numbers up to and including {10000:8}: {count:4}"
count = 0 for i in 0..100_000:
if iseban(i): inc count
echo &"\nNumber of eban numbers up to and including {100000:8}: {count:4}"
count = 0 for i in 0..1_000_000:
if iseban(i): inc count
echo &"\nNumber of eban numbers up to and including {1000000:8}: {count:4}"
count = 0 for i in 0..10_000_000:
if iseban(i): inc count
echo &"\nNumber of eban numbers up to and including {10000000:8}: {count:4}"
count = 0 for i in 0..100_000_000:
if iseban(i): inc count
echo &"\nNumber of eban numbers up to and including {100000000:8}: {count:4}"</lang>
- Output:
eban numbers up to and including 1000: 2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66 eban numbers between 1000 and 4000 (inclusive): 2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000 Number of eban numbers up to and including 10000: 79 Number of eban numbers up to and including 100000: 399 Number of eban numbers up to and including 1000000: 399 Number of eban numbers up to and including 10000000: 1599
Perl
Exhaustive search
A couple of 'e'-specific optimizations keep the running time reasonable. <lang perl>use strict; use warnings; use feature 'say'; use Lingua::EN::Numbers qw(num2en);
sub comma { reverse ((reverse shift) =~ s/(.{3})/$1,/gr) =~ s/^,//r }
sub e_ban {
my($power) = @_;
my @n;
for (1..10**$power) {
next unless 0 == $_%2;
next if $_ =~ /[789]/ or /[12].$/ or /[135]..$/ or /[135]...$/ or /[135].....$/;
push @n, $_ unless num2en($_) =~ /e/;
}
@n;
}
my @OK = e_ban(my $max = 7);
my @a = grep { $_ <= 1000 } @OK; say "Number of eban numbers up to and including 1000: @{[1+$#a]}"; say join(', ',@a); say ;
my @b = grep { $_ >= 1000 && $_ <= 4000 } @OK; say "Number of eban numbers between 1000 and 4000 (inclusive): @{[1+$#b]}"; say join(', ',@b); say ;
for my $exp (4..$max) {
my $n = + grep { $_ <= 10**$exp } @OK;
printf "Number of eban numbers and %10s: %d\n", comma(10**$exp), $n;
}</lang>
- Output:
eban numbers up to and including 1000: 2, 4, 6, 30, 32, 34, 36, 40, 42, 44, 46, 50, 52, 54, 56, 60, 62, 64, 66 eban numbers between 1000 and 4000 (inclusive): 2000, 2002, 2004, 2006, 2030, 2032, 2034, 2036, 2040, 2042, 2044, 2046, 2050, 2052, 2054, 2056, 2060, 2062, 2064, 2066, 4000 Number of eban numbers up to 10,000: 79 Number of eban numbers up to 100,000: 399 Number of eban numbers up to 1,000,000: 399 Number of eban numbers up to 10,000,000: 1599
Algorithmically generate / count
Alternately, a partial translation of Raku. Does not need to actually generate the e-ban numbers to count them. Display counts up to 10**21.
<lang perl>use strict; use warnings; use bigint; use feature 'say'; use Lingua::EN::Nums2Words 'num2word'; use List::AllUtils 'sum';
sub comma { reverse ((reverse shift) =~ s/(.{3})/$1,/gr) =~ s/^,//r }
sub nban {
my ($n, @numbers) = @_;
grep { lc(num2word($_)) !~ /[$n]/i } @numbers;
}
sub enumerate {
my ($n, $upto) = @_;
my @ban = nban($n, 1 .. 99);
my @orders;
for my $o (2 .. $upto) {
push @orders, [nban($n, map { $_ * 10**$o } 1 .. 9)];
}
for my $oom (@orders) {
next unless +@$oom;
my @these;
for my $num (@$oom) {
push @these, $num, map { $_ + $num } @ban;
}
push @ban, @these;
}
unshift @ban, 0 if nban($n, 0);
@ban
}
sub count {
my ($n, $upto) = @_;
my @orders;
for my $o (2 .. $upto) {
push @orders, [nban($n, map { $_ * 10**$o } 1 .. 9)];
}
my @count = scalar nban($n, 1 .. 99);
for my $o ( 0 .. $#orders - 1 ) {
push @count, sum(@count) * (scalar @{$orders[$o]}) + (scalar @{$orders[$o]});
}
++$count[0] if nban($n, 0);
for my $m ( 0 .. $#count - 1 ) {
next unless scalar $orders[$m];
if (nban($n, 10**($m+2))) { $count[$m]++; $count[$m + 1]-- }
}
map { sum( @count[0..$_] ) } 0..$#count;
}
for my $t ('e') {
my @bans = enumerate($t, 4); my @count = count($t, my $max = 21);
my @j = grep { $_ <= 10 } @bans;
unshift @count, @{[1+$#j]};
say "\n============= $t-ban: =============";
my @a = grep { $_ <= 1000 } @bans;
say "$t-ban numbers up to 1000: @{[1+$#a]}";
say '[', join(' ',@a), ']';
say ;
my @b = grep { $_ >= 1000 && $_ <= 4000 } @bans;
say "$t-ban numbers between 1,000 & 4,000 (inclusive): @{[1+$#b]}";
say '[', join(' ',@b), ']';
say ;
say "Counts of $t-ban numbers up to ", lc(num2word(10**$max));
for my $exp (1..$max) {
my $nu = $count[$exp-1];
printf "Up to and including %23s: %s\n", lc(num2word(10**$exp)), comma($nu);
}
}</lang>
============= e-ban: ============= e-ban numbers up to 1000: 19 [2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66] e-ban numbers between 1,000 & 4,000 (inclusive): 21 [2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000] Counts of e-ban numbers up to one sextillion Up to and including ten: 3 Up to and including one hundred: 19 Up to and including one thousand: 19 Up to and including ten thousand: 79 Up to and including one hundred thousand: 399 Up to and including one million: 399 Up to and including ten million: 1,599 Up to and including one hundred million: 7,999 Up to and including one billion: 7,999 Up to and including ten billion: 31,999 Up to and including one hundred billion: 159,999 Up to and including one trillion: 159,999 Up to and including ten trillion: 639,999 Up to and including one hundred trillion: 3,199,999 Up to and including one quadrillion: 3,199,999 Up to and including ten quadrillion: 12,799,999 Up to and including one hundred quadrillion: 63,999,999 Up to and including one quintillion: 63,999,999 Up to and including ten quintillion: 255,999,999 Up to and including one hundred quintillion: 1,279,999,999 Up to and including one sextillion: 1,279,999,999
Phix
Why count when you can calculate? <lang Phix>function count_eban(integer p10) -- returns the count of eban numbers 1..power(10,p10)
integer n = p10-floor(p10/3),
p5 = floor(n/2),
p4 = floor((n+1)/2)
return power(5,p5)*power(4,p4)-1
end function
function eban(integer n) -- returns true if n is an eban number (only fully tested to 10e9)
if n=0 then return false end if
while n do
integer thou = remainder(n,1000)
if floor(thou/100)!=0 then return false end if
if not find(floor(thou/10),{0,3,4,5,6}) then return false end if
if not find(remainder(thou,10),{0,2,4,6}) then return false end if
n = floor(n/1000)
end while
return true
end function
sequence s = {} for i=0 to 1000 do
if eban(i) then s &= i end if
end for printf(1,"eban to 1000 : %v (%d items)\n",{s,length(s)}) s = {} for i=1000 to 4000 do
if eban(i) then s &= i end if
end for printf(1,"eban 1000..4000 : %v (%d items)\n\n",{s,length(s)})
atom t0 = time() for i=0 to 21 do
printf(1,"count_eban(10^%d) : %,d\n",{i,count_eban(i)})
end for ?elapsed(time()-t0)</lang>
- Output:
eban to 1000 : {2,4,6,30,32,34,36,40,42,44,46,50,52,54,56,60,62,64,66} (19 items)
eban 1000..4000 : {2000,2002,2004,2006,2030,2032,2034,2036,2040,2042,2044,2046,2050,2052,2054,2056,2060,2062,2064,2066,4000} (21 items)
count_eban(10^0) : 0
count_eban(10^1) : 3
count_eban(10^2) : 19
count_eban(10^3) : 19
count_eban(10^4) : 79
count_eban(10^5) : 399
count_eban(10^6) : 399
count_eban(10^7) : 1,599
count_eban(10^8) : 7,999
count_eban(10^9) : 7,999
count_eban(10^10) : 31,999
count_eban(10^11) : 159,999
count_eban(10^12) : 159,999
count_eban(10^13) : 639,999
count_eban(10^14) : 3,199,999
count_eban(10^15) : 3,199,999
count_eban(10^16) : 12,799,999
count_eban(10^17) : 63,999,999
count_eban(10^18) : 63,999,999
count_eban(10^19) : 255,999,999
count_eban(10^20) : 1,279,999,999
count_eban(10^21) : 1,279,999,999
"0.0s"
PicoLisp
<lang PicoLisp>(de _eban? (N)
(let
(B (/ N 1000000000)
R (% N 1000000000)
M (/ R 1000000)
R (% N 1000000)
Z (/ R 1000)
R (% R 1000) )
(and
(>= M 30)
(<= M 66)
(setq M (% M 10)) )
(and
(>= Z 30)
(<= Z 66)
(setq Z (% Z 10)) )
(and
(>= R 30)
(<= R 66)
(setq R (% R 10)) )
(fully
'((S)
(unless (bit? 1 (val S))
(>= 6 (val S)) ) )
'(B M Z R) ) ) )
(de eban (B A)
(default A 2)
(let R (cons 0 (cons))
(for (N A (>= B N) (+ N 2))
(and
(_eban? N)
(inc R)
(push (cdr R) N) ) )
(con R (flip (cadr R)))
R ) )
(off R) (prinl "eban numbers up to an including 1000:") (setq R (eban 1000)) (println (cdr R)) (prinl "count: " (car R)) (prinl) (prinl "eban numbers between 1000 and 4000") (setq R (eban 4000 1000)) (println (cdr R)) (prinl "count: " (car R)) (prinl) (prinl "eban numbers up to an including 10000:") (prinl "count: " (car (eban 10000))) (prinl) (prinl "eban numbers up to an including 100000:") (prinl "count: " (car (eban 100000))) (prinl) (prinl "eban numbers up to an including 1000000:") (prinl "count: " (car (eban 1000000))) (prinl) (prinl "eban numbers up to an including 10000000:") (prinl "count: " (car (eban 10000000)))</lang>
- Output:
eban numbers up to an including 1000: (2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66) count: 19 eban numbers between 1000 and 4000 (2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000) count: 21 eban numbers up to an including 10000: count: 79 eban numbers up to an including 100000: count: 399 eban numbers up to an including 1000000: count: 399 eban numbers up to an including 10000000: count: 1599
Python
<lang Python>
- Use inflect
"""
show all eban numbers <= 1,000 (in a horizontal format), and a count show all eban numbers between 1,000 and 4,000 (inclusive), and a count show a count of all eban numbers up and including 10,000 show a count of all eban numbers up and including 100,000 show a count of all eban numbers up and including 1,000,000 show a count of all eban numbers up and including 10,000,000
"""
import inflect import time
before = time.perf_counter()
p = inflect.engine()
- eban numbers <= 1000
print(' ') print('eban numbers up to and including 1000:') print(' ')
count = 0
for i in range(1,1001):
if not 'e' in p.number_to_words(i):
print(str(i)+' ',end=)
count += 1
print(' ') print(' ') print('count = '+str(count)) print(' ')
- eban numbers 1000 to 4000
print(' ') print('eban numbers between 1000 and 4000 (inclusive):') print(' ')
count = 0
for i in range(1000,4001):
if not 'e' in p.number_to_words(i):
print(str(i)+' ',end=)
count += 1
print(' ') print(' ') print('count = '+str(count)) print(' ')
- eban numbers up to 10000
print(' ') print('eban numbers up to and including 10000:') print(' ')
count = 0
for i in range(1,10001):
if not 'e' in p.number_to_words(i):
count += 1
print(' ') print('count = '+str(count)) print(' ')
- eban numbers up to 100000
print(' ') print('eban numbers up to and including 100000:') print(' ')
count = 0
for i in range(1,100001):
if not 'e' in p.number_to_words(i):
count += 1
print(' ') print('count = '+str(count)) print(' ')
- eban numbers up to 1000000
print(' ') print('eban numbers up to and including 1000000:') print(' ')
count = 0
for i in range(1,1000001):
if not 'e' in p.number_to_words(i):
count += 1
print(' ') print('count = '+str(count)) print(' ')
- eban numbers up to 10000000
print(' ') print('eban numbers up to and including 10000000:') print(' ')
count = 0
for i in range(1,10000001):
if not 'e' in p.number_to_words(i):
count += 1
print(' ') print('count = '+str(count)) print(' ')
after = time.perf_counter()
print(" ") print("Run time in seconds: "+str(after - before)) </lang>
Output:
eban numbers up to and including 1000: 2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66 count = 19 eban numbers between 1000 and 4000 (inclusive): 2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000 count = 21 eban numbers up to and including 10000: count = 79 eban numbers up to and including 100000: count = 399 eban numbers up to and including 1000000: count = 399 eban numbers up to and including 10000000: count = 1599 Run time in seconds: 1134.289519125
Raku
(formerly Perl 6)
Modular approach, very little is hard coded. Change the $upto order-of-magnitude limit to adjust the search/display ranges. Change the letter(s) given to the enumerate / count subs to modify which letter(s) to disallow.
Will handle multi-character 'bans'. Demonstrate for e-ban, t-ban and subur-ban.
Directly find :
- OEIS:A006933 Numbers without e: Eban
- OEIS:A008521 Numbers without o: Oban
- OEIS:A008523 Numbers without t: Tban
- OEIS:A072954 Numbers without a, i, l, t: TALIban
- OEIS:A072955 Numbers without b, r, s, u: SUBURban
- OEIS:A072956 Numbers without r, t, u: TURban
- OEIS:A072957 Numbers without r, u: URban
- OEIS:A072958 Numbers without a, c, i, l: CALIban
- OEIS:A089589 Numbers without i: Iban
- OEIS:A089590 Numbers without u: Uban
- and so on...
Considering numbers up to 1021, as the task directions suggest.
<lang perl6>use Lingua::EN::Numbers;
sub nban ($seq, $n = 'e') { ($seq).map: { next if .&cardinal.contains(any($n.lc.comb)); $_ } }
sub enumerate ($n, $upto) {
my @ban = [nban(1 .. 99, $n)],;
my @orders;
(2 .. $upto).map: -> $o {
given $o % 3 { # Compensate for irregulars: 11 - 19
when 1 { @orders.push: [flat (10**($o - 1) X* 10 .. 19).map(*.&nban($n)), |(10**$o X* 2 .. 9).map: *.&nban($n)] }
default { @orders.push: [flat (10**$o X* 1 .. 9).map: *.&nban($n)] }
}
}
^@orders .map: -> $o {
@ban.push: [] and next unless +@orders[$o];
my @these;
@orders[$o].map: -> $m {
@these.push: $m;
for ^@ban -> $b {
next unless +@ban[$b];
@these.push: $_ for (flat @ban[$b]) »+» $m ;
}
}
@ban.push: @these;
}
@ban.unshift(0) if nban(0, $n);
flat @ban.map: *.flat;
}
sub count ($n, $upto) {
my @orders;
(2 .. $upto).map: -> $o {
given $o % 3 { # Compensate for irregulars: 11 - 19
when 1 { @orders.push: [flat (10**($o - 1) X* 10 .. 19).map(*.&nban($n)), |(10**$o X* 2 .. 9).map: *.&nban($n)] }
default { @orders.push: [flat (10**$o X* 1 .. 9).map: *.&nban($n)] }
}
}
my @count = +nban(1 .. 99, $n);
^@orders .map: -> $o {
@count.push: 0 and next unless +@orders[$o];
my $prev = so (@orders[$o].first( { $_ ~~ /^ '1' '0'+ $/ } ) // 0 );
my $sum = @count.sum;
my $these = +@orders[$o] * $sum + @orders[$o];
$these-- if $prev;
@count[1 + $o] += $these;
++@count[$o] if $prev;
}
++@count[0] if nban(0, $n);
[\+] @count;
}
- for < e o t tali subur tur ur cali i u > -> $n { # All of them
for < e t subur > -> $n { # An assortment for demonstration
my $upto = 21; # 1e21 my @bans = enumerate($n, 4); my @counts = count($n, $upto);
# DISPLAY
my @k = @bans.grep: * < 1000;
my @j = @bans.grep: 1000 <= * <= 4000;
put "\n============= {$n}-ban: =============\n" ~
"{$n}-ban numbers up to 1000: {+@k}\n[{@k».&comma}]\n\n" ~
"{$n}-ban numbers between 1,000 & 4,000: {+@j}\n[{@j».&comma}]\n" ~
"\nCounts of {$n}-ban numbers up to {cardinal 10**$upto}"
;
my $s = max (1..$upto).map: { (10**$_).&cardinal.chars };
@counts.unshift: @bans.first: * > 10, :k;
for ^$upto -> $c {
printf "Up to and including %{$s}s: %s\n", cardinal(10**($c+1)), comma(@counts[$c]);
}
}</lang>
- Output:
============= e-ban: ============= e-ban numbers up to 1000: 19 [2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66] e-ban numbers between 1,000 & 4,000: 21 [2,000 2,002 2,004 2,006 2,030 2,032 2,034 2,036 2,040 2,042 2,044 2,046 2,050 2,052 2,054 2,056 2,060 2,062 2,064 2,066 4,000] Counts of e-ban numbers up to one sextillion Up to and including ten: 3 Up to and including one hundred: 19 Up to and including one thousand: 19 Up to and including ten thousand: 79 Up to and including one hundred thousand: 399 Up to and including one million: 399 Up to and including ten million: 1,599 Up to and including one hundred million: 7,999 Up to and including one billion: 7,999 Up to and including ten billion: 31,999 Up to and including one hundred billion: 159,999 Up to and including one trillion: 159,999 Up to and including ten trillion: 639,999 Up to and including one hundred trillion: 3,199,999 Up to and including one quadrillion: 3,199,999 Up to and including ten quadrillion: 12,799,999 Up to and including one hundred quadrillion: 63,999,999 Up to and including one quintillion: 63,999,999 Up to and including ten quintillion: 255,999,999 Up to and including one hundred quintillion: 1,279,999,999 Up to and including one sextillion: 1,279,999,999 ============= t-ban: ============= t-ban numbers up to 1000: 56 [0 1 4 5 6 7 9 11 100 101 104 105 106 107 109 111 400 401 404 405 406 407 409 411 500 501 504 505 506 507 509 511 600 601 604 605 606 607 609 611 700 701 704 705 706 707 709 711 900 901 904 905 906 907 909 911] t-ban numbers between 1,000 & 4,000: 0 [] Counts of t-ban numbers up to one sextillion Up to and including ten: 7 Up to and including one hundred: 9 Up to and including one thousand: 56 Up to and including ten thousand: 56 Up to and including one hundred thousand: 56 Up to and including one million: 57 Up to and including ten million: 392 Up to and including one hundred million: 785 Up to and including one billion: 5,489 Up to and including ten billion: 38,416 Up to and including one hundred billion: 76,833 Up to and including one trillion: 537,824 Up to and including ten trillion: 537,824 Up to and including one hundred trillion: 537,824 Up to and including one quadrillion: 537,825 Up to and including ten quadrillion: 3,764,768 Up to and including one hundred quadrillion: 7,529,537 Up to and including one quintillion: 52,706,752 Up to and including ten quintillion: 52,706,752 Up to and including one hundred quintillion: 52,706,752 Up to and including one sextillion: 52,706,752 ============= subur-ban: ============= subur-ban numbers up to 1000: 35 [1 2 5 8 9 10 11 12 15 18 19 20 21 22 25 28 29 50 51 52 55 58 59 80 81 82 85 88 89 90 91 92 95 98 99] subur-ban numbers between 1,000 & 4,000: 0 [] Counts of subur-ban numbers up to one sextillion Up to and including ten: 6 Up to and including one hundred: 35 Up to and including one thousand: 35 Up to and including ten thousand: 35 Up to and including one hundred thousand: 35 Up to and including one million: 36 Up to and including ten million: 216 Up to and including one hundred million: 2,375 Up to and including one billion: 2,375 Up to and including ten billion: 2,375 Up to and including one hundred billion: 2,375 Up to and including one trillion: 2,375 Up to and including ten trillion: 2,375 Up to and including one hundred trillion: 2,375 Up to and including one quadrillion: 2,375 Up to and including ten quadrillion: 2,375 Up to and including one hundred quadrillion: 2,375 Up to and including one quintillion: 2,375 Up to and including ten quintillion: 2,375 Up to and including one hundred quintillion: 2,375 Up to and including one sextillion: 2,375
Note that the limit to one sextillion is somewhat arbitrary and is just to match the task parameters.
This will quite happily count *-bans up to one hundred centillion. (10305) It takes longer, but still on the order of seconds, not minutes.
Counts of e-ban numbers up to one hundred centillion ... Up to and including one hundred centillion: 35,184,372,088,831,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999
REXX
Programming note: REXX has no shortcuts for if statements, so the multiple if statements weren't combined into one. <lang rexx>/*REXX program to display eban numbers (those that don't have an "e" their English name)*/ numeric digits 20 /*support some gihugic numbers for pgm.*/ parse arg $ /*obtain optional arguments from the cL*/ if $= then $= '1 1000 1000 4000 1 -10000 1 -100000 1 -1000000 1 -10000000'
do k=1 by 2 to words($) /*step through the list of numbers. */
call banE word($, k), word($, k+1) /*process the numbers, from low──►high.*/
end /*k*/
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ banE: procedure; parse arg x,y,_; z= reverse(x) /*obtain the number to be examined. */
tell= y>=0 /*Is HI non-negative? Display eban #s.*/
#= 0 /*the count of eban numbers (so far).*/
do j=x to abs(y) /*probably process a range of numbers. */
if hasE(j) then iterate /*determine if the number has an "e". */
#= # + 1 /*bump the counter of eban numbers. */
if tell then _= _ j /*maybe add to a list of eban numbers. */
end /*j*/
if _\== then say strip(_) /*display the list (if there is one). */
say; say # ' eban numbers found for: ' x " " y; say copies('═', 105)
return
/*──────────────────────────────────────────────────────────────────────────────────────*/ hasE: procedure; parse arg x; z= reverse(x) /*obtain the number to be examined. */
do k=1 by 3 /*while there're dec. digit to examine.*/
@= reverse( substr(z, k, 3) ) /*obtain 3 dec. digs (a period) from Z.*/
if @==' ' then return 0 /*we have reached the "end" of the num.*/
uni= right(@, 1) /*get units dec. digit of this period. */
if uni//2==1 then return 1 /*if an odd digit, then not an eban #. */
if uni==8 then return 1 /*if an eight, " " " " " */
tens=substr(@, 2, 1) /*get tens dec. digit of this period. */
if tens==1 then return 1 /*if teens, then not an eban #. */
if tens==2 then return 1 /*if twenties, " " " " " */
if tens>6 then return 1 /*if 70s, 80s, 90s, " " " " " */
hun= left(@, 1) /*get hundreds dec. dig of this period.*/
if hun==0 then iterate /*if zero, then there is more of number*/
if hun\==' ' then return 1 /*any hundrEd (not zero) has an "e". */
end /*k*/ /*A "period" is a group of 3 dec. digs */
return 0 /*in the number, grouped from the right*/</lang>
- output when using the default inputs:
2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66 19 eban numbers found for: 1 1000 ═════════════════════════════════════════════════════════════════════════════════════════════════════════ 2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000 21 eban numbers found for: 1000 4000 ═════════════════════════════════════════════════════════════════════════════════════════════════════════ 79 eban numbers found for: 1 -10000 ═════════════════════════════════════════════════════════════════════════════════════════════════════════ 399 eban numbers found for: 1 -100000 ═════════════════════════════════════════════════════════════════════════════════════════════════════════ 399 eban numbers found for: 1 -1000000 ═════════════════════════════════════════════════════════════════════════════════════════════════════════ 1599 eban numbers found for: 1 -10000000 ═════════════════════════════════════════════════════════════════════════════════════════════════════════
Tailspin
<lang tailspin> templates isEban
def number: $; '$;' -> \(<'([246]|[3456][0246])(0[03456][0246])*'> $ !\) -> $number !
end isEban </lang>
Alternatively, if regex is not your thing, we can do it numerically, which actually runs faster <lang tailspin> templates isEban
def number: $; $ -> \(<1..> $!\) -> # when <=0> do $number ! when <?($ mod 1000 <=0|=2|=4|=6|30..66?($ mod 10 <=0|=2|=4|=6>)>)> do $ ~/ 1000 -> #
end isEban </lang>
Either version is called by the following code <lang tailspin> def small: [1..1000 -> isEban]; $small -> !OUT::write ' There are $small::length; eban numbers up to and including 1000
' -> !OUT::write
def next: [1000..4000 -> isEban]; $next -> !OUT::write ' There are $next::length; eban numbers between 1000 and 4000 (inclusive)
' -> !OUT::write ' There are $:[1..10000 -> isEban] -> $::length; eban numbers up to and including 10 000
' -> !OUT::write ' There are $:[1..100000 -> isEban] -> $::length; eban numbers up to and including 100 000
' -> !OUT::write ' There are $:[1..1000000 -> isEban] -> $::length; eban numbers up to and including 1 000 000
' -> !OUT::write ' There are $:[1..10000000 -> isEban] -> $::length; eban numbers up to and including 10 000 000
' -> !OUT::write </lang>
- Output:
[2, 4, 6, 30, 32, 34, 36, 40, 42, 44, 46, 50, 52, 54, 56, 60, 62, 64, 66] There are 19 eban numbers up to and including 1000 [2000, 2002, 2004, 2006, 2030, 2032, 2034, 2036, 2040, 2042, 2044, 2046, 2050, 2052, 2054, 2056, 2060, 2062, 2064, 2066, 4000] There are 21 eban numbers between 1000 and 4000 (inclusive) There are 79 eban numbers up to and including 10 000 There are 399 eban numbers up to and including 100 000 There are 399 eban numbers up to and including 1 000 000 There are 1599 eban numbers up to and including 10 000 000
Visual Basic .NET
<lang vbnet>Module Module1
Structure Interval
Dim start As Integer
Dim last As Integer
Dim print As Boolean
Sub New(s As Integer, l As Integer, p As Boolean)
start = s
last = l
print = p
End Sub
End Structure
Sub Main()
Dim intervals As Interval() = {
New Interval(2, 1_000, True),
New Interval(1_000, 4_000, True),
New Interval(2, 10_000, False),
New Interval(2, 100_000, False),
New Interval(2, 1_000_000, False),
New Interval(2, 10_000_000, False),
New Interval(2, 100_000_000, False),
New Interval(2, 1_000_000_000, False)
}
For Each intv In intervals
If intv.start = 2 Then
Console.WriteLine("eban numbers up to and including {0}:", intv.last)
Else
Console.WriteLine("eban numbers between {0} and {1} (inclusive):", intv.start, intv.last)
End If
Dim count = 0
For i = intv.start To intv.last Step 2
Dim b = i \ 1_000_000_000
Dim r = i Mod 1_000_000_000
Dim m = r \ 1_000_000
r = i Mod 1_000_000
Dim t = r \ 1_000
r = r Mod 1_000
If m >= 30 AndAlso m <= 66 Then
m = m Mod 10
End If
If t >= 30 AndAlso t <= 66 Then
t = t Mod 10
End If
If r >= 30 AndAlso r <= 66 Then
r = r Mod 10
End If
If b = 0 OrElse b = 2 OrElse b = 4 OrElse b = 6 Then
If m = 0 OrElse m = 2 OrElse m = 4 OrElse m = 6 Then
If t = 0 OrElse t = 2 OrElse t = 4 OrElse t = 6 Then
If r = 0 OrElse r = 2 OrElse r = 4 OrElse r = 6 Then
If intv.print Then
Console.Write("{0} ", i)
End If
count += 1
End If
End If
End If
End If
Next
If intv.print Then
Console.WriteLine()
End If
Console.WriteLine("count = {0}", count)
Console.WriteLine()
Next
End Sub
End Module</lang>
- Output:
eban numbers up to and including 1000: 2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66 count = 19 eban numbers between 1000 and 4000 (inclusive): 2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000 count = 21 eban numbers up to and including 10000: count = 79 eban numbers up to and including 100000: count = 399 eban numbers up to and including 1000000: count = 399 eban numbers up to and including 10000000: count = 1599 eban numbers up to and including 100000000: count = 7999 eban numbers up to and including 1000000000: count = 7999
Wren
<lang ecmascript>var rgs = [
[2, 1000, true], [1000, 4000, true], [2, 1e4, false], [2, 1e5, false], [2, 1e6, false], [2, 1e7, false], [2, 1e8, false], [2, 1e9, false]
] for (rg in rgs) {
if (rg[0] == 2) {
System.print("eban numbers up to and including %(rg[1])")
} else {
System.print("eban numbers between %(rg[0]) and %(rg[1]) (inclusive):")
}
var count = 0
var i = rg[0]
while (i <= rg[1]) {
var b = (i/1e9).floor
var r = i % 1e9
var m = (r/1e6).floor
r = i % 1e6
var t = (r/1000).floor
r = r % 1000
if (m >= 30 && m <= 66) m = m % 10
if (t >= 30 && t <= 66) t = t % 10
if (r >= 30 && r <= 66) r = r % 10
if (b == 0 || b == 2 || b == 4 || b == 6) {
if (m == 0 || m == 2 || m == 4 || m == 6) {
if (t == 0 || t == 2 || t == 4 || t == 6) {
if (r == 0 || r == 2 || r == 4 || r == 6) {
if (rg[2]) System.write("%(i) ")
count = count + 1
}
}
}
}
i = i + 2
}
if (rg[2]) System.print()
System.print("count = %(count)\n")
}</lang>
- Output:
eban numbers up to and including 1000 2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66 count = 19 eban numbers between 1000 and 4000 (inclusive): 2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000 count = 21 eban numbers up to and including 10000 count = 79 eban numbers up to and including 100000 count = 399 eban numbers up to and including 1000000 count = 399 eban numbers up to and including 10000000 count = 1599 eban numbers up to and including 100000000 count = 7999 eban numbers up to and including 1000000000 count = 7999
Yabasic
<lang Yabasic>data 2, 100, true data 1000, 4000, true data 2, 1e4, false data 2, 1e5, false data 2, 1e6, false data 2, 1e7, false data 2, 1e8, false REM data 2, 1e9, false // it takes a lot of time data 0, 0, false
do
read start, ended, printable
if not start break
if start = 2 then
Print "eban numbers up to and including ", ended
else
Print "eban numbers between ", start, " and ", ended, " (inclusive):"
endif
count = 0
for i = start to ended step 2
b = int(i / 1000000000)
r = mod(i, 1000000000)
m = int(r / 1000000)
r = mod(i, 1000000)
t = int(r / 1000)
r = mod(r, 1000)
if m >= 30 and m <= 66 m = mod(m, 10)
if t >= 30 and t <= 66 t = mod(t, 10)
if r >= 30 and r <= 66 r = mod(r, 10)
if b = 0 or b = 2 or b = 4 or b = 6 then
if m = 0 or m = 2 or m = 4 or m = 6 then
if t = 0 or t = 2 or t = 4 or t = 6 then
if r = 0 or r = 2 or r = 4 or r = 6 then
if printable Print i;
count = count + 1
endif
endif
endif
endif
next
if printable Print
Print "count = ", count, "\n"
loop</lang>
zkl
<lang zkl>rgs:=T( T(2, 1_000, True), // (start,end,print)
T(1_000, 4_000, True),
T(2, 1e4, False), T(2, 1e5, False), T(2, 1e6, False), T(2, 1e7, False),
T(2, 1e8, False), T(2, 1e9, False), // slow and very slow
);
foreach start,end,pr in (rgs){
if(start==2) println("eban numbers up to and including %,d:".fmt(end));
else println("eban numbers between %,d and %,d (inclusive):".fmt(start,end));
count:=0;
foreach i in ([start..end,2]){
b,r := i/100_0000_000, i%1_000_000_000;
m,r := r/1_000_000, i%1_000_000;
t,r := r/1_000, r%1_000;
if(30<=m<=66) m=m%10;
if(30<=t<=66) t=t%10;
if(30<=r<=66) r=r%10;
if(magic(b) and magic(m) and magic(t) and magic(r)){
if(pr) print(i," ");
count+=1;
}
}
if(pr) println();
println("count = %,d\n".fmt(count));
} fcn magic(z){ z.isEven and z<=6 }</lang>
- Output:
eban numbers up to and including 1,000: 2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66 count = 19 eban numbers between 1,000 and 4,000 (inclusive): 2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000 count = 21 eban numbers up to and including 10,000: count = 79 eban numbers up to and including 100,000: count = 399 eban numbers up to and including 1,000,000: count = 399 eban numbers up to and including 10,000,000: count = 1,599 eban numbers up to and including 100,000,000: count = 7,999 eban numbers up to and including 1,000,000,000: count = 7,999