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# Eban numbers

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.

•   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.

## 11l

Translation of: Nim
`F iseban(n)   I n == 0      R 0B   V (b, r) = divmod(n, 1'000'000'000)   (V m, r) = divmod(r, 1'000'000)   (V t, r) = divmod(r, 1'000)   m = I m C 30..66 {m % 10} E m   t = I t C 30..66 {t % 10} E t   r = I r C 30..66 {r % 10} E r   R Set([b, m, t, r]) <= Set([0, 2, 4, 6]) print(‘eban numbers up to and including 1000:’)L(i) 0..100   I iseban(i)      print(i, end' ‘ ’) print("\n\neban numbers between 1000 and 4000 (inclusive):")L(i) 1000..4000   I iseban(i)      print(i, end' ‘ ’) print()L(maxn) (10'000, 100'000, 1'000'000, 10'000'000)   V count = 0   L(i) 0..maxn      I iseban(i)         count++   print("\nNumber of eban numbers up to and including #8: #4".format(maxn, count))`
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
```

## AppleScript

`(*    Quickly generate all the (positive) eban numbers up to and including the    specified end number, then lose those before the start number.    0 is taken as "zero" rather than as "nought" or "nil".     WARNING: The getEbans() handler returns a potentially very long list of numbers.    Don't let such a list get assigned to a persistent variable or be the end result displayed in an editor!*) on getEbans(startNumber, endNumber)    script o        property output : {}        property listCollector : missing value        property temp : missing value    end script     if (startNumber > endNumber) then set {startNumber, endNumber} to {endNumber, startNumber}    -- The range is limited to between 0 and 10^15 to keep within AppleScript's current number precision.    -- Even so, some of the numbers may not /look/ right when displayed in a editor.    set limit to 10 ^ 15 - 1    if (endNumber > limit) then set endNumber to limit    if (startNumber > limit) then set startNumber to limit     -- Initialise the output list with 0 and the sub-1000 ebans, stopping if the end number's reached.    -- The 0's needed for rest of the process, but is removed at the end.    repeat with tens in {0, 30, 40, 50, 60}        repeat with units in {0, 2, 4, 6}            set thisEban to tens + units            if (thisEban ≤ endNumber) then set end of o's output to thisEban        end repeat    end repeat     -- Repeatedly sweep the output list, adding new ebans formed from the addition of those found previously    -- to a suitable power of 1000 times each one. Stop when the end number's reached or exceeded.    -- The output list may become very long and appending items to it individually take forever, so results are    -- collected in short, 1000-item lists, which are concatenated to the output at the end of each sweep.    set sweepLength to (count o's output)    set multiplier to 1000    repeat while (thisEban < endNumber) -- Per sweep.        set o's temp to {}        set o's listCollector to {o's temp}        repeat with i from 2 to sweepLength -- Per eban found already. (Not the 0.)            set baseAmount to (item i of o's output) * multiplier            repeat with j from 1 to sweepLength by 1000 -- Per 1000-item list.                set z to j + 999                if (z > sweepLength) then set z to sweepLength                repeat with k from j to z -- Per new eban.                    set thisEban to baseAmount + (item k of o's output)                    if (thisEban ≤ endNumber) then set end of o's temp to thisEban                    if (thisEban ≥ endNumber) then exit repeat                end repeat                if (thisEban ≥ endNumber) then exit repeat                set o's temp to {}                set end of o's listCollector to o's temp            end repeat            if (thisEban ≥ endNumber) then exit repeat        end repeat         -- Concatentate this sweep's new lists together        set listCount to (count o's listCollector)        repeat until (listCount is 1)            set o's temp to {}            repeat with i from 2 to listCount by 2                set end of o's temp to (item (i - 1) of o's listCollector) & (item i of o's listCollector)            end repeat            if (listCount mod 2 is 1) then set item -1 of o's temp to (end of o's temp) & (end of o's listCollector)            set o's listCollector to o's temp            set o's temp to {}            set listCount to (count o's listCollector)        end repeat        -- Concatenate the result to the output list and prepare to go round again.        set o's output to o's output & (beginning of o's listCollector)        set sweepLength to sweepLength * sweepLength        set multiplier to multiplier * multiplier    end repeat     -- Lose the initial 0 and any ebans before the specified start number.    set resultCount to (count o's output)    if (resultCount is 1) then        set o's output to {}    else        repeat with i from 2 to resultCount            if (item i of o's output ≥ startNumber) then exit repeat        end repeat        set o's output to items i thru resultCount of o's output    end if    if (endNumber = limit) then set end of o's output to "(Limit of AppleScript's number precision.)"     return o's outputend getEbans -- Task code:on runTask()    set output to {}    set astid to AppleScript's text item delimiters    set AppleScript's text item delimiters to ", "     set ebans to getEbans(0, 1000)    set end of output to ((count ebans) as text) & " eban numbers between 0 and 1,000:" & (linefeed & "  " & ebans)    set ebans to getEbans(1000, 4000)    set end of output to ((count ebans) as text) & " between 1,000 and 4,000:" & (linefeed & "  " & ebans)    set end of output to ((count getEbans(0, 10000)) as text) & " up to and including 10,000"    set end of output to ((count getEbans(0, 100000)) as text) & " up to and including 100,000"    set end of output to ((count getEbans(0, 1000000)) as text) & " up to and including 1,000,000"    set end of output to ((count getEbans(0, 10000000)) as text) & " up to and including 10,000,000"    set ebans to getEbans(6.606606602E+10, 1.0E+12)    set end of output to ((count ebans) as text) & " between 66,066,066,020 and 1,000,000,000,000:" & (linefeed & "  " & ebans)     set AppleScript's text item delimiters to linefeed    set output to output as text    set AppleScript's text item delimiters to astid     return outputend runTask runTask()`
Output:
`"19 eban numbers between 0 and 1,000:  2, 4, 6, 30, 32, 34, 36, 40, 42, 44, 46, 50, 52, 54, 56, 60, 62, 64, 6621 between 1,000 and 4,000:  2000, 2002, 2004, 2006, 2030, 2032, 2034, 2036, 2040, 2042, 2044, 2046, 2050, 2052, 2054, 2056, 2060, 2062, 2064, 2066, 400079 up to and including 10,000399 up to and including 100,000399 up to and including 1,000,0001599 up to and including 10,000,00016 between 66,066,066,020 and 1,000,000,000,000:  6.606606603E+10, 6.6066066032E+10, 6.6066066034E+10, 6.6066066036E+10, 6.606606604E+10, 6.6066066042E+10, 6.6066066044E+10, 6.6066066046E+10, 6.606606605E+10, 6.6066066052E+10, 6.6066066054E+10, 6.6066066056E+10, 6.606606606E+10, 6.6066066062E+10, 6.6066066064E+10, 6.6066066066E+10"`

## AutoHotkey

`eban_numbers(min, max, show:=0){	counter := 0, output := ""	i := min	while ((i+=2) <= max) 	{		b := floor(i / 1000000000)		r := Mod(i, 1000000000)		m := floor(r / 1000000)		r := Mod(i, 1000000)		t := floor(r / 1000)		r := Mod(r, 1000)				if (m >= 30 && m <= 66) 			m := Mod(m, 10)		if (t >= 30 && t <= 66)			t := Mod(t, 10)		if (r >= 30 && r <= 66)			r := Mod(r, 10)		if (b = 0 || b = 2 || b = 4 || b = 6)		&& (m = 0 || m = 2 || m = 4 || m = 6)		&& (t = 0 || t = 2 || t = 4 || t = 6)		&& (r = 0 || r = 2 || r = 4 || r = 6)			counter++, (show ? output .= i " " : "")	}	return min "-" max " : " output " Count = " counter}`
Examples:
`MsgBox, 262144, , % eban_numbers(0, 1000, 1)MsgBox, 262144, , % eban_numbers(1000, 4000, 1)MsgBox, 262144, , % eban_numbers(0, 10000)MsgBox, 262144, , % eban_numbers(0, 100000)MsgBox, 262144, , % eban_numbers(0, 1000000)MsgBox, 262144, , % eban_numbers(0, 100000000)`
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```

## AWK

` # syntax: GAWK -f EBAN_NUMBERS.AWK# converted from FreeBASICBEGIN {    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)} `
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

Translation of: D
`#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;}`
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#

Translation of: D
`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);            }        }    }}`
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++

Translation of: D
`#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;}`
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

Translation of: Kotlin
`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;    }}`
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

Translation of: Julia
`USING: arrays formatting fry io kernel math math.functionsmath.order math.ranges prettyprint sequences ; : eban? ( n -- ? )    1000000000 /mod 1000000 /mod 1000 /mod    [ dup 30 66 between? [ 10 mod ] when ] [email protected] 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 ] [email protected]4 9 [a,b] [ [ 1 10 ] dip ^ .eban-count ] each`
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

` ' 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, 1Data 1000, 4000, 1Data 2, 10000, 0Data 2, 100000, 0Data 2, 1000000, 0Data 2, 10000000, 0Data 2, 100000000, 0Data 0, 0, 0 Dim As Double tiempo = TimerDim As Integer start, ended, printable, countDim As Long i, b, r, m, tDo    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)Looptiempo = Timer - tiempoPrint "Run time: " & (tiempo) & " seconds."End `
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

Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition.

Programs in Fōrmulæ are created/edited online in its website, However they run on execution servers. By default remote servers are used, but they are limited in memory and processing power, since they are intended for demonstration and casual use. A local server can be downloaded and installed, it has no limitations (it runs in your own computer). Because of that, example programs can be fully visualized and edited, but some of them will not run if they require a moderate or heavy computation/memory resources, and no local server is being used.

## 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")    }}`
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
```

## Groovy

Translation of: Java
`class Main {    private static class Range {        int start        int end        boolean print         Range(int s, int e, boolean p) {            start = s            end = e            print = p        }    }     static void main(String[] args) {        List<Range> rgs = Arrays.asList(                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) {                println("eban numbers up to and including \$rg.end")            } else {                println("eban numbers between \$rg.start and \$rg.end")            }            int count = 0            for (int i = rg.start; i <= rg.end; ++i) {                int b = (int) (i / 1_000_000_000)                int r = i % 1_000_000_000                int m = (int) (r / 1_000_000)                r = i % 1_000_000                int t = (int) (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) {                                    print("\$i ")                                }                                count++                            }                        }                    }                }            }            if (rg.print) {                println()            }            println("count = \$count")            println()        }    }}`
Output:
`Same as the Java entry`

Translation of: Julia
`{-# LANGUAGE NumericUnderscores #-}import Data.List (intercalate)import Text.Printf (printf)import Data.List.Split (chunksOf) isEban :: Int -> BoolisEban 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 -> Stringthousands = 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`
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

` 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. 30011000 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   79100000  399   1e6  399   1e7 1599 `

## Java

Translation of: Kotlin
`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);        }    }}`
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```

## jq

Works with: jq

Works with gojq, the Go implementation of jq (*)

(*) gojq's memory requirements are rather extravagent and may prevent completion of the task beyond 1E+7; the output shown below was obtained using the C implementation of jq.

`# quotient and remainderdef quotient(\$a; \$b; f; g): f = ((\$a/\$b)|floor) | g = \$a % \$b; def tasks:    [2, 1000, true],    [1000, 4000, true],    (1e4, 1e5, 1e6, 1e7, 1e8 | [2, ., false]) ; def task:  def update(f): if (f >= 30 and f <= 66) then f %= 10 else . end;   tasks as \$rg  | if \$rg == 2    then "eban numbers up to and including \(\$rg):"    else "eban numbers between \(\$rg) and \(\$rg) (inclusive):"    end,    ( foreach (range( \$rg; 1 + \$rg; 2), null) as \$i ( { count: 0 };        .emit = false        | if \$i == null then .total = .count          else quotient(\$i; 1e9; .b; .r)          |    quotient(.r; 1e6; .m; .r)          |    quotient(.r; 1e3; .t; .r)          | update(.m) | update(.t) | update(.r)          | if all(.b, .m, .t, .r; IN(0, 2, 4, 6))            then .count += 1            | if (\$rg) then .emit=\$i else . end            else .  	    end  	  end;  	  if .emit then .emit else empty end,  	  if .total then "count = \(.count)\n" else empty end) ); task`
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 1E+4:
count = 79

eban numbers up to and including 1E+5:
count = 399

eban numbers up to and including 1E+6:
count = 399

eban numbers up to and including 1E+7:
count = 1599

eban numbers up to and including 1E+8:
count = 7999
```

## 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)) `
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

Translation of: Go
`// 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")    }}`
Output:
```Same as Go example.
```

## Lua

Translation of: lang
`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()    endend main()`
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

### Exhaustive search

Translation of: Julia
`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 = 0for i in 0..10_000:  if iseban(i):    inc countecho &"\n\nNumber of eban numbers up to and including {10000:8}: {count:4}" count = 0for i in 0..100_000:  if iseban(i):    inc countecho &"\nNumber of eban numbers up to and including {100000:8}: {count:4}" count = 0for i in 0..1_000_000:  if iseban(i):    inc countecho &"\nNumber of eban numbers up to and including {1000000:8}: {count:4}" count = 0for i in 0..10_000_000:  if iseban(i):    inc countecho &"\nNumber of eban numbers up to and including {10000000:8}: {count:4}" count = 0for i in 0..100_000_000:  if iseban(i):    inc countecho &"\nNumber of eban numbers up to and including {100000000:8}: {count:4}"`
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
```

### Algorithmic computation

Translation of: Phix
`import math, strutils, strformat #--------------------------------------------------------------------------------------------------- func ebanCount(p10: Natural): Natural =  ## Return the count of eban numbers 1..10^p10.  let    n = p10 - p10 div 3    p5 = n div 2    p4 = (n + 1) div 2  result = 5^p5 * 4^p4 - 1 #--------------------------------------------------------------------------------------------------- func eban(n: Natural): bool =  ## Return true if n is an eban number (only fully tested to 10e9).  if n == 0: return false  var n = n  while n != 0:    let thou = n mod 1000    if thou div 100 != 0: return false    if thou div 10 notin {0, 3, 4, 5, 6}: return false    if thou mod 10 notin {0, 2, 4, 6}: return false    n = n div 1000  result = true #——————————————————————————————————————————————————————————————————————————————————————————————————— var s: seq[Natural]for i in 0..1000:  if eban(i): s.add(i)echo fmt"Eban to 1000: {s.join("", "")} ({s.len} items)" s.setLen(0)for i in 1000..4000:  if eban(i): s.add(i)echo fmt"Eban 1000..4000: {s.join("", "")} ({s.len} items)" import timeslet t0 = getTime()for i in 0..21:  echo fmt"ebanCount(10^{i}): {ebanCount(i)}"echo ""echo fmt"Time: {getTime() - t0}"`
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)
ebanCount(10^0): 0
ebanCount(10^1): 3
ebanCount(10^2): 19
ebanCount(10^3): 19
ebanCount(10^4): 79
ebanCount(10^5): 399
ebanCount(10^6): 399
ebanCount(10^7): 1599
ebanCount(10^8): 7999
ebanCount(10^9): 7999
ebanCount(10^10): 31999
ebanCount(10^11): 159999
ebanCount(10^12): 159999
ebanCount(10^13): 639999
ebanCount(10^14): 3199999
ebanCount(10^15): 3199999
ebanCount(10^16): 12799999
ebanCount(10^17): 63999999
ebanCount(10^18): 63999999
ebanCount(10^19): 255999999
ebanCount(10^20): 1279999999
ebanCount(10^21): 1279999999

Time: 189 microseconds and 491 nanoseconds```

## Perl

### Exhaustive search

A couple of 'e'-specific optimizations keep the running time reasonable.

`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 \$_ =~ // or /.\$/ or /..\$/ or /...\$/ or /.....\$/;        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;}`
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.

`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 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);    }}`
```============= 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?

`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)-1end 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 trueend function sequence s = {}for i=0 to 1000 do    if eban(i) then s &= i end ifend forprintf(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 ifend forprintf(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)`
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

`(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)))`
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

` # 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 inflectimport 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)) `

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)

Works with: Rakudo version 2018.12

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 :

Considering numbers up to 1021, as the task directions suggest.

`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 if nban(0, \$n);    [\+] @count;} #for < e o t tali subur tur ur cali i u > -> \$n { # All of themfor < 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: {[email protected]}\n[{@k».&comma}]\n\n" ~        "{\$n}-ban numbers between 1,000 & 4,000: {[email protected]}\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]);    }}`
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.

`/*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*/`
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
═════════════════════════════════════════════════════════════════════════════════════════════════════════
```

## Ruby

Translation of: C#
`def main    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 intv in intervals        (start, ending, display) = intv        if start == 2 then            print "eban numbers up to and including %d:\n" % [ending]        else            print "eban numbers between %d and %d (inclusive):\n" % [start, ending]        end         count = 0        for i in (start .. ending).step(2)            b = (i / 1000000000).floor            r = (i % 1000000000)            m = (r / 1000000).floor            r = (r % 1000000)            t = (r / 1000).floor            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 display then                                print ' ', i                            end                            count = count + 1                        end                    end                end            end        end        if display then            print "\n"        end        print "count = %d\n\n" % [count]    endend main()`
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```

## Scala

Translation of: Java
`object EbanNumbers {   class ERange(s: Int, e: Int, p: Boolean) {    val start: Int = s    val end: Int = e    val print: Boolean = p  }   def main(args: Array[String]): Unit = {    val rgs = List(      new ERange(2, 1000, true),      new ERange(1000, 4000, true),      new ERange(2, 10000, false),      new ERange(2, 100000, false),      new ERange(2, 1000000, false),      new ERange(2, 10000000, false),      new ERange(2, 100000000, false),      new ERange(2, 1000000000, false)    )    for (rg <- rgs) {      if (rg.start == 2) {        println(s"eban numbers up to an including \${rg.end}")      } else {        println(s"eban numbers between \${rg.start} and \${rg.end}")      }       var count = 0      for (i <- rg.start to rg.end) {        val b = i / 1000000000        var r = i % 1000000000        var m = r / 1000000        r = i % 1000000        var 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) {                  print(s"\$i ")                }                count += 1              }            }          }        }      }      if (rg.print) {        println()      }      println(s"count = \$count")      println()    }  }}`
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

eban numbers up to an including 100000000
count = 7999

eban numbers up to an including 1000000000
count = 7999```

## Tailspin

` templates isEban  def number: \$;  '\$;' -> \(<'(|)(0)*'> \$ !\) -> \$number !end isEban `

Alternatively, if regex is not your thing, we can do it numerically, which actually runs faster

` 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 `

Either version is called by the following code

` 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 `
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

Translation of: D
`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`
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

Translation of: Go
`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 == 2) {        System.print("eban numbers up to and including %(rg)")    } else {        System.print("eban numbers between %(rg) and %(rg) (inclusive):")    }    var count = 0    var i = rg    while (i <= rg) {        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) System.write("%(i) ")                        count = count + 1                    }                }            }        }        i = i + 2    }    if (rg) System.print()    System.print("count = %(count)\n")}`
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

Translation of: Go
`data 2, 100, truedata 1000, 4000, truedata 2, 1e4, falsedata 2, 1e5, falsedata 2, 1e6, falsedata 2, 1e7, falsedata 2, 1e8, falseREM data 2, 1e9, false  // it takes a lot of timedata 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`

## zkl

Translation of: Go
`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 }`
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
```