Eban numbers

From Rosetta Code
Task
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



11l[edit]

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[edit]

(*
    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 output
end 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 output
end 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, 66
21 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, 4000
79 up to and including 10,000
399 up to and including 100,000
399 up to and including 1,000,000
1599 up to and including 10,000,000
16 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[edit]

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[edit]

# 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)
}
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[edit]

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#[edit]

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++[edit]

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

CLU[edit]

eban = cluster is numbers
    rep = null
    
    % Next valid E-ban number in the range [0..999]
    next = proc (n: int) returns (int) signals (no_more)
        if n>=66 then signal no_more end
        if n<0 then return(0) end
        
        % advance low digit (two four six)
        if     (n//10=0) then n := (n/10)*10 + 2
        elseif (n//10=2) then n := (n/10)*10 + 4 
        elseif (n//10=4) then n := (n/10)*10 + 6 
        elseif (n//10=6) then n := (n/10)*10 + 10
        end 
        
        % this might put us into tEn, twEnty which are invalid
        if     (n//100/10 =  1) then n := n + 20   % ten + 20    = 30 
        elseif (n//100/10 =  2) then n := n + 10   % twEnty + 10 = 30
        end
        return(n)
    end next
        
    % Yield all valid E_ban numbers
    % (sextillion and upwards aren't handled)
    numbers = iter () yields (int)
        parts: array[int] := array[int]$[0: 2]
        
        while true do
            num: int := 0
            for m: int in array[int]$indexes(parts) do
                num := num + parts[m] * 1000 ** m
            end
            yield(num)
            
            for m: int in array[int]$indexes(parts) do
                begin
                    parts[m] := next(parts[m])
                    break
                end except when no_more:
                    parts[m] := 0
                end
            end
            
            if array[int]$top(parts) = 0 then
                array[int]$addh(parts,2)
            end
       end
    end numbers            
end eban

start_up = proc ()
    maxmagn = 1000000000000
    po: stream := stream$primary_output()

    count: int := 0
    upto_1k: int := 0
    between_1k_4k: int := 0

    disp_1k: bool := false
    disp_4k: bool := false
    
    nextmagn: int := 10000
    
    for i: int in eban$numbers() do
        while i>nextmagn do
            stream$putl(po, int$unparse(count) 
                        || " eban numbers <= " 
                        || int$unparse(nextmagn))
            nextmagn := nextmagn * 10
        end
        
        count := count + 1

        if i<1000 then upto_1k := upto_1k + 1
        elseif i<=4000 then between_1k_4k := between_1k_4k + 1
        end 
        
        if i>1000 & ~disp_1k then
            disp_1k := true
            stream$putl(po, "\n" || int$unparse(upto_1k) 
                        || " eban numbers <= 1000\n")
        end
        
        if i<=4000 then stream$putright(po, int$unparse(i), 5) end
        
        if i>4000 & ~disp_4k then
            disp_4k := true
            stream$putl(po, "\n" || int$unparse(between_1k_4k)
                        || " eban numbers 1000 <= x <= 4000\n")
        end
        
        if nextmagn>maxmagn then break end
    end
end start_up
Output:
    2    4    6   30   32   34   36   40   42   44   46   50   52   54   56   60   62   64   66
19 eban numbers <= 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 1000 <= x <= 4000

79 eban numbers <= 10000
399 eban numbers <= 100000
399 eban numbers <= 1000000
1599 eban numbers <= 10000000
7999 eban numbers <= 100000000
7999 eban numbers <= 1000000000
31999 eban numbers <= 10000000000
159999 eban numbers <= 100000000000
159999 eban numbers <= 1000000000000

D[edit]

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[edit]

Translation of: Julia
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
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[edit]

' 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
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[edit]

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

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

In this page you can see the program(s) related to this task and their results.

Go[edit]

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[edit]

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

Haskell[edit]

Translation of: Julia
{-# 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
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[edit]

Filter =: (#~`)(`:6)

itemAmend =: (29&< *. <&67)`(,: 10&|)}
iseban =: [: *./ 0 2 4 6 e.~ [: itemAmend [: |: (4#1000)&#:


   (;~ #) iseban Filter >: i. 1000
┌──┬─────────────────────────────────────────────────────┐
192 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
┌──┬────────────────────────────────────────────────────────────────────────────────────────────────────────┐
212000 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

Java[edit]

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[edit]

Adapted from Julia and Wren

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 remainder
def 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[0] == 2
    then "eban numbers up to and including \($rg[1]):"
    else "eban numbers between \($rg[0]) and \($rg[1]) (inclusive):"
    end,
    ( foreach (range( $rg[0]; 1 + $rg[1]; 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 ($rg[2]) then .emit=$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[edit]

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[edit]

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[edit]

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()
    end
end

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

Mathematica/Wolfram Language[edit]

ClearAll[ZeroTwoFourSixQ, EbanNumbers]
ZeroTwoFourSixQ[n_Integer] := (n == 0 || n == 2 || n == 4 || n == 6)
EbanNumbers[min_, max_, show : (False | True)] := 
 Module[{counter, output, i, b, r, t, m},
  counter = 0;
  output = "";
  i = min;
  While[(i += 2) <= max,
   {b, r} = QuotientRemainder[i, 10^9];
   {m, r} = QuotientRemainder[r, 10^6];
   {t, r} = QuotientRemainder[r, 10^3];
   If[30 <= m <= 66,
    m = Mod[m, 10];
    ];
   If[30 <= t <= 66,
    t = Mod[t, 10];
    ];
   If[30 <= r <= 66,
    r = Mod[r, 10];
    ];
   If[ZeroTwoFourSixQ[b] && ZeroTwoFourSixQ[m] && ZeroTwoFourSixQ[t] &&
      ZeroTwoFourSixQ[r],
    counter++;
    If[show,
     output = output <> ToString[i] <> " ";
     ]
    ]
   ];
  Print[min, "-", max, ": ", output, " count = ", counter]
  ]
EbanNumbers[0, 1000, True]
EbanNumbers[1000, 4000, True]
EbanNumbers[0, 10^4, False]
EbanNumbers[0, 10^5, False]
EbanNumbers[0, 10^6, False]
EbanNumbers[0, 10^7, False]
Output:
0-1000: 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
0-10000:  count = 79
0-100000:  count = 399
0-1000000:  count = 399
0-10000000:  count = 1599

Nim[edit]

Exhaustive search[edit]

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 = 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}"
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[edit]

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 times
let 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[edit]

Exhaustive search[edit]

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 $_ =~ /[789]/ or /[12].$/ 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;
}
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[edit]

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[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);
    }
}
============= 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[edit]

Why count when you can calculate?

with javascript_semantics
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 = append(s,sprint(i)) end if
end for
printf(1,"eban to 1000 : %s\n\n",{join(shorten(s,"items",8))})
s = {}
for i=1000 to 4000 do
    if eban(i) then s = append(s,sprint(i)) end if
end for
printf(1,"eban 1000..4000 : %s\n\n",{join(shorten(s,"items",4))})
 
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 ... 50 52 54 56 60 62 64 66  (19 items)

eban 1000..4000 : 2000 2002 2004 2006 ... 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
"0s"

PicoLisp[edit]

(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[edit]

# 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))

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

Quackery[edit]

  [ 20 /mod
    4 /mod
    [ table 0 2 4 6 ]
    swap
    [ table 0 30 40 50 60 ]
    +
    over 0 = iff nip done
    swap recurse 1000 * + ] is n->eban ( n --> n )

  [] 1
  [ dup n->eban
    dup 10000001 < while
    swap dip join
    1+ again ]
  2drop
  dup dup
  say "eban numbers up to and including 1000:"
  cr
  findwith [ 1000 > ] [ ]
  split drop dup echo cr
  say "count: " size echo
  cr cr
  say "eban numbers between 1000 and 4000 (inclusive):"
  cr
  dup dup
  findwith [ 999 > ] [ ]
  split nip
  dup
  findwith [ 4001 > ] [ ]
  split drop dup echo cr
  say "count: " size echo
  cr cr
  say "number of eban numbers up to and including 10000: "
  dup findwith [ 10001 > ] [ ]
  echo
  cr cr
  say "number of eban numbers up to and including 100000: "
  dup findwith [ 100001 > ] [ ]
  echo
  cr cr
  say "number of eban numbers up to and including 1000000: "
  dup findwith [ 1000001 > ] [ ]
  echo
  cr cr
  say "number of eban numbers up to and including 10000000: "
  findwith [ 10000001 > ] [ ]
  echo
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

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

Raku[edit]

(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[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]);
    }
}
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[edit]

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[edit]

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]
    end
end

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[edit]

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[edit]

templates isEban
  def number: $;
  '$;' -> \(<'([246]|[3456][0246])(0[03456][0246])*'> $ !\) -> $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[edit]

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

V (Vlang)[edit]

Translation of: Go
struct Range {
    start i64
    end i64
    print bool
}
 
fn main() {
    rgs := [
        Range{2, 1000, true},
        Range{1000, 4000, true},
        Range{2, 10000, false},
        Range{2, 100000, false},
        Range{2, 1000000, false},
        Range{2, 10000000, false},
        Range{2, 100000000, false},
        Range{2, 1000000000, false}
    ]
    for rg in rgs {
        if rg.start == 2 {
            println("eban numbers up to and including $rg.end:")
        } else {
            println("eban numbers between $rg.start and $rg.end (inclusive):")
        }
        mut count := 0
        for i := rg.start; i <= rg.end; i += 2 {
            b := i / 1000000000
            mut r := i % 1000000000
            mut m := r / 1000000
            r = i % 1000000
            mut t := r / 1000
            r %= 1000
            if m >= 30 && m <= 66 {
                m %= 10
            }
            if t >= 30 && t <= 66 {
                t %= 10
            }
            if r >= 30 && r <= 66 {
                r %= 10
            }
            if b == 0 || b == 2 || b == 4 || b == 6 {            
                if m == 0 || m == 2 || m == 4 || m == 6 {
                    if t == 0 || t == 2 || t == 4 || t == 6 {
                        if r == 0 || r == 2 || r == 4 || r == 6 {
                            if rg.print {
                                print("$i ")
                            }
                            count++
                        }
                    }
                }
            }
        }
        if rg.print {
            println('')
        }
        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 

Wren[edit]

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[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")
}
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[edit]

Translation of: Go
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

zkl[edit]

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