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



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

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.

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 

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
 

Julia[edit]

I changed your program but not the output. You had used the common cut-n-paste error repeating the 10000 println("eban numbers up to and including 10000:

 
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 10000: 399
eban numbers up to and including 10000: 399
eban numbers up to and including 10000: 1599
eban numbers up to and including 10000: 7999
eban numbers up to and including 10000: 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.

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 Perl 6. 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

Perl 6[edit]

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: {[email protected]}\n[{@k».&comma}]\n\n" ~
"{$n}-ban numbers between 1,000 & 4,000: {[email protected]}\n[{@j».&comma}]\n" ~
"\nCounts of {$n}-ban numbers up to {cardinal 10**$upto}"
;
 
my $s = max (1..$upto).map: { (10**$_).&cardinal.chars };
@counts.unshift: @bans.first: * > 10, :k;
for ^$upto -> $c {
printf "Up to and including %{$s}s: %s\n", cardinal(10**($c+1)), comma(@counts[$c]);
}
}
Output:
============= e-ban: =============
e-ban numbers up to 1000: 19
[2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66]

e-ban numbers between 1,000 & 4,000: 21
[2,000 2,002 2,004 2,006 2,030 2,032 2,034 2,036 2,040 2,042 2,044 2,046 2,050 2,052 2,054 2,056 2,060 2,062 2,064 2,066 4,000]

Counts of e-ban numbers up to one sextillion
Up to and including                     ten: 3
Up to and including             one hundred: 19
Up to and including            one thousand: 19
Up to and including            ten thousand: 79
Up to and including    one hundred thousand: 399
Up to and including             one million: 399
Up to and including             ten million: 1,599
Up to and including     one hundred million: 7,999
Up to and including             one billion: 7,999
Up to and including             ten billion: 31,999
Up to and including     one hundred billion: 159,999
Up to and including            one trillion: 159,999
Up to and including            ten trillion: 639,999
Up to and including    one hundred trillion: 3,199,999
Up to and including         one quadrillion: 3,199,999
Up to and including         ten quadrillion: 12,799,999
Up to and including one hundred quadrillion: 63,999,999
Up to and including         one quintillion: 63,999,999
Up to and including         ten quintillion: 255,999,999
Up to and including one hundred quintillion: 1,279,999,999
Up to and including          one sextillion: 1,279,999,999

============= t-ban: =============
t-ban numbers up to 1000: 56
[0 1 4 5 6 7 9 11 100 101 104 105 106 107 109 111 400 401 404 405 406 407 409 411 500 501 504 505 506 507 509 511 600 601 604 605 606 607 609 611 700 701 704 705 706 707 709 711 900 901 904 905 906 907 909 911]

t-ban numbers between 1,000 & 4,000: 0
[]

Counts of t-ban numbers up to one sextillion
Up to and including                     ten: 7
Up to and including             one hundred: 9
Up to and including            one thousand: 56
Up to and including            ten thousand: 56
Up to and including    one hundred thousand: 56
Up to and including             one million: 57
Up to and including             ten million: 392
Up to and including     one hundred million: 785
Up to and including             one billion: 5,489
Up to and including             ten billion: 38,416
Up to and including     one hundred billion: 76,833
Up to and including            one trillion: 537,824
Up to and including            ten trillion: 537,824
Up to and including    one hundred trillion: 537,824
Up to and including         one quadrillion: 537,825
Up to and including         ten quadrillion: 3,764,768
Up to and including one hundred quadrillion: 7,529,537
Up to and including         one quintillion: 52,706,752
Up to and including         ten quintillion: 52,706,752
Up to and including one hundred quintillion: 52,706,752
Up to and including          one sextillion: 52,706,752

============= subur-ban: =============
subur-ban numbers up to 1000: 35
[1 2 5 8 9 10 11 12 15 18 19 20 21 22 25 28 29 50 51 52 55 58 59 80 81 82 85 88 89 90 91 92 95 98 99]

subur-ban numbers between 1,000 & 4,000: 0
[]

Counts of subur-ban numbers up to one sextillion
Up to and including                     ten: 6
Up to and including             one hundred: 35
Up to and including            one thousand: 35
Up to and including            ten thousand: 35
Up to and including    one hundred thousand: 35
Up to and including             one million: 36
Up to and including             ten million: 216
Up to and including     one hundred million: 2,375
Up to and including             one billion: 2,375
Up to and including             ten billion: 2,375
Up to and including     one hundred billion: 2,375
Up to and including            one trillion: 2,375
Up to and including            ten trillion: 2,375
Up to and including    one hundred trillion: 2,375
Up to and including         one quadrillion: 2,375
Up to and including         ten quadrillion: 2,375
Up to and including one hundred quadrillion: 2,375
Up to and including         one quintillion: 2,375
Up to and including         ten quintillion: 2,375
Up to and including one hundred quintillion: 2,375
Up to and including          one sextillion: 2,375

Note that the limit to one sextillion is somewhat arbitrary and is just to match the task parameters.

This will quite happily count *-bans up to one hundred centillion. (10305) It takes longer, but still on the order of seconds, not minutes.

Counts of e-ban numbers up to one hundred centillion
 ...
Up to and including one hundred centillion: 35,184,372,088,831,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999

Phix[edit]

Why count when you can calculate?

function count_eban(integer p10)
-- returns the count of eban numbers 1..power(10,p10)
integer n = p10-floor(p10/3),
p5 = floor(n/2),
p4 = floor((n+1)/2)
return power(5,p5)*power(4,p4)-1
end function
 
function eban(integer n)
-- returns true if n is an eban number (only fully tested to 10e9)
if n=0 then return false end if
while n do
integer thou = remainder(n,1000)
if floor(thou/100)!=0 then return false end if
if not find(floor(thou/10),{0,3,4,5,6}) then return false end if
if not find(remainder(thou,10),{0,2,4,6}) then return false end if
n = floor(n/1000)
end while
return true
end function
 
sequence s = {}
for i=0 to 1000 do
if eban(i) then s &= i end if
end for
printf(1,"eban to 1000 : %v (%d items)\n",{s,length(s)})
s = {}
for i=1000 to 4000 do
if eban(i) then s &= i end if
end for
printf(1,"eban 1000..4000 : %v (%d items)\n\n",{s,length(s)})
 
atom t0 = time()
for i=0 to 21 do
printf(1,"count_eban(10^%d) : %,d\n",{i,count_eban(i)})
end for
?elapsed(time()-t0)
Output:
eban to 1000 : {2,4,6,30,32,34,36,40,42,44,46,50,52,54,56,60,62,64,66} (19 items)
eban 1000..4000 : {2000,2002,2004,2006,2030,2032,2034,2036,2040,2042,2044,2046,2050,2052,2054,2056,2060,2062,2064,2066,4000} (21 items)

count_eban(10^0) : 0
count_eban(10^1) : 3
count_eban(10^2) : 19
count_eban(10^3) : 19
count_eban(10^4) : 79
count_eban(10^5) : 399
count_eban(10^6) : 399
count_eban(10^7) : 1,599
count_eban(10^8) : 7,999
count_eban(10^9) : 7,999
count_eban(10^10) : 31,999
count_eban(10^11) : 159,999
count_eban(10^12) : 159,999
count_eban(10^13) : 639,999
count_eban(10^14) : 3,199,999
count_eban(10^15) : 3,199,999
count_eban(10^16) : 12,799,999
count_eban(10^17) : 63,999,999
count_eban(10^18) : 63,999,999
count_eban(10^19) : 255,999,999
count_eban(10^20) : 1,279,999,999
count_eban(10^21) : 1,279,999,999
"0.0s"

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

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
═════════════════════════════════════════════════════════════════════════════════════════════════════════

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..> $!) -> #
<0> $number !
<?($ mod 1000 <0|2|4|6|30..66?($ mod 10 <0|2|4|6>)>)> $ / 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

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