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# Disarium numbers

Disarium numbers
You are encouraged to solve this task according to the task description, using any language you may know.

A Disarium number is an integer where the sum of each digit raised to the power of its position in the number, is equal to the number.

E.G.

135 is a Disarium number:

11 + 32 + 53 == 1 + 9 + 125 == 135

There are a finite number of Disarium numbers.

• Find and display the first 18 Disarium numbers.

Stretch
• Find and display all 20 Disarium numbers.

## Arturo

`disarium?: function [x][    j: 0    psum: sum map digits x 'dig [        j: j + 1        dig ^ j    ]    return psum = x] cnt: 0i: 0while [cnt < 18][    if disarium? i [        print i        cnt: cnt + 1    ]    i: i + 1]`
Output:
```0
1
2
3
4
5
6
7
8
9
89
135
175
518
598
1306
1676
2427```

## AWK

` # syntax: GAWK -f DISARIUM_NUMBERS.AWKBEGIN {    stop = 19    printf("The first %d Disarium numbers:\n",stop)    while (count < stop) {      if (is_disarium(n)) {        printf("%d ",n)        count++      }      n++    }    printf("\n")    exit(0)}function is_disarium(n,  leng,sum,x) {    x = n    leng = length(n)    while (x != 0) {      sum += (x % 10) ^ leng      leng--      x = int(x/10)    }    return((sum == n) ? 1 : 0)} `
Output:
```The first 19 Disarium numbers:
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798
```

## BASIC

### BASIC256

`function isDisarium(n)	digitos = length(string(n))	suma = 0	x = n	while x <> 0		suma += (x % 10) ^ digitos		digitos -= 1		x = x \ 10	end while	if suma = n then return True else return Falseend function limite = 19cont = 0 : n = 0print "The first"; limite; " Disarium numbers are:"while cont < limite	if isDisarium(n) then		print n; " ";		cont += 1	endif	n += 1end whileend`
Output:
```Igual que la entrada de FreeBASIC.
```

### FreeBASIC

`#define limite 19 Function isDisarium(n As Integer) As Boolean    Dim As Integer digitos = Len(Str(n))    Dim As Integer suma = 0, x = n    While x <> 0        suma += (x Mod 10) ^ digitos        digitos -= 1        x \= 10    Wend    Return Iif(suma = n, True, False)End Function Dim As Integer cont = 0, n = 0, iPrint "The first"; limite; " Disarium numbers are:"Do While cont < limite    If isDisarium(n) Then        Print n; " ";        cont += 1    End If    n += 1LoopSleep`
Output:
```Igual que la entrada de Python.
```

### PureBasic

`Procedure isDisarium(n.i)  digitos.i = Len(Str(n))  suma.i = 0  x.i = n  While x <> 0    r.i = (x % 10)    suma + Pow(r, digitos)    digitos - 1    x / 10  Wend  If suma = n     ProcedureReturn #True   Else     ProcedureReturn #False  EndIfEndProcedure OpenConsole()limite.i = 19cont.i = 0n.i = 0PrintN("The first" + Str(limite) + " Disarium numbers are:")While cont < limite  If isDisarium(n)    Print(Str(n) + #TAB\$)    cont + 1  EndIf  n + 1WendInput()CloseConsole()`
Output:
```Igual que la entrada de FreeBASIC.
```

## C++

`#include <vector>#include <iostream>#include <cmath>#include <algorithm> std::vector<int> decompose( int n ) {   std::vector<int> digits ;   while ( n != 0 ) {      digits.push_back( n % 10 ) ;      n /= 10 ;   }   std::reverse( digits.begin( ) , digits.end( ) ) ;   return digits ;} bool isDisarium( int n ) {   std::vector<int> digits( decompose( n ) ) ;   int exposum = 0 ;   for ( int i = 1 ; i < digits.size( ) + 1 ; i++ ) {      exposum += static_cast<int>( std::pow(               static_cast<double>(*(digits.begin( ) + i - 1 )) ,               static_cast<double>(i) )) ;   }   return exposum == n ;} int main( ) {   std::vector<int> disariums ;   int current = 0 ;   while ( disariums.size( ) != 18 ){      if ( isDisarium( current ) )          disariums.push_back( current ) ;      current++ ;   }   for ( int d : disariums )       std::cout << d << " " ;   std::cout << std::endl ;   return 0 ;}`
Output:
`0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427`

## Factor

Works with: Factor version 0.99 2021-06-02
`USING: io kernel lists lists.lazy math.ranges math.text.utilsmath.vectors prettyprint sequences ; : disarium? ( n -- ? )    dup 1 digit-groups dup length 1 [a,b] v^ sum = ; : disarium ( -- list ) 0 lfrom [ disarium? ] lfilter ; 19 disarium ltake [ pprint bl ] leach nl`
Output:
```0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798
```

## Go

Translation of: Wren

A translation of Version 2.

Although Go has native unsigned 64 bit arithmetic, much quicker than I was expecting at a little under a minute.

`package main import (    "fmt"    "strconv") const DMAX = 20  // maximum digitsconst LIMIT = 20 // maximum number of disariums to find func main() {    // Pre-calculated exponential and power serials    EXP := make([][]uint64, 1+DMAX)    POW := make([][]uint64, 1+DMAX)     EXP = make([]uint64, 11)    EXP = make([]uint64, 11)    POW = make([]uint64, 11)    POW = make([]uint64, 11)    for i := uint64(1); i <= 10; i++ {        EXP[i] = i    }    for i := uint64(1); i <= 9; i++ {        POW[i] = i    }    POW = 9     for i := 2; i <= DMAX; i++ {        EXP[i] = make([]uint64, 11)        POW[i] = make([]uint64, 11)    }    for i := 1; i < DMAX; i++ {        for j := 0; j <= 9; j++ {            EXP[i+1][j] = EXP[i][j] * 10            POW[i+1][j] = POW[i][j] * uint64(j)        }        EXP[i+1] = EXP[i] * 10        POW[i+1] = POW[i] + POW[i+1]    }     // Digits of candidate and values of known low bits    DIGITS := make([]int, 1+DMAX) // Digits form    Exp := make([]uint64, 1+DMAX) // Number form    Pow := make([]uint64, 1+DMAX) // Powers form     var exp, pow, min, max uint64    start := 1    final := DMAX    count := 0    for digit := start; digit <= final; digit++ {        fmt.Println("# of digits:", digit)        level := 1        DIGITS = 0        for {            // Check limits derived from already known low bit values            // to find the most possible candidates            for 0 < level && level < digit {                // Reset path to try next if checking in level is done                if DIGITS[level] > 9 {                    DIGITS[level] = 0                    level--                    DIGITS[level]++                    continue                }                 // Update known low bit values                Exp[level] = Exp[level-1] + EXP[level][DIGITS[level]]                Pow[level] = Pow[level-1] + POW[digit+1-level][DIGITS[level]]                 // Max possible value                pow = Pow[level] + POW[digit-level]                 if pow < EXP[digit] { // Try next since upper limit is invalidly low                    DIGITS[level]++                    continue                }                 max = pow % EXP[level]                pow -= max                if max < Exp[level] {                    pow -= EXP[level]                }                max = pow + Exp[level]                 if max < EXP[digit] { // Try next since upper limit is invalidly low                    DIGITS[level]++                    continue                }                 // Min possible value                exp = Exp[level] + EXP[digit]                pow = Pow[level] + 1                 if exp > max || max < pow { // Try next since upper limit is invalidly low                    DIGITS[level]++                    continue                }                 if pow > exp {                    min = pow % EXP[level]                    pow -= min                    if min > Exp[level] {                        pow += EXP[level]                    }                    min = pow + Exp[level]                } else {                    min = exp                }                 // Check limits existence                if max < min {                    DIGITS[level]++ // Try next number since current limits invalid                } else {                    level++ // Go for further level checking since limits available                }            }             // All checking is done, escape from the main check loop            if level < 1 {                break            }             // Finally check last bit of the most possible candidates            // Update known low bit values            Exp[level] = Exp[level-1] + EXP[level][DIGITS[level]]            Pow[level] = Pow[level-1] + POW[digit+1-level][DIGITS[level]]             // Loop to check all last bits of candidates            for DIGITS[level] < 10 {                // Print out new Disarium number                if Exp[level] == Pow[level] {                    s := ""                    for i := DMAX; i > 0; i-- {                        s += fmt.Sprintf("%d", DIGITS[i])                    }                    n, _ := strconv.ParseUint(s, 10, 64)                    fmt.Println(n)                    count++                    if count == LIMIT {                        fmt.Println("\nFound the first", LIMIT, "Disarium numbers.")                        return                    }                }                 // Go to followed last bit candidate                DIGITS[level]++                Exp[level] += EXP[level]                Pow[level]++            }             // Reset to try next path            DIGITS[level] = 0            level--            DIGITS[level]++        }        fmt.Println()    }}`
Output:
```# of digits: 1
0
1
2
3
4
5
6
7
8
9

# of digits: 2
89

# of digits: 3
135
175
518
598

# of digits: 4
1306
1676
2427

# of digits: 5

# of digits: 6

# of digits: 7
2646798

# of digits: 8

# of digits: 9

# of digits: 10

# of digits: 11

# of digits: 12

# of digits: 13

# of digits: 14

# of digits: 15

# of digits: 16

# of digits: 17

# of digits: 18

# of digits: 19

# of digits: 20
12157692622039623539

Found the first 20 Disarium numbers.

real	0m57.430s
user	0m57.420s
sys	0m0.105s
```

`module Disarium    whereimport Data.Char ( digitToInt) isDisarium :: Int -> BoolisDisarium n = (sum \$ map (\(c , i ) -> (digitToInt c ) ^ i ) \$ zip ( show n ) [1 , 2 ..]) == n solution :: [Int]solution = take 18 \$ filter isDisarium [0, 1 ..] `
Output:
`[0,1,2,3,4,5,6,7,8,9,89,135,175,518,598,1306,1676,2427]`

## Julia

`isdisarium(n) = sum(last(p)^first(p) for p in enumerate(reverse(digits(n)))) == n function disariums(numberwanted)    n, ret = 0, Int[]    while length(ret) < numberwanted        isdisarium(n) && push!(ret, n)        n += 1    end    return retend println(disariums(19))@time disariums(19) `
Output:
```[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 89, 135, 175, 518, 598, 1306, 1676, 2427, 2646798]
0.555962 seconds (5.29 M allocations: 562.335 MiB, 10.79% gc time)
```

## Perl

`use strict;use warnings; my (\$n,@D) = (0, 0);while (++\$n) {    my(\$m,\$sum);    map { \$sum += \$_ ** ++\$m } split '', \$n;    push @D, \$n if \$n == \$sum;    last if 19 == @D;}print "@D\n";`
Output:
`0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798`

## Phix

```with javascript_semantics
constant limit = 19
integer count = 0, n = 0
printf(1,"The first 19 Disarium numbers are:\n")
while count<limit do
atom dsum = 0
string digits = sprintf("%d",n)
for i=1 to length(digits) do
dsum += power(digits[i]-'0',i)
end for
if dsum=n then
printf(1," %d",n)
count += 1
end if
n += 1
end while
```
Output:
```The first 19 Disarium numbers are:
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798
```

### stretch

```with javascript_semantics
-- translation of https://github.com/rgxgr/Disarium-Numbers/blob/master/Disarium.c
constant DMAX = iff(machine_bits()=64?20:7)

// Pre-calculated exponential & power serials
sequence exps = repeat(repeat(0,11),1+DMAX),
pows = repeat(repeat(0,11),1+DMAX)
exps[1..2] = {{0,0,0,0,0,0,0,0,0,0,1},{0,1,2,3,4,5,6,7,8,9,10}}
pows[1..2] = {{0,0,0,0,0,0,0,0,0,0,0},{0,1,2,3,4,5,6,7,8,9, 9}}
for i=2 to DMAX do
for j=1 to 10 do
exps[i+1][j] = exps[i][j]*10
pows[i+1][j] = pows[i][j]*(j-1)
end for
exps[i+1] = exps[i]*10
pows[i+1] = pows[i] + pows[i+1]
end for

// Digits of candidate and values of known low bits
sequence digits = repeat(0,1+DMAX), // Digits form
expl = repeat(0,1+DMAX),   // Number form
powl = repeat(0,1+DMAX)    // Powers form

printf(1,"") -- (exclude console setup from timings [if pw.exe])
atom expn, powr, minn, maxx, t0 = time(), t1 = t0+1, count = 0
for digit=2 to DMAX+1 do
printf(1,"Searching %d digits (started at %s):\n", {digit-1,elapsed(time()-t0)});
integer level = 2
digits = 0
while true do
// Check limits derived from already known low bit values
// to find the most possible candidates
while 1<level and level<digit do
// Reset path to try next if checking in level is done
integer dl = digits[level]+1
if dl>10 then
digits[level] = 0;
level -= 1
digits[level] += 1
else
// Update known low bit values
expl[level] = expl[level-1] + exps[level][dl]
powl[level] = powl[level-1] + pows[digit-level+2][dl]

// Max possible value
powr = powl[level] + pows[digit-level+1]

atom ed2 = exps[digit]
if powr<ed2 then  // Try next since upper limit is invalidly low
digits[level] += 1
else
atom el11 = exps[level],
el = expl[level]
maxx = remainder(powr,el11)
powr -= maxx
if maxx<el then
powr -= el11
end if
maxx = powr + el
if maxx<ed2 then  // Try next since upper limit is invalidly low
digits[level] += 1
else
// Min possible value
expn = el + ed2
powr = powl[level] + 1

if expn>maxx or maxx<powr then // Try next since upper limit is invalidly low
digits[level] += 1
else
if powr>expn then
minn = remainder(powr,el11)
powr -= minn
if minn>el then
powr += el11
end if
minn = powr + el
else
minn = expn
end if

// Check limits existence
if maxx<minn then
digits[level] +=1   // Try next number since current limits invalid
else
level +=1   // Go for further level checking since limits available
end if
end if
end if
end if
end if
if time()>t1 and platform()!=JS then
progress("working:%v... (%s)",{digits,elapsed(time()-t0)})
t1 = time()+1
end if
end while

// All checking is done, escape from the main check loop
if level<2 then exit end if

// Final check last bit of the most possible candidates
// Update known low bit values
integer dlx = digits[level]+1
expl[level] = expl[level-1] + exps[level][dlx];
powl[level] = powl[level-1] + pows[digit+1-level][dlx];

// Loop to check all last bit of candidates
while digits[level]<10 do
// Print out new disarium number
if expl[level] == powl[level] then
if platform()!=JS then progress("") end if
integer ld = max(trim_tail(digits,0,true),2)
printf(1,"%s\n",{reverse(join(apply(digits[2..ld],sprint),""))})
count += 1
end if

// Go to followed last bit candidate
digits[level] += 1
expl[level] += exps[level]
powl[level] += 1
end while

// Reset to try next path
digits[level] = 0;
level -= 1
digits[level] += 1
end while
if platform()!=JS then progress("") end if
end for
printf(1,"%d disarium numbers found (%s)\n",{count,elapsed(time()-t0)})
```
Output:
```Searching 1 digits (started at 0s):
0
1
2
3
4
5
6
7
8
9
Searching 2 digits (started at 0s):
89
Searching 3 digits (started at 0s):
135
175
518
598
Searching 4 digits (started at 0s):
1306
1676
2427
Searching 5 digits (started at 0.0s):
Searching 6 digits (started at 0.0s):
Searching 7 digits (started at 0.0s):
2646798
Searching 8 digits (started at 0.0s):
Searching 9 digits (started at 0.0s):
Searching 10 digits (started at 0.0s):
Searching 11 digits (started at 0.1s):
Searching 12 digits (started at 0.1s):
Searching 13 digits (started at 0.3s):
Searching 14 digits (started at 0.8s):
Searching 15 digits (started at 2.5s):
Searching 16 digits (started at 6.9s):
Searching 17 digits (started at 23.2s):
Searching 18 digits (started at 1 minute and 8s):
Searching 19 digits (started at 3 minutes and 35s):
Searching 20 digits (started at 10 minutes and 8s):
12157692622039623539
20 disarium numbers found (2 hours and 7s)
```

Takes about 48min to find the 20 digit number, then trundles away for over another hour. I think that technically it should also scan for 21 and 22 digit numbers to be absolutely sure there aren't any, but that certainly exceeds my patience.

## Picat

### Iterative approach

Translation of: Python
`main =>  Limit = 19,  D = [],  N = 0,  printf("The first %d Disarium numbers are:\n",Limit),    while (D.len < Limit)     if disarium_number(N) then       D := D ++ [N]     end,     N := N + 1,     if N mod 10_000_000 == 0 then       println(test=N)     end  end,  println(D). disarium_number(N) =>  Sum = 0,  Digits = N.to_string.len,  X = N,  while (X != 0, Sum <= N)    Sum := Sum + (X mod 10) ** Digits,    Digits := Digits - 1,    X := X div 10  end,  Sum == N.`
Output:
```The first 19 Disarium numbers are:
[0,1,2,3,4,5,6,7,8,9,89,135,175,518,598,1306,1676,2427,2646798]

2.905s```

### Constraint modelling

A faster approach is to use constraint modeling. It finds the first 19 Disarium numbers in 0.591s (vs 2.905s for the iterative approach).

Note that the domain of Picat's constraint variables is -2**56..2**56 (about 10**17) which means that this approach cannot be used to handle numbers of length 20.

The cp solver and sat solvers takes about the same time for finding the 7-digits number 2646798, but the sat solver is much faster for checking longer numbers; it took almost 9 minutes to prove that there are no Disarium numbers of length 8..17.

`import sat.% import cp. main =>  D = [],  Limit = 19,  Base = 10,  foreach(Len in 1..20, D.len < Limit)     Nums = disarium_number_cp(Len,Base),     if Nums.len > 0 then       foreach(Num in Nums)         B = to_radix_string(Num,Base),         D := D ++ [B]       end     end  end,  printf("The first %d Disarius numbers in base %d:\n",D.len, Base),  println(D[1..Limit]),  nl. % Find all Disarium of a certain lengthdisarium_number_cp(Len,Base) = findall(N,disarium_number_cp(Len,Base,N)).sort. % Find a Disarium number of a certain lengthdisarium_number_cp(Len,Base,N) =>  X = new_list(Len),  X :: 0..Base-1,  N :: Base**(Len-1)-1..Base**Len-1,  N #= sum([X[I]**I : I in 1..Len]),    to_num(X,Base,N), % convert X <=> N  solve(\$[],X++[N]). % Converts a number Num to/from a list of integer List given a base Baseto_num(List, Base, Num) =>   Len = length(List),   Num #= sum([List[I]*Base**(Len-I) : I in 1..Len]).`
Output:
```The first 19 Disarius numbers in base 10:
[0,1,2,3,4,5,6,7,8,9,89,135,175,518,598,1306,1676,2427,2646798]

0.591s
```

Finding the first 23 Disarium numbers in base 11 is easier:

```The first 23 Disarius numbers in base 11:
[0,1,2,3,4,5,6,7,8,9,A,25,36,9A,105,438,488,609,85A,86A,2077,40509,43789]

0.015s```

(And finding the first 36 Disarium numbers in base 36 is even easier: 0..Z.)

## Python

`#!/usr/bin/python def isDisarium(n):    digitos = len(str(n))    suma = 0    x = n    while x != 0:        suma += (x % 10) ** digitos        digitos -= 1        x //= 10    if suma == n:        return True    else:        return False if __name__ == '__main__':    limite = 19    cont = 0    n = 0    print("The first",limite,"Disarium numbers are:")    while cont < limite:        if isDisarium(n):            print(n, end = " ")            cont += 1        n += 1`
Output:
```The first 19 Disarium numbers are:
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798```

## Raku

Not an efficient algorithm. First 18 in less than 1/4 second. 19th in around 45 seconds. Pretty much unusable for the 20th.

`my \$disarium = (^∞).hyper.map: { \$_ if \$_ == sum .polymod(10 xx *).reverse Z** 1..* }; put \$disarium[^18];put \$disarium;`
Output:
```0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427
2646798```

## Sidef

`func is_disarium(n) {    n.digits.flip.sum_kv{|k,d| d**(k+1) } == n} say 18.by(is_disarium)`
Output:
```[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 89, 135, 175, 518, 598, 1306, 1676, 2427]
```

## Wren

Library: Wren-math

### Version 1 (Brute force)

This version finds the first 19 Disarium numbers in 3.35 seconds though, clearly, finding the 20th is out of the question with this approach.

As a possible optimization, I tried caching all possible digit powers but there was no perceptible difference in running time for numbers up to 7 digits long.

`import "./math" for Int var limit = 19var count = 0var disarium = []var n = 0while (count < limit) {    var sum = 0    var digits = Int.digits(n)    for (i in 0...digits.count) sum = sum + digits[i].pow(i+1)    if (sum == n) {        disarium.add(n)        count = count + 1    }    n = n + 1}System.print("The first 19 Disarium numbers are:")System.print(disarium)`
Output:
```The first 19 Disarium numbers are:
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 89, 135, 175, 518, 598, 1306, 1676, 2427, 2646798]
```

### Version 2 (Much faster)

This is a translation of the C code referred to in the Phix entry and finds the first 19 Disarium numbers in 0.012 seconds.

Efficient though this method is, unfortunately finding the 20th is still out of reasonable reach for Wren. If we let this run until 15 digit numbers have been examined (the most that 53 bit integer math can accurately manage), then the time taken rises to 19 seconds - roughly 3 times slower than Phix.

However, we need 64 bit integer arithmetic to get up to 20 digits and this requires the use of Wren-long which (as it's written entirely in Wren, not C) needs about 7 times longer (2 minutes 16 seconds) to even reach 15 digits. Using BigInt or GMP would be even slower.

So, if the Phix example requires 48 minutes to find the 20th number, it would probably take Wren the best part of a day to do the same which is far longer than I have patience for.

`var DMAX  = 7  // maxmimum digitsvar LIMIT = 19 // maximum number of Disariums to find // Pre-calculated exponential and power serialsvar EXP = List.filled(1 + DMAX, null)var POW = List.filled(1 + DMAX, null)EXP = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1]EXP = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]POW = List.filled(11, 0)POW = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 9]for (i in 2..DMAX) {    EXP[i] = List.filled(11, 0)    POW[i] = List.filled(11, 0)}    for (i in 1...DMAX) {    for (j in 0..9) {        EXP[i+1][j] = EXP[i][j] * 10        POW[i+1][j] = POW[i][j] * j    }    EXP[i+1] = EXP[i] * 10    POW[i+1] = POW[i] + POW[i+1]} // Digits of candidate and values of known low bitsvar DIGITS = List.filled(1 + DMAX, 0)  // Digits formvar Exp    = List.filled(1 + DMAX, 0)  // Number formvar Pow    = List.filled(1 + DMAX, 0)  // Powers form var expvar powvar minvar maxvar start = 1var final = DMAXvar count = 0for (digit in start..final) {    System.print("# of digits: %(digit)")    var level = 1    DIGITS = 0    while (true) {        // Check limits derived from already known low bit values        // to find the most possible candidates        while (0 < level && level < digit) {            // Reset path to try next if checking in level is done            if (DIGITS[level] > 9) {                DIGITS[level] = 0                level = level - 1                DIGITS[level] = DIGITS[level] + 1                continue            }             // Update known low bit values            Exp[level] = Exp[level - 1] + EXP[level][DIGITS[level]]            Pow[level] = Pow[level - 1] + POW[digit + 1 - level][DIGITS[level]]             // Max possible value            pow = Pow[level] + POW[digit - level]             if (pow < EXP[digit]) {  // Try next since upper limit is invalidly low                DIGITS[level] = DIGITS[level] + 1                continue            }             max = pow % EXP[level]            pow = pow - max            if (max < Exp[level]) pow = pow - EXP[level]            max = pow + Exp[level]             if (max < EXP[digit]) {  // Try next since upper limit is invalidly low                DIGITS[level] = DIGITS[level] + 1                continue            }             // Min possible value            exp = Exp[level] + EXP[digit]            pow = Pow[level] + 1             if (exp > max || max < pow) { // Try next since upper limit is invalidly low                DIGITS[level] = DIGITS[level] + 1                continue            }             if (pow > exp ) {                min = pow % EXP[level]                pow = pow - min                 if (min > Exp[level]) {                    pow = pow + EXP[level]                }                min = pow + Exp[level]            } else {                min = exp            }             // Check limits existence            if (max < min) {                DIGITS[level] = DIGITS[level] + 1  // Try next number since current limits invalid            } else {                level= level + 1  // Go for further level checking since limits available            }        }         // All checking is done, escape from the main check loop        if (level < 1) break         // Finally check last bit of the most possible candidates        // Update known low bit values        Exp[level] = Exp[level - 1] + EXP[level][DIGITS[level]]        Pow[level] = Pow[level - 1] + POW[digit + 1 - level][DIGITS[level]]         // Loop to check all last bits of candidates        while (DIGITS[level] < 10) {            // Print out new Disarium number            if (Exp[level] == Pow[level]) {                var s = ""                for (i in DMAX...0) s = s + DIGITS[i].toString                System.print(Num.fromString(s))                count = count + 1                if (count == LIMIT) {                    System.print("\nFound the first %(LIMIT) Disarium numbers.")                    return                }            }             // Go to followed last bit candidate            DIGITS[level] = DIGITS[level] + 1            Exp[level] = Exp[level] + EXP[level]            Pow[level] = Pow[level] + 1        }         // Reset to try next path        DIGITS[level] = 0        level = level - 1        DIGITS[level] = DIGITS[level] + 1    }    System.print()}`
Output:
```# of digits: 1
0
1
2
3
4
5
6
7
8
9

# of digits: 2
89

# of digits: 3
135
175
518
598

# of digits: 4
1306
1676
2427

# of digits: 5

# of digits: 6

# of digits: 7
2646798

Found the first 19 Disarium numbers.

real	0m0.012s
user	0m0.008s
sys	0m0.004s
```

### Version 3 (Embedded)

Library: Wren-i64

An initial outing for the above module which aims to speed up 64 bit arithmetic in Wren by wrapping the corresponding C99 fixed size types.

Early indications are that this is at least 4 times faster than Wren-long as it can search up to 15 digits in about 29 seconds which in turn is about 4 times slower than the Phix entry.

This suggested that the 20th Disarium number would be found in around 3.5 hours so I thought I'd have a go. However, after taking consistently 4 times longer than Phix to search each digit length up to 19, I was pleasantly surprised when the 20th number popped up after only 81 minutes!

I haven't bothered to search all 20 digits numbers up to the unsigned 64 limit as this would take far longer and, of course, be fruitless in any case.

`import "./i64" for U64 var DMAX  = 20 // maxmimum digitsvar LIMIT = 20 // maximum number of disariums to find // Pre-calculated exponential and power serialsvar EXP = List.filled(1 + DMAX, null)var POW = List.filled(1 + DMAX, null)EXP  = List.filled(11, null)EXP  = List.filled(11, null)POW  = List.filled(11, null)POW  = List.filled(11, null)for (i in 0..9) EXP[i] = U64.zeroEXP = U64.one for (i in 0..10) EXP[i] = U64.from(i) for (i in 0..10) POW[i] = U64.zero for (i in 0..9) POW[i] = U64.from(i)POW = U64.from(9) for (i in 2..DMAX) {    EXP[i] = List.filled(11, null)    POW[i] = List.filled(11, null)    for (j in 0..10) {        EXP[i][j] = U64.zero        POW[i][j] = U64.zero    }}for (i in 1...DMAX) {    for (j in 0..9) {        EXP[i+1][j] = EXP[i][j] * 10        POW[i+1][j] = POW[i][j] * j    }    EXP[i+1] = EXP[i] * 10    POW[i+1] = POW[i] + POW[i+1]} // Digits of candidate and values of known low bitsvar DIGITS = List.filled(1 + DMAX, 0)  // Digits formvar Exp    = List.filled(1 + DMAX, null)  // Number formvar Pow    = List.filled(1 + DMAX, null)  // Powers formfor (i in 0..DMAX) {    Exp[i] = U64.zero    Pow[i] = U64.zero} var exp = U64.new()var pow = U64.new()var min = U64.new()var max = U64.new()var start = 1var final = DMAXvar count = 0for (digit in start..final) {    System.print("# of digits: %(digit)")    var level = 1    DIGITS = 0    while (true) {        // Check limits derived from already known low bit values        // to find the most possible candidates        while (0 < level && level < digit) {            // Reset path to try next if checking in level is done            if (DIGITS[level] > 9) {                DIGITS[level] = 0                level = level - 1                DIGITS[level] = DIGITS[level] + 1                continue            }             // Update known low bit values            Exp[level].add(Exp[level - 1], EXP[level][DIGITS[level]])            Pow[level].add(Pow[level - 1], POW[digit + 1 - level][DIGITS[level]])             // Max possible value            pow.add(Pow[level], POW[digit - level])             if (pow < EXP[digit]) {  // Try next since upper limit is invalidly low                DIGITS[level] = DIGITS[level] + 1                continue            }             max.rem(pow, EXP[level])            pow.sub(max)            if (max < Exp[level]) pow.sub(EXP[level])            max.add(pow, Exp[level])             if (max < EXP[digit]) {  // Try next since upper limit is invalidly low                DIGITS[level] = DIGITS[level] + 1                continue            }             // Min possible value            exp.add(Exp[level], EXP[digit])            pow.add(Pow[level], 1)             if (exp > max || max < pow) { // Try next since upper limit is invalidly low                DIGITS[level] = DIGITS[level] + 1                continue            }             if (pow > exp ) {                min.rem(pow, EXP[level])                pow.sub(min)                if (min > Exp[level]) {                    pow.add(EXP[level])                }                min.add(pow, Exp[level])            } else {                min.set(exp)            }             // Check limits existence            if (max < min) {                DIGITS[level] = DIGITS[level] + 1  // Try next number since current limits invalid            } else {                level = level + 1  // Go for further level checking since limits available            }        }         // All checking is done, escape from the main check loop        if (level < 1) break         // Final check last bit of the most possible candidates        // Update known low bit values        Exp[level].add(Exp[level - 1], EXP[level][DIGITS[level]])        Pow[level].add(Pow[level - 1], POW[digit + 1 - level][DIGITS[level]])         // Loop to check all last bit of candidates        while (DIGITS[level] < 10) {            // Print out new disarium number            if (Exp[level] == Pow[level]) {                var s = ""                for (i in DMAX...0) s = s + DIGITS[i].toString                s = s.trimStart("0")                if (s == "") s = "0"                System.print(s)                count = count + 1                if (count == LIMIT) {                    if (LIMIT < 20) {                        System.print("\nFound the first %(LIMIT) Disarium numbers.")                    } else {                        System.print("\nFound all 20 Disarium numbers.")                    }                    return                }            }             // Go to followed last bit candidate            DIGITS[level] = DIGITS[level] + 1            Exp[level].add(Exp[level], EXP[level])            Pow[level].inc        }         // Reset to try next path        DIGITS[level] = 0        level = level - 1        DIGITS[level] = DIGITS[level] + 1    }    System.print()}`
Output:
```# of digits: 1
0
1
2
3
4
5
6
7
8
9

# of digits: 2
89

# of digits: 3
135
175
518
598

# of digits: 4
1306
1676
2427

# of digits: 5

# of digits: 6

# of digits: 7
2646798

# of digits: 8

# of digits: 9

# of digits: 10

# of digits: 11

# of digits: 12

# of digits: 13

# of digits: 14

# of digits: 15

# of digits: 16

# of digits: 17

# of digits: 18

# of digits: 19

# of digits: 20
12157692622039623539

Found all 20 Disarium numbers.

real	81m16.365s
user	81m16.181s
sys	0m0.016s
```

## XPL0

1.35 seconds on Pi4.

`func Disarium(N);       \Return 'true' if N is a Disarium numberint  N, N0, D(10), A(10), I, J, Sum;[N0:= N;for J:= 0 to 10-1 do A(J):= 1;I:= 0;repeat  N:= N/10;        D(I):= rem(0);        I:= I+1;        for J:= 0 to I-1 do            A(J):= A(J) * D(J);until   N = 0;Sum:= 0;for J:= 0 to I-1 do    Sum:= Sum + A(J);return Sum = N0;]; int Cnt, N;[Cnt:= 0;  N:= 0;loop    [if Disarium(N) then            [IntOut(0, N);  ChOut(0, ^ );            Cnt:= Cnt+1;            if Cnt >= 19 then quit;            ];        N:= N+1;        ];]`
Output:
```0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798
```