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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.

## ALGOL 68

Finds the first 19 Disarium numbers - to find the 20th would require a lot more time and also the table of digit powers would need to be increased to at least 20th powers (and 64 bit integers would be required).

BEGIN # find some Disarium numbers - numbers whose digit position-power sums  #
# are equal to the number, e.g. 135 = 1^1 + 3^2 + 5^3 #
# compute the nth powers of 0-9 #
[ 1 : 9, 0 : 9 ]INT power;
FOR d FROM 0 TO 9 DO power[ 1, d ] := d OD;
FOR n FROM 2 TO 9 DO
power[ n, 0 ] := 0;
FOR d TO 9 DO
power[ n, d ] := power[ n - 1, d ] * d
OD
OD;
# print the first few Disarium numbers #
INT max disarium = 19;
INT count := 0;
INT power of ten := 10;
INT length := 1;
FOR n FROM 0 WHILE count < max disarium DO
IF n = power of ten THEN
# the number of digits just increased #
power of ten *:= 10;
length +:= 1
FI;
# form the digit power sum #
INT v := n;
INT dps := 0;
FOR p FROM length BY -1 TO 1 DO
dps +:= power[ p, v MOD 10 ];
v OVERAB 10
OD;
IF dps = n THEN
# n is Disarium #
count +:= 1;
print( ( " ", whole( n, 0 ) ) )
FI
OD
END
Output:
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798

## ALGOL W

Translation of: ALGOL 68
begin % find some Disarium numbers - numbers whose digit position-power sums  %
% are equal to the number, e.g. 135 = 1^1 + 3^2 + 5^3  %
integer array power ( 1 :: 9, 0 :: 9 );
integer MAX_DISARIUM;
integer count, powerOfTen, length, n;
% compute the nth powers of 0-9  %
for d := 0 until 9 do power( 1, d ) := d;
for n := 2 until 9 do begin
power( n, 0 ) := 0;
for d := 1 until 9 do power( n, d ) := power( n - 1, d ) * d
end for_n;
% print the first few Disarium numbers  %
MAX_DISARIUM := 19;
count := 0;
powerOfTen := 10;
length := 1;
n := 0;
while count < MAX_DISARIUM do begin
integer v, dps;
if n = powerOfTen then begin
% the number of digits just increased  %
powerOfTen := powerOfTen * 10;
length := length + 1
end if_m_eq_powerOfTen ;
% form the digit power sum  %
v := n;
dps := 0;
for p := length step -1 until 1 do begin
dps := dps + power( p, v rem 10 );
v := v div 10;
end FOR_P;
if dps = n then begin
% n is Disarium  %
count := count + 1;
writeon( i_w := 1, s_w := 0, " ", n )
end if_dps_eq_n ;
n := n + 1
end
end.
Output:
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798

## Arturo

disarium?: function [x][
j: 0
psum: sum map digits x 'dig [
j: j + 1
dig ^ j
]
return psum = x
]

cnt: 0
i: 0
while [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.AWK
BEGIN {
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 False
end function

limite = 19
cont = 0 : n = 0
print "The first"; limite; " Disarium numbers are:"
while cont < limite
if isDisarium(n) then
print n; " ";
cont += 1
endif
n += 1
end while
end
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, i
Print "The first"; limite; " Disarium numbers are:"
Do While cont < limite
If isDisarium(n) Then
Print n; " ";
cont += 1
End If
n += 1
Loop
Sleep
Output:
Igual que la entrada de Python.

### Run BASIC

function isDisarium(n)
digitos = len(str\$(n))
suma = 0 : x = n
while x <> 0
r = (x mod 10)
suma = suma + (r ^ digitos)
digitos = digitos - 1
x = int(x / 10)
wend
if suma = n then isDisarium = 1 else isDisarium = 0
end function

limite = 18 : cont = 0 : n = 0
print "The first"; limite; " Disarium numbers are:"
while cont < limite
if isDisarium(n) = 1 then
print n; " ";
cont = cont + 1
end if
n = n + 1
wend
Output:
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427

### True BASIC

FUNCTION isDisarium(n)
LET digitos = LEN(STR\$(n))
LET suma = 0
LET x = n
DO WHILE x <> 0
LET r = REMAINDER(x, 10)
LET suma = suma + (r ^ digitos)
LET digitos = digitos - 1
LET x = INT(x / 10)
LOOP
IF suma = n THEN LET isDisarium = 1 ELSE LET isDisarium = 0
END FUNCTION

LET limite = 18
LET cont = 0
LET n = 0
PRINT "The first"; limite; " Disarium numbers are:"
DO WHILE cont < limite
IF isDisarium(n) = 1 THEN
PRINT n; " ";
LET cont = cont + 1
END IF
LET n = n + 1
LOOP
END
Output:
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427

### 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
EndIf
EndProcedure

OpenConsole()
limite.i = 19
cont.i = 0
n.i = 0
PrintN("The first" + Str(limite) + " Disarium numbers are:")
While cont < limite
If isDisarium(n)
Print(Str(n) + #TAB\$)
cont + 1
EndIf
n + 1
Wend
Input()
CloseConsole()
Output:
Igual que la entrada de FreeBASIC.

### Yabasic

limite = 18 : cont = 0 : n = 0
print "The first", limite, " Disarium numbers are:"
while cont < limite
if isDisarium(n) then
print n, " ";
cont = cont + 1
fi
n = n + 1
wend
end

sub isDisarium(n)
digitos = len(str\$(n))
suma = 0 : x = n
while x <> 0
r = mod(x, 10)
suma = suma + (r ^ digitos)
digitos = digitos - 1
x = int(x / 10)
wend
if suma = n then return True else return False : fi
end sub
Output:
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427

## 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.utils
math.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 digits
const 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[0] = make([]uint64, 11)
EXP[1] = make([]uint64, 11)
POW[0] = make([]uint64, 11)
POW[1] = make([]uint64, 11)
for i := uint64(1); i <= 10; i++ {
EXP[1][i] = i
}
for i := uint64(1); i <= 9; i++ {
POW[1][i] = i
}
POW[1][10] = 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][10] = EXP[i][10] * 10
POW[i+1][10] = POW[i][10] + POW[i+1][9]
}

// 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] = 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][10]

if pow < EXP[digit][1] { // Try next since upper limit is invalidly low
DIGITS[level]++
continue
}

max = pow % EXP[level][10]
pow -= max
if max < Exp[level] {
pow -= EXP[level][10]
}
max = pow + Exp[level]

if max < EXP[digit][1] { // Try next since upper limit is invalidly low
DIGITS[level]++
continue
}

// Min possible value
exp = Exp[level] + EXP[digit][1]
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][10]
pow -= min
if min > Exp[level] {
pow += EXP[level][10]
}
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][1]
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
where
import Data.Char ( digitToInt)

isDisarium :: Int -> Bool
isDisarium 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 ret
end

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)

## Mathematica/Wolfram Language

ClearAll[DisariumQ]
DisariumQ[n_Integer] := Module[{digs},
digs = IntegerDigits[n];
digs = digs^Range[Length[digs]];
Total[digs] == n
]
i = 0;
Reap[Do[
If[DisariumQ[n],
i++;
Sow[n]
];
If[i == 19, Break[]]
,
{n, 0, \[Infinity]}
]][[2, 1]]
Output:
{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 89, 135, 175, 518, 598, 1306, 1676, 2427, 2646798}

## 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][11] = exps[i][11]*10
pows[i+1][11] = pows[i][11] + pows[i+1][10]
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[1] = 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][11]

atom ed2 = exps[digit][2]
if powr<ed2 then  // Try next since upper limit is invalidly low
digits[level] += 1
else
atom el11 = exps[level][11],
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][2]
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)
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 length
disarium_number_cp(Len,Base) = findall(N,disarium_number_cp(Len,Base,N)).sort.

% Find a Disarium number of a certain length
disarium_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 Base
to_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

## PL/M

Based on...
Translation of: ALGOL 68
... but as the original PL/M compiler only supports 8 and 16-bit unsigned integers, this stops after trying up to 9999 or finding 18 Disarium numbers. Also, PL/M only supports 1-dimensional arrays.
Works with: 8080 PL/M Compiler
... under CP/M (or an emulator)
100H: /* FIND SOME DISARIUM NUMBERS - NUMBERS WHOSE DIGIT POSITION-POWER     */
/* SUMS ARE EQUAL TO THE NUMBER, E.G. 135 = 1^1 + 3^2 + 5^3 */

/* CP/M BDOS SYSTEM CALL, IGNORE THE RETURN VALUE */
BDOS: PROCEDURE( FN, ARG ); DECLARE FN BYTE, ARG ADDRESS; GOTO 5; END;
PR\$CHAR: PROCEDURE( C ); DECLARE C BYTE; CALL BDOS( 2, C ); END;
PR\$STRING: PROCEDURE( S ); DECLARE S ADDRESS; CALL BDOS( 9, S ); END;
PR\$NUMBER: PROCEDURE( N ); /* PRINTS A NUMBER IN THE MINIMUN FIELD WIDTH */
DECLARE V ADDRESS, N\$STR ( 6 )BYTE, W BYTE;
V = N;
W = LAST( N\$STR );
N\$STR( W ) = '\$';
N\$STR( W := W - 1 ) = '0' + ( V MOD 10 );
DO WHILE( ( V := V / 10 ) > 0 );
N\$STR( W := W - 1 ) = '0' + ( V MOD 10 );
END;
CALL PR\$STRING( .N\$STR( W ) );
END PR\$NUMBER;

/* TABLE OF POWERS UP TO THE FOURTH POWER - AS WE ARE ONLY FINDING THE */
/* DISARIUM NUMBERS UP TO 9999 */
DECLARE POWER( 40 /* ( 1 : 4, 0 : 9 ) */ ) ADDRESS;
DECLARE MAX\$DISARIUM LITERALLY '9999';

DECLARE ( N, D, POWER\$OF\$TEN, COUNT, LENGTH, V, P, DPS, NSUB, NPREV )

/* COMPUTE THE NTH POWERS OF 0-9 */
DO D = 0 TO 9; POWER( D ) = D; END;
NSUB = 10;
NPREV = 0;
DO N = 2 TO 4;
POWER( NSUB ) = 0;
DO D = 1 TO 9;
POWER( NSUB + D ) = POWER( NPREV + D ) * D;
END;
NPREV = NSUB;
NSUB = NSUB + 10;
END;

/* PRINT THE DISARIUM NUMBERS UPTO 9999 OR THE 18TH, WHICHEVER IS SOONER */
POWER\$OF\$TEN = 10;
LENGTH = 1;
COUNT, N = 0;
DO WHILE( N < MAX\$DISARIUM AND COUNT < 18 );
IF N = POWER\$OF\$TEN THEN DO;
/* THE NUMBER OF DIGITS JUST INCREASED */
POWER\$OF\$TEN = POWER\$OF\$TEN * 10;
LENGTH = LENGTH + 1;
END;
/* FORM THE DIGIT POWER SUM */
V = N;
P = LENGTH * 10;
DPS = 0;
DO D = 1 TO LENGTH;
P = P - 10;
DPS = DPS + POWER( P + ( V MOD 10 ) );
V = V / 10;
END;
IF DPS = N THEN DO;
/* N IS DISARIUM */
COUNT = COUNT + 1;
CALL PR\$CHAR( ' ' );
CALL PR\$NUMBER( N );
END;
N = N + 1;
END;

EOF
Output:
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427

## 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 = (^âˆž).hyper.map: { \$_ if \$_ == sum .polymod(10 xx *).reverse Z** 1..* };

put \$disarium[^18];
put \$disarium[18];
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]

## VTL-2

Translation of: ALGOL W

Finds the first 18 Disarium numbers - computes a table of digit powers up to the fourth power.

1000 N=1
1010 D=0
1020 :N*10+D)=D
1030 D=D+1
1040 #=D<10*1020
1050 N=2
1060 :N*10)=0
1070 D=1
1080 :N*10+D)=:N-1*10+D)*D
1090 D=D+1
1100 #=D<10*1080
1120 N=N+1
1130 #=N<5*1060
2000 C=0
2010 T=10
2020 L=1
2030 N=0
2040 #=N=T=0*2070
2050 T=T*10
2060 L=L+1
2070 V=N
2080 P=L
2090 S=0
2100 V=V/10
2110 S=S+:P*10+%
2120 P=P-1
2130 #=V>1*(S-1<N)*2100
2140 #=S=N=0*2180
2150 C=C+1
2160 \$=32
2170 ?=N
2180 N=N+1
2190 #=C<18*2040

Output:
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427

## Vlang

Recommend to build first `v -prod disarium.v` and then run `./disarium`

Translation of: Go
import strconv

const dmax = 20 // maximum digits
const limit = 20 // maximum number of disariums to find

fn main() {
// Pre-calculated exponential and power serials
mut exp1 := [][]u64{len: 1+dmax, init: []u64{len: 11}}
mut pow1 := [][]u64{len: 1+dmax, init: []u64{len: 11}}

for i := u64(1); i <= 10; i++ {
exp1[1][i] = i
}
for i := u64(1); i <= 9; i++ {
pow1[1][i] = i
}
pow1[1][10] = 9

for i := 1; i < dmax; i++ {
for j := 0; j <= 9; j++ {
exp1[i+1][j] = exp1[i][j] * 10
pow1[i+1][j] = pow1[i][j] * u64(j)
}
exp1[i+1][10] = exp1[i][10] * 10
pow1[i+1][10] = pow1[i][10] + pow1[i+1][9]
}

// Digits of candidate and values of known low bits
mut digits := []int{len: 1+dmax} // Digits form
mut exp2 := []u64{len: 1+dmax} // Number form
mut pow2 := []u64{len: 1+dmax} // pow2ers form

mut exp, mut pow, mut min, mut max := u64(0),u64(0),u64(0),u64(0)
start := 1
final := dmax
mut count := 0
for digit := start; digit <= final; digit++ {
println("# of digits: \$digit")
mut level := 1
digits[0] = 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
exp2[level] = exp2[level-1] + exp1[level][digits[level]]
pow2[level] = pow2[level-1] + pow1[digit+1-level][digits[level]]

// Max possible value
pow = pow2[level] + pow1[digit-level][10]

if pow < exp1[digit][1] { // Try next since upper limit is invalidly low
digits[level]++
continue
}

max = pow % exp1[level][10]
pow -= max
if max < exp2[level] {
pow -= exp1[level][10]
}
max = pow + exp2[level]

if max < exp1[digit][1] { // Try next since upper limit is invalidly low
digits[level]++
continue
}

// Min possible value
exp = exp2[level] + exp1[digit][1]
pow = pow2[level] + 1

if exp > max || max < pow { // Try next since upper limit is invalidly low
digits[level]++
continue
}

if pow > exp {
min = pow % exp1[level][10]
pow -= min
if min > exp2[level] {
pow += exp1[level][10]
}
min = pow + exp2[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
exp2[level] = exp2[level-1] + exp1[level][digits[level]]
pow2[level] = pow2[level-1] + pow1[digit+1-level][digits[level]]

// Loop to check all last bits of candidates
for digits[level] < 10 {
// Print out new Disarium number
if exp2[level] == pow2[level] {
mut s := ""
for i := dmax; i > 0; i-- {
s += "\${digits[i]}"
}
n, _ := strconv.common_parse_uint2(s, 10, 64)
println(n)
count++
if count == limit {
println("\nFound the first \$limit Disarium numbers.")
return
}
}

// Go to followed last bit candidate
digits[level]++
exp2[level] += exp1[level][1]
pow2[level]++
}

// Reset to try next path
digits[level] = 0
level--
digits[level]++
}
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.

## 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 = 19
var count = 0
var disarium = []
var n = 0
while (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) {
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 digits
var LIMIT = 19 // maximum number of Disariums to find

// Pre-calculated exponential and power serials
var EXP = List.filled(1 + DMAX, null)
var POW = List.filled(1 + DMAX, null)
EXP[0] = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1]
EXP[1] = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
POW[0] = List.filled(11, 0)
POW[1] = [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][10] = EXP[i][10] * 10
POW[i+1][10] = POW[i][10] + POW[i+1][9]
}

// Digits of candidate and values of known low bits
var DIGITS = List.filled(1 + DMAX, 0) // Digits form
var Exp = List.filled(1 + DMAX, 0) // Number form
var Pow = List.filled(1 + DMAX, 0) // Powers form

var exp
var pow
var min
var max
var start = 1
var final = DMAX
var count = 0
for (digit in start..final) {
System.print("# of digits: %(digit)")
var level = 1
DIGITS[0] = 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][10]

if (pow < EXP[digit][1]) { // Try next since upper limit is invalidly low
DIGITS[level] = DIGITS[level] + 1
continue
}

max = pow % EXP[level][10]
pow = pow - max
if (max < Exp[level]) pow = pow - EXP[level][10]
max = pow + Exp[level]

if (max < EXP[digit][1]) { // Try next since upper limit is invalidly low
DIGITS[level] = DIGITS[level] + 1
continue
}

// Min possible value
exp = Exp[level] + EXP[digit][1]
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][10]
pow = pow - min
if (min > Exp[level]) {
pow = pow + EXP[level][10]
}
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][1]
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 digits
var LIMIT = 20 // maximum number of disariums to find

// Pre-calculated exponential and power serials
var EXP = List.filled(1 + DMAX, null)
var POW = List.filled(1 + DMAX, null)
EXP[0] = List.filled(11, null)
EXP[1] = List.filled(11, null)
POW[0] = List.filled(11, null)
POW[1] = List.filled(11, null)
for (i in 0..9) EXP[0][i] = U64.zero
EXP[0][10] = U64.one

for (i in 0..10) EXP[1][i] = U64.from(i)

for (i in 0..10) POW[0][i] = U64.zero

for (i in 0..9) POW[1][i] = U64.from(i)
POW[1][10] = 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][10] = EXP[i][10] * 10
POW[i+1][10] = POW[i][10] + POW[i+1][9]
}

// Digits of candidate and values of known low bits
var DIGITS = List.filled(1 + DMAX, 0) // Digits form
var Exp = List.filled(1 + DMAX, null) // Number form
var Pow = List.filled(1 + DMAX, null) // Powers form
for (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 = 1
var final = DMAX
var count = 0
for (digit in start..final) {
System.print("# of digits: %(digit)")
var level = 1
DIGITS[0] = 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
Pow[level].add(Pow[level - 1], POW[digit + 1 - level][DIGITS[level]])

// Max possible value

if (pow < EXP[digit][1]) { // Try next since upper limit is invalidly low
DIGITS[level] = DIGITS[level] + 1
continue
}

max.rem(pow, EXP[level][10])
pow.sub(max)
if (max < Exp[level]) pow.sub(EXP[level][10])

if (max < EXP[digit][1]) { // Try next since upper limit is invalidly low
DIGITS[level] = DIGITS[level] + 1
continue
}

// Min possible value

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][10])
pow.sub(min)
if (min > 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
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
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 number
int 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