Own digits power sum

Own digits power sum is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Description

For the purposes of this task, an own digits power sum is a decimal integer which is N digits long and is equal to the sum of its individual digits raised to the power N.

Example

The three digit integer 153 is an own digits power sum because 1³ + 5³ + 3³ = 1 + 125 + 27 = 153.

Task

Find and show here all own digits power sums for N = 3 to N = 8 inclusive.

Optionally, do the same for N = 9 which may take a while for interpreted languages.

ALGOL 68

Non-recursive, generates the possible combinations ands the own digits power sums in reverse order, without duplication.
Uses ideas from various solutions on this page, particularly the observation that the own digits power sum is independent of the order of the digits. Uses the minimum possible highest digit for the number of digits (width) and maximum number of zeros for the width to avoid some combinations. This trys 73 359 combinations, so not as optimosed as the Pascal solution, however it is much faster than the previous Algol 68 version. <lang algol68>BEGIN

   # counts of used digits, check is a copy used to check the number is an own digit power sum #
   [ 0 : 9 ]INT used, check; FOR i FROM 0 TO 9 DO check[ i ] := 0 OD;
   [ 1 : 9, 1 : 9 ]LONG INT power;     # table of digit powers       #
   FOR i TO 9 DO power[ 1, i ] := i OD;
   FOR j FROM 2 TO 9 DO
       FOR i TO 9 DO power[ j, i ] := power[ j - 1, i ] * i OD
   OD;
   # find the lowest possible first digit for each digit combination #
   # this is the roughly the low3est n where P*n^p > 10^p            #
   [ 1 : 9 ]INT lowest digit;
   lowest digit[ 2 ] := lowest digit[ 1 ] := -1;
   LONG INT p10 := 100;
   FOR i FROM 3 TO 9 DO
       FOR p FROM 2 TO 9 WHILE LONG INT np = power[ i, p ] * i;
                               np < p10 DO
           lowest digit[ i ] := p
       OD;
       p10 *:= 10
   OD;
   # find the maximum number of zeros possible for each width and max digit #
   [ 1 : 9, 1 : 9 ]INT max zeros; FOR i TO 9 DO FOR j TO 9 DO max zeros[ i, j ] := 0 OD OD;
   p10 := 1000;
   FOR w FROM 3 TO 9 DO
       FOR d FROM lowest digit[ w ] TO 9 DO
           INT nz := 9;
           WHILE IF nz < 0
                 THEN FALSE
                 ELSE LONG INT np := power[ w, d ] * nz;
                      np > p10
                 FI
           DO
               nz -:= 1
           OD;
           max zeros[ w, d ] := IF nz > w THEN 0 ELSE w - nz FI
       OD;
       p10 *:= 10
   OD;
   # find the numbers, works backeards through the possible combinations of  #
   # digits, starting from all 9s                                            #
   [ 1 : 100 ]LONG INT numbers;  # will hold the own digit power sum numbers #
   INT n count   := 0;           # count of the own digit power sums         #
   INT try count := 0;           # count of digit combinations tried         #
   [ 1 : 9 ]INT digits;          # the latest digit combination to try       #
   FOR d        TO 9 DO digits[ d ] := 9 OD;
   FOR d FROM 0 TO 8 DO used[   d ] := 0 OD; used[ 9 ] := 9;
   INT width     := 9;           # number of digits                          #
   INT last      := width;       # final digit position                      #
   p10           := 100 000 000; # min value for a width digit power sum     #
   WHILE width > 2 DO
       try count   +:= 1;
       LONG INT dps := 0;        # construct the digit power sum             #
       check        := used;
       FOR i TO 9 DO
           IF used[ i ] /= 0 THEN dps +:= used[ i ] * power[ width, i ] FI
       OD;
       # reduce the count of each digit by the number of times it appear in the digit power sum #
       LONG INT n := dps;
       WHILE check[ SHORTEN ( n MOD 10 ) ] -:= 1; # reduce the count of this digit #
             ( n OVERAB 10 ) > 0
       DO SKIP OD;
       BOOL reduce width := dps <= p10;
       IF NOT reduce width THEN
           # dps is not less than the minimum possible width number          #
           # check there are no non-zero check counts left and so result is  #
           # equal to its digit power sum                                    #
           INT z count := 0;
           FOR i FROM 0 TO 9 WHILE check[ i ] = 0 DO z count +:= 1 OD;
           IF z count = 10 THEN
               numbers[ n count +:= 1 ] := dps
           FI;
           # prepare the next digit combination: reduce the last digit #
           used[ digits[ last ] ] -:= 1;
           digits[ last ]         -:= 1;
           IF digits[ last ] = 0 THEN
               # the last digit is now zero - check this number of zeros is possible #
               IF used[ 0 ] >= max zeros[ width, digits[ 1 ] ] THEN
                   # have exceeded the maximum number of zeros for the first digit in this width #
                   digits[ last ] := -1                    
               FI
           FI;
           IF digits[ last ] >= 0 THEN
               # still processing the last digit #
               used[ digits[ last ] ] +:= 1
           ELSE
               # last digit is now -1, start processing the previous digit #
               INT prev := last;
               WHILE IF ( prev -:= 1 ) < 1
                     THEN # processed all digits #
                         FALSE
                     ELSE
                         # have another digit # 
                         used[ digits[ prev ] ] -:= 1;
                         digits[ prev ]         -:= 1;
                         digits[ prev ] < 0
                     FI
               DO SKIP OD;
               IF prev > 0 THEN
                   # still some digits to process #
                   IF prev = 1 THEN
                       IF digits[ 1 ] <= lowest digit[ width ] THEN
                           # just finished the lowest possible maximum digit for this width #
                           prev := 0
                       FI
                   FI;
                   IF prev /= 0 THEN
                       # OK to try a lower digit #
                       used[ digits[ prev ] ] +:= 1;
                       FOR i FROM prev + 1 TO width DO
                           digits[ i ] := digits[ prev ];
                           used[ digits[ prev ] ] +:= 1
                       OD
                   FI
               FI;
               IF prev <= 0 THEN
                   # processed all the digits for this width #
                   reduce width := TRUE
               FI
           FI
       FI;
       IF reduce width THEN
           # reduce the number of digits #
           width   := last -:= 1;
           IF last > 0 THEN
               # iniialise for fewer digits #
               FOR d               TO last DO digits[ d ] :=  9 OD;
               FOR d FROM last + 1 TO 9    DO digits[ d ] := -1 OD;
               FOR d FROM        0 TO 9    DO used[   d ] :=  0 OD;
               used[ 9 ] := last;
               p10   OVERAB 10
           FI
       FI
   OD;
   # show the own digit power sums #
   print( ( "Own digits power sums for N = 3 to 9 inclusive:", newline ) );
   FOR i FROM n count BY -1 TO LWB numbers DO
       print( ( whole( numbers[ i ], 0 ), newline ) )
   OD;
   print( ( "Considered ", whole( try count, 0 ), " digit combinations" ) )

END</lang>

Output:

Own digits power sums for N = 3 to 9 inclusive:
153
370
371
407
1634
8208
9474
54748
92727
93084
548834
1741725
4210818
9800817
9926315
24678050
24678051
88593477
146511208
472335975
534494836
912985153
Considered 73359 digit combinations

C

Iterative (slow)

Takes about 1.9 seconds to run (GCC 9.3.0 -O3).

Translation of: Wren

<lang c>#include <stdio.h>

1. include <math.h>

1. define MAX_DIGITS 9

int digits[MAX_DIGITS];

void getDigits(int i) {

   int ix = 0;
   while (i > 0) {
       digits[ix++] = i % 10;
       i /= 10;
   }

}

int main() {

   int n, d, i, max, lastDigit, sum, dp;
   int powers[10] = {0, 1, 4, 9, 16, 25, 36, 49, 64, 81};
   printf("Own digits power sums for N = 3 to 9 inclusive:\n");
   for (n = 3; n < 10; ++n) {
       for (d = 2; d < 10; ++d) powers[d] *= d;
       i = (int)pow(10, n-1);
       max = i * 10;
       lastDigit = 0;
       while (i < max) {
           if (!lastDigit) {
               getDigits(i);
               sum = 0;
               for (d = 0; d < n; ++d) {
                   dp = digits[d];
                   sum += powers[dp];
               }
           } else if (lastDigit == 1) {
               sum++;
           } else {
               sum += powers[lastDigit] - powers[lastDigit-1];
           }
           if (sum == i) {
               printf("%d\n", i);
               if (lastDigit == 0) printf("%d\n", i + 1);
               i += 10 - lastDigit;
               lastDigit = 0;
           } else if (sum > i) {
               i += 10 - lastDigit;
               lastDigit = 0;
           } else if (lastDigit < 9) {
               i++;
               lastDigit++;
           } else {
               i++;
               lastDigit = 0;
           }
       }
   }
   return 0;

}</lang>

Output:

Same as Wren example.

Recursive (very fast)

Translation of: Pascal

Down now to 14ms. <lang c>#include <stdio.h>

1. include <string.h>

1. define MAX_BASE 10

typedef unsigned long long ulong;

int usedDigits[MAX_BASE]; ulong powerDgt[MAX_BASE][MAX_BASE]; ulong numbers[60]; int nCount = 0;

void initPowerDgt() {

   int i, j;
   powerDgt[0][0] = 0;
   for (i = 1; i < MAX_BASE; ++i) powerDgt[0][i] = 1;
   for (j = 1; j < MAX_BASE; ++j) {
       for (i = 0; i < MAX_BASE; ++i) {
           powerDgt[j][i] = powerDgt[j-1][i] * i;
       }
   }

}

ulong calcNum(int depth, int used[MAX_BASE]) {

   int i;
   ulong result = 0, r, n;
   if (depth < 3) return 0;
   for (i = 1; i < MAX_BASE; ++i) {
       if (used[i] > 0) result += powerDgt[depth][i] * used[i];
   }
   if (result == 0) return 0;
   n = result;
   do {
       r = n / MAX_BASE;
       used[n-r*MAX_BASE]--;
       n = r;
       depth--;
   } while (r);
   if (depth) return 0;
   i = 1;
   while (i < MAX_BASE && used[i] == 0) i++;
   if (i >= MAX_BASE) numbers[nCount++] = result;
   return 0;

}

void nextDigit(int dgt, int depth) {

   int i, used[MAX_BASE];
   if (depth < MAX_BASE-1) {
       for (i = dgt; i < MAX_BASE; ++i) {
           usedDigits[dgt]++;
           nextDigit(i, depth+1);
           usedDigits[dgt]--;
       }
   }
   if (dgt == 0) dgt = 1;
   for (i = dgt; i < MAX_BASE; ++i) {
       usedDigits[i]++;
       memcpy(used, usedDigits, sizeof(usedDigits));
       calcNum(depth, used);
       usedDigits[i]--;
   }

}

int main() {

   int i, j;
   ulong t;
   initPowerDgt();
   nextDigit(0, 0);

   // sort and remove duplicates
   for (i = 0; i < nCount-1; ++i) {
       for (j = i + 1; j < nCount; ++j) {
           if (numbers[j] < numbers[i]) {
               t = numbers[i];
               numbers[i] = numbers[j];
               numbers[j] = t;
           }
       }
   }
   j = 0;
   for (i = 1; i < nCount; ++i) {
       if (numbers[i] != numbers[j]) {
           j++;
           t = numbers[i];
           numbers[i] = numbers[j];
           numbers[j] = t;
       }
   }
   printf("Own digits power sums for N = 3 to 9 inclusive:\n");
   for (i = 0; i <= j; ++i) printf("%lld\n", numbers[i]);
   return 0;

}</lang>

Output:

Same as before.

F#

<lang fsharp> // Own digits power sum. Nigel Galloway: October 2th., 2021 let fN g=let N=[|for n in 0..9->pown n g|] in let rec fN g=function n when n<10->N.[n]+g |n->fN(N.[n%10]+g)(n/10) in (fun g->fN 0 g) {3..9}|>Seq.iter(fun g->let fN=fN g in printf $"%d{g} digit are:"; {pown 10 (g-1)..(pown 10 g)-1}|>Seq.iter(fun g->if g=fN g then printf $" %d{g}"); printfn "") </lang>

Output:

3 digit are: 153 370 371 407
4 digit are: 1634 8208 9474
5 digit are: 54748 92727 93084
6 digit are: 548834
7 digit are: 1741725 4210818 9800817 9926315
8 digit are: 24678050 24678051 88593477
9 digit are: 146511208 472335975 534494836 912985153

FreeBASIC

<lang freebasic> dim as uinteger N, curr, temp, dig, sum

for N = 3 to 9

   for curr = 10^(N-1) to 10^N-1
       sum = 0
       temp = curr
       do
           dig = temp mod 10
           temp = temp \ 10
           sum += dig ^ N
       loop until temp = 0
       if sum = curr then print curr
   next curr

next N </lang>

Output:

As above.

Go

Iterative (slow)

Translation of: Wren

Library: Go-rcu

Takes about 16.5 seconds to run including compilation time. <lang go>package main

import (

   "fmt"
   "math"
   "rcu"

)

func main() {

   powers := [10]int{0, 1, 4, 9, 16, 25, 36, 49, 64, 81}
   fmt.Println("Own digits power sums for N = 3 to 9 inclusive:")
   for n := 3; n < 10; n++ {
       for d := 2; d < 10; d++ {
           powers[d] *= d
       }
       i := int(math.Pow(10, float64(n-1)))
       max := i * 10
       lastDigit := 0
       sum := 0
       var digits []int
       for i < max {
           if lastDigit == 0 {
               digits = rcu.Digits(i, 10)
               sum = 0
               for _, d := range digits {
                   sum += powers[d]
               }
           } else if lastDigit == 1 {
               sum++
           } else {
               sum += powers[lastDigit] - powers[lastDigit-1]
           }
           if sum == i {
               fmt.Println(i)
               if lastDigit == 0 {
                   fmt.Println(i + 1)
               }
               i += 10 - lastDigit
               lastDigit = 0
           } else if sum > i {
               i += 10 - lastDigit
               lastDigit = 0
           } else if lastDigit < 9 {
               i++
               lastDigit++
           } else {
               i++
               lastDigit = 0
           }
       }
   }

}</lang>

Output:

Same as Wren example.

Recursive (very fast)

Down to about 128 ms now including compilation time. Actual run time only 8 ms!

Translation of: Pascal

<lang go>package main

import "fmt"

const maxBase = 10

var usedDigits = [maxBase]int{} var powerDgt = [maxBase][maxBase]uint64{} var numbers []uint64

func initPowerDgt() {

   for i := 1; i < maxBase; i++ {
       powerDgt[0][i] = 1
   }
   for j := 1; j < maxBase; j++ {
       for i := 0; i < maxBase; i++ {
           powerDgt[j][i] = powerDgt[j-1][i] * uint64(i)
       }
   }

}

func calcNum(depth int, used [maxBase]int) uint64 {

   if depth < 3 {
       return 0
   }
   result := uint64(0)
   for i := 1; i < maxBase; i++ {
       if used[i] > 0 {
           result += uint64(used[i]) * powerDgt[depth][i]
       }
   }
   if result == 0 {
       return 0
   }
   n := result
   for {
       r := n / maxBase
       used[n-r*maxBase]--
       n = r
       depth--
       if r == 0 {
           break
       }
   }
   if depth != 0 {
       return 0
   }
   i := 1
   for i < maxBase && used[i] == 0 {
       i++
   }
   if i >= maxBase {
       numbers = append(numbers, result)
   }
   return 0

}

func nextDigit(dgt, depth int) {

   if depth < maxBase-1 {
       for i := dgt; i < maxBase; i++ {
           usedDigits[dgt]++
           nextDigit(i, depth+1)
           usedDigits[dgt]--
       }
   }
   if dgt == 0 {
       dgt = 1
   }
   for i := dgt; i < maxBase; i++ {
       usedDigits[i]++
       calcNum(depth, usedDigits)
       usedDigits[i]--
   }

}

func main() {

   initPowerDgt()
   nextDigit(0, 0)

   // sort and remove duplicates
   for i := 0; i < len(numbers)-1; i++ {
       for j := i + 1; j < len(numbers); j++ {
           if numbers[j] < numbers[i] {
               numbers[i], numbers[j] = numbers[j], numbers[i]
           }
       }
   }
   j := 0
   for i := 1; i < len(numbers); i++ {
       if numbers[i] != numbers[j] {
           j++
           numbers[i], numbers[j] = numbers[j], numbers[i]
       }
   }
   numbers = numbers[0 : j+1]
   fmt.Println("Own digits power sums for N = 3 to 9 inclusive:")
   for _, n := range numbers {
       fmt.Println(n)
   }

}</lang>

Output:

Same as before.

Haskell

Using a function from the Combinations with Repetitions task: <lang haskell>import Data.List (sort)

---

OWN DIGITS POWER SUM -----------------

ownDigitsPowerSums :: Int -> [Int] ownDigitsPowerSums n = sort (ns >>= go)

 where
   ns = combsWithRep n [0 .. 9]
   go xs
     | digitsMatch m xs = [m]
     | otherwise = []
     where
       m = foldr ((+) . (^ n)) 0 xs

digitsMatch :: Show a => a -> [Int] -> Bool digitsMatch n ds =

 sort ds == sort (digits n)

---

TEST -------------------------

main :: IO () main = do

 putStrLn "N ∈ [3 .. 8]"
 mapM_ print ([3 .. 8] >>= ownDigitsPowerSums)
 putStrLn ""
 putStrLn "N=9"
 mapM_ print $ ownDigitsPowerSums 9

---

GENERIC ------------------------

combsWithRep ::

 (Eq a) =>
 Int ->
 [a] ->
 a

combsWithRep k xs = comb k []

 where
   comb 0 ys = ys
   comb n [] = comb (pred n) (pure <$> xs)
   comb n peers = comb (pred n) (peers >>= nextLayer)
     where
       nextLayer ys@(h : _) =
         (: ys) <$> dropWhile (/= h) xs

digits :: Show a => a -> [Int] digits n = (\x -> read [x] :: Int) <$> show n</lang>

Output:

N ∈ [3 .. 8]
153
370
371
407
1634
8208
9474
54748
92727
93084
548834
1741725
4210818
9800817
9926315
24678050
24678051
88593477

N=9
146511208
472335975
534494836
912985153

Julia

<lang julia>function isowndigitspowersum(n::Integer, base=10)

   dig = digits(n, base=base)
   exponent = length(dig)
   return mapreduce(x -> x^exponent, +, dig) == n

end

for i in 10^2:10^9-1

   isowndigitspowersum(i) && println(i)

end

</lang>

Output:

153
370
371
407
1634
8208
9474
54748
92727
93084
548834
1741725
4210818
9800817
9926315
24678050
24678051
88593477
146511208
472335975
534494836
912985153

Pascal

recursive solution.Just counting the different combination of digits
See Combinations_with_repetitions
<lang pascal>program PowerOwnDigits; {$IFDEF FPC} // {$R+,O+}

 {$MODE DELPHI}{$OPTIMIZATION ON,ALL}{$COPERATORS ON}

{$ELSE}{$APPTYPE CONSOLE}{$ENDIF} uses

 SysUtils;

const

 MAXBASE = 10;//16;
 MaxDgtVal = MAXBASE - 1;

type

 tDgtVal = 0..MaxDgtVal;
 tUsedDigits = array[0..15] of Int8;
 tPower = array[tDgtVal] of Uint64; 

var

 PowerDgt: array[tDgtVal] of tPower;
 UD :tUsedDigits;
 CombIdx: array of Int8;
 Numbers : array of Uint32;
 rec_cnt : NativeInt;

 function InitCombIdx(ElemCount: Byte): pbyte;
 begin
   setlength(CombIdx, ElemCount + 1);
   Fillchar(CombIdx[0], sizeOf(CombIdx[0]) * (ElemCount + 1), #0);
   Result := @CombIdx[0];
 end;

 function NextCombWithRep(pComb: pByte; MaxVal, ElemCount: UInt32): boolean;
 var
   i, dgt: NativeInt;
 begin
   i := -1;
   repeat
     i += 1;
     dgt := pComb[i];
     if dgt < MaxVal then
       break;
   until i > ElemCount;
   Result := i >= ElemCount;
   dgt +=1;
   repeat
     pComb[i] := dgt;
     i -= 1;
   until i < 0;
 end; 
 
 function Init(ElemCount:byte):pByte;
 var
   i, j: tDgtVal;
 begin
   for i in tDgtVal do
     PowerDgt[low(tDgtVal), i] := 1;
   for j := low(tDgtVal) + 1 to High(tDgtVal) do
     for i in tDgtVal do
       PowerDgt[j, i] := PowerDgt[j - 1, i] * i;
  result := InitCombIdx(ElemCount);     
 end;
 
 function GetPowerSum(minpot:nativeInt;digits:pbyte;var UD_tmp :tUsedDigits):NativeInt;
 var
   pPower : pUint64;
   res,r,dgt : Uint64;
 begin
   dgt := minpot;
   res := 0;
   pPower := @PowerDgt[dgt,0];
   repeat
     dgt -=1;
     res += pPower[digits[dgt]];
   until dgt<=0;  
   
   result := minPot;
   repeat
     r := res DIV MAXBASE;
     UD_tmp[res-r*MAXBASE]-= 1;
     res := r;
     dec(result);
   until r = 0;
 end;
 
 procedure calcNum(digits:pbyte);
 var
   UD_tmp :tUsedDigits;
   minPot,dgt: nativeInt;
   res: Uint32;
 begin
   fillchar(UD,SizeOf(UD),#0);

   minPot := 0;
   repeat
     dgt := digits[minPot];
     if dgt = 0 then
       break;
     UD[dgt]+=1;
     inc(minPot);
   until minPot > MaxDgtVal;

   If (minPot<2) or (digits[0] = 1) then
     EXIT;

   repeat
     UD_tmp := UD;
     dgt := GetPowerSum(minpot,digits,UD_tmp);
     //number to small
     if dgt > 0 then
       break;
     if dgt=0 then
     begin
       dgt:= 1;
       while (dgt <= MaxDgtVal) and (UD_tmp[dgt] = 0) do
         dgt +=1;
       if dgt > MaxDgtVal then
       begin
         res := 0;
         for dgt := minpot-1 downto 0 do
           res += PowerDgt[minpot,digits[dgt]];
         setlength(Numbers, Length(Numbers) + 1);
         Numbers[high(Numbers)] := res;
         BREAK;
       end;
     end;
     //try one more 0
     minPot +=1;
   until minPot > MaxDgtVal;
 end;

const

 rounds = 128;

var

 digits : pByte;
 T0 : Int64;
 tmp: Uint64;
 i, j : Int32;

begin

 digits := Init(MaxDgtVal);
 //warm up
 For i := 1 to 50 do
 Begin
   setlength(numbers,0);
   digits := InitCombIdx(MaxDgtVal);
   repeat
     calcnum(digits);
   until NextCombWithRep(digits,MaxDgtVal,MaxDgtVal);
 end;
 //warm up  
 T0 := GetTickCount64;
 rec_cnt := 0;  
 For i := 1 to rounds do
 Begin
   setlength(numbers,0);
   digits := InitCombIdx(MaxDgtVal);
   repeat
     calcnum(digits);
     inc(rec_cnt);
   until NextCombWithRep(digits,MaxDgtVal,MaxDgtVal);
 end;
 T0 := GetTickCount64-T0;  
 writeln(rec_cnt DIV rounds,' recursions in runtime ',T0/rounds:5:2,' ms');

 writeln('found ',length(Numbers));
 //sort
 for i := 0 to High(Numbers) - 1 do
   for j := i + 1 to High(Numbers) do
     if Numbers[j] < Numbers[i] then
     begin
       tmp := Numbers[i];
       Numbers[i] := Numbers[j];
       Numbers[j] := tmp;
     end;

 setlength(Numbers, j + 1);
 for i := 0 to High(Numbers) do
    writeln(i+1:3,Numbers[i]:11);
 {$IFDEF WINDOWS}
 readln;
 {$ENDIF}

end.</lang>

Output:

//doing rounds = 1024 NextCombWithRep without calcnum(digits); takes: 48620 recursions in runtime  0.23 ms
TIO.RUN  CPU share: 99.04 %
48620 recursions in runtime  3.63 ms //best on TIO.RUN ..5.11 ms
found 22
  1        153
  2        370
  3        371
  4        407
  5       1634
  6       8208
  7       9474
  8      54748
  9      92727
 10      93084
 11     548834
 12    1741725
 13    4210818
 14    9800817
 15    9926315
 16   24678050
 17   24678051
 18   88593477
 19  146511208
 20  472335975
 21  534494836
 22  912985153

Perl

Brute Force

Use Parallel::ForkManager to obtain concurrency, trading some code complexity for less-than-infinite run time. Still very slow. <lang perl>use strict; use warnings; use feature 'say'; use List::Util 'sum'; use Parallel::ForkManager;

my %own_dps; my($lo,$hi) = (3,9); my $cores = 8; # configure to match hardware being used

my $start = 10**($lo-1); my $stop = 10**$hi - 1; my $step = int(1 + ($stop - $start)/ ($cores+1));

my $pm = Parallel::ForkManager->new($cores);

RUN: for my $i ( 0 .. $cores ) {

   $pm->run_on_finish (
       sub {
           my ($pid, $exit_code, $ident, $exit_signal, $core_dump, $data_ref) = @_;
           $own_dps{$ident} = $data_ref;
       }
   );

   $pm->start($i) and next RUN;

   my @values;
   for my $n ( ($start + $i*$step) .. ($start + ($i+1)*$step) ) {
       push @values, $n if $n == sum map { $_**length($n) } split , $n;
   }

   $pm->finish(0, \@values)

}

$pm->wait_all_children;

say $_ for sort { $a <=> $b } map { @$_ } values %own_dps;</lang>

Output:

153
370
371
407
1634
8208
9474
54748
92727
93084
548834
1741725
4210818
9800817
9926315
24678050
24678051
88593477
146511208
472335975
534494836
912985153

Combinatorics

Leverage the fact that all combinations of digits give same DPS. Much faster than brute force, as only non-redundant values tested. <lang perl>use strict; use warnings; use List::Util 'sum'; use Algorithm::Combinatorics qw<combinations_with_repetition>;

my @own_dps; for my $d (3..9) {

   my $iter = combinations_with_repetition([0..9], $d);
   while (my $p = $iter->next) {
       my $dps = sum map { $_**$d } @$p;
       next unless $d == length $dps and join(, @$p) == join , sort split , $dps;
       push @own_dps, $dps;
   }

}

print join "\n", sort { $a <=> $b } @own_dps;</lang>

Output:

153
370
371
407
1634
8208
9474
54748
92727
93084
548834
1741725
4210818
9800817
9926315
24678050
24678051
88593477
146511208
472335975
534494836
912985153

Python

Python :: Procedural

slower

<lang python>""" Rosetta code task: Own_digits_power_sum """

def isowndigitspowersum(integer):

   """ true if sum of (digits of number raised to number of digits) == number """
   digits = [int(c) for c in str(integer)]
   exponent = len(digits)
   return sum(x ** exponent for x in digits) == integer

print("Own digits power sums for N = 3 to 9 inclusive:") for i in range(100, 1000000000):

   if isowndigitspowersum(i):
       print(i)

</lang>

Output:

Same as Wren example. Takes over a half hour to run.

faster

Translation of: Wren

Same output.

<lang python>""" Rosetta code task: Own_digits_power_sum (recursive method)"""

MAX_BASE = 10 POWER_DIGIT = [[1 for _ in range(MAX_BASE)] for _ in range(MAX_BASE)] USED_DIGITS = [0 for _ in range(MAX_BASE)] NUMBERS = []

def calc_num(depth, used):

   """ calculate the number at a given recurse depth """
   result = 0
   if depth < 3:
       return 0
   for i in range(1, MAX_BASE):
       if used[i] > 0:
           result += used[i] * POWER_DIGIT[depth][i]
   if result != 0:
       num, rnum = result, 1
       while rnum != 0:
           rnum = num // MAX_BASE
           used[num - rnum * MAX_BASE] -= 1
           num = rnum
           depth -= 1
       if depth == 0:
           i = 1
           while i < MAX_BASE and used[i] == 0:
               i += 1
           if i >= MAX_BASE:
               NUMBERS.append(result)
   return 0

def next_digit(dgt, depth):

   """ get next digit at the given depth """
   if depth < MAX_BASE - 1:
       for i in range(dgt, MAX_BASE):
           USED_DIGITS[dgt] += 1
           next_digit(i, depth + 1)
           USED_DIGITS[dgt] -= 1

   if dgt == 0:
       dgt = 1
   for i in range(dgt, MAX_BASE):
       USED_DIGITS[i] += 1
       calc_num(depth, USED_DIGITS.copy())
       USED_DIGITS[i] -= 1

for j in range(1, MAX_BASE):

   for k in range(MAX_BASE):
       POWER_DIGIT[j][k] = POWER_DIGIT[j - 1][k] * k

next_digit(0, 0) print(NUMBERS) NUMBERS = list(set(NUMBERS)) NUMBERS.sort() print('Own digits power sums for N = 3 to 9 inclusive:') for n in NUMBERS:

   print(n)</lang>

Python :: Functional

Using a function from the Combinations with Repetitions task: <lang python>Own digit power sums

from itertools import accumulate, chain, islice, repeat from functools import reduce

1. ownDigitsPowerSums :: Int -> [Int]

def ownDigitsPowerSums(n):

   All own digit power sums of digit length N
   def go(xs):
       m = reduce(lambda a, x: a + (x ** n), xs, 0)
       return [m] if digitsMatch(m)(xs) else []

   return concatMap(go)(
       combinationsWithRepetitions(n)(range(0, 1 + 9))
   )

1. digitsMatch :: Int -> [Int] -> Bool

def digitsMatch(n):

   True if the digits in ds contain exactly
      the digits of n, in any order.
   
   def go(ds):
       return sorted(ds) == sorted(digits(n))
   return go

1. ------------------------- TEST -------------------------
2. main :: IO ()

def main():

   Own digit power sums for digit lengths 3..9
   print(
       '\n'.join([
           'N ∈ [3 .. 8]',
           *map(str, concatMap(ownDigitsPowerSums)(
               range(3, 1 + 8)
           )),
           '\nN=9',
           *map(str, ownDigitsPowerSums(9))
       ])
   )

1. ----------------------- GENERIC ------------------------

1. combinationsWithRepetitions :: Int -> [a] -> [kTuple a]

def combinationsWithRepetitions(k):

   Combinations with repetitions.
      A list of tuples, representing
      sets of cardinality k,
      with elements drawn from xs.
   
   def f(a, x):
       def go(ys, xs):
           return xs + [[x] + y for y in ys]
       return accumulate(a, go)

   def combsBySize(xs):
       return [
           tuple(x) for x in next(islice(
               reduce(
                   f, xs, chain(
                       [[[]]],
                       islice(repeat([]), k)
                   )
               ), k, None
           ))
       ]
   return combsBySize

1. concatMap :: (a -> [b]) -> [a] -> [b]

def concatMap(f):

   A concatenated list over which a function has been
      mapped.
      The list monad can be derived by using a function f
      which wraps its output in a list, (using an empty
      list to represent computational failure).
   
   def go(xs):
       return list(chain.from_iterable(map(f, xs)))
   return go

1. digits :: Int -> [Int]

def digits(n):

   The individual digits of n as integers
   return [int(c) for c in str(n)]

1. MAIN ---

if __name__ == '__main__':

   main()</lang>

Output:

N ∈ [3 .. 8]
153
370
371
407
1634
8208
9474
54748
92727
93084
548834
1741725
4210818
9800817
9926315
24678050
24678051
88593477

N=9
146511208
472335975
534494836
912985153

Raku

<lang perl6>(3..8).map: -> $p {

   my %pow = (^10).map: { $_ => $_ ** $p };
   my $start = 10 ** ($p - 1);
   my $end   = 10 ** $p;
   my @temp;
   for ^9 -> $i {
       ([X] ($i..9) xx $p).race.map: {
           next unless [<=] $_;
           my $sum = %pow{$_}.sum;
           next if $sum < $start;
           next if $sum > $end;
           @temp.push: $sum if $sum.comb.Bag eqv $_».Str.Bag
       }
   }
   .say for unique sort @temp;

}</lang>

Output:

153
370
371
407
1634
8208
9474
54748
92727
93084
548834
1741725
4210818
9800817
9926315
24678050
24678051
88593477

Combinations with repetitions

Using code from Combinations with repetitions task, a version that runs relatively quickly, and scales well. <lang perl6>proto combs_with_rep (UInt, @ ) { * } multi combs_with_rep (0, @ ) { () } multi combs_with_rep ($, []) { () } multi combs_with_rep (1, @a) { map { $_, }, @a } multi combs_with_rep ($n, [$head, *@tail]) {

   |combs_with_rep($n - 1, ($head, |@tail)).map({ $head, |@_ }),
   |combs_with_rep($n, @tail);

}

say sort gather {

   for 3..9 -> $d {
       for combs_with_rep($d, [^10]) -> @digits {
           .take if $d == .comb.elems and @digits.join == .comb.sort.join given sum @digits X** $d;
       }
   }

}</lang>

Output:

153 370 371 407 1634 8208 9474 54748 92727 93084 548834 1741725 4210818 9800817 9926315 24678050 24678051 88593477 146511208 472335975 534494836 912985153

Visual Basic .NET

<lang vbnet>Option Strict On Option Explicit On

Imports System.IO

<summary> Finds n digit numbers N such that the sum of the nth powers of their digits = N </summary> Module OwnDigitsPowerSum

   Public Sub Main
       For n As Integer = 3 To 9
           Dim fdigit As Integer =  10 - n
           Dim f(8) As Integer
           For i As Integer = 1 To 8
               f(i) = If(i = fdigit, 1, 0)
           Next i
           Dim t(8) As Integer
           For i As Integer = 1 To 8
               t(i) = If(i < fdigit, 0, 9)
           Next i
           Dim power(10) As Integer
               For i As Integer = 0 To UBound(power)
               power(i) = Cint(i ^ n)
           Next i
           Dim maxN As Integer = power(10)
           For d1 As Integer = f(1) To t(1)
               Dim p1 As Integer = power(d1)
               Dim s1 As Integer = d1 * 10
               For d2 As Integer = f(2) To t(2)
                   Dim p2 As Integer = power(d2) + p1
                   If p2 >= maxN Then Exit For
                   Dim s2 As Integer = (s1 + d2) * 10
                   For d3 As Integer = f(3) To t(3)
                       Dim p3 As Integer = power(d3) + p2
                       If p3 >= maxN Then Exit For
                       Dim s3 As Integer = (s2 + d3) * 10
                       For d4 As Integer = f(4) To t(4)
                           Dim p4 As Integer = power(d4) + p3
                           If p4 >= maxN Then Exit For
                           Dim s4 As Integer = (s3 + d4) * 10
                           For d5 As Integer = f(5) To t(5)
                               Dim p5 As Integer = power(d5) + p4
                               If p5 >= maxN Then Exit For
                               Dim s5 As Integer = (s4 + d5) * 10
                               For d6 As Integer = f(6) To t(6)
                                   Dim p6 As Integer = power(d6) + p5
                                   If p6 >= maxN Then Exit For
                                   Dim s6 As Integer = (s5 + d6) * 10
                                   For d7 As Integer = f(7) To t(7)
                                       Dim p7 As Integer = power(d7) + p6
                                       If p7 >= maxN Then Exit For
                                       Dim s7 As Integer = (s6 + d7) * 10
                                       For d8 As Integer = f(8) To t(8)
                                           Dim p8 As Integer = power(d8) + p7
                                           If p8 >= maxN Then Exit For
                                           Dim s8 As Integer = (s7 + d8) * 10
                                           If s8 = p8 Then
                                               ' found a number with 0 as the final digit
                                               ' the same number with a final digit of 1
                                               ' must also match the requirements
                                               Console.Out.WriteLine(s8)
                                               Console.Out.WriteLine(s8 + 1)
                                           End If
                                           For d9 As Integer = 2 To 9
                                               Dim p9 As Integer = power(d9) + p8
                                               If p9 >= maxN Then Exit For
                                               Dim s9 As Integer = s8 + d9
                                               If s9 = p9 Then
                                                   Console.Out.WriteLine(s9)
                                               End If
                                           Next d9
                                       Next d8
                                   Next d7
                               Next d6
                           Next d5
                       Next d4
                   Next d3
               Next d2
           Next d1
       Next n
   End Sub

End Module</lang>

Output:

153
370
371
407
1634
8208
9474
54748
92727
93084
548834
1741725
4210818
9800817
9926315
24678050
24678051
88593477
146511208
472335975
534494836
912985153

Real time: 7.553 s
User time: 7.044 s
Sys. time: 0.290 s on TIO.RUN using Visual Basic .NET (VBC)

Wren

Iterative (slow)

Library: Wren-math

Includes some simple optimizations to try and quicken up the search. However, getting up to N = 9 still took a little over 4 minutes on my machine. <lang ecmascript>import "./math" for Int

var powers = [0, 1, 4, 9, 16, 25, 36, 49, 64, 81] System.print("Own digits power sums for N = 3 to 9 inclusive:") for (n in 3..9) {

   for (d in 2..9) powers[d] = powers[d] * d
   var i = 10.pow(n-1)
   var max = i * 10
   var lastDigit = 0
   var sum = 0
   var digits = null
   while (i < max) {
       if (lastDigit == 0) {
           digits = Int.digits(i)
           sum = digits.reduce(0) { |acc, d|  acc + powers[d] }
       } else if (lastDigit == 1) {
           sum = sum + 1
       } else {
           sum = sum + powers[lastDigit] - powers[lastDigit-1]
       }
       if (sum == i) {
           System.print(i)
           if (lastDigit == 0) System.print(i + 1)
           i = i + 10 - lastDigit
           lastDigit = 0
       } else if (sum > i) {
           i = i + 10 - lastDigit
           lastDigit = 0
       } else if (lastDigit < 9) {
           i = i + 1
           lastDigit = lastDigit + 1
       } else {
           i = i + 1
           lastDigit = 0
       }
   }

}</lang>

Output:

Own digits power sums for N = 3 to 9 inclusive:
153
370
371
407
1634
8208
9474
54748
92727
93084
548834
1741725
4210818
9800817
9926315
24678050
24678051
88593477
146511208
472335975
534494836
912985153

Recursive (very fast)

Translation of: Pascal

Library: Wren-seq

Astonishing speed-up. Runtime now only 0.5 seconds! <lang ecmascript>import "./seq" for Lst

var maxBase = 10 var usedDigits = List.filled(maxBase, 0) var powerDgt = List.filled(maxBase, null) var numbers = []

var initPowerDgt = Fn.new {

   for (i in 0...maxBase) powerDgt[i] = List.filled(maxBase, 0)
   for (i in 1...maxBase) powerDgt[0][i] = 1
   for (j in 1...maxBase) {
       for (i in 0...maxBase) powerDgt[j][i] = powerDgt[j-1][i] * i
   }

}

var calcNum = Fn.new { |depth, used|

   if (depth < 3) return 0
   var result = 0
   for (i in 1...maxBase) {
       if (used[i] > 0) result = result + used[i] * powerDgt[depth][i]
   }
   if (result == 0) return 0
   var n = result
   while (true) {
       var r = (n/maxBase).floor
       used[n - r*maxBase] = used[n - r*maxBase] - 1
       n = r
       depth = depth - 1
       if (r == 0) break
   }
   if (depth != 0) return 0
   var i = 1
   while (i < maxBase && used[i] == 0) i = i + 1
   if (i >= maxBase) numbers.add(result)
   return 0

}

var nextDigit // recursive function nextDigit = Fn.new { |dgt, depth|

   if (depth < maxBase-1) {
       for (i in dgt...maxBase) {
           usedDigits[dgt] = usedDigits[dgt] + 1
           nextDigit.call(i, depth + 1)
           usedDigits[dgt] = usedDigits[dgt] - 1
       }
   }
   if (dgt == 0) dgt = 1
   for (i in dgt...maxBase) {
       usedDigits[i] = usedDigits[i] + 1
       calcNum.call(depth, usedDigits.toList)
       usedDigits[i] = usedDigits[i] - 1
   }

}

initPowerDgt.call() nextDigit.call(0, 0) numbers = Lst.distinct(numbers) numbers.sort() System.print("Own digits power sums for N = 3 to 9 inclusive:") System.print(numbers.map { |n| n.toString }.join("\n"))</lang>

Output:

Same as iterative version.