Self numbers: Difference between revisions

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Unsurprisingly, very slow compared to the Go version as Wren is interpreted and uses floating point arithmetic for all numerical work.
Unsurprisingly, very slow compared to the Go version as Wren is interpreted and uses floating point arithmetic for all numerical work.
<lang ecmascript>
<lang ecmascript>var sieve = Fn.new {
var sieve = Fn.new {
var sv = List.filled(2*1e9+9*9+1, false)
var sv = List.filled(2*1e9+9*9+1, false)
var n = 0
var n = 0

Revision as of 10:28, 9 October 2020

Self numbers 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.

A number n is a self number if there is no number g such that g + the sum of g's digits = n. So 18 is not a self number because 9+9=18, 43 is not a self number because 35+5+3=43.
The task is: 

 Display the first 50 self numbers;
 I believe that the 100000000th self number is 1022727208. You should either confirm or dispute my conjecture.

224036583-1 is a Mersenne prime, claimed to also be a self number. Extra credit to anyone proving it.

See also

C

Translation of: Go

The 'sieve based' version.

About 25% faster than Go (using GCC 7.5.0 -O3) mainly due to being able to iterate through the sieve using a pointer. <lang c>#include <stdio.h>

  1. include <stdlib.h>
  2. include <time.h>

typedef unsigned char bool;

  1. define TRUE 1
  2. define FALSE 0
  3. define MILLION 1000000
  4. define BILLION 1000 * MILLION
  5. define MAX_COUNT 2*BILLION + 9*9 + 1

void sieve(bool *sv) { 

   int n = 0, s[8], a, b, c, d, e, f, g, h, i, j;
   for (a = 0; a < 2; ++a) {
       for (b = 0; b < 10; ++b) {
           s[0] = a + b;
           for (c = 0; c < 10; ++c) {
               s[1] = s[0] + c;
               for (d = 0; d < 10; ++d) {
                   s[2] = s[1] + d;
                   for (e = 0; e < 10; ++e) {
                       s[3] = s[2] + e;
                       for (f = 0; f < 10; ++f) {
                           s[4] = s[3] + f;
                           for (g = 0; g < 10; ++g) {
                               s[5] = s[4] + g;
                               for (h = 0; h < 10; ++h) {
                                   s[6] = s[5] + h;
                                   for (i = 0; i < 10; ++i) {
                                       s[7] = s[6] + i;
                                       for (j = 0; j < 10; ++j) {
                                           sv[s[7] + j+ n++] = TRUE;
                                       }
                                   }
                               }
                           }
                       }
                   }
               }
           }
       }
   }

}

int main() { 

   int i, count = 0;
   clock_t begin = clock();
   bool *p, *sv = (bool*) calloc(MAX_COUNT, sizeof(bool));
   sieve(sv);
   printf("The first 50 self numbers are:\n");
   for (p = sv; p < sv + MAX_COUNT; ++p) {
       if (!*p) {
           if (++count <= 50) printf("%ld ", p-sv);
           if (count == 100 * MILLION) {
               printf("\n\nThe 100 millionth self number is %ld\n", p-sv);
               break;
           }
       }
   }
   clock_t end = clock();
   printf("Took %lf seconds.\n", (double)(end - begin) / CLOCKS_PER_SEC);
   return 0;

}</lang>

Output:
The first 50 self numbers are:
1 3 5 7 9 20 31 42 53 64 75 86 97 108 110 121 132 143 154 165 176 187 198 209 211 222 233 244 255 266 277 288 299 310 312 323 334 345 356 367 378 389 400 411 413 424 435 446 457 468 

The 100 millionth self number is 1022727208
Took 1.521486 seconds.

C#

Translation of: Pascal

via

Translation of: Go

(third version) Stripped down, as C# limits the size of an array to Int32.MaxValue, so the sieve isn't large enough to hit the 1,000,000,000th value.

<lang csharp>using System; using static System.Console;

class Program { 

 const int mc = 103 * 1000 * 10000 + 11 * 9 + 1;

 static bool[] sv = new bool[mc + 1];

 static void sieve() { int[] dS = new int[10000];
   for (int a = 9, i = 9999; a >= 0; a--)
     for (int b = 9; b >= 0; b--)
       for (int c = 9, s = a + b; c >= 0; c--)
         for (int d = 9, t = s + c; d >= 0; d--)
           dS[i--] = t + d;
   for (int a = 0, n = 0; a < 103; a++)
     for (int b = 0, d = dS[a]; b < 1000; b++, n += 10000)
       for (int c = 0, s = d + dS[b] + n; c < 10000; c++)
         sv[dS[c] + s++] = true; }

 static void Main() { DateTime st = DateTime.Now; sieve();
   WriteLine("Sieving took {0}s", (DateTime.Now - st).TotalSeconds); 
   WriteLine("\nThe first 50 self numbers are:");
   for (int i = 0, count = 0; count <= 50; i++) if (!sv[i]) {
       count++; if (count <= 50) Write("{0} ", i);
       else WriteLine("\n\n       Index     Self number"); }
   for (int i = 0, limit = 1, count = 0; i < mc; i++)
     if (!sv[i]) if (++count == limit) {
         WriteLine("{0,12:n0}   {1,13:n0}", count, i);
         if (limit == 1e9) break; limit *= 10; }
   WriteLine("\nOverall took {0}s", (DateTime.Now - st). TotalSeconds);
 }

}</lang>

Output:

Timing from tio.run 

Sieving took 3.4266187s

The first 50 self numbers are:
1 3 5 7 9 20 31 42 53 64 75 86 97 108 110 121 132 143 154 165 176 187 198 209 211 222 233 244 255 266 277 288 299 310 312 323 334 345 356 367 378 389 400 411 413 424 435 446 457 468 

       Index     Self number
           1               1
          10              64
         100             973
       1,000          10,188
      10,000         102,225
     100,000       1,022,675
   1,000,000      10,227,221
  10,000,000     102,272,662
 100,000,000   1,022,727,208

Overall took 7.0237244s

F#

<lang fsharp> // Self numbers. Nigel Galloway: October 6th., 2020 let fN g=let rec fG n g=match n/10 with 0->n+g |i->fG i (g+(n%10)) in fG g g let Self=let rec Self n i g=seq{let g=g@([n..i]|>List.map fN) in yield! List.except g [n..i]; yield! Self (n+100) (i+100) (List.filter(fun n->n>i) g)} in Self 0 99 []

Self |> Seq.take 50 |> Seq.iter(printf "%d "); printfn "" printfn "\n%d" (Seq.item 99999999 Self) </lang>

Output:
1 3 5 7 9 20 31 42 53 64 75 86 97 108 110 121 132 143 154 165 176 187 198 209 211 222 233 244 255 266 277 288 299 310 312 323 334 345 356 367 378 389 400 411 413 424 435 446 457 468

1022727208

Go

Low memory

Simple-minded, uses very little memory (no sieve) but slow - over 2.5 minutes. <lang go>package main

import ( 

   "fmt"
   "time"

)

func sumDigits(n int) int { 

   sum := 0
   for n > 0 {
       sum += n % 10
       n /= 10
   }
   return sum

}

func max(x, y int) int { 

   if x > y {
       return x
   }
   return y

}

func main() { 

   st := time.Now()
   count := 0
   var selfs []int
   i := 1
   pow := 10
   digits := 1
   offset := 9
   lastSelf := 0
   for count < 1e8 {
       isSelf := true
       start := max(i-offset, 0)
       sum := sumDigits(start)
       for j := start; j < i; j++ {
           if j+sum == i {
               isSelf = false
               break
           }
           if (j+1)%10 != 0 {
               sum++
           } else {
               sum = sumDigits(j + 1)
           }
       }
       if isSelf {
           count++
           lastSelf = i
           if count <= 50 {
               selfs = append(selfs, i)
               if count == 50 {
                   fmt.Println("The first 50 self numbers are:")
                   fmt.Println(selfs)
               }
           }
       }
       i++
       if i%pow == 0 {
           pow *= 10
           digits++
           offset = digits * 9
       }
   }
   fmt.Println("\nThe 100 millionth self number is", lastSelf)
   fmt.Println("Took", time.Since(st))

}</lang>

Output:
The first 50 self numbers are:
[1 3 5 7 9 20 31 42 53 64 75 86 97 108 110 121 132 143 154 165 176 187 198 209 211 222 233 244 255 266 277 288 299 310 312 323 334 345 356 367 378 389 400 411 413 424 435 446 457 468]

The 100 millionth self number is 1022727208
Took 2m35.531949399s

Sieve based

Simple sieve, requires a lot of memory but quick - around 2 seconds.

Nested 'for's used rather than a recursive function for extra speed.

Have also incorporated Enter your username's suggestion (see Talk page) of using partial sums for each loop which improves performance by about 25%. <lang go>package main

import ( 

   "fmt"
   "time"

)

func sieve() []bool { 

   sv := make([]bool, 2*1e9+9*9 + 1)
   n := 0
   var s [8]int
   for a := 0; a < 2; a++ {
       for b := 0; b < 10; b++ {
           s[0] = a + b
           for c := 0; c < 10; c++ {
               s[1] = s[0] + c
               for d := 0; d < 10; d++ {
                   s[2] = s[1] + d
                   for e := 0; e < 10; e++ {
                       s[3] = s[2] + e
                       for f := 0; f < 10; f++ {
                           s[4] = s[3] + f
                           for g := 0; g < 10; g++ {
                               s[5] = s[4] + g
                               for h := 0; h < 10; h++ {
                                   s[6] = s[5] + h 
                                   for i := 0; i < 10; i++ {
                                       s[7] = s[6] + i
                                       for j := 0; j < 10; j++ {
                                           sv[s[7]+j+n] = true
                                           n++
                                       }
                                   }
                               }
                           }
                       }
                   }
               }
           }
       }
   }
   return sv

}

func main() { 

   st := time.Now()
   sv := sieve()
   count := 0
   fmt.Println("The first 50 self numbers are:")
   for i := 0; i < len(sv); i++ {
       if !sv[i] {
           count++
           if count <= 50 {
               fmt.Printf("%d ", i)
           }
           if count == 1e8 {
               fmt.Println("\n\nThe 100 millionth self number is", i)
               break
           }
       }
   }
   fmt.Println("Took", time.Since(st))

}</lang>

Output:
The first 50 self numbers are:
1 3 5 7 9 20 31 42 53 64 75 86 97 108 110 121 132 143 154 165 176 187 198 209 211 222 233 244 255 266 277 288 299 310 312 323 334 345 356 367 378 389 400 411 413 424 435 446 457 468 

The 100 millionth self number is 1022727208
Took 1.984969034s

Extended

Translation of: Pascal

This uses horst.h's ideas (see Talk page) to find up to the 1 billionth self number in a reasonable time and using less memory than the simple 'sieve based' approach above would have needed. <lang go>package main

import ( 

   "fmt"
   "time"

)

const MAX_COUNT = 103*1e4*1e4 + 11*9 + 1

var sv = make([]bool, MAX_COUNT+1) var digitSum = make([]int, 1e4)

func init() { 

   i := 9999
   var s, t int
   for a := 9; a >= 0; a-- {
       for b := 9; b >= 0; b-- {
           s = a + b
           for c := 9; c >= 0; c-- {
               t = s + c
               for d := 9; d >= 0; d-- {
                   digitSum[i] = t + d
                   i--
               }
           }
       }
   }

}

func sieve() { 

   n := 0
   for a := 0; a < 103; a++ {
       for b := 0; b < 1e4; b++ {
           s := digitSum[a] + digitSum[b] + n
           for c := 0; c < 1e4; c++ {
               sv[digitSum[c]+s] = true
               s++
           }
           n += 1e4
       }
   }

}

func main() { 

   st := time.Now()
   sieve()
   fmt.Println("Sieving took", time.Since(st))
   count := 0
   fmt.Println("\nThe first 50 self numbers are:")
   for i := 0; i < len(sv); i++ {
       if !sv[i] {
           count++
           if count <= 50 {
               fmt.Printf("%d ", i)
           } else {
               fmt.Println("\n\n     Index  Self number")
               break
           }
       }
   }
   count = 0
   limit := 1
   for i := 0; i < len(sv); i++ {
       if !sv[i] {
           count++
           if count == limit {
               fmt.Printf("%10d  %11d\n", count, i)
               limit *= 10
               if limit == 1e10 {
                   break
               }
           }
       }
   }
   fmt.Println("\nOverall took", time.Since(st))

}</lang>

Output:
Sieving took 8.286841692s

The first 50 self numbers are:
1 3 5 7 9 20 31 42 53 64 75 86 97 108 110 121 132 143 154 165 176 187 198 209 211 222 233 244 255 266 277 288 299 310 312 323 334 345 356 367 378 389 400 411 413 424 435 446 457 468 

     Index  Self number
         1            1
        10           64
       100          973
      1000        10188
     10000       102225
    100000      1022675
   1000000     10227221
  10000000    102272662
 100000000   1022727208
1000000000  10227272649

Overall took 14.647314803s

Julia

The code first bootstraps a sliding window of size 81 and then uses this as a sieve. Note that 81 is the window size because the sum of digits of 999,999,999 (the largest digit sum of a counting number less than 1022727208) is 81. <lang julia>gsum(i) = sum(digits(i)) + i isnonself(i) = any(x -> gsum(x) == i, i-1:-1:i-max(1, ndigits(i)*9)) const last81 = filter(isnonself, 1:5000)[1:81]

function checkselfnumbers() 

   i, selfcount = 1, 0
   while selfcount <= 100_000_000 && i <= 1022727208
       if !(i in last81)
           selfcount += 1
           if selfcount < 51
               print(i, " ")
           elseif selfcount == 51
               println()
           elseif selfcount == 100_000_000
               println(i == 1022727208 ?
                   "Yes, $i is the 100,000,000th self number." :
                   "No, instead $i is the 100,000,000th self number.")
           end
       end
       popfirst!(last81)
       push!(last81, gsum(i))
       i += 1
   end

end

checkselfnumbers()

</lang>

Output:
1 3 5 7 9 20 31 42 53 64 75 86 97 108 110 121 132 143 154 165 176 187 198 209 211 222 233 244 255 266 277 288 299 310 312 323 334 345 356 367 378 389 400 411 413 424 435 446 457 468
Yes, 1022727208 is the 100,000,000th self number.

Faster version

Translation of: Pascal

Contains tweaks peculiar to the "10 to the nth" self number. Timings include compilation times. <lang julia>const MAXCOUNT = 103 * 10000 * 10000 + 11 * 9 + 1

function dosieve!(sieve, digitsum9999) 

   n = 1
   for a in 1:103, b in 1:10000
       s = digitsum9999[a] + digitsum9999[b] + n
       for c in 1:10000
           sieve[digitsum9999[c] + s] = true
           s += 1
       end
       n += 10000
   end

end

initdigitsum() = reverse!(vec([sum(k) for k in Iterators.product(9:-1:0, 9:-1:0, 9:-1:0, 9:-1:0)]))

function findselves() 

   sieve = zeros(Bool, MAXCOUNT+1)
   println("Sieve time:")
   @time begin
       digitsum = initdigitsum()
       dosieve!(sieve, digitsum)
   end
   cnt = 1
   for i in 1:MAXCOUNT+1
       if !sieve[i]
           cnt > 50 && break
           print(i, " ")
           cnt += 1
       end
   end
   println()
   limit, cnt = 1, 0
   for i in 0:MAXCOUNT
       cnt += 1 - sieve[i + 1]
       if cnt == limit
           println(lpad(cnt, 10), lpad(i, 12))
           limit *= 10
       end
   end

end

@time findselves()

</lang>

Output:
Sieve time:
  7.187635 seconds (2 allocations: 78.203 KiB)
1 3 5 7 9 20 31 42 53 64 75 86 97 108 110 121 132 143 154 165 176 187 198 209 211 222 233 244 255 266 277 288 299 310 312 323 334 345 356 367 378 389 400 411 413 424 435 446 457 468
         1           1
        10          64
       100         973
      1000       10188
     10000      102225
    100000     1022675
   1000000    10227221
  10000000   102272662
 100000000  1022727208
1000000000 10227272649
 16.999383 seconds (42.92 k allocations: 9.595 GiB, 0.01% gc time)

Pascal

Works with: Free Pascal


Just "sieving" with only one follower of every number

Translation of: Go

Extended to 10.23e9 <lang pascal>program selfnumb; {$IFDEF FPC} 

 {$MODE Delphi}
 {$Optimization ON,ALL}

{$IFEND} {$IFDEF DELPHI} {$APPTYPE CONSOLE} {$IFEND} uses 

 sysutils;

const 

 MAXCOUNT =103*10000*10000+11*9+ 1;

type 

 tDigitSum9999 = array[0..9999] of Uint8;
 tpDigitSum9999 = ^tDigitSum9999;

var 

 DigitSum9999 : tDigitSum9999;
 sieve : array of boolean;

procedure dosieve; var 

 pSieve : pBoolean;
 pDigitSum :tpDigitSum9999;
 n,c,b,a,s : NativeInt;

Begin 

 pSieve := @sieve[0];
 pDigitSum := @DigitSum9999[0];
 n := 0;
 for a := 0 to 102 do
   for b := 0 to 9999 do
   Begin
     s := pDigitSum^[a]+pDigitSum^[b]+n;
     for c := 0 to 9999 do
     Begin
       pSieve[pDigitSum^[c]+s] := true;
       s+=1;
     end;
     inc(n,10000);
   end;

end;

procedure InitDigitSum; var 

 i,d,c,b,a : NativeInt;

begin 

 i := 9999;
 for a := 9 downto 0 do
   for b := 9 downto 0 do
     for c := 9 downto 0 do
       for d := 9 downto 0 do
       Begin
         DigitSum9999[i] := a+b+c+d;
         dec(i);
       end;

end;

procedure OutPut(cnt,i:NativeUint); Begin 

 writeln(cnt:10,i:12);

end;

var 

 pSieve : pboolean;
 T0 : Uint64;
 i,cnt,limit,One: NativeUInt;

BEGIN 

 setlength(sieve,MAXCOUNT);
 pSieve := @sieve[0];
 T0 := GetTickCount64;
 InitDigitSum;
 dosieve;
 writeln('Sievetime : ',(GetTickCount64-T0 )/1000:8:3,' sec');
 //find first 50
 cnt := 0;
 for i := 0 to MAXCOUNT do
 Begin
   if NOT(pSieve[i]) then
   Begin
     inc(cnt);
     if cnt <= 50 then
       write(i:4)
     else
       BREAK;
   end;
 end;
 writeln;
 One := 1;
 limit := One;
 cnt := 0;
 for i := 0 to MAXCOUNT do
 Begin
   inc(cnt,One-Ord(pSieve[i]));
   if cnt = limit then
   Begin
     OutPut(cnt,i);
     limit := limit*10;
   end;
 end;

END.</lang>

Output:
 time ./selfnumb
Sievetime :    6.579 sec
   1   3   5   7   9  20  31  42  53  64  75  86  97 108 110 121 132 143 154 165 176 187 198 209 211 222 233 244 255 266 277 288 299 310 312 323 334 345 356 367 378 389 400 411 413 424 435 446 457 468
         1           1
        10          64
       100         973
      1000       10188
     10000      102225
    100000     1022675
   1000000    10227221
  10000000   102272662
 100000000  1022727208
1000000000 10227272649

real  0m13,252s

Phix

Translation of: Go

Replacing the problematic sv[a+b+... line with a bit of dirty inline assembly improved performance by 90%
(Of course you lose bounds checking, type checking, negative subscripts, fraction handling, and all that jazz.)
We use a string of Y/N for the sieve to force one byte per element ('\0' and 1 would be equally valid). <lang Phix>if machine_bits()=32 then crash("requires 64 bit") end if

function sieve() 

   string sv = repeat('N',2*1e9+9*9+1)
   integer n = 0
   for a=0 to 1 do
       for b=0 to 9 do
           for c=0 to 9 do
               for d=0 to 9 do                  
                  for e=0 to 9 do
                       for f=0 to 9 do                         
                           for g=0 to 9 do
                               for h=0 to 9 do
                                   for i=0 to 9 do
                                       for j=0 to 9 do

-- n += 1 -- sv[a+b+c+d+e+f+g+h+i+j+n] = 'Y'

  1. ilASM{
 [32]  -- (allows clean compilation on 32 bit, before crash as above)
 [64]
   mov rax,[a]
   mov r12,[b]
   mov r13,[c]
   mov r14,[d]
   mov r15,[e]
   add r12,r13
   add r14,r15
   add rax,r12 
   mov rdi,[sv]
   add rax,r14 
   mov r12,[f]
   mov r13,[g]
   mov r14,[h]
   mov r15,[i]
   add r12,r13
   shl rdi,2
   mov rcx,[n]
   mov r13,[j]
   add r14,r15
   add rax,r12 
   add rax,r14 
   add r13,rcx
   add rax,r13 
   add rcx,1
   mov byte[rdi+rax],'Y'
   mov [n],rcx }
                                       end for
                                   end for                                  
                               end for
                           end for  
                       end for
                   end for
               end for
           end for
           printf(1,"%d,%d\r",{a,b}) -- (show progress)
       end for
   end for
   return sv

end function

atom t0 = time() string sv = sieve() printf(1,"sieve build took %s\n",{elapsed(time()-t0)}) integer count = 0 printf(1,"The first 50 self numbers are:\n") for i=1 to length(sv) do 

   if sv[i]='N' then
       count += 1
       if count <= 50 then
           printf(1,"%3d ",i-1)
           if remainder(count,10)=0 then
               printf(1,"\n")
           end if
       end if
       if count == 1e8 then
           printf(1,"\nThe %,dth self number is %,d (%s)\n",
                    {count,i-1,elapsed(time()-t0)})
           exit
       end if
   end if

end for</lang>

Output:
sieve build took 6.6s
The first 50 self numbers are:
  1   3   5   7   9  20  31  42  53  64
 75  86  97 108 110 121 132 143 154 165
176 187 198 209 211 222 233 244 255 266
277 288 299 310 312 323 334 345 356 367
378 389 400 411 413 424 435 446 457 468

The 100,000,000th self number is 1,022,727,208 (11.0s)

generator dictionary

While this is dog-slow (see shocking estimate below), it is interesting to note that even by the time it generates the 10,000,000th number, it is only having to juggle a mere 27 generators. Just a shame that we had to push over 10,000,000 generators onto that stack, and tried to push quite a few more. Memory use is pretty low, around ~4MB.
[unlike the above, this is perfectly happy on both 32 and 64 bit]
Aside: the getd_index() check is often worth trying with phix dictionaries: if there is a high probability that the key already exists, it will yield a win, but with a low probability it will just be unhelpful overhead. <lang Phix>integer gd = new_dict(), g = 1, gdhead = 2, n = 0

function ng(integer n) 

   integer r = n
   while r do
       n += remainder(r,10)
       r = floor(r/10)
   end while
   return n

end function

function self(integer /*i*/) -- note: assumes sequentially invoked (arg i unused) integer nxt 

   n += 1
   while n=gdhead do
       while g<=gdhead do
           nxt = ng(g)
           if getd_index(nxt, gd)=NULL then -- (~25% gain)
               setd(nxt,0,gd)
           end if
           g += 1
       end while
       integer wasgdhead = gdhead
       while true do
           gdhead = pop_dict(gd)[1]
           if gdhead!=wasgdhead then exit end if

--  ?{"ding",wasgdhead} -- 2, once only... 

       end while
       nxt = ng(gdhead)

-- if getd_index(nxt, gd)=NULL then -- (~1% loss) 

           setd(nxt,0,gd)

-- end if 

       n += (n!=gdhead)
   end while
   return n

end function

atom t0 = time() printf(1,"The first 50 self numbers are:\n") pp(apply(tagset(50),self),{pp_IntFmt,"%3d",pp_IntCh,false})

constant limit = 10000000 integer chk = 100 printf(1,"\n i n size time\n") for i=51 to limit do 

   n = self(i)
   if i=chk then
       printf(1,"%,11d %,11d %6d %s\n",{i,n,dict_size(gd),elapsed(time()-t0)})
       chk *= 10
   end if

end for printf(1,"\nEstimated time for %,d :%s\n",{1e8,elapsed((time()-t0)*1e8/limit)})</lang>

Output:
The first 50 self numbers are:
{  1,  3,  5,  7,  9, 20, 31, 42, 53, 64, 75, 86, 97,108,110,121,132,143,
 154,165,176,187,198,209,211,222,233,244,255,266,277,288,299,310,312,323,
 334,345,356,367,378,389,400,411,413,424,435,446,457,468}

          i           n   size time
        100         973     18 0.1s
      1,000      10,188     13 0.2s
     10,000     102,225     10 1.0s
    100,000   1,022,675     20 9.3s
  1,000,000  10,227,221     17 1 minute and 37s
 10,000,000 102,272,662     27 16 minutes and 40s

Estimated time for 100,000,000 :2 hours, 46 minutes and 37s

For the record, I would not be at all surprised should a translation of this beat 20 minutes (for 1e8)

REXX

first 50 self numbers

<lang rexx>/*REXX program displays N self numbers, aka Colombian or Devlali numbers. OEIS A3052.*/ parse arg n . /*obtain optional argument from the CL.*/ if n== | n=="," then n= 50 /*Not specified? Then use the default.*/ @.=. /*initialize the array of self numbers.*/ 

          do j=1  for n*10                      /*scan through ten times the #s wanted.*/
          $= j                                  /*1st part of sum is the number itself.*/
               do k=1  for length(j)            /*sum the decimal digits in the number.*/
               $= $ + substr(j, k, 1)           /*add a particular digit to the sum.   */
               end   /*k*/
          @.$=                                  /*mark  J  as not being a self number. */
          end        /*j*/

list= 1 /*initialize list to the first number. */

  1. = 1 /*the count of self numbers (so far). */
          do i=3  until #==n                    /*traipse through array 'til satisfied.*/
          if @.i==  then iterate              /*Not a self number?   Then skip it.   */
          #= # + 1                              /*bump the counter of self numbers.    */
          list= list i                          /*append it to the list of self numbers*/
          end   /*i*/

say 'the first ' n " self numbers are:" /*display the title for the output list*/ say list /*display list of self numbers ──►term.*/ exit 0 /*stick a fork in it, we're all done. */</lang>

output   when using the default input:
the first  50  self numbers are:
1 3 5 7 9 20 31 42 53 64 75 86 97 108 110 121 132 143 154 165 176 187 198 209 211 222 233 244 255 266 277 288 299 310 312 323 334 345 356 367 378 389 400 411 413 424 435 446 457 468

ten millionth self number

Translation of: Go (low memory)

<lang rexx>/*REXX pgm displays the Nth self number, aka Colombian or Devlali numbers. OEIS A3052.*/ numeric digits 20 /*ensure enough decimal digits for #. */ parse arg high . /*obtain optional argument from the CL.*/ if high== | high=="," then high= 100000000 /*Not specified? Then use 100M default*/ i= 1; pow= 10; digs= 1; offset= 9; $= 0 /*$: the last self number found. */

  1. = 0 /*count of self numbers (so far). */
    do while #<high;          isSelf= 1         /*assume a self number   (so far).     */
    start= max(i-offset, 0)
    sum= sumDigs(start)

       do j=start  to i-1
       if j+sum==i  then do;  isSelf= 0         /*found a   non  self number.          */
                              iterate           /*keep looking for more self numbers.  */
                         end
       jp= j + 1                                /*shortcut variable for next statement.*/
       if jp//10==0   then sum= sumDigs(jp)
                      else sum= sum + 1
       end   /*j*/

    if isSelf  then do;  #= # + 1               /*bump the count of self numbers.      */
                         $= i                   /*save the last self number found.     */
                    end
    i= i + 1
    if i//pow==0  then do;  pow= pow * 10
                            digs= digs + 1      /*bump the number of decimal digits.   */
                            offset= digs * 9    /*bump the offset by a factor of nine. */
                       end
    end   /*while*/

say say 'the ' commas(high)th(high) " self number is: " commas($) exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ sumDigs: parse arg s 2 x; do k=1 for length(x) /*get 1st dig, & also get the rest.*/ 

                           s= s + substr(x, k, 1)  /*add a particular digit to the sum.*/
                           end  /*k*/;  return s

/*──────────────────────────────────────────────────────────────────────────────────────*/ commas: parse arg _; do c=length(_)-3 to 1 by -3; _=insert(',', _, c); end; return _ th: parse arg th; return word('th st nd rd',1+(th//10)*(th//100%10\==1)*(th//10<4))</lang>

output   when using the default input:
the  100,000,000th  self number is:  1,022,727,208

Wren

Translation of: Go

Just the sieve based version as the low memory version would take too long to run in Wren.

Note that you need a lot of memory to run this as Bools in Wren require 8 bytes of storage compared to 1 byte in Go.

Unsurprisingly, very slow compared to the Go version as Wren is interpreted and uses floating point arithmetic for all numerical work. <lang ecmascript>var sieve = Fn.new { 

   var sv = List.filled(2*1e9+9*9+1, false)
   var n = 0
   var s = [0] * 8
   for (a in 0..1) {
       for (b in 0..9) {
           s[0] = a + b
           for (c in 0..9) {
               s[1] = s[0] + c
               for (d in 0..9) { 
                  s[2] = s[1] + d                   
                  for (e in 0..9) {
                       s[3] = s[2] + e
                       for (f in 0..9) {
                           s[4] = s[3] + f                           
                           for (g in 0..9) {
                               s[5] = s[4] + g
                               for (h in 0..9) {
                                   s[6] = s[5] + h
                                   for (i in 0..9) {
                                       s[7] = s[6] + i
                                       for (j in 0..9) {                                           
                                          sv[s[7] + j + n] = true
                                          n = n + 1
                                       }
                                   }                                    
                               }
                           }  
                       }
                   }
               }
           }
       }
   }
   return sv

}

var st = System.clock var sv = sieve.call() var count = 0 System.print("The first 50 self numbers are:") for (i in 0...sv.count) { 

   if (!sv[i]) {
       count = count + 1
       if (count <= 50) System.write("%(i) ")
       if (count == 1e8) {
           System.print("\n\nThe 100 millionth self number is %(i)")
           break
       }
   }

} System.print("Took %(System.clock-st) seconds.")</lang>

Output:
The first 50 self numbers are:
1 3 5 7 9 20 31 42 53 64 75 86 97 108 110 121 132 143 154 165 176 187 198 209 211 222 233 244 255 266 277 288 299 310 312 323 334 345 356 367 378 389 400 411 413 424 435 446 457 468 

The 100 millionth self number is 1022727208
Took 222.789713 seconds.
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