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Self numbers

From Rosetta Code
Task
Self numbers
You are encouraged to solve this task according to the task description, using any language you may know.

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

AppleScript[edit]

I couldn't follow the math in the Wikipedia entry, nor the discussion and code here so far. But an initial expedient of generating a list of all the integers from 1 to just over ten times the required number of results and then deleting those that could be derived by the described method revealed the sequencing pattern on which the code below is based. On the test machine, it completes all three of the tests at the bottom in a total of around a millisecond.

(*
Base-10 self numbers by index (single or range).
Follows an observed sequence pattern whereby, after the initial single-digit odd numbers, self numbers are
grouped in runs whose members occur at numeric intervals of 11. Runs after the first one come in blocks of
ten: eight runs of ten numbers followed by two shorter runs. The numeric interval between runs is usually 2,
but that between shorter runs, and their length, depend on the highest-order digit change occurring in them.
This connection with significant digit change means every ten blocks form a higher-order block, every ten
of these a higher-order-still block, and so on.
 
The code below appears to be good up to the last self number before 10^12 — ie. 999,999,999,997, which is
returned as the 97,777,777,792nd such number. After this, instead of zero-length shorter runs, the actual
pattern apparently starts again with a single run of 10, like the one at the beginning.
*)

on selfNumbers(indexRange)
set indexRange to indexRange as list
-- Script object with subhandlers and associated properties.
script |subscript|
property startIndex : beginning of indexRange
property endIndex : end of indexRange
property counter : 0
property currentSelf : -1
property output : {}
 
-- Advance to the next self number in the sequence, append it to the output if required, indicate if finished.
on doneAfterAdding(interval)
set currentSelf to currentSelf + interval
set counter to counter + 1
if (counter < startIndex) then return false
set end of my output to currentSelf
return (counter = endIndex)
end doneAfterAdding
 
-- If necessary, fast forward to the last self number before the lowest-order block containing the first number required.
on fastForward()
if (counter ≥ startIndex) then return
-- The highest-order blocks whose ends this script handles correctly contain 9,777,777,778 self numbers.
-- The difference between equivalently positioned numbers in these blocks is 100,000,000,001.
-- The figures for successively lower-order blocks have successively fewer 7s and 0s!
set indexDiff to 9.777777778E+9
set numericDiff to 1.00000000001E+11
repeat until ((indexDiff < 98) or (counter = startIndex))
set test to counter + indexDiff
if (test < startIndex) then
set counter to test
set currentSelf to (currentSelf + numericDiff)
else
set indexDiff to (indexDiff + 2) div 10
set numericDiff to numericDiff div 10 + 1
end if
end repeat
end fastForward
 
-- Work out a shorter run length based on the most significant digit change about to happen.
on getShorterRunLength()
set shorterRunLength to 9
set temp to (|subscript|'s currentSelf) div 1000
repeat while (temp mod 10 is 9)
set shorterRunLength to shorterRunLength - 1
set temp to temp div 10
end repeat
return shorterRunLength
end getShorterRunLength
end script
 
-- Main process. Start with the single-digit odd numbers and first run.
repeat 5 times
if (|subscript|'s doneAfterAdding(2)) then return |subscript|'s output
end repeat
repeat 9 times
if (|subscript|'s doneAfterAdding(11)) then return |subscript|'s output
end repeat
-- Fast forward if the start index hasn't yet been reached.
tell |subscript| to fastForward()
 
-- Sequencing loop, per lowest-order block.
repeat
-- Eight ten-number runs, each at a numeric interval of 2 from the end of the previous one.
repeat 8 times
if (|subscript|'s doneAfterAdding(2)) then return |subscript|'s output
repeat 9 times
if (|subscript|'s doneAfterAdding(11)) then return |subscript|'s output
end repeat
end repeat
-- Two shorter runs, the second at an interval inversely related to their length.
set shorterRunLength to |subscript|'s getShorterRunLength()
repeat with interval in {2, 2 + (10 - shorterRunLength) * 13}
if (|subscript|'s doneAfterAdding(interval)) then return |subscript|'s output
repeat (shorterRunLength - 1) times
if (|subscript|'s doneAfterAdding(11)) then return |subscript|'s output
end repeat
end repeat
end repeat
end selfNumbers
 
-- Demo calls:
-- First to fiftieth self numbers.
selfNumbers({1, 50})
--> {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}
 
-- One hundred millionth:
selfNumbers(100000000)
--> {1.022727208E+9}
 
-- 97,777,777,792nd:
selfNumbers(9.7777777792E+10)
--> {9.99999999997E+11}

AWK[edit]

 
# syntax: GAWK -f SELF_NUMBERS.AWK
# converted from Go (low memory example)
BEGIN {
print("HH:MM:SS INDEX SELF")
print("-------- ---------- ----------")
count = 0
digits = 1
i = 1
last_self = 0
offset = 9
pow = 10
while (count < 1E8) {
is_self = 1
start = max(i-offset,0)
sum = sum_digits(start)
for (j=start; j<i; j++) {
if (j + sum == i) {
is_self = 0
break
}
sum = ((j+1) % 10 != 0) ? ++sum : sum_digits(j+1)
}
if (is_self) {
last_self = i
if (++count <= 50) {
selfs = selfs i " "
}
}
if (++i % pow == 0) {
pow *= 10
digits++
offset = digits * 9
}
if (count ~ /^10*$/ && arr[count]++ == 0) {
printf("%8s %10s %10s\n",strftime("%H:%M:%S"),count,last_self)
}
}
printf("\nfirst 50 self numbers:\n%s\n",selfs)
exit(0)
}
function sum_digits(x, sum,y) {
while (x) {
y = x % 10
sum += y
x = int(x/10)
}
return(sum)
}
function max(x,y) { return((x > y) ? x : y) }
 
Output:
HH:MM:SS      INDEX       SELF
-------- ---------- ----------
00:36:35          1          1
00:36:35         10         64
00:36:35        100        973
00:36:35       1000      10188
00:36:36      10000     102225
00:36:46     100000    1022675
00:38:49    1000000   10227221
01:03:01   10000000  102272662
05:27:35  100000000 1022727208

first 50 self numbers:
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

C[edit]

Sieve based[edit]

Translation of: Go

About 25% faster than Go (using GCC 7.5.0 -O3) mainly due to being able to iterate through the sieve using a pointer.

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
 
typedef unsigned char bool;
 
#define TRUE 1
#define FALSE 0
#define MILLION 1000000
#define BILLION 1000 * MILLION
#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 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;
}
}
}
free(sv);
printf("Took %lf seconds.\n", (double)(clock() - begin) / CLOCKS_PER_SEC);
return 0;
}
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.

Extended[edit]

Translation of: Pascal
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
 
typedef unsigned char bool;
 
#define TRUE 1
#define FALSE 0
#define MILLION 1000000LL
#define BILLION 1000 * MILLION
#define MAX_COUNT 103LL*10000*10000 + 11*9 + 1
 
int digitSum[10000];
 
void init() {
int i = 9999, s, t, a, b, c, d;
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;
}
}
}
}
}
 
void sieve(bool *sv) {
int a, b, c;
long long s, n = 0;
for (a = 0; a < 103; ++a) {
for (b = 0; b < 10000; ++b) {
s = digitSum[a] + digitSum[b] + n;
for (c = 0; c < 10000; ++c) {
sv[digitSum[c]+s] = TRUE;
++s;
}
n += 10000;
}
}
}
 
int main() {
long long count = 0, limit = 1;
clock_t begin = clock(), end;
bool *p, *sv = (bool*) calloc(MAX_COUNT, sizeof(bool));
init();
sieve(sv);
printf("Sieving took %lf seconds.\n", (double)(clock() - begin) / CLOCKS_PER_SEC);
printf("\nThe first 50 self numbers are:\n");
for (p = sv; p < sv + MAX_COUNT; ++p) {
if (!*p) {
if (++count <= 50) {
printf("%ld ", p-sv);
} else {
printf("\n\n Index Self number\n");
break;
}
}
}
count = 0;
for (p = sv; p < sv + MAX_COUNT; ++p) {
if (!*p) {
if (++count == limit) {
printf("%10lld  %11ld\n", count, p-sv);
limit *= 10;
if (limit == 10 * BILLION) break;
}
}
}
free(sv);
printf("\nOverall took %lf seconds.\n", (double)(clock() - begin) / CLOCKS_PER_SEC);
return 0;
}
Output:
Sieving took 7.429969 seconds.

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 11.574158 seconds.

C#[edit]

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.
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);
}
}
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

Elixir[edit]

 
defmodule SelfNums do
 
def digAndSum(number) when is_number(number) do
Integer.digits(number) |>
Enum.reduce( 0, fn(num, acc) -> num + acc end ) |>
(fn(x) -> x + number end).()
end
 
def selfFilter(list, firstNth) do
numRange = Enum.to_list 1..firstNth
numRange -- list
end
 
end
 
defmodule SelfTest do
 
import SelfNums
stop = 1000
Enum.to_list 1..stop |>
Enum.map(&digAndSum/1) |>
SelfNums.selfFilter(stop) |>
Enum.take(50) |>
IO.inspect
 
end
 
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]

F#[edit]

 
// 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=[email protected]([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)
 
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[edit]

Low memory[edit]

Simple-minded, uses very little memory (no sieve) but slow - over 2.5 minutes.

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))
}
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[edit]

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%.

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))
}
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[edit]

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.

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))
}
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[edit]

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.

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()
 
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[edit]

Translation of: Pascal

Contains tweaks peculiar to the "10 to the nth" self number. Timings include compilation times.

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()
 
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[edit]

Works with: Free Pascal

Just "sieving" with only one follower of every number
Translation of: Go

Extended to 10.23e9

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.
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[edit]

Translation of: AppleScript

Certainly puts my previous rubbish attempts (archived here) to shame.
The precise nature of the difference-pattern eludes me, I will admit.

--
-- Base-10 self numbers by index (single or range).
-- Follows an observed sequence pattern whereby, after the initial single-digit odd numbers, self numbers are
-- grouped in runs whose members occur at numeric intervals of 11. Runs after the first one come in blocks of
-- ten: eight runs of ten numbers followed by two shorter runs. The numeric interval between runs is usually 2,
-- but that between shorter runs, and their length, depend on the highest-order digit change occurring in them.
-- This connection with significant digit change means every ten blocks form a higher-order block, every ten
-- of these a higher-order-still block, and so on.
--
-- The code below appears to be good up to the last self number before 10^12 — ie. 999,999,999,997, which is
-- returned as the 97,777,777,792nd such number. After this, instead of zero-length shorter runs, the actual
-- pattern apparently starts again with a single run of 10, like the one at the beginning.
--
integer startIndex, endIndex, counter
atom currentSelf
sequence output
 
function doneAfterAdding(integer interval, n)
-- Advance to the next self number in the sequence, append it to the output if required, indicate if finished.
for i=1 to n do
currentSelf += interval
counter += 1
if counter >= startIndex then
output &= currentSelf
end if
if counter = endIndex then return true end if
end for
return false
end function
 
function selfNumbers(sequence indexRange)
startIndex = indexRange[1]
endIndex = indexRange[$]
counter = 0
currentSelf = -1
output = {}
 
-- Main process. Start with the single-digit odd numbers and first run.
if doneAfterAdding(2,5) then return output end if
if doneAfterAdding(11,9) then return output end if
 
-- If necessary, fast forward to last self number before the lowest-order block containing first number rqd.
if counter<startIndex then
-- The highest-order blocks whose ends this handles correctly contain 9,777,777,778 self numbers.
-- The difference between equivalently positioned numbers in these blocks is 100,000,000,001.
-- The figures for successively lower-order blocks have successively fewer 7s and 0s!
atom indexDiff = 9777777778,
numericDiff = 100000000001
while indexDiff>=98 and counter!=startIndex do
if counter+indexDiff < startIndex then
counter += indexDiff
currentSelf += numericDiff
else
indexDiff = (indexDiff+2)/10 -- (..78->80->8)
numericDiff = (numericDiff+9)/10 -- (..01->10->1)
end if
end while
end if
 
-- Sequencing loop, per lowest-order block.
while true do
-- Eight ten-number runs, each at a numeric interval of 2 from the end of the previous one.
for i=1 to 8 do
if doneAfterAdding(2,1) then return output end if
if doneAfterAdding(11,9) then return output end if
end for
-- Two shorter runs, the second at an interval inversely related to their length.
integer shorterRunLength = 8,
temp = floor(currentSelf/1000)
-- Work out a shorter run length based on the most significant digit change about to happen.
while remainder(temp,10)=9 do
shorterRunLength -= 1
temp = floor(temp/10)
end while
 
integer interval = 2
for i=1 to 2 do
if doneAfterAdding(interval,1) then return output end if
if doneAfterAdding(11,shorterRunLength) then return output end if
interval += (9-shorterRunLength)*13
end for
end while
end function
 
atom t0 = time()
printf(1,"The first 50 self numbers are:\n")
pp(selfNumbers({1, 50}),{pp_IntFmt,"%3d",pp_IntCh,false})
for p=8 to 9 do
integer n = power(10,p)
printf(1,"The %,dth safe number is %,d\n",{n,selfNumbers({n})[1]})
end for
printf(1,"completed in %s\n",elapsed(time()-t0))
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 100,000,000th safe number is 1,022,727,208
The 1,000,000,000th safe number is 10,227,272,649
completed in 0.1s

Raku[edit]

Translated the low memory version of the Go entry but showed only the first 50 self numbers. The machine for running this task (a Xeon E3110+8GB memory) is showing its age as, 1) took over 6 minutes to complete the Go entry, 2) not even able to run the other two Go alternative entries and 3) needed over 47 minutes to reach 1e6 iterations here. Anyway I will try this on an i5 box later to see how it goes.

Translation of: Go
# 20201127 Raku programming solution
 
my ( $st, $count, $i, $pow, $digits, $offset, $lastSelf, $done, @selfs) =
now, 0, 1, 10, 1, 9, 0, False;
 
# while ( $count < 1e8 ) {
until $done {
my $isSelf = True;
my $sum = (my $start = max ($i-$offset), 0).comb.sum;
loop ( my $j = $start; $j < $i; $j+=1 ) {
if $j+$sum == $i { $isSelf = False and last }
($j+1)%10 != 0 ?? ( $sum+=1 ) !! ( $sum = ($j+1).comb.sum ) ;
}
if $isSelf {
$count+=1;
$lastSelf = $i;
if $count50 {
@selfs.append: $i;
if $count == 50 {
say "The first 50 self numbers are:\n", @selfs;
$done = True;
}
}
}
$i+=1;
if $i % $pow == 0 {
$pow *= 10;
$digits+=1 ;
$offset = $digits * 9
}
}
 
# say "The 100 millionth self number is ", $lastSelf;
# say "Took ", now - $st, " seconds."
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]

REXX[edit]

first 50 self numbers[edit]

/*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.*/
tell = n>0; n= abs(n) /*TELL: show the self numbers if N>0 */
@.= . /*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 the list to the 1st number*/
#= 1 /*the count of self numbers (so far). */
do i=3 until #==n; if @.i=='' then iterate /*Not a self number? Then skip it. */
#= # + 1; list= list i /*bump counter of self #'s; add to list*/
end /*i*/
/*stick a fork in it, we're all done. */
say n " self numbers were found." /*display the title for the output list*/
if tell then say list /*display list of self numbers ──►term.*/
output   when using the default input:
50  self numbers were found.
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[edit]

Translation of: Go (low memory)
/*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. */
#= 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))
output   when using the default input:
the  100,000,000th  self number is:  1,022,727,208

Standard ML[edit]

 
open List;
 
val rec selfNumberNr = fn NR =>
let
val rec sumdgt = fn 0 => 0 | n => Int.rem (n, 10) + sumdgt (Int.quot(n ,10));
val rec isSelf = fn ([],l1,l2) => []
| (x::tt,l1,l2) => if exists (fn i=>i=x) l1 orelse exists (fn i=>i=x) l2
then ( isSelf (tt,l1,l2)) else x::isSelf (tt,l1,l2) ;
 
val rec partcount = fn (n, listIn , count , selfs) =>
if count >= NR then nth (selfs, length selfs + NR - count -1)
else
let
val range = tabulate (81 , fn i => 81*n +i+1) ;
val listOut = map (fn i => i + sumdgt i ) range ;
val selfs = isSelf (range,listIn,listOut)
in
partcount ( n+1 , listOut , count+length (selfs) , selfs )
end;
in
partcount (0,[],0,[])
end;
 
app ((fn s => print (s ^ " ")) o Int.toString o selfNumberNr) (tabulate (50,fn i=>i+1)) ;
selfNumberNr 100000000 ;
 

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

Wren[edit]

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.

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.")
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.