Self numbers: Difference between revisions
Line 378:
import SelfNums
stop = 1000
Enum.to_list 1..stop |>
Enum.map(&digAndSum/1) |>
|
Revision as of 20:00, 30 October 2020
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
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.
<lang applescript>(*
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}</lang>
C
Sieve based
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>
- 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;
}</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.
Extended
<lang c>#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;
}</lang>
- 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#
via
(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
Elixir
<lang elixir> 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 </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]
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
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 $i is the 100,000,000th self number.") end end popfirst!(last81) push!(last81, gsum(i)) i += 1 end
end
checkselfnumbers()
</lang>
- Output:
1 3 5 7 9 20 31 42 53 64 75 86 97 108 110 121 132 143 154 165 176 187 198 209 211 222 233 244 255 266 277 288 299 310 312 323 334 345 356 367 378 389 400 411 413 424 435 446 457 468 Yes, 1022727208 is the 100,000,000th self number.
Faster version
Contains tweaks peculiar to the "10 to the nth" self number. Timings include compilation times. <lang julia>const MAXCOUNT = 103 * 10000 * 10000 + 11 * 9 + 1
function dosieve!(sieve, digitsum9999)
n = 1 for a in 1:103, b in 1:10000 s = digitsum9999[a] + digitsum9999[b] + n for c in 1:10000 sieve[digitsum9999[c] + s] = true s += 1 end n += 10000 end
end
initdigitsum() = reverse!(vec([sum(k) for k in Iterators.product(9:-1:0, 9:-1:0, 9:-1:0, 9:-1:0)]))
function findselves()
sieve = zeros(Bool, MAXCOUNT+1) println("Sieve time:") @time begin digitsum = initdigitsum() dosieve!(sieve, digitsum) end cnt = 1 for i in 1:MAXCOUNT+1 if !sieve[i] cnt > 50 && break print(i, " ") cnt += 1 end end println() limit, cnt = 1, 0 for i in 0:MAXCOUNT cnt += 1 - sieve[i + 1] if cnt == limit println(lpad(cnt, 10), lpad(i, 12)) limit *= 10 end end
end
@time findselves()
</lang>
- Output:
Sieve time: 7.187635 seconds (2 allocations: 78.203 KiB) 1 3 5 7 9 20 31 42 53 64 75 86 97 108 110 121 132 143 154 165 176 187 198 209 211 222 233 244 255 266 277 288 299 310 312 323 334 345 356 367 378 389 400 411 413 424 435 446 457 468 1 1 10 64 100 973 1000 10188 10000 102225 100000 1022675 1000000 10227221 10000000 102272662 100000000 1022727208 1000000000 10227272649 16.999383 seconds (42.92 k allocations: 9.595 GiB, 0.01% gc time)
Pascal
Just "sieving" with only one follower of every number
Extended to 10.23e9 <lang pascal>program selfnumb; {$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
Certainly puts my previous rubbish attempts (archived here) to shame.
The precise nature of the difference-pattern eludes me, I will admit.
<lang Phix>--
-- 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))</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 100,000,000th safe number is 1,022,727,208 The 1,000,000,000th safe number is 10,227,272,649 completed in 0.1s
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.*/ 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.*/</lang>
- 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
<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. */
- = 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
Standard ML
<lang OCaml> 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;
map ((fn s => print (s ^ " ")) o Int.toString o selfNumberNr) (tabulate (50,fn i=>i+1)) ; selfNumberNr 100000000 ; </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
Wren
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.