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Smallest numbers

From Rosetta Code
Smallest 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.
Task

Smallest positive integer   k   such that the decimal expansion of   kk   contains   n,   where   n  <  51

ALGOL 68[edit]

Works with: ALGOL 68G version Any - tested with release 2.8.3.win32

Uses ALGOL 68G's LONG LONG INT which provides large integers (the default precision is sufficient for the task). Also uses the ALGOL 68G string in string procedure.

BEGIN # find the smallest k such that the decimal representation of k^k contains n for 0 <= n <= 50 #
# start with powers up to 20^20, if this proves insufficient, the kk array will be extended #
FLEX[ 1 : 20 ]STRING kk;
FOR k TO UPB kk DO kk[ k ] := whole( LONG LONG INT( k ) ^ k, 0 ) OD;
# find the numbers #
FOR i FROM 0 TO 50 DO
STRING n = whole( i, 0 );
BOOL try again := TRUE;
WHILE try again DO
try again := FALSE;
BOOL found := FALSE;
FOR k FROM LWB kk TO UPB kk WHILE NOT found DO
IF string in string( n, NIL, kk[ k ] ) THEN
found := TRUE;
print( ( " ", whole( k, -3 ) ) )
FI
OD;
IF NOT found THEN
# haven't got enough k^k values - get some more #
kk := HEAP[ 1 : UPB kk * 2 ]STRING;
FOR k TO UPB kk DO kk[ k ] := whole( LONG LONG INT( k ) ^ k, 0 ) OD;
try again := TRUE
FI
OD;
IF i MOD 10 = 9 THEN print( ( newline ) ) FI
OD
END
Output:
   9   1   3   5   2   4   4   3   7   9
  10  11   5  19  22  26   8  17  16  19
   9   8  13   7  17   4  17   3  11  18
  13   5  23  17  18   7  17  15   9  18
  16  17   9   7  12  28   6  23   9  24
  23

F#[edit]

 
// Smallest number: Nigel Galloway. April 13th., 2021
let rec fG n g=match bigint.DivRem(n,if g<10 then 10I else 100I) with (_,n) when (int n)=g->true |(n,_) when n=0I->false |(n,_)->fG n g
{0..50}|>Seq.iter(fun g->printf "%d " (1+({1..0x0FFFFFFF}|>Seq.map(fun n->(bigint n)**n)|>Seq.findIndex(fun n->fG n g)))); printfn ""
 
Output:
9 1 3 5 2 4 4 3 7 9 26 11 14 21 22 26 8 25 16 19 23 21 13 25 17 5 25 3 11 18 27 5 23 24 22 7 17 16 21 19 18 17 9 7 12 28 18 23 27 24 23
Real: 00:00:00.005

Factor[edit]

Works with: Factor version 0.99 2021-02-05
USING: formatting grouping io kernel lists lists.lazy
math.functions present sequences ;
 
: smallest ( m -- n )
present 1 lfrom [ dup ^ present subseq? ] with lfilter car ;
 
51 <iota> [ smallest ] map 10 group
[ [ "%3d" printf ] each nl ] each
Output:
  9  1  3  5  2  4  4  3  7  9
 10 11  5 19 22 26  8 17 16 19
  9  8 13  7 17  4 17  3 11 18
 13  5 23 17 18  7 17 15  9 18
 16 17  9  7 12 28  6 23  9 24
 23

FreeBASIC[edit]

Reuses some code from Arbitrary-precision_integers_(included)#FreeBASIC.

#Include once "gmp.bi"
Dim Shared As Zstring * 100000000 outtext
 
Function Power(number As String,n As Uinteger) As String'automate precision
#define dp 3321921
Dim As __mpf_struct _number,FloatAnswer
Dim As Ulongint ln=Len(number)*(n)*4
If ln>dp Then ln=dp
mpf_init2(@FloatAnswer,ln)
mpf_init2(@_number,ln)
mpf_set_str(@_number,number,10)
mpf_pow_ui(@Floatanswer,@_number,n)
gmp_sprintf( @outtext,"%." & Str(n) & "Ff",@FloatAnswer )
Var outtxt=Trim(outtext)
If Instr(outtxt,".") Then outtxt= Rtrim(outtxt,"0"):outtxt=Rtrim(outtxt,".")
Return Trim(outtxt)
End Function
 
function is_substring( s as string, j as string ) as boolean
dim as integer nj = len(j), ns = len(s)
for i as integer = 1 to ns - nj + 1
if mid(s,i,nj) = j then return true
next i
return false
end function
 
dim as integer k
 
for i as uinteger = 0 to 50
k = 0
do
k = k + 1
loop until is_substring( Power(str(k), k), str(i) )
print k;" ";
next i
Output:
 9  1  3  5  2  4  4  3  7  9  10  11  5  19  22  26  8  17  16  19  9  8  13  7  17  4  17  3  11  18  13  5  23  17  18  7  17  15  9  18  16  17  9  7  12  28  6  23  9  24  23

Go[edit]

Translation of: Wren
package main
 
import (
"fmt"
"math/big"
"strconv"
"strings"
)
 
func main() {
var res []int64
for n := 0; n <= 50; n++ {
ns := strconv.Itoa(n)
k := int64(1)
for {
bk := big.NewInt(k)
s := bk.Exp(bk, bk, nil).String()
if strings.Contains(s, ns) {
res = append(res, k)
break
}
k++
}
}
fmt.Println("The smallest positive integers K where K ^ K contains N (0..50) are:")
for i, n := range res {
fmt.Printf("%2d ", n)
if (i+1)%17 == 0 {
fmt.Println()
}
}
}
Output:
The smallest positive integers K where K ^ K contains N (0..50) are:
 9  1  3  5  2  4  4  3  7  9 10 11  5 19 22 26  8 
17 16 19  9  8 13  7 17  4 17  3 11 18 13  5 23 17 
18  7 17 15  9 18 16 17  9  7 12 28  6 23  9 24 23 

jq[edit]

Works with gojq, the Go implementation of jq

The integer precision of stedolan jq is insufficient for this task.
 
# if the input and $b are integers, then gojq will preserve precision
def power($b): . as $a | reduce range(0; $b) as $i (1; . * $a);
 
def smallest_k:
tostring as $n
| first( range(1; infinite) | select( power(.) | tostring | contains($n))) ;
 
 
# Formatting
 
def lpad($len): tostring | ($len - length) as $l | (" " * $l)[:$l] + .;
 
def nwise($n):
def n: if length <= $n then . else .[0:$n] , (.[$n:] | n) end;
n;
 
The task:
 
def task($n):
[range(0; $n) | smallest_k | lpad(3) ]
| nwise(10)
| join(" ");
 
task(51)
Output:

As for Factor, for example.

Julia[edit]

hasinktok(n) = for k in 1:100000 contains("$(BigInt(k)^k)", "$n") && return k end
 
foreach(p -> print(rpad(p[2], 4), p[1] % 17 == 0 ? "\n" : ""), enumerate(map(hasinktok, 0:50)))
 
Output:
9   1   3   5   2   4   4   3   7   9   10  11  5   19  22  26  8   
17  16  19  9   8   13  7   17  4   17  3   11  18  13  5   23  17  
18  7   17  15  9   18  16  17  9   7   12  28  6   23  9   24  23

Nim[edit]

Library: bignum
import strformat, strutils
import bignum
 
var k = 1u
var toFind = {0..50}
var results: array[0..50, uint]
while toFind.card > 0:
let str = $(pow(newInt(k), k))
for n in toFind:
if str.find($n) >= 0:
results[n] = k
toFind.excl(n)
inc k
 
echo "Smallest values of k such that k^k contains n:"
for n, k in results:
stdout.write &"{n:2} → {k:<2} ", if (n + 1) mod 9 == 0: '\n' else: ' '
echo()
Output:
Smallest values of k such that k^k contains n:
 0 → 9      1 → 1      2 → 3      3 → 5      4 → 2      5 → 4      6 → 4      7 → 3      8 → 7    
 9 → 9     10 → 10    11 → 11    12 → 5     13 → 19    14 → 22    15 → 26    16 → 8     17 → 17   
18 → 16    19 → 19    20 → 9     21 → 8     22 → 13    23 → 7     24 → 17    25 → 4     26 → 17   
27 → 3     28 → 11    29 → 18    30 → 13    31 → 5     32 → 23    33 → 17    34 → 18    35 → 7    
36 → 17    37 → 15    38 → 9     39 → 18    40 → 16    41 → 17    42 → 9     43 → 7     44 → 12   
45 → 28    46 → 6     47 → 23    48 → 9     49 → 24    50 → 23   

Pascal[edit]

Works with: Free Pascal

made like Phix but own multiplikation to BASE 1E9 here

program K_pow_K;
//First occurence of a numberstring with max DIGTIS digits in k^k
{$IFDEF FPC}
{$MODE DELPHI}
{$Optimization ON,ALL}
{$ELSE}
{$APPTYPE CONSOLE}
{$ENDIF}
 
uses
sysutils;
const
LongWordDec = 1000*1000*1000;
Digits = 6;
 
type
tMulElem = Uint32;
tMul = array of tMulElem;
tpMul = pUint32;
tFound = Uint32;
var
Pot_N_str : AnsiString;
Str_Found : array of tFound;
FirstMissing :NativeInt;
T0 : INt64;
 
procedure Out_Results(number,found:NativeInt);
var
i : NativeInt;
Begin
writeln;
writeln(#10,'Found: ',found,' at ',number,' with ',length(Pot_N_str),
' digits in Time used ',(GetTickCount64-T0)/1000:8:3,' secs');
writeln ;
writeln(' 0 1 2 3 4 5 6 7 8 9');
write(' |__________________________________________________');
For i := 0 to 99 do//decLimit-1 do
begin
if i MOD 10 = 0 then
Begin
writeln;
write((i DIV 10)*10:10,'|');
end;
number := Str_Found[i]-1;
if number > 0 then
write(number:5);
end;
writeln;
end;
 
procedure Mul_12(var Mul1,Mul2:tMul);
//Mul2 = Mul1*Mul2;
var
TmpMul : tMul;
carry,
n,prod: Uint64;
lmt1,lmt2,i,j : NativeInt;
begin
lmt1 := High(MUl1);
lmt2 := High(Mul2);
setlength(TmpMul,lmt1+lmt2+3);
For i := 0 to lmt1 do
Begin
carry := 0;
n := Mul1[i];
For j := 0 to lmt2 do
Begin
prod := n*Mul2[j]+TmpMul[i+j]+carry;
carry := prod DIV LongWordDec;
TmpMul[i+j]:=prod-carry*LongWordDec;
end;
TmpMul[i+lmt2+1] += carry;
end;
Mul2 := TmpMul;
setlength(TmpMul,0);
i := High(Mul2);
while (i>=1) AND (Mul2[i]=0) do
dec(i);
setlength(Mul2,i+1);
end;
 
procedure ConvToStr(var s:Ansistring;const Mul:tMul;i:NativeInt);
var
s9: string[9];
pS : pChar;
j,k : NativeInt;
begin
// i := High(MUL);
j := (i+1)*9;
setlength(s,j+1);
pS := pChar(s);
// fill complete with '0'
fillchar(pS[0],j,'0');
str(Mul[i],S9);
j := length(s9);
move(s9[1],pS[0],j);
k := j;
dec(i);
If i >= 0 then
repeat
str(Mul[i],S9);// no leading '0'
j := length(s9);
inc(k,9);
//move to the right place, leading '0' is already there
move(s9[1],pS[k-j],j);
dec(i);
until i<0;
setlength(s,k);
end;
 
function CheckOneString(const s:Ansistring;pow:NativeInt):NativeInt;
//check every possible number from one to DIGITS digits
var
i,k,lmt,num : NativeInt;
begin
result := 0;
 
lmt := length(s);
For i := 1 to lmt do
Begin
k := i;
num := 0;
repeat
num := num*10+ Ord(s[k])-Ord('0');
IF (num >= FirstMissing) AND (str_Found[num] = 0) then
begin
str_Found[num]:= pow+1;
// commatize only once. reference counted string
inc(result);
if num =FirstMissing then
Begin
while str_Found[FirstMissing] <> 0 do
inc(FirstMissing);
end;
end;
inc(k)
until (k>lmt) or (k-i >DIGITS-1);
end;
end;
 
var
MulErg,Square :tMUl;
number,i,j,found,decLimit: Int32;
Begin
T0 := GetTickCount64;
decLimit := 1;
For i := 1 to digits do
decLimit *= 10;
setlength(Str_Found,decLimit);
 
found := 0;
FirstMissing := 0;
number := 1;
repeat
setlength(MulErg,1);
MulErg[0] := 1;
setlength(Square,1);
Square[0]:= number;
 
If number AND 1 <> 0 then
MulErg[0] := number;
j := 2;
while j <= number do
Begin
Mul_12(Square,Square);
If number AND J <> 0 then
Mul_12(Square,MulErg);
j:= j*2;
end;
ConvToStr(Pot_N_str,MulErg,High(MulErg));
inc(found,CheckOneString(Pot_N_str,number));
inc(number);
if number AND 511 = 0 then
write(#13,number:7,' with ',length(Pot_N_str), ' digits.Found ',found);
until found >=decLimit;
Out_Results(number,found);
end.
 
Output:
TIO.RUN for 6 Digits

    512 with 1385 digits.Found 334811
   1024 with 3080 digits.Found 777542
   1536 with 4891 digits.Found 968756
   2048 with 6778 digits.Found 998285
   2560 with 8722 digits.Found 999959
   3072 with 10710 digits.Found 999999

Found: 1000000 at 3173 with 11107 digits in Time used    2.719 secs

               0    1    2    3    4    5    6    7    8    9
          |__________________________________________________
         0|    9    1    3    5    2    4    4    3    7    9
        10|   10   11    5   19   22   26    8   17   16   19
        20|    9    8   13    7   17    4   17    3   11   18
        30|   13    5   23   17   18    7   17   15    9   18
        40|   16   17    9    7   12   28    6   23    9   24
        50|   23   13   18   11    7   14    4   18   14   13
        60|   19   11   25   17   17    6    6    8   14   27
        70|   11   26    8   16    9   13   17    8   15   19
        80|   14   21    7   21   16   11   17    9   17    9
        90|   15   12   13   15   27   16   18   19   21   23

...  at home for 7 Digits
only calc k^k for 1..9604
Found: 0 at 9604 with 0 digits in Time used   45.700 secs
with ConvToStr
Found: 0 at 9604 with 38244 digits in Time used   46.406 secs
with ConvToStr and CheckOneString
Found: 10000000 at 9604 with 38244 digits in Time used   52.222 secs
   9216 with 36533 digits.Found 9999997

gmp-version[edit]

program K_pow_K_gmp;
//First occurence of a numberstring with max DIGTIS digits in k^k
{$IFDEF FPC}
{$MODE DELPHI}
{$Optimization ON,ALL}
{$ELSE}
{$APPTYPE CONSOLE}
{$ENDIF}
 
uses
sysutils,gmp;
const
LongWordDec = 1000*1000*1000;
 
Digits = 7;
 
var
Pot_N_str : AnsiString;
Str_Found : array of Uint32;
FirstMissing :NativeInt;
T0 : INt64;
 
procedure Out_Results(number,found:NativeInt);
var
i : NativeInt;
Begin
writeln;
writeln(#10,'Found: ',found,' at ',number,' with ',length(pChar(Pot_N_str)),
' digits in Time used ',(GetTickCount64-T0)/1000:8:3,' secs');
writeln ;
writeln(' 0 1 2 3 4 5 6 7 8 9');
write(' |__________________________________________________');
For i := 0 to 99 do//decLimit-1 do
begin
if i MOD 10 = 0 then
Begin
writeln;
write((i DIV 10)*10:10,'|');
end;
number := Str_Found[i]-1;
if number > 0 then
write(number:5);
end;
writeln;
end;
 
function CheckOneString(const s:Ansistring;lmt,pow:NativeInt):NativeInt;
//check every possible number from one to DIGITS digits
var
i,k,num : NativeInt;
begin
result := 0;
 
For i := 1 to lmt do
Begin
k := i;
num := 0;
repeat
num := num*10+ Ord(s[k])-Ord('0');
IF (num >= FirstMissing) AND (str_Found[num] = 0) then
begin
str_Found[num]:= pow+1;
inc(result);
if num =FirstMissing then
Begin
while str_Found[FirstMissing] <> 0 do
inc(FirstMissing);
end;
end;
inc(k)
until (k>lmt) or (k-i >DIGITS-1);
end;
end;
 
 
var
zkk: mpz_t;
number,i,found,lenS,decLimit: Int32;
Begin
T0 := GetTickCount64;
mpz_init(zkk);
 
decLimit := 1;
For i := 1 to digits do
decLimit *= 10;
setlength(Str_Found,decLimit);
 
//calc digits for max number := 10000
number:= 10000;
i := trunc(number*ln(number)/ln(10))+5;
setlength(Pot_N_str,i);
 
found := 0;
FirstMissing := 0;
number := 1;
lenS :=1;
repeat
mpz_ui_pow_ui(zkk,number,number);
mpz_get_str(pChar(Pot_N_str),10,zkk);
while Pot_N_str[lenS] <> #0 do inc(lenS);
// lenS := length(pChar(Pot_N_str));
inc(found,CheckOneString(Pot_N_str,lenS,number));
inc(number);
if number AND 511 = 0 then
write(#13,number:7,' with ',lenS, ' digits.Found ',found);
until number>9604;// found >=decLimit;
Out_Results(number,found);
end.
Output:
TIO.RUN for 7 digits  same as above
    512 with 1386 digits.Found 608645
   1024 with 3081 digits.Found 1952296
...
Found: 10000000 at 9604 with 38244 digits in Time used   13.538 secs
//only mpz_ui_pow_ui(zkk,number,number); takes <0.5s up to 9604 with string conversion 3.3s

Perl[edit]

use strict;
use warnings;
use feature 'say';
use List::Util 'first';
use Math::AnyNum 'ipow';
 
sub smallest { first { ipow($_,$_) =~ /$_[0]/ } 1..1e4 }
say join ' ', map { smallest($_) } 0..50;
Output:
9 1 3 5 2 4 4 3 7 9 10 11 5 19 22 26 8 17 16 19 9 8 13 7 17 4 17 3 11 18 13 5 23 17 18 7 17 15 9 18 16 17 9 7 12 28 6 23 9 24 23

Phix[edit]

Native numbers won't cope (14^14 exceeds a 64-bit float, 17^17 an 80-bit one), so instead of gmp I've gone with string math again. (Related recent tasks: here and here)

with javascript_semantics
constant lim = 51       -- (tested to 1,000,000)
atom t0 = time(), t1 = t0+1
sequence res = repeat(0,lim)
integer found = 0, k = 1
while found<lim do
    string kk = "1"
    for i=1 to k do
        integer carry = 0
        for j=length(kk) to 1 by -1 do
            integer digit = (kk[j]-'0')*k+carry
            kk[j] = remainder(digit,10)+'0'
            carry = floor(digit/10)
        end for
        while carry do
            kk = remainder(carry,10)+'0' & kk
            carry = floor(carry/10)
        end while
    end for
    for i=1 to length(kk) do
        integer digit = 0, j = i
        while j<=length(kk) and digit<=lim do
            digit = digit*10+kk[j]-'0'
            if digit<lim and res[digit+1]=0 then
                res[digit+1] = sprintf("%2d",k)
                found += 1
            end if
            j += 1
        end while
    end for
    if platform()!=JS and time()>t1 then
        progress("found %,d/%,d, at %d^%d which has %,d digits (%s)",
                 {found,lim,k,k,length(kk),elapsed(time()-t0)})
        t1 = time()+1
    end if
    k += 1
end while
puts(1,join_by(shorten(res,"",30),1,10))
Output:
 9    1    3    5    2    4    4    3    7    9
10   11    5   19   22   26    8   17   16   19
 9    8   13    7   17    4   17    3   11   18
13    5   23   17   18    7   17   15    9   18
16   17    9    7   12   28    6   23    9   24
23

Testing to 1,000,000 took 12mins 35s.

gmp version[edit]

with javascript_semantics
constant lim = 51       -- (tested to 1,000,000)
include mpfr.e
mpz zkk = mpz_init()
atom t0 = time(), t1 = t0+1
sequence res = repeat(0,lim)
integer found = 0, k = 1
while found<lim do
    mpz_ui_pow_ui(zkk,k,k)
    string kk = mpz_get_str(zkk)
    for i=1 to length(kk) do
        integer digit = 0, j = i
        while j<=length(kk) and digit<=lim do
            digit = digit*10+kk[j]-'0'
            if digit<lim and res[digit+1]=0 then
                res[digit+1] = sprintf("%2d",k)
                found += 1
            end if
            j += 1
        end while
    end for
    if platform()!=JS and time()>t1 then
        progress("found %,d/%,d, at %d^%d which has %,d digits (%s)",
                 {found,lim,k,k,length(kk),elapsed(time()-t0)})
        t1 = time()+1
    end if
    k += 1
end while
puts(1,join_by(shorten(res,"",30),1,10))

Same results, but nearly 30 times faster, finishing the 1,000,000 test in just 26.6s

Raku[edit]

sub smallest ( $n ) {
state @powers = '', |map { $_ ** $_ }, 1 .. *;
 
return @powers.first: :k, *.contains($n);
}
 
.say for (^51).map(&smallest).batch(10)».fmt('%2d');
Output:
( 9  1  3  5  2  4  4  3  7  9)
(10 11  5 19 22 26  8 17 16 19)
( 9  8 13  7 17  4 17  3 11 18)
(13  5 23 17 18  7 17 15  9 18)
(16 17  9  7 12 28  6 23  9 24)
(23)

REXX[edit]

Code was added to display the count of unique numbers found.

/*REXX pgm finds the  smallest positive integer  K  where   K**K   contains  N,  N < 51 */
numeric digits 200 /*ensure enough decimal digs for k**k */
parse arg hi cols . /*obtain optional argument from the CL.*/
if hi=='' | hi=="," then hi= 51 /*Not specified? Then use the default.*/
if cols=='' | cols=="," then cols= 10 /* " " " " " " */
w= 6 /*width of a number in any column. */
title=' smallest positive integer K where K**K contains N, 0 ≤ N < ' commas(hi)
say ' N │'center(title, 5 + cols*(w+1) ) /*display the title of the output. */
say '─────┼'center("" , 5 + cols*(w+1), '─') /* " " separator " " " */
u= 0;  !.= . /*number of unique #'s found; semaphore*/
$=; idx= 0 /*define $ output list; index to 0.*/
do j=0 for hi /*look for a power of K that contains N*/
do k=1 until pos(j, k**k)>0 /*calculate a bunch of powers (K**K). */
end /*k*/
if !.k==. then do; u= u+1;  !.k=; end /*Is unique? Then bump unique counter.*/
c= commas(k) /*maybe add commas to the powe of six. */
$= $ right(c, max(w, length(c) ) ) /*add a K (power) ──► list, allow big#*/
if (j+1)//cols\==0 then iterate /*have we populated a line of output? */
say center(idx, 5)'│'substr($, 2); $= /*display what we have so far (cols). */
idx= idx + cols /*bump the index count for the output*/
end /*j*/
 
if $\=='' then say center(idx, 5)"│"substr($,2) /*possible display any residual output.*/
say '─────┴'center("" , 5 + cols*(w+1), '─') /* " " separator " " " */
say
say commas(u) ' unique numbers found.'
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: parse arg ?; do jc=length(?)-3 to 1 by -3; ?=insert(',', ?, jc); end; return ?
output   when using the default inputs:
  N  │  smallest positive integer  K  where  K**K  contains  N,   0  ≤  N  <  51
─────┼───────────────────────────────────────────────────────────────────────────
  0  │     9      1      3      5      2      4      4      3      7      9
 10  │    10     11      5     19     22     26      8     17     16     19
 20  │     9      8     13      7     17      4     17      3     11     18
 30  │    13      5     23     17     18      7     17     15      9     18
 40  │    16     17      9      7     12     28      6     23      9     24
 50  │    23
─────┴───────────────────────────────────────────────────────────────────────────

23  unique numbers found.

Ring[edit]

 
load "bignumber.ring"
 
decimals(0)
see "working..." + nl
see "Smallest number k > 0 such that the decimal expansion of k^k contains n are:" + nl
 
row = 0
limit1 = 50
limit2 = 30
 
for n = 0 to limit1
strn = string(n)
for m = 1 to limit2
powm = pow(m,m)
ind = substr(powm,strn)
if ind > 0
exit
ok
next
row = row + 1
see "" + m + " "
if row%10 = 0
see nl
ok
next
 
see nl + "done..." + nl
 
func pow(num1,num2)
num1 = string(num1)
num2 = string(num2)
return FuncPower(num1,num2)
 
Output:
working...
Smallest number k > 0 such that the decimal expansion of k^k contains n are:
9 1 3 5 2 4 4 3 7 9 
10 11 5 19 22 26 8 17 16 19 
9 8 13 7 17 4 17 3 11 18 
13 5 23 17 18 7 17 15 9 18 
16 17 9 7 12 28 6 23 9 24 
23 
done...

Sidef[edit]

0..50 -> map {|n| 1..Inf -> first {|k| Str(k**k).contains(n) } }.say
Output:
[9, 1, 3, 5, 2, 4, 4, 3, 7, 9, 10, 11, 5, 19, 22, 26, 8, 17, 16, 19, 9, 8, 13, 7, 17, 4, 17, 3, 11, 18, 13, 5, 23, 17, 18, 7, 17, 15, 9, 18, 16, 17, 9, 7, 12, 28, 6, 23, 9, 24, 23]

Wren[edit]

Library: Wren-big
Library: Wren-seq
Library: Wren-fmt
import "/big" for BigInt
import "/seq" for Lst
import "/fmt" for Fmt
 
var res = []
for (n in 0..50) {
var k = 1
while (true) {
var s = BigInt.new(k).pow(k).toString
if (s.contains(n.toString)) {
res.add(k)
break
}
k = k + 1
}
}
System.print("The smallest positive integers K where K ^ K contains N (0..50) are:")
for (chunk in Lst.chunks(res, 17)) Fmt.print("$2d", chunk)
Output:
The smallest positive integers K where K ^ K contains N (0..50) are:
 9  1  3  5  2  4  4  3  7  9 10 11  5 19 22 26  8
17 16 19  9  8 13  7 17  4 17  3 11 18 13  5 23 17
18  7 17 15  9 18 16 17  9  7 12 28  6 23  9 24 23