Largest proper divisor of n

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Revision as of 17:33, 27 August 2022 by Thundergnat (talk | contribs) (syntax highlighting fixup automation)
Task
Largest proper divisor of n
You are encouraged to solve this task according to the task description, using any language you may know.
Task
a(1) = 1; for n > 1, a(n) = largest proper divisor of n, where n < 101 .



11l

Translation of: Python
F lpd(n)
   L(i) (n - 1 .< 0).step(-1)
      I n % i == 0
         R i
   R 1

L(i) 1..100
   print(‘#3’.format(lpd(i)), end' I i % 10 == 0 {"\n"} E ‘’)
Output:
  1  1  1  2  1  3  1  4  3  5
  1  6  1  7  5  8  1  9  1 10
  7 11  1 12  5 13  9 14  1 15
  1 16 11 17  7 18  1 19 13 20
  1 21  1 22 15 23  1 24  7 25
 17 26  1 27 11 28 19 29  1 30
  1 31 21 32 13 33  1 34 23 35
  1 36  1 37 25 38 11 39  1 40
 27 41  1 42 17 43 29 44  1 45
 13 46 31 47 19 48  1 49 33 50

Ada

with Ada.Text_IO;         use Ada.Text_IO;
with Ada.Integer_Text_IO; use Ada.Integer_Text_IO;

procedure Main is
   subtype param_type is Integer range 1 .. 100;
   function lpd (n : in param_type) return param_type is
      result : param_type := 1;
   begin
      for divisor in reverse 1 .. n / 2 loop
         if n rem divisor = 0 then
            result := divisor;
            exit;
         end if;
      end loop;
      return result;
   end lpd;

begin
   for I in param_type loop
      Put (Item => lpd (I), Width => 3);
      if I rem 10 = 0 then
         New_Line;
      end if;
   end loop;
end Main;
Output:
  1  1  1  2  1  3  1  4  3  5
  1  6  1  7  5  8  1  9  1 10
  7 11  1 12  5 13  9 14  1 15
  1 16 11 17  7 18  1 19 13 20
  1 21  1 22 15 23  1 24  7 25
 17 26  1 27 11 28 19 29  1 30
  1 31 21 32 13 33  1 34 23 35
  1 36  1 37 25 38 11 39  1 40
 27 41  1 42 17 43 29 44  1 45
 13 46 31 47 19 48  1 49 33 50

ALGOL 68

FOR n TO 100 DO # show the largest proper divisors for n = 1..100 #
    INT largest proper divisor := 1;
    FOR j FROM ( n OVER 2 ) BY -1 TO 2 WHILE largest proper divisor = 1 DO
        IF n MOD j = 0 THEN
            largest proper divisor := j
        FI
    OD;
    print( ( whole( largest proper divisor, -3 ) ) );
    IF n MOD 10 = 0 THEN print( ( newline ) ) FI
OD
Output:
  1  1  1  2  1  3  1  4  3  5
  1  6  1  7  5  8  1  9  1 10
  7 11  1 12  5 13  9 14  1 15
  1 16 11 17  7 18  1 19 13 20
  1 21  1 22 15 23  1 24  7 25
 17 26  1 27 11 28 19 29  1 30
  1 31 21 32 13 33  1 34 23 35
  1 36  1 37 25 38 11 39  1 40
 27 41  1 42 17 43 29 44  1 45
 13 46 31 47 19 48  1 49 33 50

ALGOL W

for n := 1 until 100 do begin % show the largest proper divisors for n = 1..100 %
    for j := n div 2 step -1 until 2 do begin
        if n rem j = 0 then begin
            writeon( i_w := 3, s_w := 0, j );
            goto foundLargestProperDivisor
        end if_n_rem_j_eq_0
    end for_j;
    writeon( i_w := 3, s_w := 0, 1 );
foundLargestProperDivisor:
    if n rem 10 = 0 then write()
end for_n.
Output:
  1  1  1  2  1  3  1  4  3  5
  1  6  1  7  5  8  1  9  1 10
  7 11  1 12  5 13  9 14  1 15
  1 16 11 17  7 18  1 19 13 20
  1 21  1 22 15 23  1 24  7 25
 17 26  1 27 11 28 19 29  1 30
  1 31 21 32 13 33  1 34 23 35
  1 36  1 37 25 38 11 39  1 40
 27 41  1 42 17 43 29 44  1 45
 13 46 31 47 19 48  1 49 33 50

APL

Works with: Dyalog APL
(/1,(0=¯1↓⍳|⊢))¨10 10⍴⍳100
Output:
 1  1  1  2  1  3  1  4  3  5
 1  6  1  7  5  8  1  9  1 10
 7 11  1 12  5 13  9 14  1 15
 1 16 11 17  7 18  1 19 13 20
 1 21  1 22 15 23  1 24  7 25
17 26  1 27 11 28 19 29  1 30
 1 31 21 32 13 33  1 34 23 35
 1 36  1 37 25 38 11 39  1 40
27 41  1 42 17 43 29 44  1 45
13 46 31 47 19 48  1 49 33 50

AppleScript

Most of this code is just to prepare the output for display. :D

on largestProperDivisor(n)
    if (n mod 2 = 0) then return n div 2
    if (n mod 3 = 0) then return n div 3
    if (n < 5) then return missing value
    repeat with i from 5 to (n ^ 0.5 div 1) by 6
        if (n mod i = 0) then return n div i
        if (n mod (i + 2) = 0) then return n div (i + 2)
    end repeat
    return 1
end largestProperDivisor

on task(max)
    script o
        property LPDs : {}
        property output : {}
    end script
    
    set w to (count (max as text)) * 2 + 3
    set padding to text 1 thru (w - 2) of "          "
    set end of o's LPDs to "1:1" & padding
    set astid to AppleScript's text item delimiters
    set AppleScript's text item delimiters to ""
    set c to 1
    repeat with n from 2 to max
        set end of o's LPDs to text 1 thru w of ((n as text) & ":" & largestProperDivisor(n) & padding)
        set c to c + 1
        if (c mod 10 is 0) then
            set end of o's output to o's LPDs as text
            set o's LPDs to {}
        end if
    end repeat
    if (c mod 10 > 0) then set end of o's output to o's LPDs as text
    set AppleScript's text item delimiters to linefeed
    set o's output to o's output as text
    set AppleScript's text item delimiters to astid
    
    return o's output
end task

task(100)
Output:
"1:1      2:1      3:1      4:2      5:1      6:3      7:1      8:4      9:3      10:5     
11:1     12:6     13:1     14:7     15:5     16:8     17:1     18:9     19:1     20:10    
21:7     22:11    23:1     24:12    25:5     26:13    27:9     28:14    29:1     30:15    
31:1     32:16    33:11    34:17    35:7     36:18    37:1     38:19    39:13    40:20    
41:1     42:21    43:1     44:22    45:15    46:23    47:1     48:24    49:7     50:25    
51:17    52:26    53:1     54:27    55:11    56:28    57:19    58:29    59:1     60:30    
61:1     62:31    63:21    64:32    65:13    66:33    67:1     68:34    69:23    70:35    
71:1     72:36    73:1     74:37    75:25    76:38    77:11    78:39    79:1     80:40    
81:27    82:41    83:1     84:42    85:17    86:43    87:29    88:44    89:1     90:45    
91:13    92:46    93:31    94:47    95:19    96:48    97:1     98:49    99:33    100:50   "


Functional

Composing functionally, for rapid drafting and refactoring, with higher levels of code reuse:

use framework "Foundation"


--------------- LARGEST PROPER DIVISOR OF N --------------

-- maxProperDivisors :: Int -> Int
on maxProperDivisors(n)
    script p
        on |λ|(x)
            0 = n mod x
        end |λ|
    end script
    
    set mb to find(p, enumFromTo(2, sqrt(n)))
    
    if missing value is mb then
        1
    else
        n div mb
    end if
end maxProperDivisors


--------------------------- TEST -------------------------
on run
    set xs to map(maxProperDivisors, enumFromTo(1, 100))
    set w to length of (maximum(xs) as string)
    
    unlines(map(unwords, ¬
        chunksOf(10, ¬
            map(compose(justifyRight(w, space), str), xs))))
end run


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

-- chunksOf :: Int -> [a] -> [[a]]
on chunksOf(k, xs)
    script
        on go(ys)
            set ab to splitAt(k, ys)
            set a to item 1 of ab
            if {}  a then
                {a} & go(item 2 of ab)
            else
                a
            end if
        end go
    end script
    result's go(xs)
end chunksOf


-- compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
on compose(f, g)
    script
        property mf : mReturn(f)
        property mg : mReturn(g)
        on |λ|(x)
            mf's |λ|(mg's |λ|(x))
        end |λ|
    end script
end compose


-- enumFromTo :: Int -> Int -> [Int]
on enumFromTo(m, n)
    if m  n then
        set lst to {}
        repeat with i from m to n
            set end of lst to i
        end repeat
        lst
    else
        {}
    end if
end enumFromTo


-- find :: (a -> Bool) -> [a] -> (a | missing value)
on find(p, xs)
    tell mReturn(p)
        set lng to length of xs
        repeat with i from 1 to lng
            if |λ|(item i of xs) then return item i of xs
        end repeat
        missing value
    end tell
end find


-- justifyRight :: Int -> Char -> String -> String
on justifyRight(n, cFiller)
    script
        on |λ|(strText)
            if n > length of strText then
                text -n thru -1 of ((replicate(n, cFiller) as text) & strText)
            else
                strText
            end if
        end |λ|
    end script
end justifyRight


-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
    -- The list obtained by applying f
    -- to each element of xs.
    tell mReturn(f)
        set lng to length of xs
        set lst to {}
        repeat with i from 1 to lng
            set end of lst to |λ|(item i of xs, i, xs)
        end repeat
        return lst
    end tell
end map


-- maximum :: Ord a => [a] -> a
on maximum(xs)
    set ca to current application
    unwrap((ca's NSArray's arrayWithArray:(xs))'s ¬
        valueForKeyPath:"@max.self")
end maximum


-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
    -- 2nd class handler function lifted into 1st class script wrapper. 
    if script is class of f then
        f
    else
        script
            property |λ| : f
        end script
    end if
end mReturn


-- Egyptian multiplication - progressively doubling a list, appending
-- stages of doubling to an accumulator where needed for binary 
-- assembly of a target length
-- replicate :: Int -> String -> String
on replicate(n, s)
    -- Egyptian multiplication - progressively doubling a list, 
    -- appending stages of doubling to an accumulator where needed 
    -- for binary assembly of a target length
    script p
        on |λ|({n})
            n  1
        end |λ|
    end script
    
    script f
        on |λ|({n, dbl, out})
            if (n mod 2) > 0 then
                set d to out & dbl
            else
                set d to out
            end if
            {n div 2, dbl & dbl, d}
        end |λ|
    end script
    
    set xs to |until|(p, f, {n, s, ""})
    item 2 of xs & item 3 of xs
end replicate


-- splitAt :: Int -> [a] -> ([a], [a])
on splitAt(n, xs)
    if n > 0 and n < length of xs then
        if class of xs is text then
            {items 1 thru n of xs as text, ¬
                items (n + 1) thru -1 of xs as text}
        else
            {items 1 thru n of xs, items (n + 1) thru -1 of xs}
        end if
    else
        if n < 1 then
            {{}, xs}
        else
            {xs, {}}
        end if
    end if
end splitAt


-- sqrt :: Num -> (Num | missing value)
on sqrt(n)
    if 0  n then
        n ^ (1 / 2)
    else
        missing value
    end if
end sqrt


-- str :: a -> String
on str(x)
    x as string
end str


-- unlines :: [String] -> String
on unlines(xs)
    -- A single string formed by the intercalation
    -- of a list of strings with the newline character.
    set {dlm, my text item delimiters} to ¬
        {my text item delimiters, linefeed}
    set s to xs as text
    set my text item delimiters to dlm
    s
end unlines


-- until :: (a -> Bool) -> (a -> a) -> a -> a
on |until|(p, f, x)
    set v to x
    set mp to mReturn(p)
    set mf to mReturn(f)
    repeat until mp's |λ|(v)
        set v to mf's |λ|(v)
    end repeat
    v
end |until|


-- unwords :: [String] -> String
on unwords(xs)
    set {dlm, my text item delimiters} to ¬
        {my text item delimiters, space}
    set s to xs as text
    set my text item delimiters to dlm
    return s
end unwords


-- unwrap :: NSValue -> a
on unwrap(nsValue)
    if nsValue is missing value then
        missing value
    else
        set ca to current application
        item 1 of ((ca's NSArray's arrayWithObject:nsValue) as list)
    end if
end unwrap
Output:
 1  1  1  2  1  3  1  4  3  5
 1  6  1  7  5  8  1  9  1 10
 7 11  1 12  5 13  9 14  1 15
 1 16 11 17  7 18  1 19 13 20
 1 21  1 22 15 23  1 24  7 25
17 26  1 27 11 28 19 29  1 30
 1 31 21 32 13 33  1 34 23 35
 1 36  1 37 25 38 11 39  1 40
27 41  1 42 17 43 29 44  1 45
13 46 31 47 19 48  1 49 33 50

Arturo

loop split.every:10 [1] ++ map 2..100 'x -> last chop factors x 'row [
    print map to [:string] row 'r -> pad r 5
]
Output:
    1     1     1     2     1     3     1     4     3     5 
    1     6     1     7     5     8     1     9     1    10 
    7    11     1    12     5    13     9    14     1    15 
    1    16    11    17     7    18     1    19    13    20 
    1    21     1    22    15    23     1    24     7    25 
   17    26     1    27    11    28    19    29     1    30 
    1    31    21    32    13    33     1    34    23    35 
    1    36     1    37    25    38    11    39     1    40 
   27    41     1    42    17    43    29    44     1    45 
   13    46    31    47    19    48     1    49    33    50

AWK

# syntax: GAWK -f LARGEST_PROPER_DIVISOR_OF_N.AWK
# converted from C
BEGIN {
    start = 1
    stop = 100
    for (i=start; i<=stop; i++) {
      printf("%3d%1s",largest_proper_divisor(i),++count%10?"":"\n")
    }
    printf("\nLargest proper divisor of n %d-%d\n",start,stop)
    exit(0)
}
function largest_proper_divisor(n,  i) {
    if (n <= 1) {
      return(1)
    }
    for (i=n-1; i>0; i--) {
      if (n % i == 0) {
        return(i)
      }
    }
}
Output:
  1   1   1   2   1   3   1   4   3   5
  1   6   1   7   5   8   1   9   1  10
  7  11   1  12   5  13   9  14   1  15
  1  16  11  17   7  18   1  19  13  20
  1  21   1  22  15  23   1  24   7  25
 17  26   1  27  11  28  19  29   1  30
  1  31  21  32  13  33   1  34  23  35
  1  36   1  37  25  38  11  39   1  40
 27  41   1  42  17  43  29  44   1  45
 13  46  31  47  19  48   1  49  33  50

Largest proper divisor of n 1-100

BASIC

10 DEFINT A-Z
20 FOR I=1 TO 100
30 IF I=1 THEN PRINT "  1";: GOTO 70
40 FOR J=I-1 TO 1 STEP -1
50 IF I MOD J=0 THEN PRINT USING "###";J;: GOTO 70
60 NEXT J
70 IF I MOD 10=0 THEN PRINT
80 NEXT I
Output:
  1  1  1  2  1  3  1  4  3  5
  1  6  1  7  5  8  1  9  1 10
  7 11  1 12  5 13  9 14  1 15
  1 16 11 17  7 18  1 19 13 20
  1 21  1 22 15 23  1 24  7 25
 17 26  1 27 11 28 19 29  1 30
  1 31 21 32 13 33  1 34 23 35
  1 36  1 37 25 38 11 39  1 40
 27 41  1 42 17 43 29 44  1 45
 13 46 31 47 19 48  1 49 33 50

BASIC256

Translation of: FreeBASIC
print "El mayor divisor propio de n es:\n"
print " 1  1 ";
for i = 3 to 100
	for j = i-1 to 1 step -1
		If i % j = 0 then
			print " "; j; " ";
			exit for
		end if
	next  j
	if i % 10 = 0 then print
next i
end


BCPL

get "libhdr"

let lpd(n) = valof
    for i = n<=1 -> 1, n-1 to 1 by -1
        if n rem i=0 resultis i

let start() be
    for i=1 to 100
    $(  writed(lpd(i), 3)
        if i rem 10=0 then wrch('*N')
    $)
Output:
  1  1  1  2  1  3  1  4  3  5
  1  6  1  7  5  8  1  9  1 10
  7 11  1 12  5 13  9 14  1 15
  1 16 11 17  7 18  1 19 13 20
  1 21  1 22 15 23  1 24  7 25
 17 26  1 27 11 28 19 29  1 30
  1 31 21 32 13 33  1 34 23 35
  1 36  1 37 25 38 11 39  1 40
 27 41  1 42 17 43 29 44  1 45
 13 46 31 47 19 48  1 49 33 50

BQN

(1⌈´↕×0=↕|⊢)¨ ∘‿10⥊1+↕100
Output:
┌─                               
╵  1  1  1  2  1  3  1  4  3  5  
   1  6  1  7  5  8  1  9  1 10  
   7 11  1 12  5 13  9 14  1 15  
   1 16 11 17  7 18  1 19 13 20  
   1 21  1 22 15 23  1 24  7 25  
  17 26  1 27 11 28 19 29  1 30  
   1 31 21 32 13 33  1 34 23 35  
   1 36  1 37 25 38 11 39  1 40  
  27 41  1 42 17 43 29 44  1 45  
  13 46 31 47 19 48  1 49 33 50  
                                ┘

C

#include <stdio.h>

unsigned int lpd(unsigned int n) {
    if (n<=1) return 1;
    int i;
    for (i=n-1; i>0; i--)
        if (n%i == 0) return i;   
}

int main() {
    int i;
    for (i=1; i<=100; i++) {
        printf("%3d", lpd(i));
        if (i % 10 == 0) printf("\n");
    }
    return 0;
}
Output:
  1  1  1  2  1  3  1  4  3  5
  1  6  1  7  5  8  1  9  1 10
  7 11  1 12  5 13  9 14  1 15
  1 16 11 17  7 18  1 19 13 20
  1 21  1 22 15 23  1 24  7 25
 17 26  1 27 11 28 19 29  1 30
  1 31 21 32 13 33  1 34 23 35
  1 36  1 37 25 38 11 39  1 40
 27 41  1 42 17 43 29 44  1 45
 13 46 31 47 19 48  1 49 33 50

C++

#include <cassert>
#include <iomanip>
#include <iostream>

int largest_proper_divisor(int n) {
    assert(n > 0);
    if ((n & 1) == 0)
        return n >> 1;
    for (int p = 3; p * p <= n; p += 2) {
        if (n % p == 0)
            return n / p;
    }
    return 1;
}

int main() {
    for (int n = 1; n < 101; ++n) {
        std::cout << std::setw(2) << largest_proper_divisor(n)
            << (n % 10 == 0 ? '\n' : ' ');
    }
}
Output:
 1  1  1  2  1  3  1  4  3  5
 1  6  1  7  5  8  1  9  1 10
 7 11  1 12  5 13  9 14  1 15
 1 16 11 17  7 18  1 19 13 20
 1 21  1 22 15 23  1 24  7 25
17 26  1 27 11 28 19 29  1 30
 1 31 21 32 13 33  1 34 23 35
 1 36  1 37 25 38 11 39  1 40
27 41  1 42 17 43 29 44  1 45
13 46 31 47 19 48  1 49 33 50

Cowgol

include "cowgol.coh";

sub print3(n: uint8) is
    print_char(' ');
    if n>9 then
        print_char('0' + n/10);
    else
        print_char(' ');
    end if;
    print_char('0' + n%10);
end sub;

sub lpd(n: uint8): (r: uint8) is
    if n <= 1 then
        r := 1;
    else
        r := n - 1;
        while r > 0 loop
            if n % r == 0 then 
                break;
            end if;
            r := r - 1;
        end loop;
    end if;
end sub;

var i: uint8 := 1;
while i <= 100 loop
    print3(lpd(i));
    if i%10 == 0 then
        print_nl();
    end if;
    i := i + 1;
end loop;
Output:
  1  1  1  2  1  3  1  4  3  5
  1  6  1  7  5  8  1  9  1 10
  7 11  1 12  5 13  9 14  1 15
  1 16 11 17  7 18  1 19 13 20
  1 21  1 22 15 23  1 24  7 25
 17 26  1 27 11 28 19 29  1 30
  1 31 21 32 13 33  1 34 23 35
  1 36  1 37 25 38 11 39  1 40
 27 41  1 42 17 43 29 44  1 45
 13 46 31 47 19 48  1 49 33 50

Draco

proc nonrec lpd(word n) word:
    word d;
    if n=1 then
        1
    else
        d := n-1;
        while n % d /= 0 do d := d-1 od;
        d
    fi
corp

proc nonrec main() void:
    word n;
    for n from 1 upto 100 do
        write(lpd(n):3);
        if n%10 = 0 then writeln() fi
    od
corp
Output:
  1  1  1  2  1  3  1  4  3  5
  1  6  1  7  5  8  1  9  1 10
  7 11  1 12  5 13  9 14  1 15
  1 16 11 17  7 18  1 19 13 20
  1 21  1 22 15 23  1 24  7 25
 17 26  1 27 11 28 19 29  1 30
  1 31 21 32 13 33  1 34 23 35
  1 36  1 37 25 38 11 39  1 40
 27 41  1 42 17 43 29 44  1 45
 13 46 31 47 19 48  1 49 33 50

F#

// Largest proper divisor of n: Nigel Galloway. June 2nd., 2021
let fN g=let rec fN n=let i=Seq.head n in match(g/i,g%i) with (1,_)->1 |(n,0)->n |_->fN(Seq.tail n) in fN(Seq.initInfinite((+)2))
seq{yield 1; yield! seq{2..100}|>Seq.map fN}|>Seq.iter(printf "%d "); printfn ""
Output:
1 1 1 2 1 3 1 4 3 5 1 6 1 7 5 8 1 9 1 10 7 11 1 12 5 13 9 14 1 15 1 16 11 17 7 18 1 19 13 20 1 21 1 22 15 23 1 24 7 25 17 26 1 27 11 28 19 29 1 30 1 31 21 32 13 33 1 34 23 35 1 36 1 37 25 38 11 39 1 40 27 41 1 42 17 43 29 44 1 45 13 46 31 47 19 48 1 49 33 50
Real: 00:00:00.015

Factor

Works with: Factor version 0.99 2021-02-05
USING: grouping kernel math math.bitwise math.functions
math.ranges prettyprint sequences ;

: odd ( m -- n )
    dup 3 /i 1 - next-odd 1 -2 <range>
    [ divisor? ] with find nip ;

: largest ( m -- n ) dup odd? [ odd ] [ 2/ ] if ;

100 [1,b] [ largest ] map 10 group simple-table.
Output:
1  1  1  2  1  3  1  4  3  5
1  6  1  7  5  8  1  9  1  10
7  11 1  12 5  13 9  14 1  15
1  16 11 17 7  18 1  19 13 20
1  21 1  22 15 23 1  24 7  25
17 26 1  27 11 28 19 29 1  30
1  31 21 32 13 33 1  34 23 35
1  36 1  37 25 38 11 39 1  40
27 41 1  42 17 43 29 44 1  45
13 46 31 47 19 48 1  49 33 50

Fermat

Func Lpd(n) =
    if n = 1 then Return(1) fi;
    for i = n\2 to 1 by -1 do
        if n|i = 0 then Return(i) fi;
    od.;

for m = 1 to 100 do
    !(Lpd(m):4);
    if m|10=0 then ! fi;
od;
Output:

1 1 1 2 1 3 1 4 3 5 1 6 1 7 5 8 1 9 1 10 7 11 1 12 5 13 9 14 1 15 1 16 11 17 7 18 1 19 13 20 1 21 1 22 15 23 1 24 7 25 17 26 1 27 11 28 19 29 1 30 1 31 21 32 13 33 1 34 23 35 1 36 1 37 25 38 11 39 1 40 27 41 1 42 17 43 29 44 1 45 13 46 31 47 19 48 1 49 33 50

FOCAL

01.10 F N=1,100;D 2;D 3
01.20 Q

02.10 S V=1
02.20 I (1-N)2.3;R
02.30 S V=N-1
02.40 S A=N/V
02.50 I (FITR(A)-A)2.6;R
02.60 S V=V-1
02.70 G 2.4

03.10 T %2,V
03.20 S A=N/10
03.30 I (FITR(A)-A)3.4;T !
03.40 R
Output:
=  1=  1=  1=  2=  1=  3=  1=  4=  3=  5
=  1=  6=  1=  7=  5=  8=  1=  9=  1= 10
=  7= 11=  1= 12=  5= 13=  9= 14=  1= 15
=  1= 16= 11= 17=  7= 18=  1= 19= 13= 20
=  1= 21=  1= 22= 15= 23=  1= 24=  7= 25
= 17= 26=  1= 27= 11= 28= 19= 29=  1= 30
=  1= 31= 21= 32= 13= 33=  1= 34= 23= 35
=  1= 36=  1= 37= 25= 38= 11= 39=  1= 40
= 27= 41=  1= 42= 17= 43= 29= 44=  1= 45
= 13= 46= 31= 47= 19= 48=  1= 49= 33= 50

Forth

Works with: Gforth
: largest-proper-divisor { n -- n }
  n 1 and 0= if n 2/ exit then 
  3
  begin
    dup dup * n <=
  while
    dup n swap /mod swap
    0= if nip exit else drop then
    2 +
  repeat drop 1 ;

: main
  101 1 do
    i largest-proper-divisor 2 .r
    i 10 mod 0= if cr else space then
  loop ;

main
bye
Output:
 1  1  1  2  1  3  1  4  3  5
 1  6  1  7  5  8  1  9  1 10
 7 11  1 12  5 13  9 14  1 15
 1 16 11 17  7 18  1 19 13 20
 1 21  1 22 15 23  1 24  7 25
17 26  1 27 11 28 19 29  1 30
 1 31 21 32 13 33  1 34 23 35
 1 36  1 37 25 38 11 39  1 40
27 41  1 42 17 43 29 44  1 45
13 46 31 47 19 48  1 49 33 50

Fortran

       program LargestProperDivisors
       implicit none
       integer i, lpd
       do 10 i=1, 100
           write (*,'(I3)',advance='no') lpd(i)
 10        if (i/10*10 .eq. i) write (*,*)       
       end program
       
       integer function lpd(n)
       implicit none
       integer n, i
       if (n .le. 1) then
           lpd = 1
       else
           do 10 i=n-1, 1, -1
 10            if (n/i*i .eq. n) goto 20
 20        lpd = i
       end if
       end function
Output:
  1  1  1  2  1  3  1  4  3  5
  1  6  1  7  5  8  1  9  1 10
  7 11  1 12  5 13  9 14  1 15
  1 16 11 17  7 18  1 19 13 20
  1 21  1 22 15 23  1 24  7 25
 17 26  1 27 11 28 19 29  1 30
  1 31 21 32 13 33  1 34 23 35
  1 36  1 37 25 38 11 39  1 40
 27 41  1 42 17 43 29 44  1 45
 13 46 31 47 19 48  1 49 33 50


FreeBASIC

Print !"El mayor divisor propio de n es:\n"
Print "  1  1"; 
For i As Byte = 3 To 100
    For j As Byte = i-1 To 1 Step -1
        If i Mod j = 0 Then Print Using "###"; j; : Exit For
    Next  j
    If i Mod 10 = 0 Then Print
Next i
Sleep
Output:
El mayor divisor propio de n es:

  1  1  1  2  1  3  1  4  3  5
  1  6  1  7  5  8  1  9  1 10
  7 11  1 12  5 13  9 14  1 15
  1 16 11 17  7 18  1 19 13 20
  1 21  1 22 15 23  1 24  7 25
 17 26  1 27 11 28 19 29  1 30
  1 31 21 32 13 33  1 34 23 35
  1 36  1 37 25 38 11 39  1 40
 27 41  1 42 17 43 29 44  1 45
 13 46 31 47 19 48  1 49 33 50

Frink

Frink contains efficient routines for factoring numbers. It uses trial division, wheel factoring, and Pollard rho factoring.

for n = 1 to 100
   println[last[allFactors[n,true,false]]]

Go

package main

import "fmt"

func largestProperDivisor(n int) int {
    for i := 2; i*i <= n; i++ {
        if n%i == 0 {
            return n / i
        }
    }
    return 1
}

func main() {
    fmt.Println("The largest proper divisors for numbers in the interval [1, 100] are:")
    fmt.Print(" 1  ")
    for n := 2; n <= 100; n++ {
        if n%2 == 0 {
            fmt.Printf("%2d  ", n/2)
        } else {
            fmt.Printf("%2d  ", largestProperDivisor(n))
        }
        if n%10 == 0 {
            fmt.Println()
        }
    }
}
Output:
The largest proper divisors for numbers in the interval [1, 100] are:
 1   1   1   2   1   3   1   4   3   5  
 1   6   1   7   5   8   1   9   1  10  
 7  11   1  12   5  13   9  14   1  15  
 1  16  11  17   7  18   1  19  13  20  
 1  21   1  22  15  23   1  24   7  25  
17  26   1  27  11  28  19  29   1  30  
 1  31  21  32  13  33   1  34  23  35  
 1  36   1  37  25  38  11  39   1  40  
27  41   1  42  17  43  29  44   1  45  
13  46  31  47  19  48   1  49  33  50  

GW-BASIC

10 PRINT 1;
20 FOR I = 1 TO 101
30 FOR D = I\2 TO 1 STEP -1
40 IF I MOD D = 0 THEN PRINT D; : GOTO 60
50 NEXT D
60 NEXT I

Haskell

import Data.List.Split (chunksOf)
import Text.Printf (printf)

lpd :: Int -> Int
lpd 1 = 1
lpd n = head [x | x <- [n -1, n -2 .. 1], n `mod` x == 0]

main :: IO ()
main =
  (putStr . unlines . map concat . chunksOf 10) $
     printf "%3d" . lpd <$> [1 .. 100]
Output:
  1  1  1  2  1  3  1  4  3  5
  1  6  1  7  5  8  1  9  1 10
  7 11  1 12  5 13  9 14  1 15
  1 16 11 17  7 18  1 19 13 20
  1 21  1 22 15 23  1 24  7 25
 17 26  1 27 11 28 19 29  1 30
  1 31 21 32 13 33  1 34 23 35
  1 36  1 37 25 38 11 39  1 40
 27 41  1 42 17 43 29 44  1 45
 13 46 31 47 19 48  1 49 33 50

Or, considering the two smallest proper divisors:

(If there are two, then the largest proper divisor will be N divided by the first proper divisor that is not 1)

(Otherwise, the largest proper divisor will be 1 itself).

import Data.List (find)
import Data.List.Split (chunksOf)
import Text.Printf (printf)

maxProperDivisors :: Int -> Int
maxProperDivisors n =
  case find ((0 ==) . rem n) [2 .. root] of
    Nothing -> 1
    Just x -> quot n x
  where
    root = (floor . sqrt) (fromIntegral n :: Double)

main :: IO ()
main =
  (putStr . unlines . map concat . chunksOf 10) $
    printf "%3d" . maxProperDivisors <$> [1 .. 100]
  1  1  1  2  1  3  1  4  3  5
  1  6  1  7  5  8  1  9  1 10
  7 11  1 12  5 13  9 14  1 15
  1 16 11 17  7 18  1 19 13 20
  1 21  1 22 15 23  1 24  7 25
 17 26  1 27 11 28 19 29  1 30
  1 31 21 32 13 33  1 34 23 35
  1 36  1 37 25 38 11 39  1 40
 27 41  1 42 17 43 29 44  1 45
 13 46 31 47 19 48  1 49 33 50

J

   lpd =:  (1 |.!.1 ])&.q:
Output:
   lpd 1+i.100
1 1 1 2 1 3 1 4 3 5 1 6 1 7 5 8 1 9 1 10 7 11 1 12 5 13 9 14 1 15 1 16 11 17 7 18 1 19 13 20 1 21 1 22 15 23 1 24 7 25 17 26 1 27 11 28 19 29 1 30 1 31 21 32 13 33 1 34 23 35 1 36 1 37 25 38 11 39 1 40 27 41 1 42 17 43 29 44 1 45 13 46 31 47 19 48 1 49 33 ...

This works by prime factorization of n, replacing the largest prime factor with 1, and then taking the product of this new list of prime factors. In J terms, we work "under" factorization, in analogy to a medical operation where the patient is "under" anaethesia: (put patient to sleep) rearrange guts (wake patient up).

The core logic only concerns itself with a list of prime factors; it is never even aware of the original input integer, nor the final integer result. In fact, note that the final multiplication is implicit and never spelled out by the programmer; product is the inverse of factorization, and we requested to work "under" factorization, thus J's algebra knows to apply the inverse of factorization (i.e. taking the product) as the final step.

jq

Works with jq

Works with gojq, the Go implementation of jq

Naive version:

# (1|largestpd) is 1 as per the task definition
def largestpd:
  if . == 1 then 1
  else . as $n
  | first( range( ($n - ($n % 2)) /2; 0; -1) | (select($n % . == 0) ))
  end;

Slightly less naive:

def largestpd:
  if . == 1 then 1
  else . as $n
  | (first(2,3,5,7 | select($n % . == 0)) // null) as $div
  | if $div then $n/$div
    else first( range( ($n - ($n % 11)) /11; 0; -1) | (select($n % . == 0) ))
    end
  end;
# For neatness
def lpad($len):
  tostring | ($len - length) as $l | (" " * $l)[:$l] + .;

def nwise($n):
  def w: if length <= $n then . else .[:$n], (.[$n:]|w) end;
  w;
  
### The task
[range(1; 101) | largestpd]
| nwise(10) | map(lpad(2)) | join(" ")
Output:
 1  1  1  2  1  3  1  4  3  5
 1  6  1  7  5  8  1  9  1 10
 7 11  1 12  5 13  9 14  1 15
 1 16 11 17  7 18  1 19 13 20
 1 21  1 22 15 23  1 24  7 25
17 26  1 27 11 28 19 29  1 30
 1 31 21 32 13 33  1 34 23 35
 1 36  1 37 25 38 11 39  1 40
27 41  1 42 17 43 29 44  1 45
13 46 31 47 19 48  1 49 33 50

Julia

largestpd(n) = (for i in n÷2:-1:1 (n % i == 0) && return i; end; 1)

foreach(n -> print(rpad(largestpd(n), 3), n % 10 == 0 ? "\n" : ""), 1:100)
Output:
1  1  1  2  1  3  1  4  3  5
1  6  1  7  5  8  1  9  1  10
7  11 1  12 5  13 9  14 1  15
1  16 11 17 7  18 1  19 13 20
1  21 1  22 15 23 1  24 7  25
17 26 1  27 11 28 19 29 1  30
1  31 21 32 13 33 1  34 23 35
1  36 1  37 25 38 11 39 1  40
27 41 1  42 17 43 29 44 1  45
13 46 31 47 19 48 1  49 33 50

MAD

            NORMAL MODE IS INTEGER
            
            INTERNAL FUNCTION(N)
            ENTRY TO LPD.
            WHENEVER N.LE.1, FUNCTION RETURN 1
            THROUGH TEST, FOR D=N-1, -1, D.L.1
TEST        WHENEVER N/D*D.E.N, FUNCTION RETURN D
            END OF FUNCTION
            
            THROUGH SHOW, FOR I=1, 10, I.GE.100
SHOW        PRINT FORMAT TABLE, 
          0     LPD.(I), LPD.(I+1), LPD.(I+2), LPD.(I+3),
          1     LPD.(I+4), LPD.(I+5), LPD.(I+6), LPD.(I+7),
          2     LPD.(I+8), LPD.(I+9)
          
            VECTOR VALUES TABLE = $10(I3)*$
            END OF PROGRAM
Output:
  1  1  1  2  1  3  1  4  3  5
  1  6  1  7  5  8  1  9  1 10
  7 11  1 12  5 13  9 14  1 15
  1 16 11 17  7 18  1 19 13 20
  1 21  1 22 15 23  1 24  7 25
 17 26  1 27 11 28 19 29  1 30
  1 31 21 32 13 33  1 34 23 35
  1 36  1 37 25 38 11 39  1 40
 27 41  1 42 17 43 29 44  1 45
 13 46 31 47 19 48  1 49 33 50

Mathematica / Wolfram Language

Last[Prepend[DeleteCases[Divisors[#], #], 1]] & /@ Range[100]
Output:
{1,1,1,2,1,3,1,4,3,5,1,6,1,7,5,8,1,9,1,10,7,11,1,12,5,13,9,14,1,15,1,16,11,17,7,18,1,19,13,20,1,21,1,22,15,23,1,24,7,25,17,26,1,27,11,28,19,29,1,30,1,31,21,32,13,33,1,34,23,35,1,36,1,37,25,38,11,39,1,40,27,41,1,42,17,43,29,44,1,45,13,46,31,47,19,48,1,49,33,50}

Modula-2

MODULE LargestProperDivisor;
FROM InOut IMPORT WriteCard, WriteLn;

VAR i: CARDINAL;

PROCEDURE lpd(n: CARDINAL): CARDINAL;
    VAR i: CARDINAL;
BEGIN
    IF n=1 THEN 
        RETURN 1;
    END;

    FOR i := n DIV 2 TO 1 BY -1 DO
        IF n MOD i = 0 THEN
            RETURN i;
        END;
    END;
END lpd;

BEGIN
    FOR i := 1 TO 100 DO
        WriteCard(lpd(i), 3);
        IF i MOD 10 = 0 THEN
            WriteLn();
        END;
    END;
END LargestProperDivisor.
Output:
  1  1  1  2  1  3  1  4  3  5
  1  6  1  7  5  8  1  9  1 10
  7 11  1 12  5 13  9 14  1 15
  1 16 11 17  7 18  1 19 13 20
  1 21  1 22 15 23  1 24  7 25
 17 26  1 27 11 28 19 29  1 30
  1 31 21 32 13 33  1 34 23 35
  1 36  1 37 25 38 11 39  1 40
 27 41  1 42 17 43 29 44  1 45
 13 46 31 47 19 48  1 49 33 50

Nim

import math, strutils

func largestProperDivisor(n: Positive): int =
  for d in 2..sqrt(float(n)).int:
    if n mod d == 0: return n div d
  result = 1

for n in 1..100:
  stdout.write ($n.largestProperDivisor).align(2), if n mod 10 == 0: '\n' else: ' '
Output:
 1  1  1  2  1  3  1  4  3  5
 1  6  1  7  5  8  1  9  1 10
 7 11  1 12  5 13  9 14  1 15
 1 16 11 17  7 18  1 19 13 20
 1 21  1 22 15 23  1 24  7 25
17 26  1 27 11 28 19 29  1 30
 1 31 21 32 13 33  1 34 23 35
 1 36  1 37 25 38 11 39  1 40
27 41  1 42 17 43 29 44  1 45
13 46 31 47 19 48  1 49 33 50

Pascal

program LarPropDiv;

function LargestProperDivisor(n:NativeInt):NativeInt;
//searching upwards to save time for example 100
//2..sqrt(n) aka 1..10 instead downwards n..sqrt(n) 100..10
var
  i,j: NativeInt;
Begin
  i := 2;
  repeat
    If n Mod i = 0 then
    Begin
      LargestProperDivisor := n DIV i;
      EXIT;
    end;
    inc(i);
  until i*i > n;
  LargestProperDivisor := 1;
end;
var
  n : Uint32;
begin
  for n := 1 to 100 do
  Begin
    write(LargestProperDivisor(n):4);
    if n mod 10 = 0 then
      Writeln;
  end;
end.
Output:
   1   1   1   2   1   3   1   4   3   5
   1   6   1   7   5   8   1   9   1  10
   7  11   1  12   5  13   9  14   1  15
   1  16  11  17   7  18   1  19  13  20
   1  21   1  22  15  23   1  24   7  25
  17  26   1  27  11  28  19  29   1  30
   1  31  21  32  13  33   1  34  23  35
   1  36   1  37  25  38  11  39   1  40
  27  41   1  42  17  43  29  44   1  45
  13  46  31  47  19  48   1  49  33  50

Perl

Library: ntheory
use strict;
use warnings;
use ntheory 'divisors';
use List::AllUtils <max natatime>;

sub proper_divisors {
  my $n = shift;
  return 1 if $n == 0;
  my @d = divisors($n);
  pop @d;
  @d;
}

my @range = 1 .. 100;
print "GPD for $range[0] through $range[-1]:\n";
my $iter = natatime 10, @range;
while( my @batch = $iter->() ) {
    printf '%3d', $_ == 1 ? 1 : max proper_divisors($_) for @batch; print "\n";
}
Output:
GPD for 1 through 100:
  1  1  1  2  1  3  1  4  3  5
  1  6  1  7  5  8  1  9  1 10
  7 11  1 12  5 13  9 14  1 15
  1 16 11 17  7 18  1 19 13 20
  1 21  1 22 15 23  1 24  7 25
 17 26  1 27 11 28 19 29  1 30
  1 31 21 32 13 33  1 34 23 35
  1 36  1 37 25 38 11 39  1 40
 27 41  1 42 17 43 29 44  1 45
 13 46 31 47 19 48  1 49 33 50

Phix

puts(1,join_by(apply(true,sprintf,{{"%3d"},vslice(apply(apply(true,factors,{tagset(100),-1}),reverse),1)}),1,10,""))
Output:
  1  1  1  2  1  3  1  4  3  5
  1  6  1  7  5  8  1  9  1 10
  7 11  1 12  5 13  9 14  1 15
  1 16 11 17  7 18  1 19 13 20
  1 21  1 22 15 23  1 24  7 25
 17 26  1 27 11 28 19 29  1 30
  1 31 21 32 13 33  1 34 23 35
  1 36  1 37 25 38 11 39  1 40
 27 41  1 42 17 43 29 44  1 45
 13 46 31 47 19 48  1 49 33 50

Alternative, same output, optimised: obviously checking for a factor from 2 up is going to be significantly faster than n-1 down... or of course as above collecting all of them and extracting/discarding all but the last, at least on much larger numbers, that is, and I suppose you could (perhaps) improve even further on this by only checking primes.

with javascript_semantics
for n=1 to 100 do
    integer p = 2,
            lim = floor(sqrt(n)),
            hf = 1
    while p<=lim do
        if remainder(n,p)=0 then
            hf = n/p
            exit
        end if
        p += 1
    end while 
    printf(1,"%3d",hf)
    if remainder(n,10)=0 then puts(1,"\n") end if
end for

Picat

main =>
  foreach(I in 1..100)
    printf("%2d%s",a(I), cond(I mod 10 == 0,"\n", " "))
  end.

a(1) = 1.
a(N) = Div =>
  a(N,N//2,Div).
a(N,I,I) :-
  N mod I == 0.
a(N,I,Div) :-
  a(N,I-1,Div).
Output:
 1  1  1  2  1  3  1  4  3  5
 1  6  1  7  5  8  1  9  1 10
 7 11  1 12  5 13  9 14  1 15
 1 16 11 17  7 18  1 19 13 20
 1 21  1 22 15 23  1 24  7 25
17 26  1 27 11 28 19 29  1 30
 1 31 21 32 13 33  1 34 23 35
 1 36  1 37 25 38 11 39  1 40
27 41  1 42 17 43 29 44  1 45
13 46 31 47 19 48  1 49 33 50


PL/I

largestProperDivisor: procedure options(main);
    lpd: procedure(n) returns(fixed);
        declare (n, i) fixed;
        if n <= 1 then return(1);
        do i=n-1 repeat(i-1) while(i>=1);
            if mod(n,i)=0 then return(i);
        end;
    end lpd;
    
    declare i fixed;
    do i=1 to 100;
        put edit(lpd(i)) (F(3));
        if mod(i,10)=0 then put skip;
    end;
end largestProperDivisor;
Output:
  1  1  1  2  1  3  1  4  3  5
  1  6  1  7  5  8  1  9  1 10
  7 11  1 12  5 13  9 14  1 15
  1 16 11 17  7 18  1 19 13 20
  1 21  1 22 15 23  1 24  7 25
 17 26  1 27 11 28 19 29  1 30
  1 31 21 32 13 33  1 34 23 35
  1 36  1 37 25 38 11 39  1 40
 27 41  1 42 17 43 29 44  1 45
 13 46 31 47 19 48  1 49 33 50

PL/M

100H:
BDOS: PROCEDURE (FN, ARG); DECLARE FN BYTE, ARG ADDRESS; GO TO 5; END BDOS;
EXIT: PROCEDURE; CALL BDOS(0,0); END EXIT;
PUT$CHAR: PROCEDURE (CH); DECLARE CH BYTE; CALL BDOS(2,CH); END PUT$CHAR;
PRINT: PROCEDURE (S); DECLARE S ADDRESS; CALL BDOS(9,S); END PRINT;

PRINT$3: PROCEDURE (N);
    DECLARE N BYTE;
    CALL PUT$CHAR(' ');
    IF N>9
        THEN CALL PUT$CHAR('0' + N/10);
        ELSE CALL PUT$CHAR(' ');
    CALL PUT$CHAR('0' + N MOD 10);
END PRINT$3;

LPD: PROCEDURE (N) BYTE;
    DECLARE (N, I) BYTE;
    IF N <= 1 THEN RETURN 1;
    I = N-1;
    DO WHILE I >= 1;
        IF N MOD I = 0 THEN RETURN I;
        I = I - 1;
    END;
END LPD;

DECLARE I BYTE;
DO I=1 TO 100;
    CALL PRINT$3(LPD(I));
    IF I MOD 10 = 0 THEN CALL PRINT(.(13,10,'$'));
END;
CALL EXIT;
EOF
Output:
  1  1  1  2  1  3  1  4  3  5
  1  6  1  7  5  8  1  9  1 10
  7 11  1 12  5 13  9 14  1 15
  1 16 11 17  7 18  1 19 13 20
  1 21  1 22 15 23  1 24  7 25
 17 26  1 27 11 28 19 29  1 30
  1 31 21 32 13 33  1 34 23 35
  1 36  1 37 25 38 11 39  1 40
 27 41  1 42 17 43 29 44  1 45
 13 46 31 47 19 48  1 49 33 50

PureBasic

Procedure.i lpd(v.i)
  For i=v/2 To 1 Step -1
    If v%i=0 : ProcedureReturn i : EndIf
  Next
  ProcedureReturn 1
EndProcedure

If OpenConsole("")
  For i=1 To 100
    Print(RSet(Str(lpd(i)),3))
    If i%10=0 : PrintN("") : EndIf
  Next
  Input()
EndIf
Output:
  1  1  1  2  1  3  1  4  3  5
  1  6  1  7  5  8  1  9  1 10
  7 11  1 12  5 13  9 14  1 15
  1 16 11 17  7 18  1 19 13 20
  1 21  1 22 15 23  1 24  7 25
 17 26  1 27 11 28 19 29  1 30
  1 31 21 32 13 33  1 34 23 35
  1 36  1 37 25 38 11 39  1 40
 27 41  1 42 17 43 29 44  1 45
 13 46 31 47 19 48  1 49 33 50

Python

def lpd(n):
    for i in range(n-1,0,-1):
        if n%i==0: return i
    return 1

for i in range(1,101):
    print("{:3}".format(lpd(i)), end=i%10==0 and '\n' or '')
Output:
  1  1  1  2  1  3  1  4  3  5
  1  6  1  7  5  8  1  9  1 10
  7 11  1 12  5 13  9 14  1 15
  1 16 11 17  7 18  1 19 13 20
  1 21  1 22 15 23  1 24  7 25
 17 26  1 27 11 28 19 29  1 30
  1 31 21 32 13 33  1 34 23 35
  1 36  1 37 25 38 11 39  1 40
 27 41  1 42 17 43 29 44  1 45
 13 46 31 47 19 48  1 49 33 50


Or, reducing the search space, formatting more flexibly (to allow for experiments with larger ranges) and composing functionally:

'''Largest proper divisor of'''

from math import isqrt


# maxProperDivisors :: Int -> Int
def maxProperDivisors(n):
    '''The largest proper divisor of n.
    '''
    secondDivisor = find(
        lambda x: 0 == (n % x)
    )(
        range(2, 1 + isqrt(n))
    )
    return 1 if None is secondDivisor else (
        n // secondDivisor
    )


# ------------------------- TEST -------------------------
# main :: IO ()
def main():
    '''Test for values of n in [1..100]'''

    xs = [
        maxProperDivisors(n) for n
        in range(1, 1 + 100)
    ]

    colWidth = 1 + len(str(max(xs)))

    print(
        '\n'.join([
            ''.join(row) for row in
            chunksOf(10)([
                str(x).rjust(colWidth, " ") for x in xs
            ])
        ])
    )


# ----------------------- GENERIC ------------------------

# chunksOf :: Int -> [a] -> [[a]]
def chunksOf(n):
    '''A series of lists of length n, subdividing the
       contents of xs. Where the length of xs is not evenly
       divisible, the final list will be shorter than n.
    '''
    def go(xs):
        return (
            xs[i:n + i] for i in range(0, len(xs), n)
        ) if 0 < n else None
    return go


# find :: (a -> Bool) -> [a] -> (a | None)
def find(p):
    '''Just the first element in the list that matches p,
       or None if no elements match.
    '''
    def go(xs):
        try:
            return next(x for x in xs if p(x))
        except StopIteration:
            return None
    return go


# MAIN ---
if __name__ == '__main__':
    main()
Output:
  1  1  1  2  1  3  1  4  3  5
  1  6  1  7  5  8  1  9  1 10
  7 11  1 12  5 13  9 14  1 15
  1 16 11 17  7 18  1 19 13 20
  1 21  1 22 15 23  1 24  7 25
 17 26  1 27 11 28 19 29  1 30
  1 31 21 32 13 33  1 34 23 35
  1 36  1 37 25 38 11 39  1 40
 27 41  1 42 17 43 29 44  1 45
 13 46 31 47 19 48  1 49 33 50

Quackery

factors is defined at Factors of an integer.

' [ 1 ] 99 times
  [ i^ 2 + factors
    -2 peek join ]
echo
Output:
[ 1 1 1 2 1 3 1 4 3 5 1 6 1 7 5 8 1 9 1 10 7 11 1 12 5 13 9 14 1 15 1 16 11 17 7 18 1 19 13 20 1 21 1 22 15 23 1 24 7 25 17 26 1 27 11 28 19 29 1 30 1 31 21 32 13 33 1 34 23 35 1 36 1 37 25 38 11 39 1 40 27 41 1 42 17 43 29 44 1 45 13 46 31 47 19 48 1 49 33 50 ]

R

largest_proper_divisor <- function(n){
  if(n == 1) return(1)
  
  lpd = 1
  
  for(i in seq(1, n-1, 1)){
    if(n %% i == 0)
      lpd = i
  }
  
  message(paste0("The largest proper divisor of ", n, " is ", lpd))
  return(lpd)
}

#Verify
for (i in 1:100){       #about 10 seconds to calculate until 10000
  largest_proper_divisor(i)
}
Output:
  The largest proper divisor of 2 is 1
The largest proper divisor of 3 is 1
The largest proper divisor of 4 is 2
The largest proper divisor of 5 is 1
The largest proper divisor of 6 is 3
The largest proper divisor of 7 is 1
The largest proper divisor of 8 is 4
The largest proper divisor of 9 is 3
.
. blah blah blah
.
The largest proper divisor of 94 is 47
The largest proper divisor of 95 is 19
The largest proper divisor of 96 is 48
The largest proper divisor of 97 is 1
The largest proper divisor of 98 is 49
The largest proper divisor of 99 is 33
The largest proper divisor of 100 is 50

Raku

A little odd to special case a(1) == 1 as technically, 1 doesn't have any proper divisors... but it matches OEIS A032742 so whatever.

Note: this example is ridiculously overpowered (and so, somewhat inefficient) for a range of 1 to 100, but will easily handle large numbers without modification.

use Prime::Factor;

for 0, 2**67 - 1 -> $add {
    my $start = now;

    my $range = $add + 1 .. $add + 100;

    say "GPD for {$range.min} through {$range.max}:";

    say ( $range.hyper(:14batch).map: {$_ == 1 ?? 1 !! $_ %% 2 ?? $_ / 2 !! .&proper-divisors.max} )\
        .batch(10)».fmt("%{$add.chars + 1}d").join: "\n";

    say (now - $start).fmt("%0.3f seconds\n");
}
Output:
GPD for 1 through 100:
 1  1  1  2  1  3  1  4  3  5
 1  6  1  7  5  8  1  9  1 10
 7 11  1 12  5 13  9 14  1 15
 1 16 11 17  7 18  1 19 13 20
 1 21  1 22 15 23  1 24  7 25
17 26  1 27 11 28 19 29  1 30
 1 31 21 32 13 33  1 34 23 35
 1 36  1 37 25 38 11 39  1 40
27 41  1 42 17 43 29 44  1 45
13 46 31 47 19 48  1 49 33 50
0.071 seconds

GPD for 147573952589676412928 through 147573952589676413027:
  73786976294838206464   49191317529892137643   73786976294838206465                      1   73786976294838206466   21081993227096630419   73786976294838206467   49191317529892137645   73786976294838206468    8680820740569200761
  73786976294838206469    5088756985850910791   73786976294838206470   49191317529892137647   73786976294838206471   13415813871788764813   73786976294838206472   29514790517935282589   73786976294838206473   49191317529892137649
  73786976294838206474    6416258808246800563   73786976294838206475     977310944302492801   73786976294838206476   49191317529892137651   73786976294838206477   29514790517935282591   73786976294838206478      87167130885810049
  73786976294838206479   49191317529892137653   73786976294838206480   21081993227096630423   73786976294838206481    7767050136298758577   73786976294838206482   49191317529892137655   73786976294838206483    2784414199805215339
  73786976294838206484   11351842506898185613   73786976294838206485   49191317529892137657   73786976294838206486     939961481462907089   73786976294838206487   29514790517935282595   73786976294838206488   49191317529892137659
  73786976294838206489      82581954443019817   73786976294838206490        147258377885867   73786976294838206491   49191317529892137661   73786976294838206492   29514790517935282597   73786976294838206493   13415813871788764817
  73786976294838206494   49191317529892137663   73786976294838206495                      1   73786976294838206496    2202596307308603179   73786976294838206497   49191317529892137665   73786976294838206498    5088756985850910793
  73786976294838206499                      1   73786976294838206500   49191317529892137667   73786976294838206501   21081993227096630429   73786976294838206502   29514790517935282601   73786976294838206503   49191317529892137669
  73786976294838206504   13415813871788764819   73786976294838206505     103270785577100359   73786976294838206506   49191317529892137671   73786976294838206507   29514790517935282603   73786976294838206508   21081993227096630431
  73786976294838206509   49191317529892137673   73786976294838206510   11351842506898185617   73786976294838206511                      1   73786976294838206512   49191317529892137675   73786976294838206513    1305964182209525779
0.440 seconds

REXX

Logic was added to the   LPDIV   function so that for   odd   numbers,   only odd divisors were used.

This addition made it about   75%   faster.

/*REXX program finds the largest proper divisors of all numbers  (up to a given limit). */
parse arg n cols .                               /*obtain optional argument from the CL.*/
if    n=='' |    n==","  then    n= 101          /*Not specified?  Then use the default.*/
if cols=='' | cols==","  then cols=  10          /* "      "         "   "   "     "    */
w= length(n)  +  1                               /*W:  the length of any output column. */
@LPD = "largest proper divisor of  N,  where  N  < "       n
idx = 1                                          /*initialize the index (output counter)*/
say ' index │'center(@LPD,  1 + cols*(w+1)     ) /*display the title for the output.    */
say '───────┼'center(""  ,  1 + cols*(w+1), '─') /*   "     "   sep   "   "     "       */
$=                                               /*a list of largest proper divs so far.*/
     do j=1  for n-1;  $= $ right(LPDIV(j), w)   /*add a largest proper divisor ──► list*/
     if j//cols\==0  then iterate                /*have we populated a line of output?  */
     say center(idx, 7)'│'  substr($, 2);   $=   /*display what we have so far  (cols). */
     idx= idx + cols                             /*bump the  index  count for the output*/
     end   /*j*/

if $\==''  then say center(idx, 7)"│"  substr($, 2)  /*possible display residual output.*/
say '───────┴'center(""  ,  1 + cols*(w+1), '─') /*display the foot separator.          */
exit 0                                           /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
LPDIV:  procedure; parse arg x;  if x<4  then return 1   /*obtain  X;  test if  X < 4.  */
        odd= x // 2;    beg= x % 2                       /*use BEG as the first divisor.*/
        if odd  then if beg//2==0  then beg= beg -  1    /*Is X odd?  Then make BEG odd.*/
                   do k=beg  by -1-odd           /*begin at halfway point and decrease. */
                   if x//k==0  then return k     /*Remainder=0?  Got largest proper div.*/
                   end   /*k*/
                                    return 1     /*If we get here,  then  X  is a prime.*/
output   when using the default inputs:
 index │  largest proper divisor of  N,  where  N  <  101
───────┼───────────────────────────────────────────────────
   1   │    1    1    1    2    1    3    1    4    3    5
  11   │    1    6    1    7    5    8    1    9    1   10
  21   │    7   11    1   12    5   13    9   14    1   15
  31   │    1   16   11   17    7   18    1   19   13   20
  41   │    1   21    1   22   15   23    1   24    7   25
  51   │   17   26    1   27   11   28   19   29    1   30
  61   │    1   31   21   32   13   33    1   34   23   35
  71   │    1   36    1   37   25   38   11   39    1   40
  81   │   27   41    1   42   17   43   29   44    1   45
  91   │   13   46   31   47   19   48    1   49   33   50
───────┴───────────────────────────────────────────────────

Ring

? "working..."
limit = 100
? "Largest proper divisor up to " + limit + " are:"
see " 1 " 
col = 1
 
for n = 2 to limit
    for m = 1 to n >> 1
        if n % m = 0 div = m ok
    next
    if div < 10 see " " ok
    see "" + div + " "
    if col++ % 10 = 0 ? "" ok
next
? "done..."
Output:
working...
Largest proper divisor up to 100 are:
 1  1  1  2  1  3  1  4  3  5 
 1  6  1  7  5  8  1  9  1 10 
 7 11  1 12  5 13  9 14  1 15 
 1 16 11 17  7 18  1 19 13 20 
 1 21  1 22 15 23  1 24  7 25 
17 26  1 27 11 28 19 29  1 30 
 1 31 21 32 13 33  1 34 23 35 
 1 36  1 37 25 38 11 39  1 40 
27 41  1 42 17 43 29 44  1 45 
13 46 31 47 19 48  1 49 33 50 
done...

Ruby

require 'prime'

def a(n)
  return 1 if n == 1 || n.prime?
  (n/2).downto(1).detect{|d| n.remainder(d) == 0}
end

(1..100).map{|n| a(n).to_s.rjust(3)}.each_slice(10){|slice| puts slice.join}
Output:
  1  1  1  2  1  3  1  4  3  5
  1  6  1  7  5  8  1  9  1 10
  7 11  1 12  5 13  9 14  1 15
  1 16 11 17  7 18  1 19 13 20
  1 21  1 22 15 23  1 24  7 25
 17 26  1 27 11 28 19 29  1 30
  1 31 21 32 13 33  1 34 23 35
  1 36  1 37 25 38 11 39  1 40
 27 41  1 42 17 43 29 44  1 45
 13 46 31 47 19 48  1 49 33 50

Seed7

$ include "seed7_05.s7i";

const func integer: largestProperDivisor (in integer: number) is func
  result
    var integer: divisor is 0;
  begin
    if not odd(number) then
      divisor := number >> 1;
    else
      divisor := number div 3 - 1;
      if odd(divisor) then
        divisor +:= 2;
      else
        incr(divisor);
      end if;
      while number rem divisor <> 0 do
        divisor -:= 2;
      end while;
    end if;
  end func;

const proc: main is func
  local
    var integer: n is 0;
  begin
    for n range 1 to 100 do
      write(largestProperDivisor(n) lpad 3);
      if n rem 10 = 0 then
        writeln;
      end if;
    end for;
  end func;
Output:
  1  1  1  2  1  3  1  4  3  5
  1  6  1  7  5  8  1  9  1 10
  7 11  1 12  5 13  9 14  1 15
  1 16 11 17  7 18  1 19 13 20
  1 21  1 22 15 23  1 24  7 25
 17 26  1 27 11 28 19 29  1 30
  1 31 21 32 13 33  1 34 23 35
  1 36  1 37 25 38 11 39  1 40
 27 41  1 42 17 43 29 44  1 45
 13 46 31 47 19 48  1 49 33 50

Swift

import Foundation

func largestProperDivisor(_ n : Int) -> Int? {
    guard n > 0 else {
        return nil
    }
    if (n & 1) == 0 {
        return n >> 1
    }
    var p = 3
    while p * p <= n {
        if n % p == 0 {
            return n / p
        }
        p += 2
    }
    return 1
}

for n in (1..<101) {
    print(String(format: "%2d", largestProperDivisor(n)!),
          terminator: n % 10 == 0 ? "\n" : " ")
}
Output:
 1  1  1  2  1  3  1  4  3  5
 1  6  1  7  5  8  1  9  1 10
 7 11  1 12  5 13  9 14  1 15
 1 16 11 17  7 18  1 19 13 20
 1 21  1 22 15 23  1 24  7 25
17 26  1 27 11 28 19 29  1 30
 1 31 21 32 13 33  1 34 23 35
 1 36  1 37 25 38 11 39  1 40
27 41  1 42 17 43 29 44  1 45
13 46 31 47 19 48  1 49 33 50


True BASIC

Translation of: FreeBASIC
PRINT "El mayor divisor propio de n es:"
PRINT
PRINT "  1  1";
FOR i = 3 To 100
    FOR j = i-1 To 1 Step -1
        IF remainder(i, j) = 0 Then
           PRINT Using$("###", j);
           EXIT FOR
        END IF
    NEXT  j
    IF remainder(i, 10) = 0 Then PRINT
NEXT i
END
Output:
Igual que la entrada de FreeBASIC.

Vlang

Translation of: go
fn largest_proper_divisor(n int) int {
    for i := 2; i*i <= n; i++ {
        if n%i == 0 {
            return n / i
        }
    }
    return 1
}
 
fn main() {
    println("The largest proper divisors for numbers in the interval [1, 100] are:")
    print(" 1  ")
    for n := 2; n <= 100; n++ {
        if n%2 == 0 {
            print("${n/2:2}  ")
        } else {
            print("${largest_proper_divisor(n):2}  ")
        }
        if n%10 == 0 {
            println('')
        }
    }
}
Output:
The largest proper divisors for numbers in the interval [1, 100] are:
 1   1   1   2   1   3   1   4   3   5  
 1   6   1   7   5   8   1   9   1  10  
 7  11   1  12   5  13   9  14   1  15  
 1  16  11  17   7  18   1  19  13  20  
 1  21   1  22  15  23   1  24   7  25  
17  26   1  27  11  28  19  29   1  30  
 1  31  21  32  13  33   1  34  23  35  
 1  36   1  37  25  38  11  39   1  40  
27  41   1  42  17  43  29  44   1  45  
13  46  31  47  19  48   1  49  33  50  


Wren

Library: Wren-math
Library: Wren-fmt
import "/math" for Int
import "/fmt" for Fmt

System.print("The largest proper divisors for numbers in the interval [1, 100] are:")
System.write(" 1  ")
for (n in 2..100) {
    if (n % 2 == 0) {
        Fmt.write("$2d  ", n / 2)
    } else {
        Fmt.write("$2d  ", Int.properDivisors(n)[-1])
    }
    if (n % 10 == 0) System.print()
}
Output:
The largest proper divisors for numbers in the interval [1, 100] are:
 1   1   1   2   1   3   1   4   3   5  
 1   6   1   7   5   8   1   9   1  10  
 7  11   1  12   5  13   9  14   1  15  
 1  16  11  17   7  18   1  19  13  20  
 1  21   1  22  15  23   1  24   7  25  
17  26   1  27  11  28  19  29   1  30  
 1  31  21  32  13  33   1  34  23  35  
 1  36   1  37  25  38  11  39   1  40  
27  41   1  42  17  43  29  44   1  45  
13  46  31  47  19  48   1  49  33  50  


Yabasic

Translation of: FreeBASIC
print "El mayor divisor propio de n es:\n"
print " 1  1 "; 
for i = 3 to 100
	for j = i-1 to 1 step -1
		If mod(i, j) = 0 then print j using "##"; : break : fi
	next  j
    if mod(i, 10) = 0 then print : fi
next i
print
end
Output:
Igual que la entrada de FreeBASIC.


X86 Assembly

      1                              ;Assemble with: tasm, tlink /t
      2 0000                                 .model  tiny
      3 0000                                 .code
      4                                      org     100h            ;.com program starts here
      5
      6                              ;bl = D = divisor
      7                              ;bh = N
      8
      9 0100  B7 01                  start:  mov     bh, 1           ;for N:= 1 to 100 do
     10 0102  8A DF                  d10:    mov     bl, bh          ;D:= if N=1 then 1 else N/2
     11 0104  D0 EB                          shr     bl, 1
     12 0106  75 02                          jne     d15
     13 0108  FE C3                           inc    bl
     14 010A                         d15:
     15 010A  8A C7                  d20:    mov     al, bh          ;while rem(N/D) # 0 do
     16 010C  98                             cbw                     ; ah:= 0 (extend sign of al into ah)
     17 010D  F6 FB                          idiv    bl              ; al:= ax/bl; ah:= remainder
     18 010F  84 E4                          test    ah, ah
     19 0111  74 04                          je      d30
     20 0113  FE CB                           dec    bl              ; D--
     21 0115  EB F3                           jmp    d20
     22 0117                         d30:
     23 0117  8A C3                          mov     al, bl          ;output number in D
     24 0119  D4 0A                          aam     10              ;ah:= al/10; al:= remainder
     25 011B  50                             push    ax              ;save low digit in remainder
     26 011C  8A C4                          mov     al, ah          ;get high digit
     27 011E  04 20                          add     al, 20h         ;if zero make it a space char
     28 0120  84 E4                          test    ah, ah
     29 0122  74 02                          je      d50
     30 0124  04 10                           add    al, 10h         ;else make it into ASCII digit
     31 0126  CD 29                  d50:    int     29h             ;output high digit or space
     32 0128  58                             pop     ax              ;get low digit
     33 0129  04 30                          add     al, 30h         ;make it ASCII
     34 012B  CD 29                          int     29h             ;output low digit
     35
     36 012D  B0 20                          mov     al, 20h         ;output space char
     37 012F  CD 29                          int     29h
     38
     39 0131  8A C7                          mov     al, bh          ;if remainder(N/10) = 0 then CR LF
     40 0133  D4 0A                          aam     10              ;ah:= al/10; al:= remainder
     41 0135  3C 00                          cmp     al, 0
     42 0137  75 08                          jne     next
     43 0139  B0 0D                           mov    al, 0Dh         ;CR
     44 013B  CD 29                           int    29h
     45 013D  B0 0A                           mov    al, 0Ah         ;LF
     46 013F  CD 29                           int    29h
     47
     48 0141  FE C7                  next:   inc     bh              ;next N
     49 0143  80 FF 64                       cmp     bh, 100
     50 0146  7E BA                          jle     d10
     51 0148  C3                             ret
     52
     53                                      end     start
Output:
 1  1  1  2  1  3  1  4  3  5
 1  6  1  7  5  8  1  9  1 10
 7 11  1 12  5 13  9 14  1 15
 1 16 11 17  7 18  1 19 13 20
 1 21  1 22 15 23  1 24  7 25
17 26  1 27 11 28 19 29  1 30
 1 31 21 32 13 33  1 34 23 35
 1 36  1 37 25 38 11 39  1 40
27 41  1 42 17 43 29 44  1 45
13 46 31 47 19 48  1 49 33 50

XPL0

int N, D;
[for N:= 1 to 100 do
    [D:= if N=1 then 1 else N/2;
    while rem(N/D) do D:= D-1;
    if D<10 then ChOut(0, ^ );
    IntOut(0, D);
    if rem(N/10) = 0 then CrLf(0) else ChOut(0, ^ );
    ];
]
Output:
 1  1  1  2  1  3  1  4  3  5
 1  6  1  7  5  8  1  9  1 10
 7 11  1 12  5 13  9 14  1 15
 1 16 11 17  7 18  1 19 13 20
 1 21  1 22 15 23  1 24  7 25
17 26  1 27 11 28 19 29  1 30
 1 31 21 32 13 33  1 34 23 35
 1 36  1 37 25 38 11 39  1 40
27 41  1 42 17 43 29 44  1 45
13 46 31 47 19 48  1 49 33 50