Proper divisors: Difference between revisions
m →version 3: updated the parameter parsing for the Pdivs function. |
m →version 3: extended the notes on the speed improvements. |
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===version 3=== |
===version 3=== |
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When factoring 20,000 integers, this REXX version is |
When factoring 20,000 integers, this REXX version is about '''10%''' faster than the REXX version 2. |
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<br>When factoring 200,000 integers, this REXX version is |
<br>When factoring 200,000 integers, this REXX version is about '''30%''' faster. |
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<br>When factoring 2,000,000 integers, this REXX version is about '''40%''' faster. |
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It accomplishes a faster speed by incorporating the calculation of an ''integer square root'' of an integer (without using any floating point arithmetic). |
It accomplishes a faster speed by incorporating the calculation of an ''integer square root'' of an integer (without using any floating point arithmetic). |
Revision as of 00:45, 8 October 2017
The proper divisors of a positive integer N are those numbers, other than N itself, that divide N without remainder.
For N > 1 they will always include 1, but for N == 1 there are no proper divisors.
- Examples
The proper divisors of 6 are 1, 2, and 3.
The proper divisors of 100 are 1, 2, 4, 5, 10, 20, 25, and 50.
- Task
- Create a routine to generate all the proper divisors of a number.
- use it to show the proper divisors of the numbers 1 to 10 inclusive.
- Find a number in the range 1 to 20,000 with the most proper divisors. Show the number and just the count of how many proper divisors it has.
Show all output here.
- Related tasks
- Amicable pairs
- Abundant, deficient and perfect number classifications
- Aliquot sequence classifications
- Factors of an integer
- Prime decomposition
360 Assembly
This program uses two ASSIST macros (XDECO, XPRNT) to keep the code as short as possible. <lang 360asm>* Proper divisors 14/06/2016 PROPDIV CSECT
USING PROPDIV,R13 base register B 72(R15) skip savearea DC 17F'0' savearea STM R14,R12,12(R13) prolog ST R13,4(R15) " ST R15,8(R13) " LR R13,R15 " LA R10,1 n=1
LOOPN1 C R10,=F'10' do n=1 to 10
BH ELOOPN1 LR R1,R10 n BAL R14,PDIV pdiv(n) ST R0,NN nn=pdiv(n) MVC PG,PGT init buffer LA R11,PG pgi=0 XDECO R10,XDEC edit n MVC 0(3,R11),XDEC+9 output n LA R11,7(R11) pgi=pgi+7 L R1,NN nn XDECO R1,XDEC edit nn MVC 0(3,R11),XDEC+9 output nn LA R11,20(R11) pgi=pgi+20 LA R5,1 i=1
LOOPNI C R5,NN do i=1 to nn
BH ELOOPNI LR R1,R5 i SLA R1,2 *4 L R2,TDIV-4(R1) tdiv(i) XDECO R2,XDEC edit tdiv(i) MVC 0(3,R11),XDEC+9 output tdiv(i) LA R11,3(R11) pgi=pgi+3 LA R5,1(R5) i=i+1 B LOOPNI
ELOOPNI XPRNT PG,80 print buffer
LA R10,1(R10) n=n+1 B LOOPN1
ELOOPN1 SR R0,R0 0
ST R0,M m=0 LA R10,1 n=1
LOOPN2 C R10,=F'20000' do n=1 to 20000
BH ELOOPN2 LR R1,R10 n BAL R14,PDIV nn=pdiv(n) C R0,M if nn>m BNH NNNHM ST R10,II ii=n ST R0,M m=nn
NNNHM LA R10,1(R10) n=n+1
B LOOPN2
ELOOPN2 MVC PG,PGR init buffer
L R1,II ii XDECO R1,XDEC edit ii MVC PG(5),XDEC+7 output ii L R1,M m XDECO R1,XDEC edit m MVC PG+9(4),XDEC+8 output m XPRNT PG,80 print buffer L R13,4(0,R13) epilog LM R14,R12,12(R13) " XR R15,R15 " BR R14 exit
- ------- pdiv --function(x)----->number of divisors---
PDIV ST R1,X x
C R1,=F'1' if x=1 BNE NOTONE LA R0,0 return(0) BR R14
NOTONE LR R4,R1 x
N R4,=X'00000001' mod(x,2) LA R4,1(R4) +1 ST R4,ODD odd=mod(x,2)+1 LA R8,1 ia=1 LA R0,1 1 ST R0,TDIV tdiv(1)=1 SR R9,R9 ib=0 L R7,ODD odd LA R7,1(R7) j=odd+1
LOOPJ LR R5,R7 do j=odd+1 by odd
MR R4,R7 j*j C R5,X while j*j<x BNL ELOOPJ L R4,X x SRDA R4,32 . DR R4,R7 /j LTR R4,R4 if mod(x,j)=0 BNZ ITERJ LA R8,1(R8) ia=ia+1 LR R1,R8 ia SLA R1,2 *4 (F) ST R7,TDIV-4(R1) tdiv(ia)=j LA R9,1(R9) ib=ib+1 L R4,X x SRDA R4,32 . DR R4,R7 j LR R2,R9 ib SLA R2,2 *4 (F) ST R5,TDIVB-4(R2) tdivb(ib)=x/j
ITERJ A R7,ODD j=j+odd
B LOOPJ
ELOOPJ LR R5,R7 j
MR R4,R7 j*j C R5,X if j*j=x BNE JTJNEX LA R8,1(R8) ia=ia+1 LR R1,R8 ia SLA R1,2 *4 (F) ST R7,TDIV-4(R1) tdiv(ia)=j
JTJNEX LA R1,TDIV(R1) @tdiv(ia+1)
LA R2,TDIVB-4(R2) @tdivb(ib) LTR R6,R9 do i=ib to 1 by -1 BZ ELOOPI
LOOPI MVC 0(4,R1),0(R2) tdiv(ia)=tdivb(i)
LA R8,1(R8) ia=ia+1 LA R1,4(R1) r1+=4 SH R2,=H'4' r2-=4 BCT R6,LOOPI i=i-1
ELOOPI LR R0,R8 return(ia)
BR R14 return to caller
- ---- ----------------------------------------
TDIV DS 80F TDIVB DS 40F M DS F NN DS F II DS F X DS F ODD DS F PGT DC CL80'... has .. proper divisors:' PGR DC CL80'..... has ... proper divisors.' PG DC CL80' ' XDEC DS CL12
YREGS END PROPDIV</lang>
- Output:
1 has 0 proper divisors: 2 has 1 proper divisors: 1 3 has 1 proper divisors: 1 4 has 2 proper divisors: 1 2 5 has 1 proper divisors: 1 6 has 3 proper divisors: 1 2 3 7 has 1 proper divisors: 1 8 has 3 proper divisors: 1 2 4 9 has 2 proper divisors: 1 3 10 has 3 proper divisors: 1 2 5 15120 has 79 proper divisors.
ALGOL 68
<lang algol68># MODE to hold an element of a list of proper divisors # MODE DIVISORLIST = STRUCT( INT divisor, REF DIVISORLIST next );
- end of divisor list value #
REF DIVISORLIST nil divisor list = REF DIVISORLIST(NIL);
- resturns a DIVISORLIST containing the proper divisors of n #
- if n = 1, 0 or -1, we return no divisors #
PROC proper divisors = ( INT n )REF DIVISORLIST:
BEGIN REF DIVISORLIST result := nil divisor list; REF DIVISORLIST end list := result; INT abs n = ABS n; IF abs n > 1 THEN # build the list of divisors backeards, so they are # # returned in ascending order # INT root n = ENTIER sqrt( abs n ); FOR d FROM root n BY -1 TO 2 DO IF abs n MOD d = 0 THEN # found another divisor # result := HEAP DIVISORLIST := DIVISORLIST( d, result ); IF end list IS nil divisor list THEN # first result # end list := result FI; IF d * d /= n THEN # add the other divisor to the end of # # the list # next OF end list := HEAP DIVISORLIST := DIVISORLIST( abs n OVER d, nil divisor list ); end list := next OF end list FI FI OD; # 1 is always a proper divisor of numbers > 1 # result := HEAP DIVISORLIST := DIVISORLIST( 1, result ) FI; result END # proper divisors # ;
- returns the number of divisors in a DIVISORLIST #
PROC count divisors = ( REF DIVISORLIST list )INT:
BEGIN INT result := 0; REF DIVISORLIST divisors := list; WHILE divisors ISNT nil divisor list DO result +:= 1; divisors := next OF divisors OD; result END # count divisors # ;
- find the proper divisors of 1 : 10 #
FOR n TO 10 DO
REF DIVISORLIST divisors := proper divisors( n ); print( ( "Proper divisors of: ", whole( n, -2 ), ": " ) ); WHILE divisors ISNT nil divisor list DO print( ( " ", whole( divisor OF divisors, 0 ) ) ); divisors := next OF divisors OD; print( ( newline ) )
OD;
- find the first/only number in 1 : 20 000 with the most divisors #
INT max number = 20 000; INT max divisors := 0; INT has max divisors := 0; INT with max divisors := 0; FOR d TO max number DO
INT divisor count = count divisors( proper divisors( d ) ); IF divisor count > max divisors THEN # found a number with more divisors than the previous max # max divisors := divisor count; has max divisors := d; with max divisors := 1 ELIF divisor count = max divisors THEN # found another number with that many divisors # with max divisors +:= 1 FI
OD; print( ( whole( has max divisors, 0 )
, " is the " , IF with max divisors < 2 THEN "only" ELSE "first" FI , " number upto " , whole( max number, 0 ) , " with " , whole( max divisors, 0 ) , " divisors" , newline ) )</lang>
- Output:
Proper divisors of: 1: Proper divisors of: 2: 1 Proper divisors of: 3: 1 Proper divisors of: 4: 1 2 Proper divisors of: 5: 1 Proper divisors of: 6: 1 2 3 Proper divisors of: 7: 1 Proper divisors of: 8: 1 2 4 Proper divisors of: 9: 1 3 Proper divisors of: 10: 1 2 5 15120 is the first number upto 20000 with 79 divisors
AppleScript
<lang AppleScript>-- PROPER DIVISORS -----------------------------------------------------------
-- properDivisors :: Int -> [Int] on properDivisors(n)
if n = 1 then {1} else set realRoot to n ^ (1 / 2) set intRoot to realRoot as integer set blnPerfectSquare to intRoot = realRoot -- isFactor :: Int -> Bool script isFactor on |λ|(x) n mod x = 0 end |λ| end script -- Factors up to square root of n, set lows to filter(isFactor, enumFromTo(1, intRoot)) -- and quotients of these factors beyond the square root, -- integerQuotient :: Int -> Int script integerQuotient on |λ|(x) (n / x) as integer end |λ| end script -- excluding n itself (last item) items 1 thru -2 of (lows & map(integerQuotient, ¬ items (1 + (blnPerfectSquare as integer)) thru -1 of reverse of lows)) end if
end properDivisors
-- TEST ----------------------------------------------------------------------
on run
-- numberAndDivisors :: Int -> [Int] script numberAndDivisors on |λ|(n) {num:n, divisors:properDivisors(n)} end |λ| end script -- maxDivisorCount :: Record -> Int -> Record script maxDivisorCount on |λ|(a, n) set intDivisors to length of properDivisors(n) if intDivisors ≥ divisors of a then {num:n, divisors:intDivisors} else a end if end |λ| end script {oneToTen:map(numberAndDivisors, ¬ enumFromTo(1, 10)), mostDivisors:foldl(maxDivisorCount, ¬ {num:0, divisors:0}, enumFromTo(1, 20000))} ¬
end run
-- GENERIC FUNCTIONS ---------------------------------------------------------
-- enumFromTo :: Int -> Int -> [Int] on enumFromTo(m, n)
if m > n then set d to -1 else set d to 1 end if set lst to {} repeat with i from m to n by d set end of lst to i end repeat return lst
end enumFromTo
-- filter :: (a -> Bool) -> [a] -> [a] on filter(f, xs)
tell mReturn(f) set lst to {} set lng to length of xs repeat with i from 1 to lng set v to item i of xs if |λ|(v, i, xs) then set end of lst to v end repeat return lst end tell
end filter
-- foldl :: (a -> b -> a) -> a -> [b] -> a on foldl(f, startValue, xs)
tell mReturn(f) set v to startValue set lng to length of xs repeat with i from 1 to lng set v to |λ|(v, item i of xs, i, xs) end repeat return v end tell
end foldl
-- map :: (a -> b) -> [a] -> [b] on map(f, 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
-- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Script on mReturn(f)
if class of f is script then f else script property |λ| : f end script end if
end mReturn</lang>
- Output:
<lang AppleScript>{oneToTen:{{num:1, divisors:{1}}, {num:2, divisors:{1}}, {num:3, divisors:{1}}, {num:4, divisors:{1, 2}}, {num:5, divisors:{1}}, {num:6, divisors:{1, 2, 3}}, {num:7, divisors:{1}}, {num:8, divisors:{1, 2, 4}}, {num:9, divisors:{1, 3}}, {num:10, divisors:{1, 2, 5}}}, mostDivisors:{num:18480, divisors:79}}</lang>
AWK
<lang AWK>
- syntax: GAWK -f PROPER_DIVISORS.AWK
BEGIN {
show = 0 # show divisors: 0=no, 1=yes print(" N cnt DIVISORS") for (i=1; i<=20000; i++) { divisors(i) if (i <= 10 || i == 100) { # including 100 as it was an example in task description printf("%5d %3d %s\n",i,Dcnt,Dstr) } if (Dcnt < max_cnt) { continue } if (Dcnt > max_cnt) { rec = "" max_cnt = Dcnt } rec = sprintf("%s%5d %3d %s\n",rec,i,Dcnt,show?Dstr:"divisors not shown") } printf("%s",rec) exit(0)
} function divisors(n, i) {
if (n == 1) { Dcnt = 0 Dstr = "" return } Dcnt = Dstr = 1 for (i=2; i<n; i++) { if (n % i == 0) { Dcnt++ Dstr = sprintf("%s %s",Dstr,i) } } return
} </lang>
output:
N cnt DIVISORS 1 0 2 1 1 3 1 1 4 2 1 2 5 1 1 6 3 1 2 3 7 1 1 8 3 1 2 4 9 2 1 3 10 3 1 2 5 100 8 1 2 4 5 10 20 25 50 15120 79 divisors not shown 18480 79 divisors not shown
BaCon
<lang qbasic> FUNCTION ProperDivisor(nr, show)
LOCAL probe, total
FOR probe = 1 TO nr-1 IF MOD(nr, probe) = 0 THEN IF show THEN PRINT " ", probe; INCR total END IF NEXT
RETURN total
END FUNCTION
FOR x = 1 TO 10
PRINT x, ":"; IF ProperDivisor(x, 1) = 0 THEN PRINT " 0"; PRINT
NEXT
FOR x = 1 TO 20000
DivisorCount = ProperDivisor(x, 0) IF DivisorCount > MaxDivisors THEN MaxDivisors = DivisorCount MagicNumber = x END IF
NEXT
PRINT "Most proper divisors for number in the range 1-20000: ", MagicNumber, " with ", MaxDivisors, " divisors." </lang>
- Output:
1: 0 2: 1 3: 1 4: 1 2 5: 1 6: 1 2 3 7: 1 8: 1 2 4 9: 1 3 10: 1 2 5 Most proper divisors for number in the range 1-20000: 15120 with 79 divisors.
C
C has tedious boilerplate related to allocating memory for dynamic arrays, so we just skip the problem of storing values altogether. <lang c>
- include <stdio.h>
- include <stdbool.h>
int proper_divisors(const int n, bool print_flag) {
int count = 0;
for (int i = 1; i < n; ++i) { if (n % i == 0) { count++; if (print_flag) printf("%d ", i); } }
if (print_flag) printf("\n");
return count;
}
int main(void) {
for (int i = 1; i <= 10; ++i) { printf("%d: ", i); proper_divisors(i, true); }
int max = 0; int max_i = 1;
for (int i = 1; i <= 20000; ++i) { int v = proper_divisors(i, false); if (v >= max) { max = v; max_i = i; } }
printf("%d with %d divisors\n", max_i, max); return 0;
} </lang>
- Output:
1: 2: 1 3: 1 4: 1 2 5: 1 6: 1 2 3 7: 1 8: 1 2 4 9: 1 3 10: 1 2 5 18480 with 79 divisors
C#
<lang csharp>namespace RosettaCode.ProperDivisors {
using System; using System.Collections.Generic; using System.Linq;
internal static class Program { private static IEnumerable<int> ProperDivisors(int number) { return Enumerable.Range(1, number / 2) .Where(divisor => number % divisor == 0); }
private static void Main() { foreach (var number in Enumerable.Range(1, 10)) { Console.WriteLine("{0}: {{{1}}}", number, string.Join(", ", ProperDivisors(number))); }
var record = Enumerable.Range(1, 20000).Select(number => new { Number = number, Count = ProperDivisors(number).Count() }).OrderByDescending(currentRecord => currentRecord.Count).First(); Console.WriteLine("{0}: {1}", record.Number, record.Count); } }
}</lang>
- Output:
1: {} 2: {1} 3: {1} 4: {1, 2} 5: {1} 6: {1, 2, 3} 7: {1} 8: {1, 2, 4} 9: {1, 3} 10: {1, 2, 5} 15120: 79
C++
<lang cpp>#include <vector>
- include <iostream>
- include <algorithm>
std::vector<int> properDivisors ( int number ) {
std::vector<int> divisors ; for ( int i = 1 ; i < number / 2 + 1 ; i++ ) if ( number % i == 0 )
divisors.push_back( i ) ;
return divisors ;
}
int main( ) {
std::vector<int> divisors ; unsigned int maxdivisors = 0 ; int corresponding_number = 0 ; for ( int i = 1 ; i < 11 ; i++ ) { divisors = properDivisors ( i ) ; std::cout << "Proper divisors of " << i << ":\n" ; for ( int number : divisors ) {
std::cout << number << " " ;
} std::cout << std::endl ; divisors.clear( ) ; } for ( int i = 11 ; i < 20001 ; i++ ) { divisors = properDivisors ( i ) ; if ( divisors.size( ) > maxdivisors ) {
maxdivisors = divisors.size( ) ; corresponding_number = i ;
} divisors.clear( ) ; }
std::cout << "Most divisors has " << corresponding_number << " , it has " << maxdivisors << " divisors!\n" ; return 0 ;
} </lang>
- Output:
Proper divisors of 1: Proper divisors of 2: 1 Proper divisors of 3: 1 Proper divisors of 4: 1 2 Proper divisors of 5: 1 Proper divisors of 6: 1 2 3 Proper divisors of 7: 1 Proper divisors of 8: 1 2 4 Proper divisors of 9: 1 3 Proper divisors of 10: 1 2 5 Most divisors has 15120 , it has 79 divisors!
Ceylon
<lang ceylon>shared void run() {
function divisors(Integer int) => if(int <= 1) then {} else (1..int / 2).filter((Integer element) => element.divides(int));
for(i in 1..10) { print("``i`` => ``divisors(i)``"); }
value start = 1; value end = 20k;
value mostDivisors = map {for(i in start..end) i->divisors(i).size} .inverse() .max(byKey(byIncreasing(Integer.magnitude)));
print("the number(s) with the most divisors between ``start`` and ``end`` is/are: ``mostDivisors?.item else "nothing"`` with ``mostDivisors?.key else "no"`` divisors"); }</lang>
- Output:
1 => [] 2 => { 1 } 3 => { 1 } 4 => { 1, 2 } 5 => { 1 } 6 => { 1, 2, 3 } 7 => { 1 } 8 => { 1, 2, 4 } 9 => { 1, 3 } 10 => { 1, 2, 5 } the number(s) with the most divisors between 1 and 20000 is/are: [15120, 18480] with 79 divisors
Clojure
<lang lisp>(ns properdivisors
(:gen-class))
(defn proper-divisors [n]
" Proper divisors of n" (if (= n 1) [] (filter #(= 0 (rem n %)) (range 1 n))))
- Property divisors of numbers 1 to 20,000 inclusive
(def data (for [n (range 1 (inc 20000))]
[n (proper-divisors n)]))
- Find Max
(defn maximal-key [k x & xs]
" Normal max-key only finds one key that produces maximum, while this function finds them all " (reduce (fn [ys x] (let [c (compare (k x) (k (peek ys)))] (cond (pos? c) [x] (neg? c) ys :else (conj ys x)))) [x] xs))
(println "n\tcnt\tPROPER DIVISORS") (doseq [n (range 1 11)]
(let [factors (proper-divisors n)] (println n "\t" (count factors) "\t" factors)))
(def max-data (apply maximal-key (fn i pd (count pd)) data))
(doseq [[n factors] max-data]
(println n " has " (count factors) " divisors"))
</lang>
- Output:
n cnt PROPER DIVISORS 1 0 [] 2 1 (1) 3 1 (1) 4 2 (1 2) 5 1 (1) 6 3 (1 2 3) 7 1 (1) 8 3 (1 2 4) 9 2 (1 3) 10 3 (1 2 5) 15120 has 79 divisors 18480 has 79 divisors
Common Lisp
Ideally, the smallest-divisor function would only try prime numbers instead of odd numbers. <lang lisp>(defun proper-divisors-recursive (product &optional (results '(1)))
"(int,list)->list::Function to find all proper divisors of a +ve integer." (defun smallest-divisor (x) "int->int::Find the smallest divisor of an integer > 1." (if (evenp x) 2 (do ((lim (truncate (sqrt x))) (sd 3 (+ sd 2))) ((or (integerp (/ x sd)) (> sd lim)) (if (> sd lim) x sd))))) (defun pd-rec (fac) "(int,int)->nil::Recursive function to find proper divisors of a +ve integer" (when (not (member fac results)) (push fac results) (let ((hifac (/ fac (smallest-divisor fac)))) (pd-rec hifac) (pd-rec (/ product hifac))))) (pd-rec product) (butlast (sort (copy-list results) #'<)))
(defun task (method &optional (n 1) (most-pds '(0)))
(dotimes (i 19999) (let ((npds (length (funcall method (incf n)))) (hiest (car most-pds))) (when (>= npds hiest) (if (> npds hiest) (setf most-pds (list npds (list n))) (setf most-pds (list npds (cons n (second most-pds)))))))) most-pds)
(defun main ()
(format t "Task 1:Proper Divisors of [1,10]:~%") (dotimes (i 10) (format t "~A:~A~%" (1+ i) (proper-divisors-recursive (1+ i)))) (format t "Task 2:Count & list of numbers <=20,000 with the most Proper Divisors:~%~A~%" (task #'proper-divisors-recursive)))</lang>
- Output:
CL-USER(10): (main) Task 1:Proper Divisors of [1,10]: 1:NIL 2:(1) 3:(1) 4:(1 2) 5:(1) 6:(1 2 3) 7:(1) 8:(1 2 4) 9:(1 3) 10:(1 2 5) Task 2:Count & list of numbers <=20,000 with the most Proper Divisors: (79 (18480 15120)) NIL
Component Pascal
<lang oberon2> MODULE RosettaProperDivisor; IMPORT StdLog;
PROCEDURE Pd*(n: LONGINT;OUT r: ARRAY OF LONGINT):LONGINT; VAR i,j: LONGINT; BEGIN i := 1;j := 0; IF n > 1 THEN WHILE (i < n) DO IF (n MOD i) = 0 THEN IF (j < LEN(r)) THEN r[j] := i END; INC(j) END; INC(i) END; END; RETURN j END Pd;
PROCEDURE Do*; VAR r: ARRAY 128 OF LONGINT; i,j,found,max,idxMx: LONGINT; mx: ARRAY 128 OF LONGINT; BEGIN FOR i := 1 TO 10 DO found := Pd(i,r); IF found > LEN(r) THEN (* Error. more pd than r can admit *) HALT(1) END; StdLog.Int(i);StdLog.String("[");StdLog.Int(found);StdLog.String("]:> "); FOR j := 0 TO found - 1 DO StdLog.Int(r[j]);StdLog.Char(' '); END; StdLog.Ln END;
max := 0;idxMx := 0;
FOR i := 1 TO 20000 DO found := Pd(i,r); IF found > max THEN idxMx:= 0;mx[idxMx] := i;max := found
ELSIF found = max THEN
INC(idxMx);mx[idxMx] := i END; END;
StdLog.String("Found: ");StdLog.Int(idxMx + 1);
StdLog.String(" Numbers with the longest proper divisors [");
StdLog.Int(max);StdLog.String("]: ");StdLog.Ln; FOR i := 0 TO idxMx DO
StdLog.Int(mx[i]);StdLog.Ln
END END Do;
END RosettaProperDivisor.
^Q RosettaProperDivisor.Do~ </lang>
- Output:
1[ 0]:> 2[ 1]:> 1 3[ 1]:> 1 4[ 2]:> 1 2 5[ 1]:> 1 6[ 3]:> 1 2 3 7[ 1]:> 1 8[ 3]:> 1 2 4 9[ 2]:> 1 3 10[ 3]:> 1 2 5 Found: 2 Numbers with the longest proper divisors [ 79]: 15120 18480
D
Currently the lambda of the filter allocates a closure on the GC-managed heap. <lang d>void main() /*@safe*/ {
import std.stdio, std.algorithm, std.range, std.typecons;
immutable properDivs = (in uint n) pure nothrow @safe /*@nogc*/ => iota(1, (n + 1) / 2 + 1).filter!(x => n % x == 0 && n != x);
iota(1, 11).map!properDivs.writeln; iota(1, 20_001).map!(n => tuple(properDivs(n).count, n)).reduce!max.writeln;
}</lang>
- Output:
[[], [1], [1], [1, 2], [1], [1, 2, 3], [1], [1, 2, 4], [1, 3], [1, 2, 5]] Tuple!(uint, int)(79, 18480)
The Run-time is about 0.67 seconds with the ldc2 compiler.
EchoLisp
<lang scheme> (lib 'list) ;; list-delete
- let n = product p_i^a_i , p_i prime
- number of divisors = product (a_i + 1) - 1
(define (numdivs n)
(1- (apply * (map (lambda(g) (1+ (length g))) (group (prime-factors n))))))
(remember 'numdivs)
- prime powers
- input
- a list g of grouped prime factors ( 3 3 3 ..)
- returns (1 3 9 27 ...)
(define (ppows g (mult 1)) (for/fold (ppows '(1)) ((a g)) (set! mult (* mult a)) (cons mult ppows)))
- proper divisors
- decomp n into ((2 2 ..) ( 3 3 ..) ) prime factors groups
- combines (1 2 4 8 ..) (1 3 9 ..) lists
- remove n from the list
(define (divs n)
(if (<= n 1) null (list-delete (for/fold (divs'(1)) ((g (map ppows (group (prime-factors n)))))
(for*/list ((a divs) (b g)) (* a b)))
n )))
- find number(s) with max # of proper divisors
- returns list of (n . maxdivs) for n in range 2..N
(define (most-proper N)
(define maxdivs 1) (define ndivs 0) (for/fold (most-proper null) ((n (in-range 2 N))) (set! ndivs (numdivs n)) #:continue (< ndivs maxdivs) (when (> ndivs maxdivs) (set!-values (most-proper maxdivs) (values null ndivs))) (cons (cons n maxdivs) most-proper)))
</lang>
- Output:
<lang scheme> (for ((i (in-range 1 11))) (writeln i (divs i))) 1 null 2 (1) 3 (1) 4 (2 1) 5 (1) 6 (2 3 1) 7 (1) 8 (4 2 1) 9 (3 1) 10 (2 5 1)
(most-proper 20000)
→ ((18480 . 79) (15120 . 79))
(most-proper 1_000_000)
→ ((997920 . 239) (982800 . 239) (942480 . 239) (831600 . 239) (720720 . 239))
(lib 'bigint) (numdivs 95952222101012742144) → 666 ;; 🎩 </lang>
Eiffel
<lang Eiffel> class APPLICATION
create make
feature
make -- Test the feature proper_divisors. local list: LINKED_LIST [INTEGER] count, number: INTEGER do across 1 |..| 10 as c loop list := proper_divisors (c.item) io.put_string (c.item.out + ": ") across list as l loop io.put_string (l.item.out + " ") end io.new_line end across 1 |..| 20000 as c loop list := proper_divisors (c.item) if list.count > count then count := list.count number := c.item end end io.put_string (number.out + " has with " + count.out + " divisors the highest number of proper divisors.") end
proper_divisors (n: INTEGER): LINKED_LIST [INTEGER] -- Proper divisors of 'n'. do create Result.make across 1 |..| (n - 1) as c loop if n \\ c.item = 0 then Result.extend (c.item) end end end
end </lang>
- Output:
1: 2: 1 3: 1 4: 1 2 5: 1 6: 1 2 3 7: 1 8: 1 2 4 9: 1 3 10: 1 2 5 15120 has with 79 divisors the highest number of proper divisors.
Elixir
<lang elixir>defmodule Proper do
def divisors(1), do: [] def divisors(n), do: [1 | divisors(2,n,:math.sqrt(n))] |> Enum.sort defp divisors(k,_n,q) when k>q, do: [] defp divisors(k,n,q) when rem(n,k)>0, do: divisors(k+1,n,q) defp divisors(k,n,q) when k * k == n, do: [k | divisors(k+1,n,q)] defp divisors(k,n,q) , do: [k,div(n,k) | divisors(k+1,n,q)] def most_divisors(limit) do {length,nums} = Enum.group_by(1..limit, fn n -> length(divisors(n)) end) |> Enum.max_by(fn {length,_nums} -> length end) IO.puts "With #{length}, Number #{inspect nums} has the most divisors" end
end
Enum.each(1..10, fn n ->
IO.puts "#{n}: #{inspect Proper.divisors(n)}"
end) Proper.most_divisors(20000)</lang>
- Output:
1: [] 2: [1] 3: [1] 4: [1, 2] 5: [1] 6: [1, 2, 3] 7: [1] 8: [1, 2, 4] 9: [1, 3] 10: [1, 2, 5] With 79, Number [18480, 15120] has the most divisors
Erlang
<lang erlang>-module(properdivs). -export([divs/1,sumdivs/1,longest/1]).
divs(0) -> []; divs(1) -> []; divs(N) -> lists:sort([1] ++ divisors(2,N,math:sqrt(N))).
divisors(K,_N,Q) when K > Q -> []; divisors(K,N,Q) when N rem K =/= 0 ->
divisors(K+1,N,Q);
divisors(K,N,Q) when K * K == N ->
[K] ++ divisors(K+1,N,Q);
divisors(K,N,Q) ->
[K, N div K] ++ divisors(K+1,N,Q).
sumdivs(N) -> lists:sum(divs(N)).
longest(Limit) -> longest(Limit,0,0,1).
longest(L,Current,CurLeng,Acc) when Acc >= L ->
io:format("With ~w, Number ~w has the most divisors~n", [CurLeng,Current]);
longest(L,Current,CurLeng,Acc) ->
A = length(divs(Acc)), if A > CurLeng -> longest(L,Acc,A,Acc+1); true -> longest(L,Current,CurLeng,Acc+1) end.</lang>
- Output:
1> [io:format("X: ~w, N: ~w~n", [N,properdivs:divs(N)]) || N <- lists:seq(1,10)]. X: 1, N: [] X: 2, N: [1] X: 3, N: [1] X: 4, N: [1,2] X: 5, N: [1] X: 6, N: [1,2,3] X: 7, N: [1] X: 8, N: [1,2,4] X: 9, N: [1,3] X: 10, N: [1,2,5] [ok,ok,ok,ok,ok,ok,ok,ok,ok,ok] 2> properdivs:longest(20000). With 79, Number 15120 has the most divisors
F#
<lang F#> let mutable a=0 let mutable b=0 let mutable c=0 let mutable d=0 let mutable e=0 let mutable f=0 for k=1 to 10 do
b <- 0 f <- k/2 printf "divisor " for l=1 to f do if k%l=0 then b <- b+1 printf " %i," l printf "no of divisor %i" b printfn ""
for i=1 to 20000 do
b <- 0 f <- i/2 for j=1 to f do if i%j=0 then b <- b+1 if b=c then d <- 0 d <- i if c<b then c <- b
printfn "%i has %i divisor" d c
</lang>
A purely functional approach. <lang fsharp> // the simple function with the answer let propDivs n = [1..n/2] |> List.filter (fun x->n % x = 0)
// to cache the result length; helpful for a long search let propDivDat n = propDivs n |> fun xs -> n, xs.Length, xs
// UI: always the longest and messiest let show (n,count,divs) =
let showCount = count |> function | 0-> "no proper divisors" | 1->"1 proper divisor" | _-> sprintf "%d proper divisors" count let showDiv = divs |> function | []->"" | x::[]->sprintf ": %d" x | _->divs |> Seq.map string |> String.concat "," |> sprintf ": %s" printfn "%d has %s%s" n showCount showDiv
// generate output [1..10] |> List.iter (propDivDat >> show)
// use a sequence: we don't really need to hold this data, just iterate over it Seq.init 20000 ( ((+) 1) >> propDivDat) |> Seq.fold (fun a b ->match a,b with | (_,c1,_),(_,c2,_) when c2 > c1 -> b | _-> a) (0,0,[]) |> fun (n,count,_) -> (n,count,[]) |> show </lang>
- Output:
1 has no proper divisors 2 has 1 proper divisor: 1 3 has 1 proper divisor: 1 4 has 2 proper divisors: 1,2 5 has 1 proper divisor: 1 6 has 3 proper divisors: 1,2,3 7 has 1 proper divisor: 1 8 has 3 proper divisors: 1,2,4 9 has 2 proper divisors: 1,3 10 has 3 proper divisors: 1,2,5 15120 has 79 proper divisors
Fortran
Compiled using G95 compiler, run on x86 system under Puppy Linux <lang Fortran>
function icntprop(num ) icnt=0 do i=1 , num-1 if (mod(num , i) .eq. 0) then icnt = icnt + 1 if (num .lt. 11) print *,' ',i end if end do icntprop = icnt end function limit = 20000 maxcnt = 0 print *,'N divisors' do j=1,limit,1 if (j .lt. 11) print *,j icnt = icntprop(j) if (icnt .gt. maxcnt) then maxcnt = icnt maxj = j end if end do print *,' ' print *,' from 1 to ',limit print *,maxj,' has max proper divisors: ',maxcnt end
</lang>
- Output:
N divisors 1 2 1 3 1 4 1 2 5 1 6 1 2 3 7 1 8 1 2 4 9 1 3 10 1 2 5 from 1 to 20000 15120 has max proper divisors: 79
FreeBASIC
<lang freebasic> ' FreeBASIC v1.05.0 win64
Sub ListProperDivisors(limit As Integer)
If limit < 1 Then Return For i As Integer = 1 To limit Print Using "##"; i; Print " ->"; If i = 1 Then Print " (None)" Continue For End if For j As Integer = 1 To i \ 2 If i Mod j = 0 Then Print " "; j; Next j Print Next i
End Sub
Function CountProperDivisors(number As Integer) As Integer
If number < 2 Then Return 0 Dim count As Integer = 0 For i As Integer = 1 To number \ 2 If number Mod i = 0 Then count += 1 Next Return count
End Function
Dim As Integer n, count, most = 1, maxCount = 0
Print "The proper divisors of the following numbers are :" Print ListProperDivisors(10)
For n As Integer = 2 To 20000
count = CountProperDivisors(n) If count > maxCount Then maxCount = count most = n EndIf
Next
Print Print Str(most); " has the most proper divisors, namely"; maxCount Print Print "Press any key to exit the program" Sleep End </lang>
- Output:
The proper divisors of the following numbers are : 1 -> (None) 2 -> 1 3 -> 1 4 -> 1 2 5 -> 1 6 -> 1 2 3 7 -> 1 8 -> 1 2 4 9 -> 1 3 10 -> 1 2 5 15120 has the most proper divisors, namely 79
GFA Basic
<lang> OPENW 1 CLEARW 1 ' ' Array f% is used to hold the divisors DIM f%(SQR(20000)) ! cannot redim arrays, so set size to largest needed ' ' 1. Show proper divisors of 1 to 10, inclusive ' FOR i%=1 TO 10
num%=@proper_divisors(i%) PRINT "Divisors for ";i%;":"; FOR j%=1 TO num% PRINT " ";f%(j%); NEXT j% PRINT
NEXT i% ' ' 2. Find (smallest) number <= 20000 with largest number of proper divisors ' result%=1 ! largest so far number%=0 ! its number of divisors FOR i%=1 TO 20000
num%=@proper_divisors(i%) IF num%>number% result%=i% number%=num% ENDIF
NEXT i% PRINT "Largest number of divisors is ";number%;" for ";result% ' ~INP(2) CLOSEW 1 ' ' find the proper divisors of n%, placing results in f% ' and return the number found ' FUNCTION proper_divisors(n%)
LOCAL i%,root%,count% ' ARRAYFILL f%(),0 count%=1 ! index of next slot in f% to fill ' IF n%>1 f%(count%)=1 count%=count%+1 root%=SQR(n%) FOR i%=2 TO root% IF n% MOD i%=0 f%(count%)=i% count%=count%+1 IF i%*i%<>n% ! root% is an integer, so check if i% is actual squa- lists:seq(1,10)].
X: 1, N: [] X: 2, N: [1] X: 3, N: [1] X: 4, N: [1,2] X: 5, N: [1] X: 6, N: [1,2,3] X: 7, N: [1] X: 8, N: [1,2,4] X: 9, N: [1,3] X: 10, N: [1,2,5] [ok,ok,ok,ok,ok,ok,ok,ok,ok,ok]
2> properdivs:longest(20000). With 79, Number 15120 has the most divisors re root of n%
f%(count%)=n%/i% count%=count%+1 ENDIF ENDIF NEXT i% ENDIF ' RETURN count%-1
ENDFUNC </lang>
Output is:
Divisors for 1: Divisors for 2: 1 Divisors for 3: 1 Divisors for 4: 1 2 Divisors for 5: 1 Divisors for 6: 1 2 3 Divisors for 7: 1 Divisors for 8: 1 2 4 Divisors for 9: 1 3 Divisors for 10: 1 2 5 Largest number of divisors is 79 for 15120
Haskell
<lang Haskell>import Data.Ord import Data.List
divisors :: (Integral a) => a -> [a] divisors n = filter ((0 ==) . (n `mod`)) [1 .. (n `div` 2)]
main :: IO () main = do
putStrLn "divisors of 1 to 10:" mapM_ (print . divisors) [1 .. 10] putStrLn "a number with the most divisors within 1 to 20000 (number, count):" print $ maximumBy (comparing snd) [(n, length $ divisors n) | n <- [1 .. 20000]]</lang>
- Output:
divisors of 1 to 10: [] [1] [1] [1,2] [1] [1,2,3] [1] [1,2,4] [1,3] [1,2,5] a number with the most divisors within 1 to 20000 (number, count): (18480,79)
Or, for a little more efficiency, filtering only up to the root, and deriving the higher proper divisors from the lower ones, as quotients:
<lang haskell>import Data.List (maximumBy) import Data.Ord (comparing) import Control.Monad (ap)
properDivisors
:: Integral a => a -> [a]
properDivisors n =
let root = (floor . sqrt . fromIntegral) n lows = filter ((0 ==) . rem n) [1 .. root] in init (lows ++ (if root * root == n then tail else id) (reverse (quot n <$> lows)))
main :: IO () main = do
putStrLn "Proper divisors of 1 to 10:" mapM_ (print . properDivisors) [1 .. 10] mapM_ putStrLn [ "" , "A number in the range 1 to 20,000 with the most proper divisors," , "as (number, count of proper divisors):" , "" ] print $ maximumBy (comparing snd) $ ap (,) (length . properDivisors) <$> [1 .. 20000]</lang>
- Output:
Proper divisors of 1 to 10: [] [1] [1] [1,2] [1] [1,2,3] [1] [1,2,4] [1,3] [1,2,5] A number in the range 1 to 20,000 with the most proper divisors, as (number, count of proper divisors): (18480,79)
J
The proper divisors of an integer are the Factors of an integer without the integer itself.
So, borrowing from the J implementation of that related task:
<lang J>factors=: [: /:~@, */&>@{@((^ i.@>:)&.>/)@q:~&__ properDivisors=: factors -. ]</lang>
Proper divisors of numbers 1 through 10:
<lang J> (,&": ' -- ' ,&": properDivisors)&>1+i.10 1 -- 2 -- 1 3 -- 1 4 -- 1 2 5 -- 1 6 -- 1 2 3 7 -- 1 8 -- 1 2 4 9 -- 1 3 10 -- 1 2 5</lang>
Number(s) not exceeding 20000 with largest number of proper divisors (and the count of those divisors):
<lang J> (, #@properDivisors)&> 1+I.(= >./) #@properDivisors@> 1+i.20000 15120 79 18480 79</lang>
Note that it's a bit more efficient to simply count factors here, when selecting the candidate numbers.
<lang J> (, #@properDivisors)&> 1+I.(= >./) #@factors@> 1+i.20000 15120 79 18480 79</lang>
We could also arbitrarily toss either 15120 or 18480 (keeping the other number), if it were important that we produce only one result.
Java
<lang java5>import java.util.Collections; import java.util.LinkedList; import java.util.List;
public class Proper{
public static List<Integer> properDivs(int n){ List<Integer> divs = new LinkedList<Integer>(); if(n == 1) return divs; divs.add(1); for(int x = 2; x < n; x++){ if(n % x == 0) divs.add(x); } Collections.sort(divs); return divs; } public static void main(String[] args){ for(int x = 1; x <= 10; x++){ System.out.println(x + ": " + properDivs(x)); } int x = 0, count = 0; for(int n = 1; n <= 20000; n++){ if(properDivs(n).size() > count){ x = n; count = properDivs(n).size(); } } System.out.println(x + ": " + count); }
}</lang>
- Output:
1: [] 2: [1] 3: [1] 4: [1, 2] 5: [1] 6: [1, 2, 3] 7: [1] 8: [1, 2, 4] 9: [1, 3] 10: [1, 2, 5] 15120: 79
JavaScript
ES5
<lang JavaScript>(function () {
// Proper divisors function properDivisors(n) { if (n < 2) return []; else { var rRoot = Math.sqrt(n), intRoot = Math.floor(rRoot),
lows = range(1, intRoot).filter(function (x) { return (n % x) === 0; });
return lows.concat(lows.slice(1).map(function (x) { return n / x; }).reverse().slice((rRoot === intRoot) | 0)); } }
// [m..n] function range(m, n) { var a = Array(n - m + 1), i = n + 1; while (i--) a[i - 1] = i; return a; }
var tblOneToTen = [ ['Number', 'Proper Divisors', 'Count'] ].concat(range(1, 10).map(function (x) { var ds = properDivisors(x);
return [x, ds.join(', '), ds.length]; })),
dctMostBelow20k = range(1, 20000).reduce(function (a, x) { var lng = properDivisors(x).length;
return lng > a.divisorCount ? { n: x, divisorCount: lng } : a; }, { n: 0, divisorCount: 0 });
// a -> bool -> s -> s function wikiTable(lstRows, blnHeaderRow, strStyle) { return '{| class="wikitable" ' + ( strStyle ? 'style="' + strStyle + '"' : ) + lstRows.map(function (lstRow, iRow) { var strDelim = ((blnHeaderRow && !iRow) ? '!' : '|');
return '\n|-\n' + strDelim + ' ' + lstRow.map(function (v) { return typeof v === 'undefined' ? ' ' : v; }).join(' ' + strDelim + strDelim + ' '); }).join() + '\n|}'; }
return wikiTable( tblOneToTen, true ) + '\n\nMost proper divisors below 20,000:\n\n ' + JSON.stringify( dctMostBelow20k );
})();</lang>
- Output:
Number | Proper Divisors | Count |
---|---|---|
1 | 0 | |
2 | 1 | 1 |
3 | 1 | 1 |
4 | 1, 2 | 2 |
5 | 1 | 1 |
6 | 1, 2, 3 | 3 |
7 | 1 | 1 |
8 | 1, 2, 4 | 3 |
9 | 1, 3 | 2 |
10 | 1, 2, 5 | 3 |
Most proper divisors below 20,000:
{"n":15120,"divisorCount":79}
ES6
<lang JavaScript>(() => {
// properDivisors :: Int -> [Int] let properDivisors = n => { let rRoot = Math.sqrt(n), intRoot = Math.floor(rRoot), blnPerfectSquare = rRoot === intRoot,
lows = range(1, intRoot) .filter(x => (n % x) === 0);
// for perfect squares, we can drop // the head of the 'highs' list return lows.concat(lows .map(x => n / x) .reverse() .slice(blnPerfectSquare | 0) ) .slice(0, -1); // except n itself },
// range :: Int -> Int -> [Int] range = (m, n) => Array.from({ length: (n - m) + 1 }, (_, i) => m + i);
return { properDivisorsOf1to10: range(1, 10) .reduce((a, x) => ( a[x.toString()] = properDivisors(x), a ), {}),
intMaxDivisorsUnder20k: range(1, 20000) .reduce((a, x) => { let intDivisors = properDivisors(x) .length;
return intDivisors >= a.divisors ? { max: x, divisors: intDivisors } : a;
}, { max: 0, divisors: 0 }) };
})();</lang>
- Output:
<lang JavaScript>{
"properDivisorsOf1to10":{ "1":[], "2":[1], "3":[1], "4":[1, 2], "5":[1], "6":[1, 2, 3], "7":[1], "8":[1, 2, 4], "9":[1, 3], "10":[1, 2, 5] }, "intMaxDivisorsUnder20k":{"max":18480, "divisors":79}
}</lang>
jq
In the following, proper_divisors returns a stream. In order to count the number of items in the stream economically, we first define "count(stream)": <lang jq>def count(stream): reduce stream as $i (0; . + 1);
- unordered
def proper_divisors:
. as $n | if $n > 1 then 1, ( range(2; 1 + (sqrt|floor)) as $i | if ($n % $i) == 0 then $i, (($n / $i) | if . == $i then empty else . end) else empty
end)
else empty end;
- The first integer in 1 .. n inclusive
- with the maximal number of proper divisors in that range:
def most_proper_divisors(n):
reduce range(1; n+1) as $i ( [null, 0]; count( $i | proper_divisors ) as $count | if $count > .[1] then [$i, $count] else . end);</lang>
The tasks: <lang jq>"The proper divisors of the numbers 1 to 10 inclusive are:", (range(1;11) as $i | "\($i): \( [ $i | proper_divisors] )"), "", "The pair consisting of the least number in the range 1 to 20,000 with", "the maximal number proper divisors together with the corresponding", "count of proper divisors is:", most_proper_divisors(20000) </lang>
- Output:
<lang sh>$ jq -n -c -r -f /Users/peter/jq/proper_divisors.jq The proper divisors of the numbers 1 to 10 inclusive are: 1: [] 2: [1] 3: [1] 4: [1,2] 5: [1] 6: [1,2,3] 7: [1] 8: [1,2,4] 9: [1,3] 10: [1,2,5]
The pair consisting of the least number in the range 1 to 20,000 with the maximal number proper divisors together with the corresponding count of proper divisors is: [15120,79]</lang>
Julia
Use factor
to obtain the prime factorization of the target number. I adopted the argument handling style of factor
in my properdivisors
function.
<lang Julia>
function properdivisors{T<:Integer}(n::T)
0 < n || throw(ArgumentError("number to be factored must be ≥ 0, got $n")) 1 < n || return T[] !isprime(n) || return T[one(T), n] f = factor(n) d = T[one(T)] for (k, v) in f c = T[k^i for i in 0:v] d = d*c' d = reshape(d, length(d)) end sort!(d) return d[1:end-1]
end
lo = 1 hi = 10 println("List the proper divisors for ", lo, " through ", hi, ".") for i in lo:hi
println(@sprintf("%4d", i), " ", properdivisors(i))
end
hi = 2*10^4 println("\nFind the numbers within [", lo, ",", hi, "] having the most divisors.")
maxdiv = 0 nlst = Int[]
for i in lo:hi
ndiv = length(properdivisors(i)) if ndiv > maxdiv maxdiv = ndiv nlst = [i] elseif ndiv == maxdiv push!(nlst, i) end
end
println(nlst, " have the maximum proper divisor count of ", maxdiv, ".") </lang>
- Output:
List the proper divisors for 1 through 10. 1 [] 2 [1,2] 3 [1,3] 4 [1,2] 5 [1,5] 6 [1,2,3] 7 [1,7] 8 [1,2,4] 9 [1,3] 10 [1,2,5] Find the numbers within [1,20000] having the most divisors. [15120,18480] have the maximum proper divisor count of 79.
Kotlin
<lang scala>// version 1.0.5-2
fun listProperDivisors(limit: Int) {
if (limit < 1) return for(i in 1..limit) { print(i.toString().padStart(2) + " -> ") if (i == 1) { println("(None)") continue } (1..i/2).filter{ i % it == 0 }.forEach { print(" $it") } println() }
}
fun countProperDivisors(n: Int): Int {
if (n < 2) return 0 return (1..n/2).count { (n % it) == 0 }
}
fun main(args: Array<String>) {
println("The proper divisors of the following numbers are :\n") listProperDivisors(10) println() var count: Int var maxCount = 0 val most: MutableList<Int> = mutableListOf(1) for (n in 2..20000) { count = countProperDivisors(n) if (count == maxCount) most.add(n) else if (count > maxCount) { maxCount = count most.clear() most.add(n) } } println("The following number(s) have the most proper divisors, namely " + maxCount + "\n") for (n in most) println(n)
}</lang>
- Output:
The proper divisors of the following numbers are : 1 -> (None) 2 -> 1 3 -> 1 4 -> 1 2 5 -> 1 6 -> 1 2 3 7 -> 1 8 -> 1 2 4 9 -> 1 3 10 -> 1 2 5 The following number(s) have the most proper divisors, namely 79 15120 18480
Lua
<lang Lua>-- Return a table of the proper divisors of n function propDivs (n)
if n < 2 then return {} end local divs, sqr = {1}, math.sqrt(n) for d = 2, sqr do if n % d == 0 then table.insert(divs, d) if d ~= sqr then table.insert(divs, n/d) end end end table.sort(divs) return divs
end
-- Show n followed by all values in t function show (n, t)
io.write(n .. ":\t") for _, v in pairs(t) do io.write(v .. " ") end print()
end
-- Main procedure local mostDivs, numDivs, answer = 0 for i = 1, 10 do show(i, propDivs(i)) end for i = 1, 20000 do
numDivs = #propDivs(i) if numDivs > mostDivs then mostDivs = numDivs answer = i end
end print(answer .. " has " .. mostDivs .. " proper divisors.")</lang>
- Output:
1: 2: 1 3: 1 4: 1 2 5: 1 6: 1 2 3 7: 1 8: 1 2 4 9: 1 3 10: 1 2 5 15120 has 79 proper divisors.
Mathematica / Wolfram Language
A Function that yields the proper divisors of an integer n: <lang Mathematica>ProperDivisors[n_Integer /; n > 0] := Most@Divisors@n;</lang>
Proper divisors of n from 1 to 10: <lang Mathematica>Grid@Table[{n, ProperDivisors[n]}, {n, 1, 10}]</lang>
- Output:
1 {} 2 {1} 3 {1} 4 {1,2} 5 {1} 6 {1,2,3} 7 {1} 8 {1,2,4} 9 {1,3} 10 {1,2,5}
The number with the most divisors between 1 and 20,000: <lang Mathematica>Fold[
Last[SortBy[{#1, {#2, Length@ProperDivisors[#2]}}, Last]] &, {0, 0}, Range[20000]]</lang>
- Output:
{18480, 79}
An alternate way to find the number with the most divisors between 1 and 20,000: <lang Mathematica>Last@SortBy[
Table[ {n, Length@ProperDivisors[n]}, {n, 1, 20000}], Last]</lang>
- Output:
{15120, 79}
Oberon-2
<lang oberon2> MODULE ProperDivisors; IMPORT
Out;
CONST
initialSize = 128;
TYPE
Result* = POINTER TO ResultDesc; ResultDesc = RECORD found-: LONGINT; (* number of slots in pd *) pd-: POINTER TO ARRAY OF LONGINT; cap: LONGINT; (* Capacity *) END;
VAR
i,found,max,idxMx: LONGINT; mx: ARRAY 32 OF LONGINT; rs: Result;
PROCEDURE (r: Result) Init(size: LONGINT); BEGIN r.found := 0; r.cap := size; NEW(r.pd,r.cap); END Init;
PROCEDURE (r: Result) Add(n: LONGINT); BEGIN (* Out.String("--->");Out.LongInt(n,0);Out.String(" At: ");Out.LongInt(r.found,0);Out.Ln; *) IF (r.found < LEN(r.pd^) - 1) THEN r.pd[r.found] := n; ELSE (* expand pd for more room *) END; INC(r.found); END Add;
PROCEDURE (r:Result) Show(); VAR i: LONGINT; BEGIN Out.String("(Result:");Out.LongInt(r.found + 1,0);(* Out.String("/");Out.LongInt(r.cap,0);*) Out.String("-"); IF r.found > 0 THEN FOR i:= 0 TO r.found - 1 DO Out.LongInt(r.pd[i],0); IF i = r.found - 1 THEN Out.Char(')') ELSE Out.Char(',') END END END; Out.Ln END Show;
PROCEDURE (r:Result) Reset(); BEGIN r.found := 0; END Reset;
PROCEDURE GetFor(n: LONGINT;VAR rs: Result); VAR i: LONGINT; BEGIN IF n > 1 THEN rs.Add(1);i := 2; WHILE (i < n) DO IF (n MOD i) = 0 THEN rs.Add(i) END; INC(i) END END; END GetFor;
BEGIN
NEW(rs);rs.Init(initialSize); FOR i := 1 TO 10 DO Out.LongInt(i,4);Out.Char(':'); GetFor(i,rs); rs.Show(); rs.Reset(); END; Out.LongInt(100,4);Out.Char(':');GetFor(100,rs);rs.Show();rs.Reset(); max := 0;idxMx := 0;found := 0; FOR i := 1 TO 20000 DO GetFor(i,rs); IF rs.found > max THEN idxMx:= 0;mx[idxMx] := i;max := rs.found ELSIF rs.found = max THEN INC(idxMx);mx[idxMx] := i END; rs.Reset() END; Out.String("Found: ");Out.LongInt(idxMx + 1,0); Out.String(" Numbers with most proper divisors "); Out.LongInt(max,0);Out.String(": ");Out.Ln; FOR i := 0 TO idxMx DO Out.LongInt(mx[i],0);Out.Ln END
END ProperDivisors. </lang>
- Output:
1:(Result:1- 2:(Result:2-1) 3:(Result:2-1) 4:(Result:3-1,2) 5:(Result:2-1) 6:(Result:4-1,2,3) 7:(Result:2-1) 8:(Result:4-1,2,4) 9:(Result:3-1,3) 10:(Result:4-1,2,5) 100:(Result:9-1,2,4,5,10,20,25,50) Found: 2 Numbers with most proper divisors 79: 15120 18480
Oforth
<lang Oforth>Integer method: properDivs self 2 / seq filter(#[ self swap mod 0 == ]) }
10 seq apply(#[ dup print " : " print properDivs println ]) 20000 seq map(#[ dup properDivs size Pair new ]) reduce(#maxKey) println</lang>
- Output:
1 : [] 2 : [1] 3 : [1] 4 : [1, 2] 5 : [1] 6 : [1, 2, 3] 7 : [1] 8 : [1, 2, 4] 9 : [1, 3] 10 : [1, 2, 5] [79, 15120]
PARI/GP
<lang parigp>proper(n)=if(n==1, [], my(d=divisors(n)); d[2..#d]); apply(proper, [1..10]) r=at=0; for(n=1,20000, t=numdiv(n); if(t>r, r=t; at=n)); [at, numdiv(t)-1]</lang>
- Output:
%1 = [[], [2], [3], [2, 4], [5], [2, 3, 6], [7], [2, 4, 8], [3, 9], [2, 5, 10]] %2 = [15120, 7]
Pascal
Using prime factorisation <lang pascal>{$IFDEF FPC}{$MODE DELPHI}{$ELSE}{$APPTYPE CONSOLE}{$ENDIF} uses
sysutils;
const
MAXPROPERDIVS = 1920;
type
tRes = array[0..MAXPROPERDIVS] of LongWord; tPot = record potPrim, potMax :LongWord; end;
tprimeFac = record pfPrims : array[1..10] of tPot; pfCnt, pfNum : LongWord; end; tSmallPrimes = array[0..6541] of longWord;
var
SmallPrimes: tSmallPrimes;
procedure InitSmallPrimes; var
pr,testPr,j,maxprimidx: Longword; isPrime : boolean;
Begin
maxprimidx := 0; SmallPrimes[0] := 2; pr := 3; repeat isprime := true; j := 0; repeat testPr := SmallPrimes[j]; IF testPr*testPr > pr then break; If pr mod testPr = 0 then Begin isprime := false; break; end; inc(j); until false;
if isprime then Begin inc(maxprimidx); SmallPrimes[maxprimidx]:= pr; end; inc(pr,2); until pr > 1 shl 16 -1;
end;
procedure PrimeFacOut(primeDecomp:tprimeFac); var
i : LongWord;
begin
with primeDecomp do Begin write(pfNum,' = '); For i := 1 to pfCnt-1 do with pfPrims[i] do If potMax = 1 then write(potPrim,'*') else write(potPrim,'^',potMax,'*'); with pfPrims[pfCnt] do If potMax = 1 then write(potPrim) else write(potPrim,'^',potMax); end;
end;
procedure PrimeDecomposition(n:LongWord;var res:tprimeFac); var
i,pr,cnt,quot{to minimize divisions} : LongWord;
Begin
res.pfNum := n; res.pfCnt:= 0; i := 0; cnt := 0; repeat pr := SmallPrimes[i]; IF pr*pr>n then Break;
quot := n div pr; IF pr*quot = n then with res do Begin inc(pfCnt); with pfPrims[pfCnt] do Begin potPrim := pr; potMax := 0; repeat n := quot; quot := quot div pr; inc(potMax); until pr*quot <> n; end; end; inc(i); until false; //a big prime left over? IF n <> 1 then with res do Begin inc(pfCnt); with pfPrims[pfCnt] do Begin potPrim := n; potMax := 1; end; end;
end;
function CntProperDivs(const primeDecomp:tprimeFac):LongWord; //count of proper divisors var
i: LongWord;
begin
result := 1; with primeDecomp do For i := 1 to pfCnt do result := result*(pfPrims[i].potMax+1); //remove dec(result);
end;
function findProperdivs(n:LongWord;var res:TRes):LongWord; //simple trial division to get a sorted list of all proper divisors var
i,j: LongWord;
Begin
result := 0; i := 1; j := n; while j>i do begin j := n DIV i; IF i*j = n then Begin //smaller factor part at the beginning upwards res[result]:= i; IF i <> j then //bigger factor at the end downwards res[MAXPROPERDIVS-result]:= j else //n is square number res[MAXPROPERDIVS-result]:= 0; inc(result); end; inc(i); end;
If result>0 then Begin //move close together i := result; j := MAXPROPERDIVS-result+1; result := 2*result-1; repeat res[i] := res[j]; inc(j); inc(i); until i > result;
if res[result-1] = 0 then dec(result); end;
end;
procedure AllFacsOut(n: Longword); var
res:TRes; i,k,j:LongInt;
Begin
j := findProperdivs(n,res); write(n:5,' : '); For k := 0 to j-2 do write(res[k],','); IF j>=1 then write(res[j-1]); writeln;
end;
var
primeDecomp: tprimeFac; rs : tRes; i,j,max,maxcnt: LongWord;
BEGIN
InitSmallPrimes; For i := 1 to 10 do AllFacsOut(i); writeln; max := 0; maxCnt := 0; For i := 1 to 20*1000 do Begin PrimeDecomposition(i,primeDecomp); j := CntProperDivs(primeDecomp); IF j> maxCnt then Begin maxcnt := j; max := i; end; end; PrimeDecomposition(max,primeDecomp); j := CntProperDivs(primeDecomp);
PrimeFacOut(primeDecomp);writeln(' ',j:10,' factors'); writeln; //https://en.wikipedia.org/wiki/Highly_composite_number <= HCN //http://wwwhomes.uni-bielefeld.de/achim/highly.txt the first 1200 HCN max := 3491888400; PrimeDecomposition(max,primeDecomp); j := CntProperDivs(primeDecomp); PrimeFacOut(primeDecomp);writeln(' ',j:10,' factors'); writeln;
END.</lang>
- Output:
1 : 2 : 1 3 : 1 4 : 1,2 5 : 1 6 : 1,2,3 7 : 1 8 : 1,2,4 9 : 1,3 10 : 1,2,515120 = 2^4*3^3*5*7 79 factors
3491888400 = 2^4*3^3*5^2*7*11*13*17*19 1919 factors
real 0m0.004s
Perl
Using a module for divisors
<lang perl>use ntheory qw/divisors/; sub proper_divisors {
my $n = shift; # Like Pari/GP, divisors(0) = (0,1) and divisors(1) = () return 1 if $n == 0; my @d = divisors($n); pop @d; # divisors are in sorted order, so last entry is the input @d;
} say "$_: ", join " ", proper_divisors($_) for 1..10;
- 1. For the max, we can do a traditional loop.
my($max,$ind) = (0,0); for (1..20000) {
my $nd = scalar proper_divisors($_); ($max,$ind) = ($nd,$_) if $nd > $max;
} say "$max $ind";
- 2. Or we can use List::Util's max with decoration (this exploits its implementation)
{
use List::Util qw/max/; no warnings 'numeric'; say max(map { scalar(proper_divisors($_)) . " $_" } 1..20000);
}</lang>
- Output:
1: 2: 1 3: 1 4: 1 2 5: 1 6: 1 2 3 7: 1 8: 1 2 4 9: 1 3 10: 1 2 5 79 15120 79 18480
Note that the first code will choose the first max, while the second chooses the last.
Perl 6
<lang perl6>sub propdiv (\x) {
my @l =(1 if x > 1), gather for 2 .. x.sqrt.floor -> \d { my \y = x div d; if y * d == x { take d; take y unless y == d } } gather @l.deepmap(*.take);
}
say "$_ ", propdiv($_) for 1..10;
my $max = 0; my @candidates; for 1..20000 {
my @pd = propdiv($_); my $pd = @pd.elems; if $pd > $max { @candidates = (); $max = $pd; } push @candidates, $_ if $pd == $max;
} say "max = $max, candidates = @candidates[]";</lang>
- Output:
1 Nil 2 1 3 1 4 1 2 5 1 6 1 2 3 7 1 8 1 2 4 9 1 3 10 1 2 5 max = 79, candidates = 15120 18480
Phix
The factors routine is an auto-include. The actual implementation of it, from builtins\pfactors.e is <lang Phix>global function factors(atom n, integer include1=0) -- returns a list of all integer factors of n -- if include1 is 0 (the default), result does not contain either 1 or n -- if include1 is 1, and n>1, the result contains 1 and n -- if include1 is -1, and n>1, the result contains 1 but not n sequence lfactors = {}, hfactors = {} atom hfactor integer p = 2,
lim = floor(sqrt(n))
if n!=1 and include1!=0 then lfactors = {1} if include1=1 then hfactors = {n} end if end if while p<=lim do if remainder(n,p)=0 then lfactors = append(lfactors,p) hfactor = n/p if hfactor=p then exit end if hfactors = prepend(hfactors,hfactor) end if p += 1 end while return lfactors & hfactors
end function</lang> The compiler knows where to find that, so the main program is just: <lang Phix>for i=1 to 10 do
?{i,factors(i,-1)}
end for
integer maxd = 0, k sequence candidates = {}
for i=1 to 20000 do k = length(factors(i,-1)) if k>=maxd then if k=maxd then candidates &= i else candidates = {i} maxd = k end if end if end for printf(1,"%d divisors:", maxd) ?candidates {} = wait_key()</lang>
- Output:
{1,{}} {2,{1}} {3,{1}} {4,{1,2}} {5,{1}} {6,{1,2,3}} {7,{1}} {8,{1,2,4}} {9,{1,3}} {10,{1,2,5}} 79 divisors:{15120,18480}
PHP
<lang php><?php function ProperDivisors($n) {
yield 1; $large_divisors = []; for ($i = 2; $i <= sqrt($n); $i++) { if ($n % $i == 0) { yield $i; if ($i*$i != $n) { $large_divisors[] = $n / $i; } } } foreach (array_reverse($large_divisors) as $i) { yield $i; }
}
assert([1, 2, 4, 5, 10, 20, 25, 50] ==
iterator_to_array(ProperDivisors(100)));
foreach (range(1, 10) as $n) {
echo "$n =>"; foreach (ProperDivisors($n) as $divisor) { echo " $divisor"; } echo "\n";
}
$divisorsCount = []; for ($i = 1; $i < 20000; $i++) {
$divisorsCount[sizeof(iterator_to_array(ProperDivisors($i)))][] = $i;
} ksort($divisorsCount);
echo "Numbers with most divisors: ", implode(", ", end($divisorsCount)), ".\n"; echo "They have ", key($divisorsCount), " divisors.\n";
</lang>
Outputs:
1 => 1 2 => 1 3 => 1 4 => 1 2 5 => 1 6 => 1 2 3 7 => 1 8 => 1 2 4 9 => 1 3 10 => 1 2 5 Numbers with most divisors: 15120, 18480. They have 79 divisors.
PicoLisp
<lang PicoLisp># Generate all proper divisors. (de propdiv (N)
(head -1 (filter '((X) (=0 (% N X))) (range 1 N) )) )
- Obtaining the values from 1 to 10 inclusive.
(mapcar propdiv (range 1 10))
- Output:
- (NIL (1) (1) (1 2) (1) (1 2 3) (1) (1 2 4) (1 3) (1 2 5))</lang>
Brute-force
<lang PicoLisp>(de propdiv (N)
(cdr (rot (make (for I N (and (=0 (% N I)) (link I)) ) ) ) ) )
(de countdiv (N)
(let C -1 (for I N (and (=0 (% N I)) (inc 'C)) ) C ) )
(let F (-5 -8)
(tab F "N" "LIST") (for I 10 (tab F I (glue " + " (propdiv I)) ) ) )
(println
(maxi countdiv (range 1 20000) ) )</lang>
Factorization
<lang PicoLisp>(de accu1 (Var Key)
(if (assoc Key (val Var)) (con @ (inc (cdr @))) (push Var (cons Key 2)) ) Key )
(de factor (N)
(let (R NIL D 2 L (1 2 2 . (4 2 4 2 4 6 2 6 .)) M (sqrt N) ) (while (>= M D) (if (=0 (% N D)) (setq M (sqrt (setq N (/ N (accu1 'R D)))) ) (inc 'D (pop 'L)) ) ) (accu1 'R N) (dec (apply * (mapcar cdr R))) ) )
(bench
(println (maxi factor (range 1 20000) ) @@ ) )</lang>
Output:
15120 79 0.081 sec
PL/I
<lang pli>*process source xref;
(subrg): cpd: Proc Options(main); p9a=time(); Dcl (p9a,p9b) Pic'(9)9'; Dcl cnt(3) Bin Fixed(31) Init((3)0); Dcl x Bin Fixed(31); Dcl pd(300) Bin Fixed(31); Dcl sumpd Bin Fixed(31); Dcl npd Bin Fixed(31); Dcl hi Bin Fixed(31) Init(0); Dcl (xl(10),xi) Bin Fixed(31); Dcl i Bin Fixed(31); Do x=1 To 10; Call proper_divisors(x,pd,npd); Put Edit(x,' -> ',(pd(i) Do i=1 To npd))(Skip,f(2),a,10(f(2))); End; xi=0; Do x=1 To 20000; Call proper_divisors(x,pd,npd); Select; When(npd>hi) Do; xi=1; xl(1)=x; hi=npd; End; When(npd=hi) Do; xi+=1; xl(xi)=x; End; Otherwise; End; End; Put Edit(hi,' -> ',(xl(i) Do i=1 To xi))(Skip,f(3),a,10(f(6)));
x= 166320; Call proper_divisors(x,pd,npd); Put Edit(x,' -> ',npd)(Skip,f(8),a,f(4)); x=1441440; Call proper_divisors(x,pd,npd); Put Edit(x,' -> ',npd)(Skip,f(8),a,f(4));
p9b=time(); Put Edit((p9b-p9a)/1000,' seconds elapsed')(Skip,f(6,3),a); Return;
proper_divisors: Proc(n,pd,npd); Dcl (n,pd(300),npd) Bin Fixed(31); Dcl (d,delta) Bin Fixed(31); npd=0; If n>1 Then Do; If mod(n,2)=1 Then /* odd number */ delta=2; Else /* even number */ delta=1; Do d=1 To n/2 By delta; If mod(n,d)=0 Then Do; npd+=1; pd(npd)=d; End; End; End; End;
End;</lang>
- Output:
1 -> 2 -> 1 3 -> 1 4 -> 1 2 5 -> 1 6 -> 1 2 3 7 -> 1 8 -> 1 2 4 9 -> 1 3 10 -> 1 2 5 79 -> 15120 18480 166320 -> 159 1441440 -> 287 0.530 seconds elapsed
PowerShell
version 1
<lang PowerShell> function proper-divisor ($n) {
if($n -ge 2) { $lim = [Math]::Floor([Math]::Sqrt($n)) $less, $greater = @(1), @() for($i = 2; $i -lt $lim; $i++){ if($n%$i -eq 0) { $less += @($i) $greater = @($n/$i) + $greater } } if(($lim -ne 1) -and ($n%$lim -eq 0)) {$less += @($lim)} $($less + $greater) } else {@()}
} "$(proper-divisor 100)" "$(proper-divisor 496)" "$(proper-divisor 2048)" </lang>
version 2
<lang PowerShell> function proper-divisor ($n) {
if($n -ge 2) { $lim = [Math]::Floor($n/2)+1 $proper = @(1) for($i = 2; $i -lt $lim; $i++){ if($n%$i -eq 0) { $proper += @($i) } } $proper } else {@()}
} "$(proper-divisor 100)" "$(proper-divisor 496)" "$(proper-divisor 2048)" </lang>
version 3
<lang PowerShell> function eratosthenes ($n) {
if($n -gt 1){ $prime = @(0..$n| foreach{$true}) $m = [Math]::Floor([Math]::Sqrt($n)) function multiple($i) { for($j = $i*$i; $j -le $n; $j += $i) { $prime[$j] = $false } } multiple 2 for($i = 3; $i -le $m; $i += 2) { if($prime[$i]) {multiple $i} } 2 for($i = 3; $i -le $n; $i += 2) { if($prime[$i]) {$i} } } else { Write-Error "$n is not greater than 1" }
} function prime-decomposition ($n) {
$array = eratosthenes $n $prime = @() foreach($p in $array) { while($n%$p -eq 0) { $n /= $p $prime += @($p) } } $prime
} function proper-divisor ($n) {
if($n -ge 2) { $array = prime-decomposition $n $lim = $array.Count function state($res, $i){ if($i -lt $lim) { state ($res) ($i + 1) state ($res*$array[$i]) ($i + 1) } elseif($res -lt $n) {$res} } state 1 0 | sort -Unique } else {@()}
} "$(proper-divisor 100)" "$(proper-divisor 496)" "$(proper-divisor 2048)" </lang> Output:
1 2 4 5 10 20 25 50 1 2 4 8 16 31 62 124 248 1 2 4 8 16 32 64 128 256 512 1024
Prolog
Taking a cue from an SO answer:
<lang prolog>divisor(N, Divisor) :-
UpperBound is round(sqrt(N)), between(1, UpperBound, D), 0 is N mod D, ( Divisor = D ; LargerDivisor is N/D, LargerDivisor =\= D, Divisor = LargerDivisor ).
proper_divisor(N, D) :-
divisor(N, D), D =\= N.
%% Task 1
%
proper_divisors(N, Ds) :-
setof(D, proper_divisor(N, D), Ds).
%% Task 2
%
show_proper_divisors_of_range(Low, High) :-
findall( N:Ds, ( between(Low, High, N), proper_divisors(N, Ds) ), Results ), maplist(writeln, Results).
%% Task 3
%
proper_divisor_count(N, Count) :-
proper_divisors(N, Ds), length(Ds, Count).
find_most_proper_divisors_in_range(Low, High, Result) :-
aggregate_all( max(Count, N), ( between(Low, High, N), proper_divisor_count(N, Count) ), max(MaxCount, Num) ), Result = (num(Num)-divisor_count(MaxCount)).</lang>
Output:
<lang prolog>?- show_proper_divisors_of_range(1,10). 2:[1] 3:[1] 4:[1,2] 5:[1] 6:[1,2,3] 7:[1] 8:[1,2,4] 9:[1,3] 10:[1,2,5] true.
?- find_most_proper_divisors_in_range(1,20000,Result). Result = num(15120)-divisor_count(79). </lang>
PureBasic
<lang PureBasic> EnableExplicit
Procedure ListProperDivisors(Number, List Lst())
If Number < 2 : ProcedureReturn : EndIf Protected i For i = 1 To Number / 2 If Number % i = 0 AddElement(Lst()) Lst() = i EndIf Next
EndProcedure
Procedure.i CountProperDivisors(Number)
If Number < 2 : ProcedureReturn 0 : EndIf Protected i, count = 0 For i = 1 To Number / 2 If Number % i = 0 count + 1 EndIf Next ProcedureReturn count
EndProcedure
Define n, count, most = 1, maxCount = 0 If OpenConsole()
PrintN("The proper divisors of the following numbers are : ") PrintN("") NewList lst() For n = 1 To 10 ListProperDivisors(n, lst()) Print(RSet(Str(n), 3) + " -> ") If ListSize(lst()) = 0 Print("(None)") Else ForEach lst() Print(Str(lst()) + " ") Next EndIf ClearList(lst()) PrintN("") Next For n = 2 To 20000 count = CountProperDivisors(n) If count > maxCount maxCount = count most = n EndIf Next PrintN("") PrintN(Str(most) + " has the most proper divisors, namely " + Str(maxCount)) PrintN("") PrintN("Press any key to close the console") Repeat: Delay(10) : Until Inkey() <> "" CloseConsole()
EndIf </lang>
- Output:
The proper divisors of the following numbers are : 1 -> (None) 2 -> 1 3 -> 1 4 -> 1 2 5 -> 1 6 -> 1 2 3 7 -> 1 8 -> 1 2 4 9 -> 1 3 10 -> 1 2 5 15120 has the most proper divisors, namely 79
Python
Python: Literal
A very literal interpretation <lang python>>>> def proper_divs2(n): ... return {x for x in range(1, (n + 1) // 2 + 1) if n % x == 0 and n != x} ... >>> [proper_divs2(n) for n in range(1, 11)] [set(), {1}, {1}, {1, 2}, {1}, {1, 2, 3}, {1}, {1, 2, 4}, {1, 3}, {1, 2, 5}] >>> >>> n, length = max(((n, len(proper_divs2(n))) for n in range(1, 20001)), key=lambda pd: pd[1]) >>> n 15120 >>> length 79 >>> </lang>
Python: From prime factors
I found a reference on how to generate factors from all the prime factors and the number of times each prime factor goes into N - its multiplicity.
For example, given N having prime factors P(i) with associated multiplicity M(i}) then the factors are given by:
for m[0] in range(M(0) + 1): for m[1] in range(M[1] + 1): ... for m[i - 1] in range(M[i - 1] + 1): mult = 1 for j in range(i): mult *= P[j] ** m[j] yield mult
This version is over an order of magnitude faster for generating the proper divisors of the first 20,000 integers; at the expense of simplicity. <lang python>from math import sqrt from functools import lru_cache, reduce from collections import Counter from itertools import product
MUL = int.__mul__
def prime_factors(n):
'Map prime factors to their multiplicity for n' d = _divs(n) d = [] if d == [n] else (d[:-1] if d[-1] == d else d) pf = Counter(d) return dict(pf)
@lru_cache(maxsize=None) def _divs(n):
'Memoized recursive function returning prime factors of n as a list' for i in range(2, int(sqrt(n)+1)): d, m = divmod(n, i) if not m: return [i] + _divs(d) return [n]
def proper_divs(n):
Return the set of proper divisors of n. pf = prime_factors(n) pfactors, occurrences = pf.keys(), pf.values() multiplicities = product(*(range(oc + 1) for oc in occurrences)) divs = {reduce(MUL, (pf**m for pf, m in zip(pfactors, multis)), 1) for multis in multiplicities} try: divs.remove(n) except KeyError: pass return divs or ({1} if n != 1 else set())
if __name__ == '__main__':
rangemax = 20000 print([proper_divs(n) for n in range(1, 11)]) print(*max(((n, len(proper_divs(n))) for n in range(1, 20001)), key=lambda pd: pd[1]))</lang>
- Output:
[set(), {1}, {1}, {1, 2}, {1}, {1, 2, 3}, {1}, {1, 2, 4}, {1, 3}, {1, 2, 5}] 15120 79
R
<lang r>
- Proper divisors. 12/10/16 aev
require(numbers); V <- sapply(1:20000, Sigma, k = 0, proper = TRUE); ind <- which(V==max(V)); cat(" *** max number of divisors:", max(V), "\n"," *** for the following indices:",ind, "\n"); </lang>
- Output:
Loading required package: numbers *** max number of divisors: 79 *** for the following indices: 15120 18480
Racket
Short version
<lang racket>#lang racket (require math) (define (proper-divisors n) (drop-right (divisors n) 1)) (for ([n (in-range 1 (add1 10))])
(printf "proper divisors of: ~a\t~a\n" n (proper-divisors n)))
(define most-under-20000
(for/fold ([best '(1)]) ([n (in-range 2 (add1 20000))]) (define divs (proper-divisors n)) (if (< (length (cdr best)) (length divs)) (cons n divs) best)))
(printf "~a has ~a proper divisors\n"
(car most-under-20000) (length (cdr most-under-20000)))</lang>
- Output:
proper divisors of: 1 () proper divisors of: 2 (1) proper divisors of: 3 (1) proper divisors of: 4 (1 2) proper divisors of: 5 (1) proper divisors of: 6 (1 2 3) proper divisors of: 7 (1) proper divisors of: 8 (1 2 4) proper divisors of: 9 (1 3) proper divisors of: 10 (1 2 5) 15120 has 79 proper divisors
Long version
The main module will only be executed when this file is executed. When used as a library, it will not be used. <lang racket>#lang racket/base (provide fold-divisors ; name as per "Abundant..."
proper-divisors)
(define (fold-divisors v n k0 kons)
(define n1 (add1 n)) (cond [(>= n1 (vector-length v)) (define rv (make-vector n1 k0)) (for* ([n (in-range 1 n1)] [m (in-range (* 2 n) n1 n)]) (vector-set! rv m (kons n (vector-ref rv m)))) rv] [else v]))
(define proper-divisors
(let ([p.d-v (vector)]) (λ (n) (set! p.d-v (reverse (fold-divisors p.d-v n null cons))) (vector-ref p.d-v n))))
(module+ main
(for ([n (in-range 1 (add1 10))]) (printf "proper divisors of: ~a\t~a\n" n (proper-divisors n)))
(define count-proper-divisors (let ([p.d-v (vector)]) (λ(n) (set! p.d-v (fold-divisors p.d-v n 0 (λ (d n) (add1 n)))) (vector-ref p.d-v n))))
(void (count-proper-divisors 20000))
(define-values [C I] (for*/fold ([C 0] [I 1]) ([i (in-range 1 (add1 20000))] [c (in-value (count-proper-divisors i))] #:when [> c C]) (values c i))) (printf "~a has ~a proper divisors\n" I C))</lang>
The output is the same as the short version above.
REXX
version 1
<lang rexx>Call time 'R' Do x=1 To 10
Say x '->' proper_divisors(x) End
hi=1 Do x=1 To 20000
/* If x//1000=0 Then Say x */ npd=count_proper_divisors(x) Select When npd>hi Then Do list.npd=x hi=npd End When npd=hi Then list.hi=list.hi x Otherwise Nop End End
Say hi '->' list.hi
Say ' 166320 ->' count_proper_divisors(166320) Say '1441440 ->' count_proper_divisors(1441440)
Say time('E') 'seconds elapsed' Exit
proper_divisors: Procedure Parse Arg n If n=1 Then Return pd= /* Optimization reduces 37 seconds to 28 seconds */ If n//2=1 Then /* odd number */
delta=2
Else /* even number */
delta=1
Do d=1 To n%2 By delta
If n//d=0 Then pd=pd d End
Return space(pd)
count_proper_divisors: Procedure Parse Arg n Return words(proper_divisors(n))</lang>
- Output:
1 -> 2 -> 1 3 -> 1 4 -> 1 2 5 -> 1 6 -> 1 2 3 7 -> 1 8 -> 1 2 4 9 -> 1 3 10 -> 1 2 5 79 -> 15120 18480 166320 -> 159 1441440 -> 287 28.342000 seconds elapsed
version 2
The following REXX version is an adaptation of the optimized version for the REXX language example for Factors of an integer.
This REXX version handles all integers (negative, zero, positive) and automatically adjusts the precision (digits).
It also allows the specification of the ranges (for display and for finding the maximum), and allows for extra numbers.
With the (subroutine) optimization, it's over twenty times faster. <lang rexx>/*REXX program finds proper divisors (and count) of integer ranges; finds the max count.*/ parse arg bot top inc range xtra /*obtain optional arguments from the CL*/ if bot== | bot=="," then bot= 1 /*Not specified? Then use the default.*/ if top== | top=="," then top= 10 /* " " " " " " */ if inc== | inc=="," then inc= 1 /* " " " " " " */ if range== | range=="," then range=20000 /* " " " " " " */ w=1+max(length(top), length(bot), length(range)) /*determine the biggest number of these*/ numeric digits max(9, w) /*have enough digits for // operator.*/ @.= 'and' /*a literal used to separate #s in list*/
do n=bot to top by inc /*process the first range specified. */ q=Pdivs(n); #=words(q) /*get proper divs; get number of Pdivs.*/ if q=='∞' then #=q /*adjust number of Pdivisors for zero. */ say right(n, max(20, w) ) 'has' center(#, 4) "proper divisors: " q end /*n*/
say /* [↑] process 1st range of integers.*/ m=0 /*M ≡ maximum number of Pdivs (so far).*/
do r=1 for range /*process the second range specified. */ q=Pdivs(r); #=words(q) /*get proper divs; get number of Pdivs.*/ if #<m then iterate /*Less then max? Then ignore this #. */ @.#=@.# @. r; m=# /*add this Pdiv to max list; set new M.*/ end /*r*/ /* [↑] process 2nd range of integers.*/
say m ' is the highest number of proper divisors in range 1──►'range,
", and it's for: " subword(@.m, 3)
say /* [↓] handle any given extra numbers.*/
do i=1 for words(xtra); n=word(xtra, i) /*obtain an extra number from XTRA list*/ w=max(w, 1 + length(n) ) /*use maximum width for aligned output.*/ numeric digits max(9, 1 + length(n) ) /*have enough digits for // operator.*/ q=Pdivs(n); #=words(q) /*get proper divs; get number of Pdivs.*/ say right(n, max(20, w) ) 'has' center(#, 4) "proper divisors." end /*i*/ /* [↑] support extra specified integers*/
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ Pdivs: procedure; parse arg x,b; x=abs(x); if x==1 then return /*unity?*/
odd=x // 2; if x==0 then return '∞' /*zero ?*/ r=0 /* [↓] ══integer square root══ ___ */ q=1; do while q<=z; q=q*4; end /*R: an integer which will be √ X */ do while q>1; q=q%4; _=z-r-q; r=r%2; if _>=0 then do; z=_; r=r+q; end end /*while q>1*/ /* [↑] compute the integer sqrt of X.*/ a=1 /* [↓] use all, or only odd #s. ___*/ do j=2 +odd by 1 +odd to r -(r*r==x) /*divide by some integers up to √ X */ if x//j==0 then do; a=a j; b=x%j b /*if ÷, add both divisors to α & ß. */ end end /*j*/ /* [↑] % is the REXX integer division*/ /* [↓] adjust for a square. ___*/ if j*j==x then return a j b /*Was X a square? If so, add √ X */ return a b /*return the divisors (both lists). */</lang>
- output when using the following input: 0 10 1 20000 166320 1441440 11796480000
0 has ∞ proper divisors: ∞ 1 has 0 proper divisors: 2 has 1 proper divisors: 1 3 has 1 proper divisors: 1 4 has 2 proper divisors: 1 2 5 has 1 proper divisors: 1 6 has 3 proper divisors: 1 2 3 7 has 1 proper divisors: 1 8 has 3 proper divisors: 1 2 4 9 has 2 proper divisors: 1 3 10 has 3 proper divisors: 1 2 5 79 is the highest number of proper divisors in range 1──►20000, and it's for: 15120 and 18480 166320 has 159 proper divisors. 1441440 has 287 proper divisors. 11796480000 has 329 proper divisors.
version 3
When factoring 20,000 integers, this REXX version is about 10% faster than the REXX version 2.
When factoring 200,000 integers, this REXX version is about 30% faster.
When factoring 2,000,000 integers, this REXX version is about 40% faster.
It accomplishes a faster speed by incorporating the calculation of an integer square root of an integer (without using any floating point arithmetic). <lang rexx>/*REXX program finds proper divisors (and count) of integer ranges; finds the max count.*/ parse arg bot top inc range xtra /*obtain optional arguments from the CL*/ if bot== | bot=="," then bot= 1 /*Not specified? Then use the default.*/ if top== | top=="," then top= 10 /* " " " " " " */ if inc== | inc=="," then inc= 1 /* " " " " " " */ if range== | range=="," then range=20000 /* " " " " " " */ w=1+max(length(top), length(bot), length(range)) /*determine the biggest number of these*/ numeric digits max(9, w) /*have enough digits for // operator.*/ @.= 'and' /*a literal used to separate #s in list*/
do n=bot to top by inc /*process the first range specified. */ q=Pdivs(n); #=words(q) /*get proper divs; get number of Pdivs.*/ if q=='∞' then #=q /*adjust number of Pdivisors for zero. */ say right(n, max(20, w) ) 'has' center(#, 4) "proper divisors: " q end /*n*/
say /* [↑] process 1st range of integers.*/ m=0 /*M ≡ maximum number of Pdivs (so far).*/
do r=1 for range /*process the second range specified. */ q=Pdivs(r); #=words(q) /*get proper divs; get number of Pdivs.*/ if #<m then iterate /*Less then max? Then ignore this #. */ @.#=@.# @. r; m=# /*add this Pdiv to max list; set new M.*/ end /*r*/ /* [↑] process 2nd range of integers.*/
say m ' is the highest number of proper divisors in range 1──►'range,
", and it's for: " subword(@.m, 3)
say /* [↓] handle any given extra numbers.*/
do i=1 for words(xtra); n=word(xtra, i) /*obtain an extra number from XTRA list*/ w=max(w, 1 + length(n) ) /*use maximum width for aligned output.*/ numeric digits max(9, 1 + length(n) ) /*have enough digits for // operator.*/ q=Pdivs(n); #=words(q) /*get proper divs; get number of Pdivs.*/ say right(n, max(20, w) ) 'has' center(#, 4) "proper divisors." end /*i*/ /* [↑] support extra specified integers*/
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ Pdivs: procedure; parse arg x 1 z,b; x=abs(x); if x==1 then return /*unity?*/
odd=x // 2; if x==0 then return '∞' /*zero ?*/ r=0 /* [↓] ══integer square root══ ___ */ q=1; do while q<=z; q=q*4; end /*R: an integer which will be √ X */ do while q>1; q=q%4; _=z-r-q; r=r%2; if _>=0 then do; z=_; r=r+q; end end /*while q>1*/ /* [↑] compute the integer sqrt of X.*/ a=1 /* [↓] use all, or only odd #s. ___*/ do j=2 +odd by 1 +odd to r -(r*r==x) /*divide by some integers up to √ X */ if x//j==0 then do; a=a j; b=x%j b /*if ÷, add both divisors to α & ß. */ end end /*j*/ /* [↑] % is the REXX integer division*/ /* [↓] adjust for a square. ___*/ if j*j==x then return a j b /*Was X a square? If so, add √ X */ return a b /*return the divisors (both lists). */</lang>
- output is identical to the 2nd REXX version when using the same inputs.
Ring
<lang ring>
- Project : Proper divisors
- Date : 2017/09/22
- Author : Gal Zsolt (~ CalmoSoft ~)
- Email : <calmosoft@gmail.com>
limit = 10 for n=1 to limit
if n=1 see "" + 1 + " -> (None)" + nl loop ok see "" + n + " -> " for m=1 to n- 1 if n%m = 0 see " " + m ok next see nl
next </lang> Output:
1 -> (None) 2 -> 1 3 -> 1 4 -> 1 2 5 -> 1 6 -> 1 2 3 7 -> 1 8 -> 1 2 4 9 -> 1 3 10 -> 1 2 5
Ruby
<lang ruby>require "prime"
class Integer
def proper_divisors return [] if self == 1 primes = prime_division.flat_map{|prime, freq| [prime] * freq} (1...primes.size).each_with_object([1]) do |n, res| primes.combination(n).map{|combi| res << combi.inject(:*)} end.flatten.uniq end
end
(1..10).map{|n| puts "#{n}: #{n.proper_divisors}"}
size, select = (1..20_000).group_by{|n| n.proper_divisors.size}.max select.each do |n|
puts "#{n} has #{size} divisors"
end</lang>
- Output:
1: [] 2: [1] 3: [1] 4: [1, 2] 5: [1] 6: [1, 2, 3] 7: [1] 8: [1, 2, 4] 9: [1, 3] 10: [1, 2, 5] 15120 has 79 divisors 18480 has 79 divisors
An Alternative Approach
<lang ruby>#Determine the integer within a range of integers that has the most proper divisors
- Nigel Galloway: December 23rd., 2014
require "prime" n, g = 0 (1..20000).each{|i| e = i.prime_division.inject(1){|n,g| n * (g[1]+1)}
n, g = e, i if e > n}
puts "#{g} has #{n-1} proper divisors"</lang>
- Output:
In the range 1..200000
15120 has 79 proper divisors
and in the ranges 1..2000000 & 1..20000000
166320 has 159 proper divisors 1441440 has 287 proper divisors
Rust
<lang rust>trait ProperDivisors {
fn proper_divisors(&self) -> Option<Vec<u64>>;
}
impl ProperDivisors for u64 {
fn proper_divisors(&self) -> Option<Vec<u64>> { if self.le(&1) { return None; } let mut divisors: Vec<u64> = Vec::new();
for i in 1..*self { if *self % i == 0 { divisors.push(i); } } Option::from(divisors) }
}
fn main() {
for i in 1..11 { println!("Proper divisors of {:2}: {:?}", i, i.proper_divisors().unwrap_or(vec![])); }
let mut most_idx: u64 = 0; let mut most_divisors: Vec<u64> = Vec::new(); for i in 1..20_001 { let divs = i.proper_divisors().unwrap_or(vec![]); if divs.len() > most_divisors.len() { most_divisors = divs; most_idx = i; } } println!("In 1 to 20000, {} has the most proper divisors at {}", most_idx, most_divisors.len());
} </lang>
- Output:
Proper divisors of 1: [] Proper divisors of 2: [1] Proper divisors of 3: [1] Proper divisors of 4: [1, 2] Proper divisors of 5: [1] Proper divisors of 6: [1, 2, 3] Proper divisors of 7: [1] Proper divisors of 8: [1, 2, 4] Proper divisors of 9: [1, 3] Proper divisors of 10: [1, 2, 5] In 1 to 20000, 15120 has the most proper divisors at 79
Scala
Simple proper divisors
<lang Scala>def properDivisors(n: Int) = (1 to n/2).filter(i => n % i == 0) def format(i: Int, divisors: Seq[Int]) = f"$i%5d ${divisors.length}%2d ${divisors mkString " "}"
println(f" n cnt PROPER DIVISORS") val (count, list) = (1 to 20000).foldLeft( (0, List[Int]()) ) { (max, i) =>
val divisors = properDivisors(i) if (i <= 10 || i == 100) println( format(i, divisors) ) if (max._1 < divisors.length) (divisors.length, List(i)) else if (max._1 == divisors.length) (divisors.length, max._2 ::: List(i)) else max
}
list.foreach( number => println(f"$number%5d ${properDivisors(number).length}") )</lang>
- Output:
n cnt PROPER DIVISORS 1 0 2 1 1 3 1 1 4 2 1 2 5 1 1 6 3 1 2 3 7 1 1 8 3 1 2 4 9 2 1 3 10 3 1 2 5 100 8 1 2 4 5 10 20 25 50 15120 79 18480 79
Proper divisors for big (long) numbers
<lang Scala>import annotation.tailrec import collection.parallel.mutable.ParSeq
def factorize(n: Long): List[Long] = {
@tailrec def factors(tuple: (Long, Long, List[Long], Int)): List[Long] = { tuple match { case (1, _, acc, _) => acc case (n, k, acc, _) if (n % k == 0) => factors((n / k, k, acc ++ ParSeq(k), Math.sqrt(n / k).toInt)) case (n, k, acc, sqr) if (k < sqr) => factors(n, k + 1, acc, sqr) case (n, k, acc, sqr) if (k >= sqr) => factors((1, k, acc ++ ParSeq(n), 0)) } } factors((n, 2, List[Long](), Math.sqrt(n).toInt))
} def properDivisors(n: Long) = {
val factors = factorize(n) val products = (1 until factors.length).map(i => factors.combinations(i).map(_.product).toList).flatten (1L +: products).filter(_ < n)
}</lang>
Sidef
<lang ruby>func propdiv (n) {
gather { { |d| if (d `divides` n) { take(d, n//d) } } << 1..n.isqrt }.grep{ _ != n }.uniq.sort
}
var max = 0 var candidates = [] 20_000.times { |i| var divs = propdiv(i).len if (divs > max) { candidates = [] max = divs } candidates << i if (divs == max) } say "max = #{max}, candidates = #{candidates}"</lang>- Output:
1 [] 2 [1] 3 [1] 4 [1, 2] 5 [1] 6 [1, 2, 3] 7 [1] 8 [1, 2, 4] 9 [1, 3] 10 [1, 2, 5] max = 79, candidates = [15120, 18480]
Swift
Simple function: <lang Swift>func properDivs1(n: Int) -> [Int] {
return filter (1 ..< n) { n % $0 == 0 }
}</lang> More efficient function: <lang Swift>import func Darwin.sqrt
func sqrt(x:Int) -> Int { return Int(sqrt(Double(x))) }
func properDivs(n: Int) -> [Int] {
if n == 1 { return [] } var result = [Int]() for div in filter (1 ... sqrt(n), { n % $0 == 0 }) { result.append(div)
if n/div != div && n/div != n { result.append(n/div) } } return sorted(result)
}</lang> Rest of the task: <lang Swift>for i in 1...10 {
println("\(i): \(properDivs(i))")
}
var (num, max) = (0,0)
for i in 1...20_000 {
let count = properDivs(i).count if (count > max) { (num, max) = (i, count) }
}
println("\(num): \(max)")</lang>
- Output:
1: [] 2: [1] 3: [1] 4: [1, 2] 5: [1] 6: [1, 2, 3] 7: [1] 8: [1, 2, 4] 9: [1, 3] 10: [1, 2, 5] 15120: 79
Tcl
Note that if a number, , greater than 1 divides exactly, both and are proper divisors. (The raw answers are not sorted; the pretty-printer code sorts.) <lang tcl>proc properDivisors {n} {
if {$n == 1} return set divs 1 for {set i 2} {$i*$i <= $n} {incr i} {
if {!($n % $i)} { lappend divs $i if {$i*$i < $n} { lappend divs [expr {$n / $i}] } }
} return $divs
}
for {set i 1} {$i <= 10} {incr i} {
puts "$i => {[join [lsort -int [properDivisors $i]] ,]}"
} set maxI [set maxC 0] for {set i 1} {$i <= 20000} {incr i} {
set c [llength [properDivisors $i]] if {$c > $maxC} {
set maxI $i set maxC $c
}
} puts "max: $maxI => (...$maxC…)"</lang>
- Output:
1 => {} 2 => {1} 3 => {1} 4 => {1,2} 5 => {1} 6 => {1,2,3} 7 => {1} 8 => {1,2,4} 9 => {1,3} 10 => {1,2,5} max: 15120 => (...79...)
zkl
<lang zkl>fcn properDivs(n){ [1.. (n + 1)/2 + 1].filter('wrap(x){ n%x==0 and n!=x }) } [1..10].apply(properDivs).println(); [1..20_001].apply('wrap(n){ T(properDivs(n).len(),n) })
.reduce(fcn([(a,_)]ab, [(c,_)]cd){ a>c and ab or cd },T(0,0)) .println();</lang>
- Output:
L(L(),L(1),L(1),L(1,2),L(1),L(1,2,3),L(1),L(1,2,4),L(1,3),L(1,2,5)) L(79,18480)
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