Aliquot sequence classifications
You are encouraged to solve this task according to the task description, using any language you may know.
An aliquot sequence of a positive integer K is defined recursively as the first member being K and subsequent members being the sum of the Proper divisors of the previous term.
- If the terms eventually reach 0 then the series for K is said to terminate.
There are several classifications for non termination:- If the second term is K then all future terms are also K and so the sequence repeats from the first term with period 1 and K is called perfect.
- If the third term would be repeating K then the sequence repeats with period 2 and K is called amicable.
- If the Nth term would be repeating K for the first time, with N > 3 then the sequence repeats with period N - 1 and K is called sociable.
Perfect, amicable and sociable numbers eventually repeat the original number K; there are other repetitions...- Some K have a sequence that eventually forms a periodic repetition of period 1 but of a number other than K, for example 95 which forms the sequence
95, 25, 6, 6, 6, ...such K are called aspiring. - K that have a sequence that eventually forms a periodic repetition of period >= 2 but of a number other than K, for example 562 which forms the sequence
562, 284, 220, 284, 220, ...such K are called cyclic.
- Some K have a sequence that eventually forms a periodic repetition of period 1 but of a number other than K, for example 95 which forms the sequence
And finally:- Some K form aliquot sequences that are not known to be either terminating or periodic; these K are to be called non-terminating.
For the purposes of this task, K is to be classed as non-terminating if it has not been otherwise classed after generating 16 terms or if any term of the sequence is greater than 2**47 = 140,737,488,355,328.
- Some K form aliquot sequences that are not known to be either terminating or periodic; these K are to be called non-terminating.
- Task
- Create routine(s) to generate the aliquot sequence of a positive integer enough to classify it according to the classifications given above.
- Use it to display the classification and sequences of the numbers one to ten inclusive.
- Use it to show the classification and sequences of the following integers, in order:
- 11, 12, 28, 496, 220, 1184, 12496, 1264460, 790, 909, 562, 1064, 1488, and optionally 15355717786080.
Show all output on this page.
- Related tasks
- Abundant, deficient and perfect number classifications. (Classifications from only the first two members of the whole sequence).
- Proper divisors
- Amicable pairs
ALGOL 68
Assumes LONG INT is at least 64 bits, as in Algol 68G. <lang algol68>BEGIN
# aliquot sequence classification #
# maximum sequence length we consider #
INT max sequence length = 16;
# possible classifications #
STRING perfect classification = "perfect ";
STRING amicable classification = "amicable ";
STRING sociable classification = "sociable ";
STRING aspiring classification = "aspiring ";
STRING cyclic classification = "cyclic ";
STRING terminating classification = "terminating ";
STRING non terminating classification = "non terminating";
# structure to hold an aliquot sequence and its classification #
MODE ALIQUOT = STRUCT( STRING classification
, [ 1 : max sequence length ]LONG INT sequence
, INT length
);
# maximum value for sequence elements - if any element is more than this, #
# we assume it is non-teriminating #
LONG INT max element = 140 737 488 355 328;
# returns the sum of the proper divisors of n #
OP DIVISORSUM = ( LONG INT n )LONG INT:
BEGIN
LONG INT abs n = ABS n;
IF abs n < 2 THEN
0 # -1, 0 and 1 have no proper divisors #
ELSE
# have a number with possible divisors #
LONG INT result := 1; # 1 is always a divisor #
# a FOR loop counter can only be an INT, hence the WHILE loop #
LONG INT d := ENTIER long sqrt( abs n );
WHILE d > 1 DO
IF abs n MOD d = 0 THEN
# found another divisor #
result +:= d;
IF d * d /= abs n THEN
# add the other divisor #
result +:= abs n OVER d
FI
FI;
d -:= 1
OD;
result
FI
END # DIVISORSUM # ;
# generates the aliquot sequence of the number k and its classification #
# at most max elements of the sequence are considered #
OP CLASSIFY = ( LONG INT k )ALIQUOT :
BEGIN
ALIQUOT result;
classification OF result := "non-terminating";
INT lb = LWB sequence OF result;
INT ub = UPB sequence OF result;
( sequence OF result )[ lb ] := k; # the first element is always k #
length OF result := 1;
FOR i FROM lb + 1 TO ub DO
( sequence OF result )[ i ] := 0
OD;
BOOL classified := FALSE;
LONG INT prev k := k;
FOR i FROM lb + 1 TO ub WHILE NOT classified DO
length OF result +:= 1;
LONG INT next k := ( sequence OF result )[ i ] := DIVISORSUM prev k;
classified := TRUE;
IF next k = 0 THEN # the sequence terminates #
classification OF result := terminating classification
ELIF next k > max element THEN # the sequence gets too large #
classification OF result := non terminating classification
ELIF next k = k THEN # the sequence that returns to k #
classification OF result
:= IF i = lb + 1 THEN perfect classification
ELIF i = lb + 2 THEN amicable classification
ELSE sociable classification
FI
ELIF next k = prev k THEN # the sequence repeats with non-k #
classification OF result := aspiring classification
ELSE # check for repeating sequence with a period more than 1 #
classified := FALSE;
FOR prev pos FROM lb TO i - 2 WHILE NOT classified DO
IF classified := ( sequence OF result )[ prev pos ] = next k THEN
# found a repeatition #
classification OF result := cyclic classification
FI
OD
FI;
prev k := next k
OD;
result
END # CLASSIFY # ;
# test cases as per the task #
[]LONG INT test cases =
( 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
, 11, 12, 28, 496, 220, 1184, 12496, 1264460, 790, 909
, 562, 1064, 1488
, 15355717786080
);
FOR i FROM LWB test cases TO UPB test cases DO
LONG INT k := test cases[ i ];
ALIQUOT seq = CLASSIFY k;
print( ( whole( k, -14 ), ": ", classification OF seq, ":" ) );
FOR e FROM LWB sequence OF seq + 1 TO length OF seq DO
print( ( " ", whole( ( sequence OF seq )[ e ], 0 ) ) )
OD;
print( ( newline ) )
OD
END</lang>
- Output:
1: terminating : 0
2: terminating : 1 0
3: terminating : 1 0
4: terminating : 3 1 0
5: terminating : 1 0
6: perfect : 6
7: terminating : 1 0
8: terminating : 7 1 0
9: terminating : 4 3 1 0
10: terminating : 8 7 1 0
11: terminating : 1 0
12: terminating : 16 15 9 4 3 1 0
28: perfect : 28
496: perfect : 496
220: amicable : 284 220
1184: amicable : 1210 1184
12496: sociable : 14288 15472 14536 14264 12496
1264460: sociable : 1547860 1727636 1305184 1264460
790: aspiring : 650 652 496 496
909: aspiring : 417 143 25 6 6
562: cyclic : 284 220 284
1064: cyclic : 1336 1184 1210 1184
1488: non-terminating: 2480 3472 4464 8432 9424 10416 21328 22320 55056 95728 96720 236592 459792 881392 882384
15355717786080: non terminating: 44534663601120 144940087464480
AppleScript
<lang applescript>on aliquotSum(n)
set sum to (n > 1) as integer
if (n > 1) then
set sqrt to n ^ 0.5
if (sqrt mod 1 is 0) then
set sum to sum + sqrt as integer
set sqrt to sqrt - 1
end if
repeat with i from (sqrt div 1) to 2 by -1
if (n mod i is 0) then set sum to sum + i + n div i
end repeat
end if
return sum
end aliquotSum
on aliquotSequence(k, maxLength, maxN)
-- Generate the sequence within the specified limitations.
set sequence to {k}
set n to k
repeat (maxLength - 1) times
set n to aliquotSum(n)
set repetition to (sequence contains n)
if (repetition) then exit repeat
set end of sequence to n
if ((n = 0) or (n > maxN)) then exit repeat
end repeat
-- Analyse it.
set sequenceLength to (count sequence)
if (sequenceLength is 1) then
set classification to "perfect"
else if (sequence ends with 0) then
set classification to "terminating"
else if (n = k) then
if (sequenceLength is 2) then
set classification to "amicable"
else
set classification to "sociable"
end if
else if (repetition) then
if (sequence ends with n) then
set classification to "aspiring"
else
set classification to "cyclic"
end if
else
set classification to "non-terminating"
end if
return {sequence:sequence, classification:classification}
end aliquotSequence
-- Task code: local output, maxLength, maxN, spacing, astid, k set output to {""} set {maxLength, maxN} to {16, 2 ^ 47} set spacing to " " set astid to AppleScript's text item delimiters set AppleScript's text item delimiters to ", " repeat with k in {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ¬
11, 12, 28, 496, 220, 1184, 12496, 1264460, 790, 909, 562, 1064, 1488, 1.535571778608E+13}
set thisResult to aliquotSequence(k's contents, maxLength, maxN)
set end of output to text -18 thru -1 of (spacing & k) & ": " & ¬
text 1 thru 17 of (thisResult's classification & spacing) & thisResult's sequence
end repeat set AppleScript's text item delimiters to linefeed set output to output as text set AppleScript's text item delimiters to astid return output</lang>
- Output:
<lang applescript>"
1: terminating 1, 0
2: terminating 2, 1, 0
3: terminating 3, 1, 0
4: terminating 4, 3, 1, 0
5: terminating 5, 1, 0
6: perfect 6
7: terminating 7, 1, 0
8: terminating 8, 7, 1, 0
9: terminating 9, 4, 3, 1, 0
10: terminating 10, 8, 7, 1, 0
11: terminating 11, 1, 0
12: terminating 12, 16, 15, 9, 4, 3, 1, 0
28: perfect 28
496: perfect 496
220: amicable 220, 284
1184: amicable 1184, 1210
12496: sociable 12496, 14288, 15472, 14536, 14264
1264460: sociable 1264460, 1547860, 1727636, 1305184
790: aspiring 790, 650, 652, 496
909: aspiring 909, 417, 143, 25, 6
562: cyclic 562, 284, 220
1064: cyclic 1064, 1336, 1184, 1210
1488: non-terminating 1488, 2480, 3472, 4464, 8432, 9424, 10416, 21328, 22320, 55056, 95728, 96720, 236592, 459792, 881392, 882384
1.535571778608E+13: non-terminating 1.535571778608E+13, 4.453466360112E+13, 1.449400874645E+14"</lang>
AWK
<lang awk>
- !/bin/gawk -f
function sumprop(num, i,sum,root) { if (num == 1) return 0 sum=1 root=sqrt(num) for ( i=2; i < root; i++) {
if (num % i == 0 )
{
sum = sum + i + num/i
}
}
if (num % root == 0)
{
sum = sum + root
}
return sum } function class(k, oldk,newk,seq){
- first term
oldk = k seq = " "
- second term
newk = sumprop(oldk) oldk = newk seq = seq " " newk if (newk == 0) return "terminating " seq if (newk == k) return "perfect " seq
- third term
newk = sumprop(oldk) oldk = newk seq = seq " " newk if (newk == 0) return "terminating " seq if (newk == k) return "amicable " seq for (t=4; t<17; t++) { newk = sumprop(oldk) seq = seq " " newk if (newk == 0) return "terminating " seq if (newk == k) return "sociable (period " t-1 ") "seq if (newk == oldk) return "aspiring " seq if (index(seq," " newk " ") > 0) return "cyclic (at " newk ") " seq if (newk > 140737488355328) return "non-terminating (term > 140737488355328) " seq oldk = newk } return "non-terminating (after 16 terms) " seq } BEGIN{ print "Number classification sequence" for (j=1; j < 11; j++)
{
print j,class(j)}
print 11,class(11)
print 12,class(12)
print 28,class(28)
print 496,class(496)
print 220,class(220)
print 1184,class(1184)
print 12496,class(12496)
print 1264460,class(1264460)
print 790,class(790)
print 909,class(909)
print 562,class(562)
print 1064,class(1064)
print 1488,class(1488)
print 15355717786080,class(15355717786080)
}
</lang>
- Output:
Number classification sequence 1 terminating 0 2 terminating 1 0 3 terminating 1 0 4 terminating 3 1 0 5 terminating 1 0 6 perfect 6 7 terminating 1 0 8 terminating 7 1 0 9 terminating 4 3 1 0 10 terminating 8 7 1 0 11 terminating 1 0 12 terminating 16 15 9 4 3 1 0 28 perfect 28 496 perfect 496 220 amicable 284 220 1184 amicable 1210 1184 12496 sociable (period 5) 14288 15472 14536 14264 12496 1264460 sociable (period 4) 1547860 1727636 1305184 1264460 790 aspiring 650 652 496 496 909 aspiring 417 143 25 6 6 562 cyclic (at 284) 284 220 284 1064 cyclic (at 1184) 1336 1184 1210 1184 1488 non-terminating (after 16 terms) 2480 3472 4464 8432 9424 10416 21328 22320 55056 95728 96720 236592 459792 881392 882384 1.53557e+13 non-terminating (term > 140737488355328) 4.45347e+13 1.4494e+14 4.71714e+14
C
Both implementations can process integers or a file containing all the integers from the command line.
Brute Force
The following implementation is a brute force method which takes a very, very long time for 15355717786080. To be fair to C, that's also true for many of the other implementations on this page which also implement the brute force method. See the next implementation for the best solution. <lang C>
- include<stdlib.h>
- include<string.h>
- include<stdio.h>
unsigned long long bruteForceProperDivisorSum(unsigned long long n){ unsigned long long i,sum = 0;
for(i=1;i<(n+1)/2;i++) if(n%i==0 && n!=i) sum += i;
return sum; }
void printSeries(unsigned long long* arr,int size,char* type){ int i;
printf("\nInteger : %llu, Type : %s, Series : ",arr[0],type);
for(i=0;i<size-1;i++) printf("%llu, ",arr[i]); printf("%llu",arr[i]); }
void aliquotClassifier(unsigned long long n){ unsigned long long arr[16]; int i,j;
arr[0] = n;
for(i=1;i<16;i++){ arr[i] = bruteForceProperDivisorSum(arr[i-1]);
if(arr[i]==0||arr[i]==n||(arr[i]==arr[i-1] && arr[i]!=n)){ printSeries(arr,i+1,(arr[i]==0)?"Terminating":(arr[i]==n && i==1)?"Perfect":(arr[i]==n && i==2)?"Amicable":(arr[i]==arr[i-1] && arr[i]!=n)?"Aspiring":"Sociable"); return; }
for(j=1;j<i;j++){ if(arr[j]==arr[i]){ printSeries(arr,i+1,"Cyclic"); return; } } }
printSeries(arr,i+1,"Non-Terminating"); }
void processFile(char* fileName){ FILE* fp = fopen(fileName,"r"); char str[21];
while(fgets(str,21,fp)!=NULL) aliquotClassifier(strtoull(str,(char**)NULL,10));
fclose(fp); }
int main(int argC,char* argV[]) {
if(argC!=2)
printf("Usage : %s <positive integer>",argV[0]); else{ if(strchr(argV[1],'.')!=NULL) processFile(argV[1]); else aliquotClassifier(strtoull(argV[1],(char**)NULL,10)); } return 0; } </lang> Input file, you can include 15355717786080 or similar numbers in this list but be prepared to wait for a very, very long time.:
1 2 3 4 5 6 7 8 9 10 11 12 28 496 220 1184 12496 1264460 790 909 562 1064 1488
Invocation and output for both individual number and input file:
C:\rosettaCode>bruteAliquot.exe 10 Integer : 10, Type : Terminating, Series : 10, 8, 7, 1, 0 C:\rosettaCode>bruteAliquot.exe aliquotData.txt Integer : 1, Type : Terminating, Series : 1, 0 Integer : 2, Type : Terminating, Series : 2, 1, 0 Integer : 3, Type : Terminating, Series : 3, 1, 0 Integer : 4, Type : Terminating, Series : 4, 3, 1, 0 Integer : 5, Type : Terminating, Series : 5, 1, 0 Integer : 6, Type : Perfect, Series : 6, 6 Integer : 7, Type : Terminating, Series : 7, 1, 0 Integer : 8, Type : Terminating, Series : 8, 7, 1, 0 Integer : 9, Type : Terminating, Series : 9, 4, 3, 1, 0 Integer : 10, Type : Terminating, Series : 10, 8, 7, 1, 0 Integer : 11, Type : Terminating, Series : 11, 1, 0 Integer : 12, Type : Terminating, Series : 12, 16, 15, 9, 4, 3, 1, 0 Integer : 28, Type : Perfect, Series : 28, 28 Integer : 496, Type : Perfect, Series : 496, 496 Integer : 220, Type : Amicable, Series : 220, 284, 220 Integer : 1184, Type : Amicable, Series : 1184, 1210, 1184 Integer : 12496, Type : Sociable, Series : 12496, 14288, 15472, 14536, 14264, 12496 Integer : 1264460, Type : Sociable, Series : 1264460, 1547860, 1727636, 1305184, 1264460 Integer : 790, Type : Aspiring, Series : 790, 650, 652, 496, 496 Integer : 909, Type : Aspiring, Series : 909, 417, 143, 25, 6, 6 Integer : 562, Type : Cyclic, Series : 562, 284, 220, 284 Integer : 1064, Type : Cyclic, Series : 1064, 1336, 1184, 1210, 1184 Integer : 1488, Type : Non-Terminating, Series : 1488, 2480, 3472, 4464, 8432, 9424, 10416, 21328, 22320, 55056, 95728, 96720, 236592, 459792, 881392, 882384, 68719476751
Number Theoretic
The following implementation, based on Number Theory, is the best solution for such a problem. All cases are handled, including 15355717786080, with all the numbers being processed and the output written to console practically instantaneously. The above brute force implementation is the original one and it remains to serve as a comparison of the phenomenal difference the right approach can make to a problem. <lang C>
- include<string.h>
- include<stdlib.h>
- include<stdio.h>
unsigned long long raiseTo(unsigned long long base, unsigned long long power){
unsigned long long result = 1,i;
for (i=0; i<power;i++) {
result*=base;
}
return result;
}
unsigned long long properDivisorSum(unsigned long long n){ unsigned long long prod = 1; unsigned long long temp = n,i,count = 0;
while(n%2 == 0){ count++; n /= 2; }
if(count!=0) prod *= (raiseTo(2,count + 1) - 1);
for(i=3;i*i<=n;i+=2){ count = 0;
while(n%i == 0){ count++; n /= i; }
if(count==1) prod *= (i+1); else if(count > 1) prod *= ((raiseTo(i,count + 1) - 1)/(i-1)); }
if(n>2) prod *= (n+1);
return prod - temp; }
void printSeries(unsigned long long* arr,int size,char* type){ int i;
printf("\nInteger : %llu, Type : %s, Series : ",arr[0],type);
for(i=0;i<size-1;i++) printf("%llu, ",arr[i]); printf("%llu",arr[i]); }
void aliquotClassifier(unsigned long long n){ unsigned long long arr[16]; int i,j;
arr[0] = n;
for(i=1;i<16;i++){ arr[i] = properDivisorSum(arr[i-1]);
if(arr[i]==0||arr[i]==n||(arr[i]==arr[i-1] && arr[i]!=n)){ printSeries(arr,i+1,(arr[i]==0)?"Terminating":(arr[i]==n && i==1)?"Perfect":(arr[i]==n && i==2)?"Amicable":(arr[i]==arr[i-1] && arr[i]!=n)?"Aspiring":"Sociable"); return; }
for(j=1;j<i;j++){ if(arr[j]==arr[i]){ printSeries(arr,i+1,"Cyclic"); return; } } }
printSeries(arr,i+1,"Non-Terminating"); }
void processFile(char* fileName){ FILE* fp = fopen(fileName,"r"); char str[21];
while(fgets(str,21,fp)!=NULL) aliquotClassifier(strtoull(str,(char**)NULL,10));
fclose(fp); }
int main(int argC,char* argV[]) {
if(argC!=2)
printf("Usage : %s <positive integer>",argV[0]); else{ if(strchr(argV[1],'.')!=NULL) processFile(argV[1]); else aliquotClassifier(strtoull(argV[1],(char**)NULL,10)); } return 0; } </lang> Input file, to emphasize the effectiveness of this approach, the last number in the file is 153557177860800, 10 times the special case mentioned in the task.
1 2 3 4 5 6 7 8 9 10 11 12 28 496 220 1184 12496 1264460 790 909 562 1064 1488 15355717786080 153557177860800
Invocation and output for both individual number and input file:
C:\rosettaCode>bruteAliquot.exe 10 Integer : 10, Type : Terminating, Series : 10, 8, 7, 1, 0 C:\rosettaCode>aliquotProper.exe aliquotData.txt Integer : 1, Type : Terminating, Series : 1, 0 Integer : 2, Type : Terminating, Series : 2, 1, 0 Integer : 3, Type : Terminating, Series : 3, 1, 0 Integer : 4, Type : Terminating, Series : 4, 3, 1, 0 Integer : 5, Type : Terminating, Series : 5, 1, 0 Integer : 6, Type : Perfect, Series : 6, 6 Integer : 7, Type : Terminating, Series : 7, 1, 0 Integer : 8, Type : Terminating, Series : 8, 7, 1, 0 Integer : 9, Type : Terminating, Series : 9, 4, 3, 1, 0 Integer : 10, Type : Terminating, Series : 10, 8, 7, 1, 0 Integer : 11, Type : Terminating, Series : 11, 1, 0 Integer : 12, Type : Terminating, Series : 12, 16, 15, 9, 4, 3, 1, 0 Integer : 28, Type : Perfect, Series : 28, 28 Integer : 496, Type : Perfect, Series : 496, 496 Integer : 220, Type : Amicable, Series : 220, 284, 220 Integer : 1184, Type : Amicable, Series : 1184, 1210, 1184 Integer : 12496, Type : Sociable, Series : 12496, 14288, 15472, 14536, 14264, 12496 Integer : 1264460, Type : Sociable, Series : 1264460, 1547860, 1727636, 1305184, 1264460 Integer : 790, Type : Aspiring, Series : 790, 650, 652, 496, 496 Integer : 909, Type : Aspiring, Series : 909, 417, 143, 25, 6, 6 Integer : 562, Type : Cyclic, Series : 562, 284, 220, 284 Integer : 1064, Type : Cyclic, Series : 1064, 1336, 1184, 1210, 1184 Integer : 1488, Type : Non-Terminating, Series : 1488, 2480, 3472, 4464, 8432, 9424, 10416, 21328, 22320, 55056, 95728, 96720, 236592, 459792, 881392, 882384, 68719476751 Integer : 15355717786080, Type : Non-Terminating, Series : 15355717786080, 44534663601120, 144940087464480, 471714103310688, 1130798979186912, 2688948041357088, 6050151708497568, 13613157922 102611548462968, 1977286128289819992, 3415126495450394808, 68719476751 Integer : 153557177860800, Type : Non-Terminating, Series : 153557177860800, 470221741508000, 685337334283120, 908681172226160, 1276860840159280, 1867115442105104, 1751034184622896, 16436297 336056, 1405725265675144, 1230017019320456, 68719476751
C++
This one follows the trail blazed by the "Number Theoretic" C example above. <lang cpp>#include <cstdint>
- include <iostream>
- include <string>
using integer = uint64_t;
// See https://en.wikipedia.org/wiki/Divisor_function integer divisor_sum(integer n) {
integer total = 1, power = 2;
// Deal with powers of 2 first
for (; n % 2 == 0; power *= 2, n /= 2)
total += power;
// Odd prime factors up to the square root
for (integer p = 3; p * p <= n; p += 2) {
integer sum = 1;
for (power = p; n % p == 0; power *= p, n /= p)
sum += power;
total *= sum;
}
// If n > 1 then it's prime
if (n > 1)
total *= n + 1;
return total;
}
// See https://en.wikipedia.org/wiki/Aliquot_sequence void classify_aliquot_sequence(integer n) {
constexpr int limit = 16;
integer terms[limit];
terms[0] = n;
std::string classification("non-terminating");
int length = 1;
for (int i = 1; i < limit; ++i) {
++length;
terms[i] = divisor_sum(terms[i - 1]) - terms[i - 1];
if (terms[i] == n) {
classification =
(i == 1 ? "perfect" : (i == 2 ? "amicable" : "sociable"));
break;
}
int j = 1;
for (; j < i; ++j) {
if (terms[i] == terms[i - j])
break;
}
if (j < i) {
classification = (j == 1 ? "aspiring" : "cyclic");
break;
}
if (terms[i] == 0) {
classification = "terminating";
break;
}
}
std::cout << n << ": " << classification << ", sequence: " << terms[0];
for (int i = 1; i < length && terms[i] != terms[i - 1]; ++i)
std::cout << ' ' << terms[i];
std::cout << '\n';
}
int main() {
for (integer i = 1; i <= 10; ++i)
classify_aliquot_sequence(i);
for (integer i : {11, 12, 28, 496, 220, 1184, 12496, 1264460, 790, 909, 562,
1064, 1488})
classify_aliquot_sequence(i);
classify_aliquot_sequence(15355717786080);
classify_aliquot_sequence(153557177860800);
return 0;
}</lang>
- Output:
1: terminating, sequence: 1 0 2: terminating, sequence: 2 1 0 3: terminating, sequence: 3 1 0 4: terminating, sequence: 4 3 1 0 5: terminating, sequence: 5 1 0 6: perfect, sequence: 6 7: terminating, sequence: 7 1 0 8: terminating, sequence: 8 7 1 0 9: terminating, sequence: 9 4 3 1 0 10: terminating, sequence: 10 8 7 1 0 11: terminating, sequence: 11 1 0 12: terminating, sequence: 12 16 15 9 4 3 1 0 28: perfect, sequence: 28 496: perfect, sequence: 496 220: amicable, sequence: 220 284 220 1184: amicable, sequence: 1184 1210 1184 12496: sociable, sequence: 12496 14288 15472 14536 14264 12496 1264460: sociable, sequence: 1264460 1547860 1727636 1305184 1264460 790: aspiring, sequence: 790 650 652 496 909: aspiring, sequence: 909 417 143 25 6 562: cyclic, sequence: 562 284 220 284 1064: cyclic, sequence: 1064 1336 1184 1210 1184 1488: non-terminating, sequence: 1488 2480 3472 4464 8432 9424 10416 21328 22320 55056 95728 96720 236592 459792 881392 882384 15355717786080: non-terminating, sequence: 15355717786080 44534663601120 144940087464480 471714103310688 1130798979186912 2688948041357088 6050151708497568 13613157922639968 35513546724070632 74727605255142168 162658586225561832 353930992506879768 642678347124409032 1125102611548462968 1977286128289819992 3415126495450394808 153557177860800: non-terminating, sequence: 153557177860800 470221741508000 685337334283120 908681172226160 1276860840159280 1867115442105104 1751034184622896 1643629718341256 1441432897905784 1647351883321016 1557892692704584 1363939602434936 1194001297910344 1597170567336056 1405725265675144 1230017019320456
Common Lisp
Uses the Lisp function proper-divisors-recursive from Task:Proper Divisors. <lang lisp>(defparameter *nlimit* 16) (defparameter *klimit* (expt 2 47)) (defparameter *asht* (make-hash-table)) (load "proper-divisors")
(defun ht-insert (v n)
(setf (gethash v *asht*) n))
(defun ht-find (v n)
(let ((nprev (gethash v *asht*))) (if nprev (- n nprev) nil)))
(defun ht-list ()
(defun sort-keys (&optional (res '())) (maphash #'(lambda (k v) (push (cons k v) res)) *asht*) (sort (copy-list res) #'< :key (lambda (p) (cdr p)))) (let ((sorted (sort-keys))) (dotimes (i (length sorted)) (format t "~A " (car (nth i sorted))))))
(defun aliquot-generator (K1)
"integer->function::fn to generate aliquot sequence" (let ((Kn K1)) #'(lambda () (setf Kn (reduce #'+ (proper-divisors-recursive Kn) :initial-value 0)))))
(defun aliquot (K1)
"integer->symbol|nil::classify aliquot sequence"
(defun aliquot-sym (Kn n)
(let* ((period (ht-find Kn n))
(sym (if period
(cond ; period event
((= Kn K1)
(case period (1 'PERF) (2 'AMIC) (otherwise 'SOCI)))
((= period 1) 'ASPI)
(t 'CYCL))
(cond ; else check for limit event
((= Kn 0) 'TERM)
((> Kn *klimit*) 'TLIM)
((= n *nlimit*) 'NLIM)
(t nil)))))
;; if period event store the period, if no event insert the value
(if sym (when period (setf (symbol-plist sym) (list period)))
(ht-insert Kn n))
sym))
(defun aliquot-str (sym &optional (period 0))
(case sym (TERM "terminating") (PERF "perfect") (AMIC "amicable") (ASPI "aspiring")
(SOCI (format nil "sociable (period ~A)" (car (symbol-plist sym))))
(CYCL (format nil "cyclic (period ~A)" (car (symbol-plist sym))))
(NLIM (format nil "non-terminating (no classification before added term limit of ~A)" *nlimit*))
(TLIM (format nil "non-terminating (term threshold of ~A exceeded)" *klimit*))
(otherwise "unknown")))
(clrhash *asht*)
(let ((fgen (aliquot-generator K1)))
(setf (symbol-function 'aliseq) #'(lambda () (funcall fgen))))
(ht-insert K1 0)
(do* ((n 1 (1+ n))
(Kn (aliseq) (aliseq))
(alisym (aliquot-sym Kn n) (aliquot-sym Kn n)))
(alisym (format t "~A:" (aliquot-str alisym)) (ht-list) (format t "~A~%" Kn) alisym)))
(defun main ()
(princ "The last item in each sequence triggers classification.") (terpri) (dotimes (k 10) (aliquot (+ k 1))) (dolist (k '(11 12 28 496 220 1184 12496 1264460 790 909 562 1064 1488 15355717786080)) (aliquot k)))</lang>
- Output:
CL-USER(45): (main) The last item in each sequence triggers classification. terminating:1 0 terminating:2 1 0 terminating:3 1 0 terminating:4 3 1 0 terminating:5 1 0 perfect:6 6 terminating:7 1 0 terminating:8 7 1 0 terminating:9 4 3 1 0 terminating:10 8 7 1 0 terminating:11 1 0 terminating:12 16 15 9 4 3 1 0 perfect:28 28 perfect:496 496 amicable:220 284 220 amicable:1184 1210 1184 sociable (period 5):12496 14288 15472 14536 14264 12496 sociable (period 4):1264460 1547860 1727636 1305184 1264460 aspiring:790 650 652 496 496 aspiring:909 417 143 25 6 6 cyclic (period 2):562 284 220 284 cyclic (period 2):1064 1336 1184 1210 1184 non-terminating (no classification before added term limit of 16):1488 2480 3472 4464 8432 9424 10416 21328 22320 55056 95728 96720 236592 459792 881392 882384 1474608 non-terminating (term threshold of 140737488355328 exceeded):15355717786080 44534663601120 144940087464480 NIL
D
<lang d>import std.stdio, std.range, std.algorithm, std.typecons, std.conv;
auto properDivisors(in ulong n) pure nothrow @safe /*@nogc*/ {
return iota(1UL, (n + 1) / 2 + 1).filter!(x => n % x == 0 && n != x);
}
enum pDivsSum = (in ulong n) pure nothrow @safe /*@nogc*/ =>
n.properDivisors.sum;
auto aliquot(in ulong n,
in size_t maxLen=16,
in ulong maxTerm=2UL^^47) pure nothrow @safe {
if (n == 0)
return tuple("Terminating", [0UL]);
ulong[] s = [n];
size_t sLen = 1;
ulong newN = n;
while (sLen <= maxLen && newN < maxTerm) {
newN = s.back.pDivsSum;
if (s.canFind(newN)) {
if (s[0] == newN) {
if (sLen == 1) {
return tuple("Perfect", s);
} else if (sLen == 2) {
return tuple("Amicable", s);
} else
return tuple(text("Sociable of length ", sLen), s);
} else if (s.back == newN) {
return tuple("Aspiring", s);
} else
return tuple(text("Cyclic back to ", newN), s);
} else if (newN == 0) {
return tuple("Terminating", s ~ 0);
} else {
s ~= newN;
sLen++;
}
}
return tuple("Non-terminating", s);
}
void main() {
foreach (immutable n; 1 .. 11)
writefln("%s: %s", n.aliquot[]);
writeln;
foreach (immutable n; [11, 12, 28, 496, 220, 1184, 12496, 1264460,
790, 909, 562, 1064, 1488])
writefln("%s: %s", n.aliquot[]);
}</lang>
- Output:
Terminating: [1, 0] Terminating: [2, 1, 0] Terminating: [3, 1, 0] Terminating: [4, 3, 1, 0] Terminating: [5, 1, 0] Perfect: [6] Terminating: [7, 1, 0] Terminating: [8, 7, 1, 0] Terminating: [9, 4, 3, 1, 0] Terminating: [10, 8, 7, 1, 0] Terminating: [11, 1, 0] Terminating: [12, 16, 15, 9, 4, 3, 1, 0] Perfect: [28] Perfect: [496] Amicable: [220, 284] Amicable: [1184, 1210] Sociable of length 5: [12496, 14288, 15472, 14536, 14264] Sociable of length 4: [1264460, 1547860, 1727636, 1305184] Aspiring: [790, 650, 652, 496] Aspiring: [909, 417, 143, 25, 6] Cyclic back to 284: [562, 284, 220] Cyclic back to 1184: [1064, 1336, 1184, 1210] Non-terminating: [1488, 2480, 3472, 4464, 8432, 9424, 10416, 21328, 22320, 55056, 95728, 96720, 236592, 459792, 881392, 882384, 1474608]
EchoLisp
<lang scheme>
- implementation of Floyd algorithm to find cycles in a graph
- see Wikipedia https://en.wikipedia.org/wiki/Cycle_detection
- returns (cycle-length cycle-starter steps)
- steps = 0 if no cycle found
- it's all about a tortoise 🐢 running at speed f(x) after a hare 🐰 at speed f(f (x))
- when they meet, a cycle is found
(define (floyd f x0 steps maxvalue) (define lam 1) ; cycle length (define tortoise (f x0)) (define hare (f (f x0)))
;; cyclic ? yes if steps > 0 (while (and (!= tortoise hare) (> steps 0)) (set!-values (tortoise hare) (values (f tortoise) (f (f hare)))) #:break (and (> hare maxvalue) (set! steps 0)) (set! steps (1- steps)))
;; first repetition = cycle starter (set! tortoise x0) (while (and (!= tortoise hare) (> steps 0)) (set!-values (tortoise hare) (values (f tortoise) (f hare))))
;; length of shortest cycle (set! hare (f tortoise)) (while (and (!= tortoise hare) (> steps 0)) (set! hare (f hare)) (set! lam (1+ lam))) (values lam tortoise steps))
- find cycle and classify
(define (taxonomy n (steps 16) (maxvalue 140737488355328)) (define-values (cycle starter steps) (floyd sum-divisors n steps maxvalue)) (write n (cond (( = steps 0) 'non-terminating) (( = starter 0) 'terminating) ((and (= starter n) (= cycle 1)) 'perfect) ((and (= starter n) (= cycle 2)) 'amicable)
((= starter n) 'sociable )
((= cycle 1) 'aspiring ) (else 'cyclic)))
(aliquote n starter) )
- print sequence
(define (aliquote x0 (starter -1) (end -1 )(n 8))
(for ((i n)) (write x0) (set! x0 (sum-divisors x0)) #:break (and (= x0 end) (write x0)) (when (= x0 starter) (set! end starter))) (writeln ...))
</lang>
- Output:
<lang scheme> (lib 'math) (lib 'bigint)
(for-each taxonomy (range 1 13))
1 terminating 1 0 0 ... 2 terminating 2 1 0 0 ... 3 terminating 3 1 0 0 ... 4 terminating 4 3 1 0 0 ... 5 terminating 5 1 0 0 ... 6 perfect 6 6 6 ... 7 terminating 7 1 0 0 ... 8 terminating 8 7 1 0 0 ... 9 terminating 9 4 3 1 0 0 ... 10 terminating 10 8 7 1 0 0 ... 11 terminating 11 1 0 0 ... 12 terminating 12 16 15 9 4 3 1 0 0 ...
(for-each taxonomy '( 28 496 220 1184 12496 1264460 790 909 562 1064 1488 15355717786080))
28 perfect 28 28 28 ... 496 perfect 496 496 496 ... 220 amicable 220 284 220 284 220 ... 1184 amicable 1184 1210 1184 1210 1184 ... 12496 sociable 12496 14288 15472 14536 14264 12496 14288 15472 ... 1264460 sociable 1264460 1547860 1727636 1305184 1264460 1547860 1727636 1305184 1264460 ... 790 aspiring 790 650 652 496 496 ... 909 aspiring 909 417 143 25 6 6 ... 562 cyclic 562 284 220 284 ... 1064 cyclic 1064 1336 1184 1210 1184 ... 1488 non-terminating 1488 2480 3472 4464 8432 9424 10416 21328 ... 15355717786080 non-terminating 15355717786080 44534663601120 144940087464480 471714103310688 1130798979186912 2688948041357088 6050151708497568 13613157922639968 ...
(taxonomy 1000) ;; 1000 non-terminating after 16 steps 1000 non-terminating 1000 1340 1516 1144 1376 1396 1054 674 ...
(taxonomy 1000 32) ;; but terminating if we increase the number of steps 1000 terminating 1000 1340 1516 1144 1376 1396 1054 674 340 416 466 236 184 176 196 203 37 1 0 0 ... </lang>
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)]
end
defmodule Aliquot do
def sequence(n, maxlen\\16, maxterm\\140737488355328)
def sequence(0, _maxlen, _maxterm), do: "terminating"
def sequence(n, maxlen, maxterm) do
{msg, s} = sequence(n, maxlen, maxterm, [n])
{msg, Enum.reverse(s)}
end
defp sequence(n, maxlen, maxterm, s) when length(s) < maxlen and n < maxterm do
m = Proper.divisors(n) |> Enum.sum
cond do
m in s ->
case {m, List.last(s), hd(s)} do
{x,x,_} ->
case length(s) do
1 -> {"perfect", s}
2 -> {"amicable", s}
_ -> {"sociable of length #{length(s)}", s}
end
{x,_,x} -> {"aspiring", [m | s]}
_ -> {"cyclic back to #{m}", [m | s]}
end
m == 0 -> {"terminating", [0 | s]}
true -> sequence(m, maxlen, maxterm, [m | s])
end
end
defp sequence(_, _, _, s), do: {"non-terminating", s}
end
Enum.each(1..10, fn n ->
{msg, s} = Aliquot.sequence(n)
:io.fwrite("~7w:~21s: ~p~n", [n, msg, s])
end) IO.puts "" [11, 12, 28, 496, 220, 1184, 12496, 1264460, 790, 909, 562, 1064, 1488, 15355717786080] |> Enum.each(fn n ->
{msg, s} = Aliquot.sequence(n)
if n<10000000, do: :io.fwrite("~7w:~21s: ~p~n", [n, msg, s]),
else: :io.fwrite("~w: ~s: ~p~n", [n, msg, s])
end)</lang>
- Output:
1: terminating: [1,0]
2: terminating: [2,1,0]
3: terminating: [3,1,0]
4: terminating: [4,3,1,0]
5: terminating: [5,1,0]
6: perfect: [6]
7: terminating: [7,1,0]
8: terminating: [8,7,1,0]
9: terminating: [9,4,3,1,0]
10: terminating: [10,8,7,1,0]
11: terminating: [11,1,0]
12: terminating: [12,16,15,9,4,3,1,0]
28: perfect: [28]
496: perfect: [496]
220: amicable: [220,284]
1184: amicable: [1184,1210]
12496: sociable of length 5: [12496,14288,15472,14536,14264]
1264460: sociable of length 4: [1264460,1547860,1727636,1305184]
790: aspiring: [790,650,652,496,496]
909: aspiring: [909,417,143,25,6,6]
562: cyclic back to 284: [562,284,220,284]
1064: cyclic back to 1184: [1064,1336,1184,1210,1184]
1488: non-terminating: [1488,2480,3472,4464,8432,9424,10416,21328,
22320,55056,95728,96720,236592,459792,881392,
882384]
15355717786080: non-terminating: [15355717786080,44534663601120,
144940087464480]
Factor
For convenience, the term that caused termination is always included in the output sequence. <lang factor>USING: combinators combinators.short-circuit formatting kernel literals locals math math.functions math.primes.factors math.ranges namespaces pair-rocket sequences sets ; FROM: namespaces => set ; IN: rosetta-code.aliquot
SYMBOL: terms CONSTANT: 2^47 $[ 2 47 ^ ] CONSTANT: test-cases {
11 12 28 496 220 1184 12496 1264460 790 909 562 1064 1488 15355717786080
}
- next-term ( n -- m ) dup divisors sum swap - ;
- continue-aliquot? ( hs term -- hs term ? )
{
[ terms get 15 < ]
[ swap in? not ]
[ nip zero? not ]
[ nip 2^47 < ]
} 2&& ;
- next-aliquot ( hs term -- hs next-term term )
[ swap [ adjoin ] keep ] [ dup [ next-term ] dip ] bi terms inc ;
- aliquot ( k -- seq )
0 terms set HS{ } clone swap
[ continue-aliquot? ] [ next-aliquot ] produce
[ drop ] 2dip swap suffix ;
- non-terminating? ( seq -- ? )
{ [ length 15 > ] [ [ 2^47 > ] any? ] } 1|| ;
- classify ( seq -- classification-str )
{
[ seq non-terminating? ] => [ "non-terminating" ]
[ seq last zero? ] => [ "terminating" ]
[ seq length 2 = ] => [ "perfect" ]
[ seq length 3 = ] => [ "amicable" ]
[ seq first seq last = ] => [ "sociable" ]
[ seq 2 tail* first2 = ] => [ "aspiring" ]
[ "cyclic" ]
} cond ;
- .classify ( k -- )
dup aliquot [ classify ] keep "%14u: %15s: %[%d, %]\n" printf ;
- main ( -- )
10 [1,b] test-cases append [ .classify ] each ;
MAIN: main</lang>
- Output:
1: terminating: { 1, 0 }
2: terminating: { 2, 1, 0 }
3: terminating: { 3, 1, 0 }
4: terminating: { 4, 3, 1, 0 }
5: terminating: { 5, 1, 0 }
6: perfect: { 6, 6 }
7: terminating: { 7, 1, 0 }
8: terminating: { 8, 7, 1, 0 }
9: terminating: { 9, 4, 3, 1, 0 }
10: terminating: { 10, 8, 7, 1, 0 }
11: terminating: { 11, 1, 0 }
12: terminating: { 12, 16, 15, 9, 4, 3, 1, 0 }
28: perfect: { 28, 28 }
496: perfect: { 496, 496 }
220: amicable: { 220, 284, 220 }
1184: amicable: { 1184, 1210, 1184 }
12496: sociable: { 12496, 14288, 15472, 14536, 14264, 12496 }
1264460: sociable: { 1264460, 1547860, 1727636, 1305184, 1264460 }
790: aspiring: { 790, 650, 652, 496, 496 }
909: aspiring: { 909, 417, 143, 25, 6, 6 }
562: cyclic: { 562, 284, 220, 284 }
1064: cyclic: { 1064, 1336, 1184, 1210, 1184 }
1488: non-terminating: { 1488, 2480, 3472, 4464, 8432, 9424, 10416, 21328, 22320, 55056, 95728, 96720, 236592, 459792, 881392, 882384 }
15355717786080: non-terminating: { 15355717786080, 44534663601120, 144940087464480 }
Fortran
This is straightforward for Fortran compilers that allow 64-bit integers, as with INTEGER*8 - though one must have faith in the correct functioning of the computer for such large numbers....
Output:
After 1, terminates! 1
After 2, terminates! 2,1
After 2, terminates! 3,1
After 3, terminates! 4,3,1
After 2, terminates! 5,1
Perfect! 6
After 2, terminates! 7,1
After 3, terminates! 8,7,1
After 4, terminates! 9,4,3,1
After 4, terminates! 10,8,7,1
After 2, terminates! 11,1
After 7, terminates! 12,16,15,9,4,3,1
Perfect! 28
Perfect! 496
Amicable: 220,284
Amicable: 1184,1210
Sociable 5: 12496,14288,15472,14536,14264
Sociable 4: 1264460,1547860,1727636,1305184
Aspiring: 790,650,652,496
Aspiring: 909,417,143,25,6
Cyclic end 2, to 284: 562,284,220
Cyclic end 2, to 1184: 1064,1336,1184,1210
After 16, non-terminating? 1488,2480,3472,4464,8432,9424,10416,21328,22320,55056,95728,96720,
236592,459792,881392,882384
After 2, overflows! 15355717786080,44534663601120
Allowing more rope leads 1488 to overflow after the 83'rd value. Extending TOOBIG to 2**48 produces overflow from step 88, and the monster test value manages one more step, to 144940087464480 and confirmed via the Mathematica example. Because the task involves only a few numbers to test, there is not so much advantage to be gained by pre-calculating a set of sums of proper divisors, but it does mean that no special tests are needed for N = 1 in function SUMF.
A more flexible syntax (such as Algol's) would enable the double scan of the TRAIL array to be avoided, as in if TRAIL[I:=MinLoc(Abs(TRAIL(1:L) - SF))] = SF then... That is, find the first index of array TRAIL such that ABS(TRAIL(1:L) - SF) is minimal, save that index in I, then access that element of TRAIL and test if it is equal to SF. The INDEX function could be use to find the first match, except that it is defined only for character variables. Alternatively, use an explicit DO-loop to search for equality, thus not employing fancy syntax, and not having to wonder if the ANY function will stop on the first match rather than wastefully continue the testing for all array elements. The modern style in manual writing is to employ vaguely general talk about arrays and omit specific details.
<lang Fortran>
MODULE FACTORSTUFF !This protocol evades the need for multiple parameters, or COMMON, or one shapeless main line...
Concocted by R.N.McLean, MMXV. c INTEGER*4 I4LIMIT c PARAMETER (I4LIMIT = 2147483647)
INTEGER*8 TOOBIG !Some bounds.
PARAMETER (TOOBIG = 2**47) !Computer arithmetic is not with real numbers.
INTEGER LOTS !Nor is computer storage infinite.
PARAMETER (LOTS = 10000) !So there can't be all that many of these.
INTEGER*8 KNOWNSUM(LOTS) !If multiple references are expected, it is worthwhile calculating these.
CONTAINS !Assistants.
INTEGER*8 FUNCTION SUMF(N) !Sum of the proper divisors of N.
INTEGER*8 N !The number in question.
INTEGER*8 F,F2 !Candidate factor, and its square.
INTEGER*8 S,INC,BOOST !Assistants.
IF (N.LE.LOTS) THEN !If we're within reach,
SUMF = KNOWNSUM(N) !The result is to hand.
ELSE !Otherwise, some on-the-spot effort ensues.
Could use SUMF in place of S, but some compilers have been confused by such usage.
S = 1 !1 is always a factor of N, but N is deemed not proper.
F = 1 !Prepare a crude search for factors.
INC = 1 !One by plodding one.
IF (MOD(N,2) .EQ. 1) INC = 2!Ah, but an odd number cannot have an even number as a divisor.
1 F = F + INC !So half the time we can doubleplod.
F2 = F*F !Up to F2 < N rather than F < SQRT(N) and worries over inexact arithmetic.
IF (F2 .LT. N) THEN !F2 = N handled below.
IF (MOD(N,F) .EQ. 0) THEN !Does F divide N?
BOOST = F + N/F !Yes. The divisor and its counterpart.
IF (S .GT. TOOBIG - BOOST) GO TO 666 !Would their augmentation cause an overflow?
S = S + BOOST !No, so count in the two divisors just discovered.
END IF !So much for a divisor discovered.
GO TO 1 !Try for another.
END IF !So much for N = p*q style factors.
IF (F2 .EQ. N) THEN !Special case: N may be a perfect square, not necessarily of a prime number.
IF (S .GT. TOOBIG - F) GO TO 666 !It is. And it too might cause overflow.
S = S + F !But if not, count F once only.
END IF !All done.
SUMF = S !This is the result.
END IF !Whichever way obtained,
RETURN !Done.
Cannot calculate the sum, because it exceeds the INTEGER*8 limit.
666 SUMF = -666 !An expression of dismay that the caller will notice.
END FUNCTION SUMF !Alternatively, find the prime factors, and combine them...
SUBROUTINE PREPARESUMF !Initialise the KNOWNSUM array.
Convert the Sieve of Eratoshenes to have each slot contain the sum of the proper divisors of its slot number. Changes to instead count the number of factors, or prime factors, etc. would be simple enough.
INTEGER*8 F !A factor for numbers such as 2F, 3F, 4F, 5F, ...
KNOWNSUM(1) = 0 !Proper divisors of N do not include N.
KNOWNSUM(2:LOTS) = 1 !So, although 1 divides all N without remainder, 1 is excluded for itself.
DO F = 2,LOTS/2 !Step through all the possible divisors of numbers not exceeding LOTS.
FORALL(I = F + F:LOTS:F) KNOWNSUM(I) = KNOWNSUM(I) + F !And augment each corresponding slot.
END DO !Different divisors can hit the same slot. For instance, 6 by 2 and also by 3.
END SUBROUTINE PREPARESUMF !Could alternatively generate all products of prime numbers.
SUBROUTINE CLASSIFY(N) !Traipse along the SumF trail.
INTEGER*8 N !The starter.
INTEGER ROPE !The size of my memory is not so great..
PARAMETER(ROPE = 16) !Indeed, this is strictly limited.
INTEGER*8 TRAIL(ROPE) !But the numbers can be large.
INTEGER*8 SF !The working sum of proper divisors.
INTEGER I,L !Indices, merely.
CHARACTER*28 THIS !A perfect scratchpad for remarks.
L = 1 !Every journey starts with its first step.
TRAIL(1) = N !Which is this.
SF = N !Syncopation.
10 SF = SUMF(SF) !Step onwards.
IF (SF .LT. 0) THEN !Trouble?
WRITE (THIS,11) L,"overflows!" !Yes. Too big a number.
11 FORMAT ("After ",I0,", ",A) !Describe the situation.
CALL REPORT(ADJUSTR(THIS)) !And give the report.
ELSE IF (SF .EQ. 0) THEN !Otherwise, a finish?
WRITE (THIS,11) L,"terminates!" !Yay!
CALL REPORT(ADJUSTR(THIS)) !This sequence is finished.
ELSE IF (ANY(TRAIL(1:L) .EQ. SF)) THEN !Otherwise, is there an echo somewhere?
IF (L .EQ. 1) THEN !Yes!
CALL REPORT("Perfect!") !Are we at the start?
ELSE IF (L .EQ. 2) THEN !Or perhaps not far along.
CALL REPORT("Amicable:") !These are held special.
ELSE !Otherwise, we've wandered further along.
I = MINLOC(ABS(TRAIL(1:L) - SF),DIM=1) !Damnit, re-scan the array to finger the first matching element.
IF (I .EQ. 1) THEN !If all the way back to the start,
WRITE (THIS,12) L !Then there are this many elements in the sociable ring.
12 FORMAT ("Sociable ",I0,":") !Computers are good at counting.
CALL REPORT(ADJUSTR(THIS)) !So, perform an added service.
ELSE IF (I .EQ. L) THEN !Perhaps we've hit a perfect number!
CALL REPORT("Aspiring:") !A cycle of length one.
ELSE !But otherwise,
WRITE (THIS,13) L - I + 1,SF !A longer cycle. Amicable, or sociable.
13 FORMAT ("Cyclic end ",I0,", to ",I0,":") !Name the flashback value too.
CALL REPORT(ADJUSTR(THIS)) !Thus.
END IF !So much for cycles.
END IF !So much for finding an echo.
ELSE !Otherwise, nothing special has happened.
IF (L .GE. ROPE) THEN !So, how long is a piece of string?
WRITE (THIS,11) L,"non-terminating?" !Not long enough!
CALL REPORT(ADJUSTR(THIS)) !So we give up.
ELSE !But if there is more scope,
L = L + 1 !Advance one more step.
TRAIL(L) = SF !Save the latest result.
GO TO 10 !And try for the next.
END IF !So much for continuing.
END IF !So much for the classification.
RETURN !Finished.
CONTAINS !Not quite.
SUBROUTINE REPORT(WHAT) !There is this service routine.
CHARACTER*(*) WHAT !Whatever the length of the text, the FORMAT's A28 shows 28 characters, right-aligned.
WRITE (6,1) WHAT,TRAIL(1:L)!Mysteriously, a fresh line after every twelve elements.
1 FORMAT (A28,1X,12(I0:",")) !And obviously, the : signifies "do not print what follows unless there is another number to go.
END SUBROUTINE REPORT !That was easy.
END SUBROUTINE CLASSIFY !Enough.
END MODULE FACTORSTUFF !Enough assistants.
PROGRAM CLASSIFYTHEM !Report on the nature of the sequence N, Sumf(N), Sumf(Sumf(N)), etc.
USE FACTORSTUFF !This should help.
INTEGER*8 I,N !Steppers.
INTEGER*8 THIS(14) !A testing collection.
DATA THIS/11,12,28,496,220,1184,12496,1264460,790,909, !Old-style continuation character in column six.
1 562,1064,1488,15355717786080/ !Monster value far exceeds the INTEGER*4 limit
CALL PREPARESUMF !Prepare for 1:LOTS, even though this test run will use only a few.
DO I = 1,10 !As specified, the first ten integers.
CALL CLASSIFY(I)
END DO
DO I = 1,SIZE(THIS) !Now for the specified list.
CALL CLASSIFY(THIS(I))
END DO
END !Done.
</lang>
FreeBASIC
<lang freebasic>function raiseTo( bas as ulongint, power as ulongint ) as ulongint
dim as ulongint result = 1, i
for i = 1 to power
result*=bas
next i
return result
end function
function properDivisorSum( n as ulongint ) as ulongint dim as ulongint prod = 1, temp = n, i = 3, count = 0 while n mod 2 = 0 count += 1 n /= 2 wend if count<>0 then prod *= (raiseTo(2,count + 1) - 1) while i*i <= n count = 0 while n mod i = 0 count += 1 n /= i wend if count = 1 then prod *= (i+1) elseif count > 1 then prod *= ((raiseTo(i,count + 1) - 1)/(i-1)) end if
i += 2
wend if n>2 then prod *= (n+1) return prod - temp end function
sub printSeries( arr() as ulongint ptr, size as integer, ty as string) dim as integer i dim as string outstr = "Integer: "+str(arr(0))+", Type: "+ty+", Series: " for i=0 to size-2 outstr = outstr + str(arr(i))+", "
next i
outstr = outstr + str(arr(i)) print outstr end sub
sub aliquotClassifier(n as ulongint) dim as ulongint arr(0 to 15) dim as integer i, j dim as string ty = "Sociable"
arr(0) = n for i = 1 to 15
arr(i) = properDivisorSum(arr(i-1)) if arr(i)=0 orelse arr(i)=n orelse (arr(i) = arr(i-1) and arr(i)<>n) then if arr(i) = 0 then ty = "Terminating" elseif arr(i) = n and i = 1 then ty = "Perfect" elseif arr(i) = n and i = 2 then ty = "Amicable" elseif arr(i) = arr(i-1) and arr(i)<>n then ty = "Aspiring" end if printSeries(arr(),i+1,ty) return end if for j = 1 to i-1 if arr(j) = arr(i) then printSeries(arr(),i+1,"Cyclic") return end if next j next i printSeries(arr(),i+1,"Non-Terminating") end sub
dim as ulongint nums(0 to 22) = {_
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 28, 496, 220, 1184,_ 12496, 1264460, 790, 909, 562, 1064, 1488}
for n as ubyte = 0 to 22
aliquotClassifier(nums(n))
next n</lang>
Go
<lang go>package main
import (
"fmt" "math" "strings"
)
const threshold = uint64(1) << 47
func indexOf(s []uint64, search uint64) int {
for i, e := range s {
if e == search {
return i
}
}
return -1
}
func contains(s []uint64, search uint64) bool {
return indexOf(s, search) > -1
}
func maxOf(i1, i2 int) int {
if i1 > i2 {
return i1
}
return i2
}
func sumProperDivisors(n uint64) uint64 {
if n < 2 {
return 0
}
sqrt := uint64(math.Sqrt(float64(n)))
sum := uint64(1)
for i := uint64(2); i <= sqrt; i++ {
if n % i != 0 {
continue
}
sum += i + n / i
}
if sqrt * sqrt == n {
sum -= sqrt
}
return sum
}
func classifySequence(k uint64) ([]uint64, string) {
if k == 0 {
panic("Argument must be positive.")
}
last := k
var seq []uint64
seq = append(seq, k)
for {
last = sumProperDivisors(last)
seq = append(seq, last)
n := len(seq)
aliquot := ""
switch {
case last == 0:
aliquot = "Terminating"
case n == 2 && last == k:
aliquot = "Perfect"
case n == 3 && last == k:
aliquot = "Amicable"
case n >= 4 && last == k:
aliquot = fmt.Sprintf("Sociable[%d]", n - 1)
case last == seq[n - 2]:
aliquot = "Aspiring"
case contains(seq[1 : maxOf(1, n - 2)], last):
aliquot = fmt.Sprintf("Cyclic[%d]", n - 1 - indexOf(seq[:], last))
case n == 16 || last > threshold:
aliquot = "Non-Terminating"
}
if aliquot != "" {
return seq, aliquot
}
}
}
func joinWithCommas(seq []uint64) string {
res := fmt.Sprint(seq) res = strings.Replace(res, " ", ", ", -1) return res
}
func main() {
fmt.Println("Aliquot classifications - periods for Sociable/Cyclic in square brackets:\n")
for k := uint64(1); k <= 10; k++ {
seq, aliquot := classifySequence(k)
fmt.Printf("%2d: %-15s %s\n", k, aliquot, joinWithCommas(seq))
}
fmt.Println()
s := []uint64{
11, 12, 28, 496, 220, 1184, 12496, 1264460, 790, 909, 562, 1064, 1488,
}
for _, k := range s {
seq, aliquot := classifySequence(k)
fmt.Printf("%7d: %-15s %s\n", k, aliquot, joinWithCommas(seq))
}
fmt.Println()
k := uint64(15355717786080)
seq, aliquot := classifySequence(k)
fmt.Printf("%d: %-15s %s\n", k, aliquot, joinWithCommas(seq))
}</lang>
- Output:
Aliquot classifications - periods for Sociable/Cyclic in square brackets:
1: Terminating [1, 0]
2: Terminating [2, 1, 0]
3: Terminating [3, 1, 0]
4: Terminating [4, 3, 1, 0]
5: Terminating [5, 1, 0]
6: Perfect [6, 6]
7: Terminating [7, 1, 0]
8: Terminating [8, 7, 1, 0]
9: Terminating [9, 4, 3, 1, 0]
10: Terminating [10, 8, 7, 1, 0]
11: Terminating [11, 1, 0]
12: Terminating [12, 16, 15, 9, 4, 3, 1, 0]
28: Perfect [28, 28]
496: Perfect [496, 496]
220: Amicable [220, 284, 220]
1184: Amicable [1184, 1210, 1184]
12496: Sociable[5] [12496, 14288, 15472, 14536, 14264, 12496]
1264460: Sociable[4] [1264460, 1547860, 1727636, 1305184, 1264460]
790: Aspiring [790, 650, 652, 496, 496]
909: Aspiring [909, 417, 143, 25, 6, 6]
562: Cyclic[2] [562, 284, 220, 284]
1064: Cyclic[2] [1064, 1336, 1184, 1210, 1184]
1488: Non-Terminating [1488, 2480, 3472, 4464, 8432, 9424, 10416, 21328, 22320, 55056, 95728, 96720, 236592, 459792, 881392, 882384]
15355717786080: Non-Terminating [15355717786080, 44534663601120, 144940087464480]
Haskell
<lang Haskell>divisors :: (Integral a) => a -> [a] divisors n = filter ((0 ==) . (n `mod`)) [1 .. (n `div` 2)]
data Class
= Terminating | Perfect | Amicable | Sociable | Aspiring | Cyclic | Nonterminating deriving (Show)
aliquot :: (Integral a) => a -> [a] aliquot 0 = [0] aliquot n = n : (aliquot $ sum $ divisors n)
classify :: (Num a, Eq a) => [a] -> Class classify [] = Nonterminating classify [0] = Terminating classify [_] = Nonterminating classify [a,b]
| a == b = Perfect | b == 0 = Terminating | otherwise = Nonterminating
classify x@(a:b:c:_)
| a == b = Perfect
| a == c = Amicable
| a `elem` (drop 1 x) = Sociable
| otherwise =
case classify (drop 1 x) of
Perfect -> Aspiring
Amicable -> Cyclic
Sociable -> Cyclic
d -> d
main :: IO () main = do
let cls n = let ali = take 16 $ aliquot n in (classify ali, ali) mapM_ (print . cls) $ [1..10] ++ [11, 12, 28, 496, 220, 1184, 12496, 1264460, 790, 909, 562, 1064, 1488]</lang>
- Output:
(Terminating,[1,0]) (Terminating,[2,1,0]) (Terminating,[3,1,0]) (Terminating,[4,3,1,0]) (Terminating,[5,1,0]) (Perfect,[6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6]) (Terminating,[7,1,0]) (Terminating,[8,7,1,0]) (Terminating,[9,4,3,1,0]) (Terminating,[10,8,7,1,0]) (Terminating,[11,1,0]) (Terminating,[12,16,15,9,4,3,1,0]) (Perfect,[28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28]) (Perfect,[496,496,496,496,496,496,496,496,496,496,496,496,496,496,496,496]) (Amicable,[220,284,220,284,220,284,220,284,220,284,220,284,220,284,220,284]) (Amicable,[1184,1210,1184,1210,1184,1210,1184,1210,1184,1210,1184,1210,1184,1210,1184,1210]) (Sociable,[12496,14288,15472,14536,14264,12496,14288,15472,14536,14264,12496,14288,15472,14536,14264,12496]) (Sociable,[1264460,1547860,1727636,1305184,1264460,1547860,1727636,1305184,1264460,1547860,1727636,1305184,1264460,1547860,1727636,1305184]) (Aspiring,[790,650,652,496,496,496,496,496,496,496,496,496,496,496,496,496]) (Aspiring,[909,417,143,25,6,6,6,6,6,6,6,6,6,6,6,6]) (Cyclic,[562,284,220,284,220,284,220,284,220,284,220,284,220,284,220,284]) (Cyclic,[1064,1336,1184,1210,1184,1210,1184,1210,1184,1210,1184,1210,1184,1210,1184,1210]) (Nonterminating,[1488,2480,3472,4464,8432,9424,10416,21328,22320,55056,95728,96720,236592,459792,881392,882384])
J
Implementation: <lang J>proper_divisors=: [: */@>@}:@,@{ [: (^ i.@>:)&.>/ 2 p: x: aliquot=: +/@proper_divisors ::0: rc_aliquot_sequence=: aliquot^:(i.16)&> rc_classify=: 3 :0
if. 16 ~:# y do. ' invalid '
elseif. 6 > {: y do. ' terminate '
elseif. (+./y>2^47) +. 16 = #~.y do. ' non-terminating'
elseif. 1=#~. y do. ' perfect '
elseif. 8= st=. {.#/.~ y do. ' amicable '
elseif. 1 < st do. ' sociable '
elseif. =/_2{. y do. ' aspiring '
elseif. 1 do. ' cyclic '
end.
) rc_display_aliquot_sequence=: (rc_classify,' ',":)@:rc_aliquot_sequence</lang>
Task example: <lang J> rc_display_aliquot_sequence&> >: i.10
terminate 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 terminate 2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 terminate 3 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 terminate 4 3 1 0 0 0 0 0 0 0 0 0 0 0 0 0 terminate 5 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 perfect 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 terminate 7 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 terminate 8 7 1 0 0 0 0 0 0 0 0 0 0 0 0 0 terminate 9 4 3 1 0 0 0 0 0 0 0 0 0 0 0 0 terminate 10 8 7 1 0 0 0 0 0 0 0 0 0 0 0 0
rc_display_aliquot_sequence&>11, 12, 28, 496, 220, 1184, 12496, 1264460, 790, 909, 562, 1064, 1488, 15355717786080x terminate 11 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 terminate 12 16 15 9 4 3 1 0 0 0 0 0 0 0 0 0 perfect 28 28 28 28 28 28 28 28 28 28 28 28 28 28 28 28 perfect 496 496 496 496 496 496 496 496 496 496 496 496 496 496 496 496 amicable 220 284 220 284 220 284 220 284 220 284 220 284 220 284 220 284 amicable 1184 1210 1184 1210 1184 1210 1184 1210 1184 1210 1184 1210 1184 1210 1184 1210 sociable 12496 14288 15472 14536 14264 12496 14288 15472 14536 14264 12496 14288 15472 14536 14264 12496 sociable 1264460 1547860 1727636 1305184 1264460 1547860 1727636 1305184 1264460 1547860 1727636 1305184 1264460 1547860 1727636 1305184 aspiring 790 650 652 496 496 496 496 496 496 496 496 496 496 496 496 496 aspiring 909 417 143 25 6 6 6 6 6 6 6 6 6 6 6 6 cyclic 562 284 220 284 220 284 220 284 220 284 220 284 220 284 220 284 cyclic 1064 1336 1184 1210 1184 1210 1184 1210 1184 1210 1184 1210 1184 1210 1184 1210 non-terminating 1488 2480 3472 4464 8432 9424 10416 21328 22320 55056 95728 96720 236592 459792 881392 882384 non-terminating 15355717786080 44534663601120 144940087464480 471714103310688 1130798979186912 2688948041357088 6050151708497568 13613157922639968 35513546724070632 74727605255142168 162658586225561832 353930992506879768 642678347124409032 1125102611548462968 1977286128289819992 3415126495450394808</lang>
Java
<lang java>import java.util.ArrayList; import java.util.Arrays; import java.util.List; import java.util.stream.LongStream;
public class AliquotSequenceClassifications {
private static Long properDivsSum(long n) {
return LongStream.rangeClosed(1, (n + 1) / 2).filter(i -> n % i == 0 && n != i).sum();
}
static boolean aliquot(long n, int maxLen, long maxTerm) {
List<Long> s = new ArrayList<>(maxLen);
s.add(n);
long newN = n;
while (s.size() <= maxLen && newN < maxTerm) {
newN = properDivsSum(s.get(s.size() - 1));
if (s.contains(newN)) {
if (s.get(0) == newN) {
switch (s.size()) {
case 1:
return report("Perfect", s);
case 2:
return report("Amicable", s);
default:
return report("Sociable of length " + s.size(), s);
}
} else if (s.get(s.size() - 1) == newN) {
return report("Aspiring", s);
} else
return report("Cyclic back to " + newN, s);
} else {
s.add(newN);
if (newN == 0)
return report("Terminating", s);
}
}
return report("Non-terminating", s);
}
static boolean report(String msg, List<Long> result) {
System.out.println(msg + ": " + result);
return false;
}
public static void main(String[] args) {
long[] arr = {
11, 12, 28, 496, 220, 1184, 12496, 1264460,
790, 909, 562, 1064, 1488};
LongStream.rangeClosed(1, 10).forEach(n -> aliquot(n, 16, 1L << 47));
System.out.println();
Arrays.stream(arr).forEach(n -> aliquot(n, 16, 1L << 47));
}
}</lang>
Terminating: [1, 0] Terminating: [2, 1, 0] Terminating: [3, 1, 0] Terminating: [4, 3, 1, 0] Terminating: [5, 1, 0] Perfect: [6] Terminating: [7, 1, 0] Terminating: [8, 7, 1, 0] Terminating: [9, 4, 3, 1, 0] Terminating: [10, 8, 7, 1, 0] Terminating: [11, 1, 0] Terminating: [12, 16, 15, 9, 4, 3, 1, 0] Perfect: [28] Perfect: [496] Amicable: [220, 284] Amicable: [1184, 1210] Sociable of length 5: [12496, 14288, 15472, 14536, 14264] Sociable of length 4: [1264460, 1547860, 1727636, 1305184] Aspiring: [790, 650, 652, 496] Aspiring: [909, 417, 143, 25, 6] Cyclic back to 284: [562, 284, 220] Cyclic back to 1184: [1064, 1336, 1184, 1210] Non-terminating: [1488, 2480, 3472, 4464, 8432, 9424, 10416, 21328, 22320, 55056, 95728, 96720, 236592, 459792, 881392, 882384, 1474608]
jq
<lang jq># "until" is available in more recent versions of jq
- than jq 1.4
def until(cond; next):
def _until: if cond then . else (next|_until) end; _until;
- 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;
- sum of proper divisors, or 0
def pdsum:
[proper_divisors] | add // 0;
- input is n
- maxlen defaults to 16;
- maxterm defaults to 2^47
def aliquot(maxlen; maxterm):
(maxlen // 15) as $maxlen
| (maxterm // 40737488355328) as $maxterm
| if . == 0 then "terminating at 0"
else
# [s, slen, new] = [[n], 1, n]
[ [.], 1, .]
| until( type == "string" or .[1] > $maxlen or .[2] > $maxterm;
.[0] as $s | .[1] as $slen
| ($s | .[length-1] | pdsum) as $new
| if ($s|index($new)) then
if $s[0] == $new then
if $slen == 1 then "perfect \($s)"
elif $slen == 2 then "amicable: \($s)"
else "sociable of length \($slen): \($s)"
end
elif ($s | .[length-1]) == $new then "aspiring: \($s)"
else "cyclic back to \($new): \($s)"
end
elif $new == 0 then "terminating: \($s + [0])"
else [ ($s + [$new]), ($slen + 1), $new ]
end )
| if type == "string" then . else "non-terminating: \(.[0])" end
end;
def task:
def pp: "\(.): \(aliquot(null;null))";
(range(1; 11) | pp),
"",
((11, 12, 28, 496, 220, 1184, 12496, 1264460,
790, 909, 562, 1064, 1488, 15355717786080) | pp);
task</lang>
- Output:
<lang sh>$ jq -n -r -f aliquot.jq 1: terminating: [1,0] 2: terminating: [2,1,0] 3: terminating: [3,1,0] 4: terminating: [4,3,1,0] 5: terminating: [5,1,0] 6: perfect [6] 7: terminating: [7,1,0] 8: terminating: [8,7,1,0] 9: terminating: [9,4,3,1,0] 10: terminating: [10,8,7,1,0]
11: terminating: [11,1,0] 12: terminating: [12,16,15,9,4,3,1,0] 28: perfect [28] 496: perfect [496] 220: amicable: [220,284] 1184: amicable: [1184,1210] 12496: sociable of length 5: [12496,14288,15472,14536,14264] 1264460: sociable of length 4: [1264460,1547860,1727636,1305184] 790: aspiring: [790,650,652,496] 909: aspiring: [909,417,143,25,6] 562: cyclic back to 284: [562,284,220] 1064: cyclic back to 1184: [1064,1336,1184,1210] 1488: non-terminating: [1488,2480,3472,4464,8432,9424,10416,21328,22320,55056,95728,96720,236592,459792,881392,882384] 15355717786080: non-terminating: [15355717786080,44534663601120]</lang>
Julia
Core Function <lang Julia> function aliquotclassifier{T<:Integer}(n::T)
a = T[n]
b = divisorsum(a[end])
len = 1
while len < 17 && !(b in a) && 0 < b && b < 2^47+1
push!(a, b)
b = divisorsum(a[end])
len += 1
end
if b in a
1 < len || return ("Perfect", a)
if b == a[1]
2 < len || return ("Amicable", a)
return ("Sociable", a)
elseif b == a[end]
return ("Aspiring", a)
else
return ("Cyclic", push!(a, b))
end
end
push!(a, b)
b != 0 || return ("Terminating", a)
return ("Non-terminating", a)
end </lang>
Supporting Functions <lang Julia> function pcontrib{T<:Integer}(p::T, a::T)
n = one(T)
pcon = one(T)
for i in 1:a
n *= p
pcon += n
end
return pcon
end
function divisorsum{T<:Integer}(n::T)
dsum = one(T)
for (p, a) in factor(n)
dsum *= pcontrib(p, a)
end
dsum -= n
end </lang>
Main <lang Julia>using Printf
println("Classification Tests:") tests = [1:12, 28, 496, 220, 1184, 12496, 1264460, 790, 909, 562, 1064, 1488] for i in tests
(class, a) = aliquotclassifier(i)
println(@sprintf("%8d => ", i), @sprintf("%16s, ", class), a)
end </lang>
- Output:
Classification Tests:
1 => Terminating, [1,0]
2 => Terminating, [2,1,0]
3 => Terminating, [3,1,0]
4 => Terminating, [4,3,1,0]
5 => Terminating, [5,1,0]
6 => Perfect, [6]
7 => Terminating, [7,1,0]
8 => Terminating, [8,7,1,0]
9 => Terminating, [9,4,3,1,0]
10 => Terminating, [10,8,7,1,0]
11 => Terminating, [11,1,0]
12 => Terminating, [12,16,15,9,4,3,1,0]
28 => Perfect, [28]
496 => Perfect, [496]
220 => Amicable, [220,284]
1184 => Amicable, [1184,1210]
12496 => Sociable, [12496,14288,15472,14536,14264]
1264460 => Sociable, [1264460,1547860,1727636,1305184]
790 => Aspiring, [790,650,652,496]
909 => Aspiring, [909,417,143,25,6]
562 => Cyclic, [562,284,220,284]
1064 => Cyclic, [1064,1336,1184,1210,1184]
1488 => Non-terminating, [1488,2480,3472,4464,8432,9424,10416,21328,22320,55056,95728,96720,236592,459792,881392,882384,1474608,2461648]
Kotlin
<lang scala>// version 1.1.3
data class Classification(val sequence: List<Long>, val aliquot: String)
const val THRESHOLD = 1L shl 47
fun sumProperDivisors(n: Long): Long {
if (n < 2L) return 0L
val sqrt = Math.sqrt(n.toDouble()).toLong()
var sum = 1L + (2L..sqrt)
.filter { n % it == 0L }
.map { it + n / it }
.sum()
if (sqrt * sqrt == n) sum -= sqrt
return sum
}
fun classifySequence(k: Long): Classification {
require(k > 0)
var last = k
val seq = mutableListOf(k)
while (true) {
last = sumProperDivisors(last)
seq.add(last)
val n = seq.size
val aliquot = when {
last == 0L -> "Terminating"
n == 2 && last == k -> "Perfect"
n == 3 && last == k -> "Amicable"
n >= 4 && last == k -> "Sociable[${n - 1}]"
last == seq[n - 2] -> "Aspiring"
last in seq.slice(1..n - 3) -> "Cyclic[${n - 1 - seq.indexOf(last)}]"
n == 16 || last > THRESHOLD -> "Non-Terminating"
else -> ""
}
if (aliquot != "") return Classification(seq, aliquot)
}
}
fun main(args: Array<String>) {
println("Aliqot classifications - periods for Sociable/Cyclic in square brackets:\n")
for (k in 1L..10) {
val (seq, aliquot) = classifySequence(k)
println("${"%2d".format(k)}: ${aliquot.padEnd(15)} $seq")
}
val la = longArrayOf(
11, 12, 28, 496, 220, 1184, 12496, 1264460, 790, 909, 562, 1064, 1488
)
println()
for (k in la) {
val (seq, aliquot) = classifySequence(k)
println("${"%7d".format(k)}: ${aliquot.padEnd(15)} $seq")
}
println()
val k = 15355717786080L
val (seq, aliquot) = classifySequence(k)
println("$k: ${aliquot.padEnd(15)} $seq")
}</lang>
- Output:
Aliqot classifications - periods for Sociable/Cyclic in square brackets:
1: Terminating [1, 0]
2: Terminating [2, 1, 0]
3: Terminating [3, 1, 0]
4: Terminating [4, 3, 1, 0]
5: Terminating [5, 1, 0]
6: Perfect [6, 6]
7: Terminating [7, 1, 0]
8: Terminating [8, 7, 1, 0]
9: Terminating [9, 4, 3, 1, 0]
10: Terminating [10, 8, 7, 1, 0]
11: Terminating [11, 1, 0]
12: Terminating [12, 16, 15, 9, 4, 3, 1, 0]
28: Perfect [28, 28]
496: Perfect [496, 496]
220: Amicable [220, 284, 220]
1184: Amicable [1184, 1210, 1184]
12496: Sociable[5] [12496, 14288, 15472, 14536, 14264, 12496]
1264460: Sociable[4] [1264460, 1547860, 1727636, 1305184, 1264460]
790: Aspiring [790, 650, 652, 496, 496]
909: Aspiring [909, 417, 143, 25, 6, 6]
562: Cyclic[2] [562, 284, 220, 284]
1064: Cyclic[2] [1064, 1336, 1184, 1210, 1184]
1488: Non-Terminating [1488, 2480, 3472, 4464, 8432, 9424, 10416, 21328, 22320, 55056, 95728, 96720, 236592, 459792, 881392, 882384]
15355717786080: Non-Terminating [15355717786080, 44534663601120, 144940087464480]
Liberty BASIC
Based on my analysis of integers up to 10,000 I have revised the criteria for non-termination as follows: 52 elements, or 11 consecutive increases of elements, or an element greater than 30 million. This is not a perfect algorithm, but seems to me to be a reasonable compromise between accuracy and speed. I'll stay away from the really large numbers - at least for now.
Of integers below 10,000--
4004 is the longest non-terminating integer by the revised criteria. The elements range from a minimum of 2,440 to a maximum of 302,666. I suspect that if the sequence were run out far enough, it would terminate in some fashion.
4344 has the longest terminating sequence.
6672 has the longest aspiring sequence.
6420 has the longest cyclic sequence.
8128 is the largest perfect integer.
There are no sociable sequences. <lang lb> print "ROSETTA CODE - Aliquot sequence classifications" [Start] input "Enter an integer: "; K K=abs(int(K)): if K=0 then goto [Quit] call PrintAS K goto [Start]
[Quit] print "Program complete." end
sub PrintAS K
Length=52
dim Aseq(Length)
n=K: class=0
for element=2 to Length
Aseq(element)=PDtotal(n)
print Aseq(element); " ";
select case
case Aseq(element)=0
print " terminating": class=1: exit for
case Aseq(element)=K and element=2
print " perfect": class=2: exit for
case Aseq(element)=K and element=3
print " amicable": class=3: exit for
case Aseq(element)=K and element>3
print " sociable": class=4: exit for
case Aseq(element)<>K and Aseq(element-1)=Aseq(element)
print " aspiring": class=5: exit for
case Aseq(element)<>K and Aseq(element-2)= Aseq(element)
print " cyclic": class=6: exit for
end select
n=Aseq(element)
if n>priorn then priorn=n: inc=inc+1 else inc=0: priorn=0
if inc=11 or n>30000000 then exit for
next element
if class=0 then print " non-terminating"
end sub
function PDtotal(n)
for y=2 to n
if (n mod y)=0 then PDtotal=PDtotal+(n/y)
next
end function </lang>
- Output:
ROSETTA CODE - Aliquot sequence classifications Enter an integer: 1 0 terminating Enter an integer: 2 1 0 terminating Enter an integer: 3 1 0 terminating Enter an integer: 4 3 1 0 terminating Enter an integer: 5 1 0 terminating Enter an integer: 6 6 perfect Enter an integer: 7 1 0 terminating Enter an integer: 8 7 1 0 terminating Enter an integer: 9 4 3 1 0 terminating Enter an integer: 10 8 7 1 0 terminating Enter an integer: 11 1 0 terminating Enter an integer: 12 16 15 9 4 3 1 0 terminating Enter an integer: 28 28 perfect Enter an integer: 496 496 perfect Enter an integer: 220 284 220 amicable Enter an integer: 1184 1210 1184 amicable Enter an integer: 12496 14288 15472 14536 14264 12496 sociable Enter an integer: 1264460 1547860 1727636 1305184 1264460 sociable Enter an integer: 790 650 652 496 496 aspiring Enter an integer: 909 417 143 25 6 6 aspiring Enter an integer: 562 284 220 284 cyclic Enter an integer: 1064 1336 1184 1210 1184 cyclic Enter an integer: 1488 2480 3472 4464 8432 9424 10416 21328 22320 55056 95728 96720 non-terminating - - - - - - - - - - - - Enter an integer: 4004 5404 5460 13356 25956 49756 49812 83244 138964 144326 127978 67322 36250 34040 48040 60140 71572 58208 64264 60836 47692 35776 42456 69144 110376 244824 373356 594884 446170 356954 219706 118874 88720 117740 174916 174972 291844 302666 2564 38 217322 185014 92510 95626 49274 25894 17198 8602 6950 6070 4874 2440 3140 non-terminating Enter an integer: 4344 6576 10536 15864 23856 47568 75440 112048 111152 104236 105428 79078 45842 22924 20924 15700 18586 9296 11536 14256 30756 47868 63852 94404 125900 147520 204524 153400 237200 333634 238334 121306 62438 31222 16514 9406 4706 2938 1850 1684 1 270 1034 694 350 394 200 265 59 1 0 terminating Enter an integer: 6672 10688 10648 11312 13984 16256 16384 16383 6145 1235 445 95 25 6 6 aspiring Enter an integer: 6420 11724 15660 34740 71184 112832 121864 106646 53326 45458 37486 18746 16198 14042 11878 5942 2974 1490 1210 1184 1210 cyclic Enter an integer: 8128 8128 perfect Enter an integer: Program complete.
Mathematica / Wolfram Language
<lang Mathematica>seq[n_] :=
NestList[If[# == 0, 0,
DivisorSum[#, # &, Function[div, div != #]]] &, n, 16];
class[seq_] :=
Which[Length[seq] < 2, "Non-terminating", MemberQ[seq, 0], "Terminating", seq1 == seq2, "Perfect", Length[seq] > 2 && seq1 == seq3, "Amicable", Length[seq] > 3 && MemberQ[seq4 ;;, seq1], "Sociable", MatchQ[class[Rest[seq]], "Perfect" | "Aspiring"], "Aspiring", MatchQ[class[Rest[seq]], "Amicable" | "Sociable" | "Cyclic"], "Cyclic", True, "Non-terminating"];
notate[seq_] :=
Which[seq == {}, {},
MemberQ[Rest[seq],
seq1], {Prepend[TakeWhile[Rest[seq], # != seq1 &],
seq1]}, True, Prepend[notate[Rest[seq]], seq1]];
Print[{#, class[seq[#]], notate[seq[#]] /. {0} -> 0}] & /@ {1, 2, 3, 4, 5, 6, 7,
8, 9, 10, 11, 12, 28, 496, 220, 1184, 12496, 1264460, 790, 909, 562, 1064, 1488, 15355717786080};</lang>
- Output:
{1, Terminating, {1, 0}}
{2, Terminating, {2, 1, 0}}
{3, Terminating, {3, 1, 0}}
{4, Terminating, {4, 3, 1, 0}}
{5, Terminating, {5, 1, 0}}
{6, Perfect, {{6}}}
{7, Terminating, {7, 1, 0}}
{8, Terminating, {8, 7, 1, 0}}
{9, Terminating, {9, 4, 3, 1, 0}}
{10, Terminating, {10, 8, 7, 1, 0}}
{11, Terminating, {11, 1, 0}}
{12, Terminating, {12, 16, 15, 9, 4, 3, 1, 0}}
{28, Perfect, {{28}}}
{496, Perfect, {{496}}}
{220, Amicable, {{220, 284}}}
{1184, Amicable, {{1184, 1210}}}
{12496, Sociable, {{12496, 14288, 15472, 14536, 14264}}}
{1264460, Sociable, {{1264460, 1547860, 1727636, 1305184}}}
{790, Aspiring, {790, 650, 652, {496}}}
{909, Aspiring, {909, 417, 143, 25, {6}}}
{562, Cyclic, {562, {284, 220}}}
{1064, Cyclic, {1064, 1336, {1184, 1210}}}
{1488, Non-terminating, {1488, 2480, 3472, 4464, 8432, 9424, 10416, 21328, 22320, 55056, 95728, 96720, 236592, 459792, 881392, 882384, 1474608}}
{15355717786080, Non-terminating, {15355717786080, 44534663601120, 144940087464480, 471714103310688, 1130798979186912, 2688948041357088, 6050151708497568, 13613157922639968, 35513546724070632, 74727605255142168, 162658586225561832, 353930992506879768, 642678347124409032, 1125102611548462968, 1977286128289819992, 3415126495450394808, 7156435369823219592}}
Nim
<lang Nim> import math import strformat from strutils import addSep import times
type
# Classification categories. Category = enum Unknown Terminating = "terminating" Perfect = "perfect" Amicable = "amicable" Sociable = "sociable" Aspiring = "aspiring" Cyclic = "cyclic" NonTerminating = "non-terminating"
# Aliquot sequence. AliquotSeq = seq[int]
const Limit = 2^47 # Limit beyond which the category is considered to be "NonTerminating".
- ---------------------------------------------------------------------------------------------------
proc sumProperDivisors(n: int): int =
## Compute the sum of proper divisors.*
if n == 1: return 0
result = 1
for d in 2..sqrt(n.toFloat).int:
if n mod d == 0:
inc result, d
if n div d != d:
inc result, n div d
- ---------------------------------------------------------------------------------------------------
iterator aliquotSeq(n: int): int =
## Yield the elements of the aliquot sequence of "n". ## Stopped if the current value is null or equal to "n".
var k = n while true: k = sumProperDivisors(k) yield k
- ---------------------------------------------------------------------------------------------------
proc `$`(a: AliquotSeq): string =
## Return the representation of an allquot sequence.
for n in a:
result.addSep(", ", 0)
result.addInt(n)
- ---------------------------------------------------------------------------------------------------
proc classification(n: int): tuple[cat: Category, values: AliquotSeq] =
## Return the category of the aliquot sequence of a number "n" and the sequence itself.
var count = 0 # Number of elements currently generated.
var prev = n # Previous element in the sequence.
result.cat = Unknown
for k in aliquotSeq(n):
inc count
if k == 0:
result.cat = Terminating
elif k == n:
result.cat = case count
of 1: Perfect
of 2: Amicable
else: Sociable
elif k > Limit or count > 16:
result.cat = NonTerminating
elif k == prev:
result.cat = Aspiring
elif k in result.values:
result.cat = Cyclic
prev = k
result.values.add(k)
if result.cat != Unknown:
break
- ---------------------------------------------------------------------------------------------------
let t0 = getTime()
for n in 1..10:
let (cat, aseq) = classification(n)
echo fmt"{n:14}: {cat:<20} {aseq}"
echo "" for n in [11, 12, 28, 496, 220, 1184, 12496, 1264460,
790, 909, 562, 1064, 1488, 15355717786080.int]:
let (cat, aseq) = classification(n)
echo fmt"{n:14}: {cat:<20} {aseq}"
echo "" echo fmt"Processed in {(getTime() - t0).inMilliseconds} ms." </lang>
- Output:
1: terminating 0
2: terminating 1, 0
3: terminating 1, 0
4: terminating 3, 1, 0
5: terminating 1, 0
6: perfect 6
7: terminating 1, 0
8: terminating 7, 1, 0
9: terminating 4, 3, 1, 0
10: terminating 8, 7, 1, 0
11: terminating 1, 0
12: terminating 16, 15, 9, 4, 3, 1, 0
28: perfect 28
496: perfect 496
220: amicable 284, 220
1184: amicable 1210, 1184
12496: sociable 14288, 15472, 14536, 14264, 12496
1264460: sociable 1547860, 1727636, 1305184, 1264460
790: aspiring 650, 652, 496, 496
909: aspiring 417, 143, 25, 6, 6
562: cyclic 284, 220, 284
1064: cyclic 1336, 1184, 1210, 1184
1488: non-terminating 2480, 3472, 4464, 8432, 9424, 10416, 21328, 22320, 55056, 95728, 96720, 236592, 459792, 881392, 882384, 1474608, 2461648
15355717786080: non-terminating 44534663601120, 144940087464480
Processed in 105 ms.
Oforth
<lang oforth>import: mapping import: quicksort import: math
Object method: sum ( coll -- m )
#+ self reduce dup ifNull: [ drop 0 ] ;
Integer method: properDivs | i l |
Array new dup 1 over add ->l 2 self nsqrt tuck for: i [ self i mod ifFalse: [ i l add self i / l add ] ] sq self == ifTrue: [ l pop drop ] dup sort
- aliquot( n -- [] ) \ Returns aliquot sequence of n
| end l |
2 47 pow ->end Array new dup n over add ->l while ( l size 16 < l last 0 <> and l last end <= and ) [ l last properDivs sum l add ]
- aliquotClass( n -- [] s ) \ Returns aliquot sequence and classification
| l i j |
n aliquot dup ->l
l last 0 == ifTrue: [ "terminate" return ]
l second n == ifTrue: [ "perfect" return ]
3 l at n == ifTrue: [ "amicable" return ]
l indexOfFrom(n, 2) ifNotNull: [ "sociable" return ]
l size loop: i [
l indexOfFrom(l at(i), i 1+ ) -> j
j i 1+ == ifTrue: [ "aspiring" return ]
j ifNotNull: [ "cyclic" return ]
]
"non-terminating"
- </lang>
- Output:
>#[ dup . aliquotClass . ":" . . printcr ] 10 each 1 terminate : [1, 0] 2 terminate : [2, 1, 0] 3 terminate : [3, 1, 0] 4 terminate : [4, 3, 1, 0] 5 terminate : [5, 1, 0] 6 perfect : [6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6] 7 terminate : [7, 1, 0] 8 terminate : [8, 7, 1, 0] 9 terminate : [9, 4, 3, 1, 0] 10 terminate : [10, 8, 7, 1, 0] ok
>#[ dup . aliquotClass . ":" . . printcr ] [ 11, 12, 28, 496, 220, 1184, 12496, 1264460, 790, 909, 562, 1064, 1488, 15355717786080 ] apply 11 terminate : [11, 1, 0] 12 terminate : [12, 16, 15, 9, 4, 3, 1, 0] 28 perfect : [28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28] 496 perfect : [496, 496, 496, 496, 496, 496, 496, 496, 496, 496, 496, 496, 496, 496, 496, 496] 220 amicable : [220, 284, 220, 284, 220, 284, 220, 284, 220, 284, 220, 284, 220, 284, 220, 284] 1184 amicable : [1184, 1210, 1184, 1210, 1184, 1210, 1184, 1210, 1184, 1210, 1184, 1210, 1184, 1210, 1184, 1210] 12496 sociable : [12496, 14288, 15472, 14536, 14264, 12496, 14288, 15472, 14536, 14264, 12496, 14288, 15472, 14536, 14264, 12496] 1264460 sociable : [1264460, 1547860, 1727636, 1305184, 1264460, 1547860, 1727636, 1305184, 1264460, 1547860, 1727636, 1305184, 1264460, 1547860, 1727636, 1305184] 790 aspiring : [790, 650, 652, 496, 496, 496, 496, 496, 496, 496, 496, 496, 496, 496, 496, 496] 909 aspiring : [909, 417, 143, 25, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6] 562 cyclic : [562, 284, 220, 284, 220, 284, 220, 284, 220, 284, 220, 284, 220, 284, 220, 284] 1064 cyclic : [1064, 1336, 1184, 1210, 1184, 1210, 1184, 1210, 1184, 1210, 1184, 1210, 1184, 1210, 1184, 1210] 1488 non-terminating : [1488, 2480, 3472, 4464, 8432, 9424, 10416, 21328, 22320, 55056, 95728, 96720, 236592, 459792, 881392, 882384] 15355717786080 non-terminating : [15355717786080, 44534663601120, 144940087464480] ok >
PARI/GP
Define function aliquot(). Works with recent versions of PARI/GP >= 2.8: <lang parigp>aliquot(x) = {
my (L = List(x), M = Map(Mat([x,1])), k, m = "non-term.", n = x);
for (i = 2, 16, n = vecsum(divisors(n)) - n;
if (n > 2^47, break,
n == 0, m = "terminates"; break,
mapisdefined(M, n, &k),
m = if (k == 1,
if (i == 2, "perfect",
i == 3, "amicable",
i > 3, concat("sociable-",i-1)),
k < i-1, concat("cyclic-",i-k),
"aspiring"); break,
mapput(M, n, i); listput(L, n));
);
printf("%16d: %10s, %s\n", x, m, Vec(L));
}</lang>
Output:
gp > apply(aliquot, concat([1..10],[11,12,28,496,220,1184,12496,1264460,790,909,562,1064,1488,15355717786080]));
1: terminates, [1]
2: terminates, [2, 1]
3: terminates, [3, 1]
4: terminates, [4, 3, 1]
5: terminates, [5, 1]
6: perfect, [6]
7: terminates, [7, 1]
8: terminates, [8, 7, 1]
9: terminates, [9, 4, 3, 1]
10: terminates, [10, 8, 7, 1]
11: terminates, [11, 1]
12: terminates, [12, 16, 15, 9, 4, 3, 1]
28: perfect, [28]
496: perfect, [496]
220: amicable, [220, 284]
1184: amicable, [1184, 1210]
12496: sociable-5, [12496, 14288, 15472, 14536, 14264]
1264460: sociable-4, [1264460, 1547860, 1727636, 1305184]
790: aspiring, [790, 650, 652, 496]
909: aspiring, [909, 417, 143, 25, 6]
562: cyclic-2, [562, 284, 220]
1064: cyclic-2, [1064, 1336, 1184, 1210]
1488: non-term., [1488, 2480, 3472, 4464, 8432, 9424, 10416, 21328, 22320, 55056, 95728, 96720, 236592, 459792, 881392, 882384]
15355717786080: non-term., [15355717786080, 44534663601120]
Perl
<lang perl>use ntheory qw/divisor_sum/;
sub aliquot {
my($n, $maxterms, $maxn) = @_; $maxterms = 16 unless defined $maxterms; $maxn = 2**47 unless defined $maxn;
my %terms = ($n => 1);
my @allterms = ($n);
for my $term (2 .. $maxterms) {
$n = divisor_sum($n)-$n;
# push onto allterms here if we want the cyclic term to display
last if $n > $maxn;
return ("terminates",@allterms, 0) if $n == 0;
if (defined $terms{$n}) {
return ("perfect",@allterms) if $term == 2 && $terms{$n} == 1;
return ("amicible",@allterms) if $term == 3 && $terms{$n} == 1;
return ("sociable-".($term-1),@allterms) if $term > 3 && $terms{$n} == 1;
return ("aspiring",@allterms) if $terms{$n} == $term-1;
return ("cyclic-".($term-$terms{$n}),@allterms) if $terms{$n} < $term-1;
}
$terms{$n} = $term;
push @allterms, $n;
}
("non-term",@allterms);
}
for my $n (1..10) {
my($class, @seq) = aliquot($n); printf "%14d %10s [@seq]\n", $n, $class;
} print "\n"; for my $n (qw/11 12 28 496 220 1184 12496 1264460 790 909 562 1064 1488 15355717786080/) {
my($class, @seq) = aliquot($n); printf "%14d %10s [@seq]\n", $n, $class;
}</lang>
- Output:
1 terminates [1 0]
2 terminates [2 1 0]
3 terminates [3 1 0]
4 terminates [4 3 1 0]
5 terminates [5 1 0]
6 perfect [6]
7 terminates [7 1 0]
8 terminates [8 7 1 0]
9 terminates [9 4 3 1 0]
10 terminates [10 8 7 1 0]
11 terminates [11 1 0]
12 terminates [12 16 15 9 4 3 1 0]
28 perfect [28]
496 perfect [496]
220 amicible [220 284]
1184 amicible [1184 1210]
12496 sociable-5 [12496 14288 15472 14536 14264]
1264460 sociable-4 [1264460 1547860 1727636 1305184]
790 aspiring [790 650 652 496]
909 aspiring [909 417 143 25 6]
562 cyclic-2 [562 284 220]
1064 cyclic-2 [1064 1336 1184 1210]
1488 non-term [1488 2480 3472 4464 8432 9424 10416 21328 22320 55056 95728 96720 236592 459792 881392 882384]
15355717786080 non-term [15355717786080 44534663601120]
Phix
Translated from the Python example <lang Phix>function aliquot(atom n) sequence s = {n} integer k
if n=0 then return {"terminating",{0}} end if
while length(s)<16
and n<140737488355328 do
n = sum(factors(n,-1))
k = find(n,s)
if k then
if k=1 then
if length(s)=1 then return {"perfect",s}
elsif length(s)=2 then return {"amicable",s}
end if return {"sociable",s}
elsif k=length(s) then return {"aspiring",s}
end if return {"cyclic",append(s,n)}
elsif n=0 then return {"terminating",s}
end if
s = append(s,n)
end while
return {"non-terminating",s}
end function
function flat_d(sequence s)
for i=1 to length(s) do s[i] = sprintf("%d",s[i]) end for
return join(s,",")
end function
constant n = tagset(12)&{28, 496, 220, 1184, 12496, 1264460, 790, 909, 562, 1064, 1488, 15355717786080} sequence class, dseq for i=1 to length(n) do
{class, dseq} = aliquot(n[i])
printf(1,"%14d => %15s, {%s}\n",{n[i],class,flat_d(dseq)})
end for</lang>
- Output:
1 => terminating, {1}
2 => terminating, {2,1}
3 => terminating, {3,1}
4 => terminating, {4,3,1}
5 => terminating, {5,1}
6 => perfect, {6}
7 => terminating, {7,1}
8 => terminating, {8,7,1}
9 => terminating, {9,4,3,1}
10 => terminating, {10,8,7,1}
11 => terminating, {11,1}
12 => terminating, {12,16,15,9,4,3,1}
28 => perfect, {28}
496 => perfect, {496}
220 => amicable, {220,284}
1184 => amicable, {1184,1210}
12496 => sociable, {12496,14288,15472,14536,14264}
1264460 => sociable, {1264460,1547860,1727636,1305184}
790 => aspiring, {790,650,652,496}
909 => aspiring, {909,417,143,25,6}
562 => cyclic, {562,284,220,284}
1064 => cyclic, {1064,1336,1184,1210,1184}
1488 => non-terminating, {1488,2480,3472,4464,8432,9424,10416,21328,22320,55056,95728,96720,236592,459792,881392,882384}
15355717786080 => non-terminating, {15355717786080,44534663601120,144940087464480}
PowerShell
To make the PowerShell 4.0 code below work with PowerShell 2.0:
Replace any instances of ".Where{...}" with " | Where {...}"
Replace any instances of ".ForEach{...}" with " | ForEach {...}"
To make the PowerShell 4.0 code below work with PowerShell 3.0:
Replace any instances of ".Where{...}" with ".Where({...})"
Replace any instances of ".ForEach{...}" with ".ForEach({...})"
Simple <lang powershell>function Get-NextAliquot ( [int]$X )
{
If ( $X -gt 1 )
{
$NextAliquot = 0
(1..($X/2)).Where{ $x % $_ -eq 0 }.ForEach{ $NextAliquot += $_ }.Where{ $_ }
return $NextAliquot
}
}
function Get-AliquotSequence ( [int]$K, [int]$N )
{
$X = $K
$X
(1..($N-1)).ForEach{ $X = Get-NextAliquot $X; $X }
}
function Classify-AlliquotSequence ( [int[]]$Sequence )
{
$K = $Sequence[0]
$LastN = $Sequence.Count
If ( $Sequence[-1] -eq 0 ) { return "terminating" }
If ( $Sequence[-1] -eq 1 ) { return "terminating" }
If ( $Sequence[1] -eq $K ) { return "perfect" }
If ( $Sequence[2] -eq $K ) { return "amicable" }
If ( $Sequence[3..($Sequence.Count-1)] -contains $K ) { return "sociable" }
If ( $Sequence[-1] -eq $Sequence[-2] ) { return "aspiring" }
If ( $Sequence.Count -gt ( $Sequence | Select -Unique ).Count ) { return "cyclic" }
return "non-terminating and non-repeating through N = $($Sequence.Count)"
}
(1..10).ForEach{ [string]$_ + " is " + ( Classify-AlliquotSequence -Sequence ( Get-AliquotSequence -K $_ -N 16 ) ) }
( 11, 12, 28, 496, 220, 1184, 790, 909, 562, 1064, 1488 ).ForEach{ [string]$_ + " is " + ( Classify-AlliquotSequence -Sequence ( Get-AliquotSequence -K $_ -N 16 ) ) }</lang> Optimized <lang powershell>function Get-NextAliquot ( [int]$X )
{
If ( $X -gt 1 )
{
$NextAliquot = 1
If ( $X -gt 2 )
{
$XSquareRoot = [math]::Sqrt( $X )
(2..$XSquareRoot).Where{ $X % $_ -eq 0 }.ForEach{ $NextAliquot += $_ + $x / $_ }
If ( $XSquareRoot % 1 -eq 0 ) { $NextAliquot -= $XSquareRoot }
}
return $NextAliquot
}
}
function Get-AliquotSequence ( [int]$K, [int]$N )
{
$X = $K
$X
$i = 1
While ( $X -and $i -lt $N )
{
$i++
$Next = Get-NextAliquot $X
If ( $Next )
{
If ( $X -eq $Next )
{
($i..$N).ForEach{ $X }
$i = $N
}
Else
{
$X = $Next
$X
}
}
Else
{
$i = $N
}
}
}
function Classify-AlliquotSequence ( [int[]]$Sequence )
{
$K = $Sequence[0]
$LastN = $Sequence.Count
If ( $Sequence[-1] -eq 0 ) { return "terminating" }
If ( $Sequence[-1] -eq 1 ) { return "terminating" }
If ( $Sequence[1] -eq $K ) { return "perfect" }
If ( $Sequence[2] -eq $K ) { return "amicable" }
If ( $Sequence[3..($Sequence.Count-1)] -contains $K ) { return "sociable" }
If ( $Sequence[-1] -eq $Sequence[-2] ) { return "aspiring" }
If ( $Sequence.Count -gt ( $Sequence | Select -Unique ).Count ) { return "cyclic" }
return "non-terminating and non-repeating through N = $($Sequence.Count)"
}
(1..10).ForEach{ [string]$_ + " is " + ( Classify-AlliquotSequence -Sequence ( Get-AliquotSequence -K $_ -N 16 ) ) }
( 11, 12, 28, 496, 220, 1184, 12496, 1264460, 790, 909, 562, 1064, 1488 ).ForEach{ [string]$_ + " is " + ( Classify-AlliquotSequence -Sequence ( Get-AliquotSequence -K $_ -N 16 ) ) }</lang>
- Output:
1 is terminating 2 is terminating 3 is terminating 4 is terminating 5 is terminating 6 is perfect 7 is terminating 8 is terminating 9 is terminating 10 is terminating 11 is terminating 12 is terminating 28 is perfect 496 is perfect 220 is amicable 1184 is amicable 12496 is sociable 1264460 is sociable 790 is aspiring 909 is aspiring 562 is cyclic 1064 is cyclic 1488 is non-terminating and non-repeating through N = 16
Version 3.0
<lang PowerShell> function Get-Aliquot {
[CmdletBinding()]
[OutputType([PScustomObject])]
Param
(
[Parameter(Mandatory=$true,
ValueFromPipeline=$true,
ValueFromPipelineByPropertyName=$true)]
[int]
$InputObject
)
Begin
{
function Get-NextAliquot ([int]$X)
{
if ($X -gt 1)
{
$nextAliquot = 1
if ($X -gt 2)
{
$xSquareRoot = [Math]::Sqrt($X)
2..$xSquareRoot | Where-Object {$X % $_ -eq 0} | ForEach-Object {$nextAliquot += $_ + $x / $_}
if ($xSquareRoot % 1 -eq 0) {$nextAliquot -= $xSquareRoot}
}
$nextAliquot
}
}
function Get-AliquotSequence ([int]$K, [int]$N)
{
$X = $K
$X
$i = 1
while ($X -and $i -lt $N)
{
$i++
$next = Get-NextAliquot $X
if ($next)
{
if ($X -eq $next)
{
$i..$N | ForEach-Object {$X}
$i = $N
}
else
{
$X = $next
$X
}
}
else
{
$i = $N
}
}
}
function Classify-AlliquotSequence ([int[]]$Sequence)
{
$k = $Sequence[0]
if ($Sequence[-1] -eq 0) {return "terminating"}
if ($Sequence[-1] -eq 1) {return "terminating"}
if ($Sequence[1] -eq $k) {return "perfect" }
if ($Sequence[2] -eq $k) {return "amicable" }
if ($Sequence[3..($Sequence.Count-1)] -contains $k) {return "sociable" }
if ($Sequence[-1] -eq $Sequence[-2] ) {return "aspiring" }
if ($Sequence.Count -gt ($Sequence | Select -Unique).Count ) {return "cyclic" }
return "non-terminating and non-repeating through N = $($Sequence.Count)"
}
}
Process
{
$_ | ForEach-Object {
[PSCustomObject]@{
Number = $_
Classification = (Classify-AlliquotSequence -Sequence (Get-AliquotSequence -K $_ -N 16))
}
}
}
} </lang> <lang PowerShell> $oneToTen = 1..10 | Get-Aliquot $selected = 11, 12, 28, 496, 220, 1184, 12496, 1264460, 790, 909, 562, 1064, 1488 | Get-Aliquot
$numbers = $oneToTen, $selected $numbers </lang>
- Output:
Number Classification
------ --------------
1 terminating
2 terminating
3 terminating
4 terminating
5 terminating
6 perfect
7 terminating
8 terminating
9 terminating
10 terminating
11 terminating
12 terminating
28 perfect
496 perfect
220 amicable
1184 amicable
12496 sociable
1264460 sociable
790 aspiring
909 aspiring
562 cyclic
1064 cyclic
1488 non-terminating and non-repeating through N = 16
Prolog
<lang prolog>% See https://en.wikipedia.org/wiki/Divisor_function divisor_sum(N, Total):-
divisor_sum_prime(N, 2, 2, Total1, 1, N1), divisor_sum(N1, 3, Total, Total1).
divisor_sum(1, _, Total, Total):-
!.
divisor_sum(N, Prime, Total, Running_total):-
Prime * Prime =< N, !, divisor_sum_prime(N, Prime, Prime, P, 1, M), Next_prime is Prime + 2, Running_total1 is P * Running_total, divisor_sum(M, Next_prime, Total, Running_total1).
divisor_sum(N, _, Total, Running_total):-
Total is (N + 1) * Running_total.
divisor_sum_prime(N, Prime, Power, Total, Running_total, M):-
0 is N mod Prime, !, Running_total1 is Running_total + Power, Power1 is Power * Prime, N1 is N // Prime, divisor_sum_prime(N1, Prime, Power1, Total, Running_total1, M).
divisor_sum_prime(N, _, _, Total, Total, N).
% See https://en.wikipedia.org/wiki/Aliquot_sequence aliquot_sequence(N, Limit, Sequence, Class):-
aliquot_sequence(N, Limit, [N], Sequence, Class).
aliquot_sequence(_, 0, _, [], 'non-terminating'):-!. aliquot_sequence(_, _, [0|_], [0], terminating):-!. aliquot_sequence(N, _, [N, N|_], [], perfect):-!. aliquot_sequence(N, _, [N, _, N|_], [N], amicable):-!. aliquot_sequence(N, _, [N|S], [N], sociable):-
memberchk(N, S), !.
aliquot_sequence(_, _, [Term, Term|_], [], aspiring):-!. aliquot_sequence(_, _, [Term|S], [Term], cyclic):-
memberchk(Term, S), !.
aliquot_sequence(N, Limit, [Term|S], [Term|Rest], Class):-
divisor_sum(Term, Sum), Term1 is Sum - Term, L1 is Limit - 1, aliquot_sequence(N, L1, [Term1, Term|S], Rest, Class).
write_aliquot_sequence(N, Sequence, Class):-
writef('%w: %w, sequence:', [N, Class]),
write_aliquot_sequence(Sequence).
write_aliquot_sequence([]):-
nl, !.
write_aliquot_sequence([Term|Rest]):-
writef(' %w', [Term]),
write_aliquot_sequence(Rest).
main:-
between(1, 10, N), aliquot_sequence(N, 16, Sequence, Class), write_aliquot_sequence(N, Sequence, Class), fail.
main:-
member(N, [11, 12, 28, 496, 220, 1184, 12496, 1264460, 790, 909, 562, 1064, 1488]), aliquot_sequence(N, 16, Sequence, Class), write_aliquot_sequence(N, Sequence, Class), fail.
main.</lang>
- Output:
1: terminating, sequence: 1 0 2: terminating, sequence: 2 1 0 3: terminating, sequence: 3 1 0 4: terminating, sequence: 4 3 1 0 5: terminating, sequence: 5 1 0 6: perfect, sequence: 6 7: terminating, sequence: 7 1 0 8: terminating, sequence: 8 7 1 0 9: terminating, sequence: 9 4 3 1 0 10: terminating, sequence: 10 8 7 1 0 11: terminating, sequence: 11 1 0 12: terminating, sequence: 12 16 15 9 4 3 1 0 28: perfect, sequence: 28 496: perfect, sequence: 496 220: amicable, sequence: 220 284 220 1184: amicable, sequence: 1184 1210 1184 12496: sociable, sequence: 12496 14288 15472 14536 14264 12496 1264460: sociable, sequence: 1264460 1547860 1727636 1305184 1264460 790: aspiring, sequence: 790 650 652 496 909: aspiring, sequence: 909 417 143 25 6 562: cyclic, sequence: 562 284 220 284 1064: cyclic, sequence: 1064 1336 1184 1210 1184 1488: non-terminating, sequence: 1488 2480 3472 4464 8432 9424 10416 21328 22320 55056 95728 96720 236592 459792 881392 882384
Python
Importing Proper divisors from prime factors:
<lang python>from proper_divisors import proper_divs from functools import lru_cache
@lru_cache()
def pdsum(n):
return sum(proper_divs(n))
def aliquot(n, maxlen=16, maxterm=2**47):
if n == 0:
return 'terminating', [0]
s, slen, new = [n], 1, n
while slen <= maxlen and new < maxterm:
new = pdsum(s[-1])
if new in s:
if s[0] == new:
if slen == 1:
return 'perfect', s
elif slen == 2:
return 'amicable', s
else:
return 'sociable of length %i' % slen, s
elif s[-1] == new:
return 'aspiring', s
else:
return 'cyclic back to %i' % new, s
elif new == 0:
return 'terminating', s + [0]
else:
s.append(new)
slen += 1
else:
return 'non-terminating', s
if __name__ == '__main__':
for n in range(1, 11):
print('%s: %r' % aliquot(n))
print()
for n in [11, 12, 28, 496, 220, 1184, 12496, 1264460, 790, 909, 562, 1064, 1488, 15355717786080]:
print('%s: %r' % aliquot(n))</lang>
- Output:
terminating: [1, 0] terminating: [2, 1, 0] terminating: [3, 1, 0] terminating: [4, 3, 1, 0] terminating: [5, 1, 0] perfect: [6] terminating: [7, 1, 0] terminating: [8, 7, 1, 0] terminating: [9, 4, 3, 1, 0] terminating: [10, 8, 7, 1, 0] terminating: [11, 1, 0] terminating: [12, 16, 15, 9, 4, 3, 1, 0] perfect: [28] perfect: [496] amicable: [220, 284] amicable: [1184, 1210] sociable of length 5: [12496, 14288, 15472, 14536, 14264] sociable of length 4: [1264460, 1547860, 1727636, 1305184] aspiring: [790, 650, 652, 496] aspiring: [909, 417, 143, 25, 6] cyclic back to 284: [562, 284, 220] cyclic back to 1184: [1064, 1336, 1184, 1210] non-terminating: [1488, 2480, 3472, 4464, 8432, 9424, 10416, 21328, 22320, 55056, 95728, 96720, 236592, 459792, 881392, 882384, 1474608] non-terminating: [15355717786080, 44534663601120, 144940087464480]
Racket
fold-divisors is used from Proper_divisors#Racket, but for the truly big numbers, we use divisors from math/number-theory.
<lang racket>#lang racket (require "proper-divisors.rkt" math/number-theory)
(define SCOPE 20000)
(define P
(let ((P-v (vector)))
(λ (n)
(cond
[(> n SCOPE)
(apply + (drop-right (divisors n) 1))]
[else
(set! P-v (fold-divisors P-v n 0 +))
(vector-ref P-v n)]))))
- initialise P-v
(void (P SCOPE))
(define (aliquot-sequence-class K)
;; note that seq is reversed as a list, since we're consing
(define (inr-asc seq)
(match seq
[(list 0 _ ...)
(values "terminating" seq)]
[(list (== K) (== K) _ ...)
(values "perfect" seq)]
[(list n n _ ...)
(values (format "aspiring to ~a" n) seq)]
[(list (== K) ami (== K) _ ...)
(values (format "amicable with ~a" ami) seq)]
[(list (== K) cycle ... (== K))
(values (format "sociable length ~a" (add1 (length cycle))) seq)]
[(list n cycle ... n _ ...)
(values (format "cyclic on ~a length ~a" n (add1 (length cycle))) seq)]
[(list X _ ...)
#:when (> X 140737488355328)
(values "non-terminating big number" seq)]
[(list seq ...)
#:when (> (length seq) 16)
(values "non-terminating long sequence" seq)]
[(list seq1 seq ...) (inr-asc (list* (P seq1) seq1 seq))]))
(inr-asc (list K)))
(define (report-aliquot-sequence-class n)
(define-values (c s) (aliquot-sequence-class n)) (printf "~a:\t~a\t~a~%" n c (reverse s)))
(for ((i (in-range 1 10)))
(report-aliquot-sequence-class i))
(newline)
(for ((i (in-list '(11 12 28 496 220 1184 12496 1264460 790 909 562 1064 1488 15355717786080))))
(report-aliquot-sequence-class i))</lang>
- Output:
1: terminating (1 0) 2: terminating (2 1 0) 3: terminating (3 1 0) 4: terminating (4 3 1 0) 5: terminating (5 1 0) 6: perfect (6 6) 7: terminating (7 1 0) 8: terminating (8 7 1 0) 9: terminating (9 4 3 1 0) 11: terminating (11 1 0) 12: terminating (12 16 15 9 4 3 1 0) 28: perfect (28 28) 496: perfect (496 496) 220: amicable with 284 (220 284 220) 1184: amicable with 1210 (1184 1210 1184) 12496: sociable length 5 (12496 14288 15472 14536 14264 12496) 1264460: sociable length 4 (1264460 1547860 1727636 1305184 1264460) 790: aspiring to 496 (790 650 652 496 496) 909: aspiring to 6 (909 417 143 25 6 6) 562: cyclic on 284 length 2 (562 284 220 284) 1064: cyclic on 1184 length 2 (1064 1336 1184 1210 1184) 1488: non-terminating long sequence (1488 2480 3472 4464 8432 9424 10416 21328 22320 55056 95728 96720 236592 459792 881392 882384 1474608) 15355717786080: non-terminating big number (15355717786080 44534663601120 144940087464480)
Raku
(formerly Perl 6)
<lang perl6>sub propdivsum (\x) {
my @l = x > 1;
(2 .. x.sqrt.floor).map: -> \d {
unless x % d { my \y = x div d; y == d ?? @l.push: d !! @l.append: d,y }
}
sum @l;
}
multi quality (0,1) { 'perfect ' } multi quality (0,2) { 'amicable' } multi quality (0,$n) { "sociable-$n" } multi quality ($,1) { 'aspiring' } multi quality ($,$n) { "cyclic-$n" }
sub aliquotidian ($x) {
my %seen;
my @seq = $x, &propdivsum ... *;
for 0..16 -> $to {
my $this = @seq[$to] or return "$x\tterminating\t[@seq[^$to]]";
last if $this > 140737488355328;
if %seen{$this}:exists {
my $from = %seen{$this};
return "$x\t&quality($from, $to-$from)\t[@seq[^$to]]";
}
%seen{$this} = $to;
}
"$x non-terminating\t[{@seq}]";
}
aliquotidian($_).say for flat
1..10, 11, 12, 28, 496, 220, 1184, 12496, 1264460, 790, 909, 562, 1064, 1488, 15355717786080;</lang>
- Output:
1 terminating [1] 2 terminating [2 1] 3 terminating [3 1] 4 terminating [4 3 1] 5 terminating [5 1] 6 perfect [6] 7 terminating [7 1] 8 terminating [8 7 1] 9 terminating [9 4 3 1] 10 terminating [10 8 7 1] 11 terminating [11 1] 12 terminating [12 16 15 9 4 3 1] 28 perfect [28] 496 perfect [496] 220 amicable [220 284] 1184 amicable [1184 1210] 12496 sociable-5 [12496 14288 15472 14536 14264] 1264460 sociable-4 [1264460 1547860 1727636 1305184] 790 aspiring [790 650 652 496] 909 aspiring [909 417 143 25 6] 562 cyclic-2 [562 284 220] 1064 cyclic-2 [1064 1336 1184 1210] 1488 non-terminating [1488 2480 3472 4464 8432 9424 10416 21328 22320 55056 95728 96720 236592 459792 881392 882384 1474608 ...] 15355717786080 non-terminating [15355717786080 44534663601120 144940087464480 ...]
REXX
Programming notes:
This REXX version uses memoization.
Two versions of classifications of non-terminating are used:
- (lowercase) non-terminating ─── due to more than sixteen cyclic numbers
- (uppercase) NON-TERMINATING ─── due to a cyclic number that is larger than 247
Both of the above limitations are imposed by this Rosetta Code task's restriction requirements: For the purposes of this task, ···. <lang rexx>/*REXX program classifies various positive integers for types of aliquot sequences. */ parse arg low high $L /*obtain optional arguments from the CL*/ high= word(high low 10,1); low= word(low 1,1) /*obtain the LOW and HIGH (range). */ if $L= then $L=11 12 28 496 220 1184 12496 1264460 790 909 562 1064 1488 15355717786080 numeric digits 100 /*be able to compute the number: BIG */ big= 2**47; NTlimit= 16 + 1 /*limits for a non─terminating sequence*/ numeric digits max(9, length(big) ) /*be able to handle big numbers for // */ digs= digits() /*used for align numbers for the output*/
- .= .; #.0= 0; #.1= 0 /*#. are the proper divisor sums. */
say center('numbers from ' low " ───► " high ' (inclusive)', 153, "═")
do n=low to high; call classify n /*call a subroutine to classify number.*/
end /*n*/ /* [↑] process a range of integers. */
say say center('first numbers for each classification', 153, "═") class.= 0 /* [↓] ensure one number of each class*/
do q=1 until class.sociable\==0 /*the only one that has to be counted. */
call classify -q /*minus (-) sign indicates don't tell. */
_= what; upper _ /*obtain the class and uppercase it. */
class._= class._ + 1 /*bump counter for this class sequence.*/
if class._==1 then say right(q, digs)':' center(what, digs) $
end /*q*/ /* [↑] only display the 1st occurrence*/
say /* [↑] process until all classes found*/ say center('classifications for specific numbers', 153, "═")
do i=1 for words($L) /*$L: is a list of "special numbers".*/
call classify word($L, i) /*call a subroutine to classify number.*/
end /*i*/ /* [↑] process a list of integers. */
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ classify: parse arg a 1 aa; a= abs(a) /*obtain number that's to be classified*/
if #.a\==. then s= #.a /*Was this number been summed before?*/
else s= sigma(a) /*No, then classify number the hard way*/
#.a= s /*define sum of the proper divisors. */
$= s /*define the start of integer sequence.*/
what= 'terminating' /*assume this kind of classification. */
c.= 0 /*clear all cyclic sequences (to zero).*/
c.s= 1 /*set the first cyclic sequence. */
if $==a then what= 'perfect' /*check for a "perfect" number. */
else do t=1 while s>0 /*loop until sum isn't 0 or > big.*/
m= s /*obtain the last number in sequence. */
if #.m==. then s= sigma(m) /*Not defined? Then sum proper divisors*/
else s= #.m /*use the previously found integer. */
if m==s then if m>=0 then do; what= 'aspiring'; leave; end
parse var $ . word2 . /*obtain the 2nd number in sequence. */
if word2==a then do; what= 'amicable'; leave; end
$= $ s /*append a sum to the integer sequence.*/
if s==a then if t>3 then do; what= 'sociable'; leave; end
if c.s then if m>0 then do; what= 'cyclic' ; leave; end
c.s= 1 /*assign another possible cyclic number*/
/* [↓] Rosetta Code task's limit: >16 */
if t>NTlimit then do; what= 'non─terminating'; leave; end
if s>big then do; what= 'NON─TERMINATING'; leave; end
end /*t*/ /* [↑] only permit within reason. */
if aa>0 then say right(a, digs)':' center(what, digs) $
return /* [↑] only display if AA is positive*/
/*──────────────────────────────────────────────────────────────────────────────────────*/ sigma: procedure expose #. !.; parse arg x; if 11<2 then return 0; odd= x // 2
s= 1 /* [↓] use EVEN or ODD integers. ___*/
do j=2+odd by 1+odd while j*j<x /*divide by all the integers up to √ X */
if x//j==0 then s= s + j + x % j /*add the two divisors to the sum. */
end /*j*/ /* [↓] adjust for square. ___*/
if j*j==x then s= s + j /*Was X a square? If so, add √ X */
#.x= s /*memoize division sum for argument X.*/
return s /*return " " " " " */</lang>
- output when using the default input:
(Shown at three-quarter size.)
═════════════════════════════════════════════════════════numbers from 1 ───► 10 (inclusive)══════════════════════════════════════════════════════════
1: terminating 0
2: terminating 1 0
3: terminating 1 0
4: terminating 3 1 0
5: terminating 1 0
6: perfect 6
7: terminating 1 0
8: terminating 7 1 0
9: terminating 4 3 1 0
10: terminating 8 7 1 0
══════════════════════════════════════════════════════════first numbers for each classification══════════════════════════════════════════════════════════
1: terminating 0
6: perfect 6
25: aspiring 6
138: non─terminating 150 222 234 312 528 960 2088 3762 5598 6570 10746 13254 13830 19434 20886 21606 25098 26742 26754
220: amicable 284 220
562: cyclic 284 220 284
12496: sociable 14288 15472 14536 14264 12496
══════════════════════════════════════════════════════════classifications for specific numbers═══════════════════════════════════════════════════════════
11: terminating 1 0
12: terminating 16 15 9 4 3 1 0
28: perfect 28
496: perfect 496
220: amicable 284 220
1184: amicable 1210 1184
12496: sociable 14288 15472 14536 14264 12496
1264460: cyclic 1547860 1727636 1305184 1264460 1547860
790: aspiring 650 652 496
909: aspiring 417 143 25 6
562: cyclic 284 220 284
1064: cyclic 1336 1184 1210 1184
1488: non─terminating 2480 3472 4464 8432 9424 10416 21328 22320 55056 95728 96720 236592 459792 881392 882384 1474608 2461648 3172912 3173904
15355717786080: NON─TERMINATING 44534663601120 144940087464480
Ring
<lang ring>
- Project : Aliquot sequence classnifications
see "Rosetta Code - aliquot sequence classnifications" + nl while true
see "enter an integer: "
give k
k=fabs(floor(number(k)))
if k=0
exit
ok
printas(k)
end see "program complete."
func printas(k)
length=52
aseq = list(length)
n=k
classn=0
priorn = 0
inc = 0
for element=2 to length
aseq[element]=pdtotal(n)
see aseq[element] + " " + nl
if aseq[element]=0
see " terminating" + nl
classn=1
exit
ok
if aseq[element]=k and element=2
see " perfect" + nl
classn=2
exit
ok
if aseq[element]=k and element=3
see " amicable" + nl
classn=3
exit
ok
if aseq[element]=k and element>3
see " sociable" + nl
classn=4
exit
ok
if aseq[element]!=k and aseq[element-1]=aseq[element]
see " aspiring" + nl
classn=5
exit
ok
if aseq[element]!=k and element>2 and aseq[element-2]= aseq[element]
see " cyclic" + nl
classn=6
exit
ok
n=aseq[element]
if n>priorn
priorn=n
inc=inc+1
but n<=priorn
inc=0
priorn=0
ok
if inc=11 or n>30000000
exit
ok
next
if classn=0
see " non-terminating" + nl
ok
func pdtotal(n)
pdtotal = 0
for y=2 to n
if (n % y)=0
pdtotal=pdtotal+(n/y)
ok
next
return pdtotal
</lang> Output:
ROSETTA CODE - Aliquot sequence classifications Enter an integer: 1 0 terminating Enter an integer: 2 1 0 terminating Enter an integer: 3 1 0 terminating Enter an integer: 4 3 1 0 terminating Enter an integer: 5 1 0 terminating Enter an integer: 6 6 perfect Enter an integer: 7 1 0 terminating Enter an integer: 8 7 1 0 terminating Enter an integer: 9 4 3 1 0 terminating Enter an integer: 10 8 7 1 0 terminating Enter an integer: 11 1 0 terminating Enter an integer: 12 16 15 9 4 3 1 0 terminating Enter an integer: 28 28 perfect Enter an integer: 496 496 perfect Enter an integer: 220 284 220 amicable Enter an integer: 1184 1210 1184 amicable Enter an integer: 12496 14288 15472 14536 14264 12496 sociable Enter an integer: 1264460 1547860 1727636 1305184 1264460 sociable Enter an integer: 790 650 652 496 496 aspiring Enter an integer: 909 417 143 25 6 6 aspiring Enter an integer: 562 284 220 284 cyclic Enter an integer: 1064 1336 1184 1210 1184 cyclic Enter an integer: 1488 2480 3472 4464 8432 9424 10416 21328 22320 55056 95728 96720 non-terminating - - - - - - - - - - - - Enter an integer: 4004 5404 5460 13356 25956 49756 49812 83244 138964 144326 127978 67322 36250 34040 48040 60140 71572 58208 64264 60836 47692 35776 42456 69144 110376 244824 373356 594884 446170 356954 219706 118874 88720 117740 174916 174972 291844 302666 2564 38 217322 185014 92510 95626 49274 25894 17198 8602 6950 6070 4874 2440 3140 non-terminating Enter an integer: 4344 6576 10536 15864 23856 47568 75440 112048 111152 104236 105428 79078 45842 22924 20924 15700 18586 9296 11536 14256 30756 47868 63852 94404 125900 147520 204524 153400 237200 333634 238334 121306 62438 31222 16514 9406 4706 2938 1850 1684 1 270 1034 694 350 394 200 265 59 1 0 terminating Enter an integer: 6672 10688 10648 11312 13984 16256 16384 16383 6145 1235 445 95 25 6 6 aspiring Enter an integer: 6420 11724 15660 34740 71184 112832 121864 106646 53326 45458 37486 18746 16198 14042 11878 5942 2974 1490 1210 1184 1210 cyclic Enter an integer: 8128 8128 perfect Enter an integer: Program complete.
Ruby
With proper_divisors#Ruby in place:
<lang ruby>def aliquot(n, maxlen=16, maxterm=2**47)
return "terminating", [0] if n == 0
s = []
while (s << n).size <= maxlen and n < maxterm
n = n.proper_divisors.inject(0, :+)
if s.include?(n)
case n
when s[0]
case s.size
when 1 then return "perfect", s
when 2 then return "amicable", s
else return "sociable of length #{s.size}", s
end
when s[-1] then return "aspiring", s
else return "cyclic back to #{n}", s
end
elsif n == 0 then return "terminating", s << 0
end
end
return "non-terminating", s
end
for n in 1..10
puts "%20s: %p" % aliquot(n)
end puts for n in [11, 12, 28, 496, 220, 1184, 12496, 1264460, 790, 909, 562, 1064, 1488, 15355717786080]
puts "%20s: %p" % aliquot(n)
end</lang>
- Output:
terminating: [1, 0]
terminating: [2, 1, 0]
terminating: [3, 1, 0]
terminating: [4, 3, 1, 0]
terminating: [5, 1, 0]
perfect: [6]
terminating: [7, 1, 0]
terminating: [8, 7, 1, 0]
terminating: [9, 4, 3, 1, 0]
terminating: [10, 8, 7, 1, 0]
terminating: [11, 1, 0]
terminating: [12, 16, 15, 9, 4, 3, 1, 0]
perfect: [28]
perfect: [496]
amicable: [220, 284]
amicable: [1184, 1210]
sociable of length 5: [12496, 14288, 15472, 14536, 14264]
sociable of length 4: [1264460, 1547860, 1727636, 1305184]
aspiring: [790, 650, 652, 496]
aspiring: [909, 417, 143, 25, 6]
cyclic back to 284: [562, 284, 220]
cyclic back to 1184: [1064, 1336, 1184, 1210]
non-terminating: [1488, 2480, 3472, 4464, 8432, 9424, 10416, 21328, 22320, 55056, 95728, 96720, 236592, 459792, 881392, 882384, 1474608]
non-terminating: [15355717786080, 44534663601120, 144940087464480]
Rust
<lang rust>#[derive(Debug)] enum AliquotType { Terminating, Perfect, Amicable, Sociable, Aspiring, Cyclic, NonTerminating }
fn classify_aliquot(num: i64) -> (AliquotType, Vec<i64>) {
let limit = 1i64 << 47; //140737488355328
let mut terms = Some(num).into_iter().collect::<Vec<_>>();
for i in 0..16 {
let n = terms[i];
let divsum = (1..(n + 1) / 2 + 1).filter(|&x| n % x == 0 && n != x).fold(0, |sum, x| sum + x);
let classification = if divsum == 0 {
Some(AliquotType::Terminating)
}
else if divsum > limit {
Some(AliquotType::NonTerminating)
}
else if let Some(prev_idx) = terms.iter().position(|&x| x == divsum) {
let cycle_len = terms.len() - prev_idx;
Some(if prev_idx == 0 {
match cycle_len {
1 => AliquotType::Perfect,
2 => AliquotType::Amicable,
_ => AliquotType::Sociable
}
}
else {
if cycle_len == 1 {AliquotType::Aspiring} else {AliquotType::Cyclic}
})
}
else {
None
};
terms.push(divsum);
if let Some(result) = classification {
return (result, terms);
}
}
(AliquotType::NonTerminating, terms)
}
fn main() {
let nums = [1i64, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 28, 496, 220, 1184, 12496, 1264460, 790, 909, 562, 1064, 1488/*, 15355717786080*/];
for num in &nums {
println!("{} {:?}", num, classify_aliquot(*num));
}
}</lang>
- Output:
1 (Terminating, [1, 0]) 2 (Terminating, [2, 1, 0]) 3 (Terminating, [3, 1, 0]) 4 (Terminating, [4, 3, 1, 0]) 5 (Terminating, [5, 1, 0]) 6 (Perfect, [6, 6]) 7 (Terminating, [7, 1, 0]) 8 (Terminating, [8, 7, 1, 0]) 9 (Terminating, [9, 4, 3, 1, 0]) 10 (Terminating, [10, 8, 7, 1, 0]) 11 (Terminating, [11, 1, 0]) 12 (Terminating, [12, 16, 15, 9, 4, 3, 1, 0]) 28 (Perfect, [28, 28]) 496 (Perfect, [496, 496]) 220 (Amicable, [220, 284, 220]) 1184 (Amicable, [1184, 1210, 1184]) 12496 (Sociable, [12496, 14288, 15472, 14536, 14264, 12496]) 1264460 (Sociable, [1264460, 1547860, 1727636, 1305184, 1264460]) 790 (Aspiring, [790, 650, 652, 496, 496]) 909 (Aspiring, [909, 417, 143, 25, 6, 6]) 562 (Cyclic, [562, 284, 220, 284]) 1064 (Cyclic, [1064, 1336, 1184, 1210, 1184]) 1488 (NonTerminating, [1488, 2480, 3472, 4464, 8432, 9424, 10416, 21328, 22320, 55056, 95728, 96720, 236592, 459792, 881392, 882384, 1474608])
Scala
Put proper_divisors#Scala the full /Proper divisors for big (long) numbers/ section to the beginning: <lang Scala>def createAliquotSeq(n: Long, step: Int, list: List[Long]): (String, List[Long]) = {
val sum = properDivisors(n).sum
if (sum == 0) ("terminate", list ::: List(sum))
else if (step >= 16 || sum > 140737488355328L) ("non-term", list)
else {
list.indexOf(sum) match {
case -1 => createAliquotSeq(sum, step + 1, list ::: List(sum))
case 0 => if (step == 0) ("perfect", list ::: List(sum))
else if (step == 1) ("amicable", list ::: List(sum))
else ("sociable-" + (step + 1), list ::: List(sum))
case index => if (step == index) ("aspiring", list ::: List(sum))
else ("cyclic-" + (step - index + 1), list ::: List(sum))
}
}
} val numbers = List(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 28, 496, 220, 1184,
12496, 1264460, 790, 909, 562, 1064, 1488, 15355717786080L)
val result = numbers.map(i => createAliquotSeq(i, 0, List(i)))
result foreach { v => println(f"${v._2.head}%14d ${v._1}%10s [${v._2 mkString " "}]" ) }</lang>
- Output:
1 terminate [1 0]
2 terminate [2 1 0]
3 terminate [3 1 0]
4 terminate [4 3 1 0]
5 terminate [5 1 0]
6 perfect [6 6]
7 terminate [7 1 0]
8 terminate [8 7 1 0]
9 terminate [9 4 3 1 0]
10 terminate [10 8 7 1 0]
11 terminate [11 1 0]
12 terminate [12 16 15 9 4 3 1 0]
28 perfect [28 28]
496 perfect [496 496]
220 amicable [220 284 220]
1184 amicable [1184 1210 1184]
12496 sociable-5 [12496 14288 15472 14536 14264 12496]
1264460 sociable-4 [1264460 1547860 1727636 1305184 1264460]
790 aspiring [790 650 652 496 496]
909 aspiring [909 417 143 25 6 6]
562 cyclic-2 [562 284 220 284]
1064 cyclic-2 [1064 1336 1184 1210 1184]
1488 non-term [1488 2480 3472 4464 8432 9424 10416 21328 22320 55056 95728 96720 236592 459792 881392 882384 1474608]
15355717786080 non-term [15355717786080 44534663601120]
Swift
<lang swift>extension BinaryInteger {
@inlinable
public func factors(sorted: Bool = true) -> [Self] {
let maxN = Self(Double(self).squareRoot())
var res = Set<Self>()
for factor in stride(from: 1, through: maxN, by: 1) where self % factor == 0 {
res.insert(factor)
res.insert(self / factor)
}
return sorted ? res.sorted() : Array(res) }
}
struct SeqClass: CustomStringConvertible {
var seq: [Int] var desc: String
var description: String {
return "\(desc): \(seq)"
}
}
func classifySequence(k: Int, threshold: Int = 1 << 47) -> SeqClass {
var last = k var seq = [k]
while true {
last = last.factors().dropLast().reduce(0, +)
seq.append(last)
let n = seq.count
if last == 0 {
return SeqClass(seq: seq, desc: "Terminating")
} else if n == 2 && last == k {
return SeqClass(seq: seq, desc: "Perfect")
} else if n == 3 && last == k {
return SeqClass(seq: seq, desc: "Amicable")
} else if n >= 4 && last == k {
return SeqClass(seq: seq, desc: "Sociable[\(n - 1)]")
} else if last == seq[n - 2] {
return SeqClass(seq: seq, desc: "Aspiring")
} else if seq.dropFirst().dropLast(2).contains(last) {
return SeqClass(seq: seq, desc: "Cyclic[\(n - 1 - seq.firstIndex(of: last)!)]")
} else if n == 16 || last > threshold {
return SeqClass(seq: seq, desc: "Non-terminating")
}
}
fatalError()
}
for i in 1...10 {
print("\(i): \(classifySequence(k: i))")
}
print()
for i in [11, 12, 28, 496, 220, 1184, 12496, 1264460, 790, 909, 562, 1064, 1488] {
print("\(i): \(classifySequence(k: i))")
}
print()
print("\(15355717786080): \(classifySequence(k: 15355717786080))")</lang>
- Output:
1: Terminating: [1, 0] 2: Terminating: [2, 1, 0] 3: Terminating: [3, 1, 0] 4: Terminating: [4, 3, 1, 0] 5: Terminating: [5, 1, 0] 6: Perfect: [6, 6] 7: Terminating: [7, 1, 0] 8: Terminating: [8, 7, 1, 0] 9: Terminating: [9, 4, 3, 1, 0] 10: Terminating: [10, 8, 7, 1, 0] 11: Terminating: [11, 1, 0] 12: Terminating: [12, 16, 15, 9, 4, 3, 1, 0] 28: Perfect: [28, 28] 496: Perfect: [496, 496] 220: Amicable: [220, 284, 220] 1184: Amicable: [1184, 1210, 1184] 12496: Sociable[5]: [12496, 14288, 15472, 14536, 14264, 12496] 1264460: Sociable[4]: [1264460, 1547860, 1727636, 1305184, 1264460] 790: Aspiring: [790, 650, 652, 496, 496] 909: Aspiring: [909, 417, 143, 25, 6, 6] 562: Cyclic[2]: [562, 284, 220, 284] 1064: Cyclic[2]: [1064, 1336, 1184, 1210, 1184] 1488: Non-terminating: [1488, 2480, 3472, 4464, 8432, 9424, 10416, 21328, 22320, 55056, 95728, 96720, 236592, 459792, 881392, 882384] 15355717786080: Non-terminating: [15355717786080, 44534663601120, 144940087464480]
Tcl
This solution creates an iterator from a coroutine to generate aliquot sequences. al_classify uses a "RESULT" exception to achieve some unusual control flow.
<lang Tcl>proc ProperDivisors {n} {
if {$n == 1} {return 0}
set divs 1
set sum 1
for {set i 2} {$i*$i <= $n} {incr i} {
if {! ($n % $i)} {
lappend divs $i
incr sum $i
if {$i*$i<$n} {
lappend divs [set d [expr {$n / $i}]]
incr sum $d
}
}
}
list $sum $divs
}
proc al_iter {n} {
yield [info coroutine]
while {$n} {
yield $n
lassign [ProperDivisors $n] n
}
yield 0
return -code break
}
proc al_classify {n} {
coroutine iter al_iter $n
set items {}
try {
set type "non-terminating"
while {[llength $items] < 16} {
set i [iter]
if {$i == 0} {
set type "terminating"
}
set ix [lsearch -exact $items $i]
set items [linsert $items 0 $i]
switch $ix {
-1 { continue }
0 { throw RESULT "perfect" }
1 { throw RESULT "amicable" }
default { throw RESULT "sociable" }
}
}
} trap {RESULT} {type} {
rename iter {}
set map {
perfect aspiring
amicable cyclic
sociable cyclic
}
if {$ix != [llength $items]-2} {
set type [dict get $map $type]
}
}
list $type [lreverse $items]
}
for {set i 1} {$i <= 10} {incr i} {
puts [format "%8d -> %-16s : %s" $i {*}[al_classify $i]]
}
foreach i {11 12 28 496 220 1184 12496 1264460 790 909 562 1064 1488 } {
puts [format "%8d -> %-16s : %s" $i {*}[al_classify $i]]
}
- stretch goal .. let's time it
set i 15355717786080 puts [time {
puts [format "%8d -> %-16s : %s" $i {*}[al_classify $i]]
}]</lang>
- Output:
1 -> terminating : 1 0
2 -> terminating : 2 1 0
3 -> terminating : 3 1 0
4 -> terminating : 4 3 1 0
5 -> terminating : 5 1 0
6 -> perfect : 6 6
7 -> terminating : 7 1 0
8 -> terminating : 8 7 1 0
9 -> terminating : 9 4 3 1 0
10 -> terminating : 10 8 7 1 0
11 -> terminating : 11 1 0
12 -> terminating : 12 16 15 9 4 3 1 0
28 -> perfect : 28 28
496 -> perfect : 496 496
220 -> amicable : 220 284 220
1184 -> amicable : 1184 1210 1184
12496 -> sociable : 12496 14288 15472 14536 14264 12496
1264460 -> sociable : 1264460 1547860 1727636 1305184 1264460
790 -> aspiring : 790 650 652 496 496
909 -> aspiring : 909 417 143 25 6 6
562 -> cyclic : 562 284 220 284
1064 -> cyclic : 1064 1336 1184 1210 1184
1488 -> non-terminating : 1488 2480 3472 4464 8432 9424 10416 21328 22320 55056 95728 96720 236592 459792 881392 882384
15355717786080 -> non-terminating : 15355717786080 44534663601120 144940087464480 471714103310688 1130798979186912 2688948041357088 6050151708497568 13613157922639968 35513546724070632 74727605255142168 162658586225561832 353930992506879768 642678347124409032 1125102611548462968 1977286128289819992 3415126495450394808
556214046 microseconds per iteration
The large number finished (notice native bignums), but it took over 500 seconds ...
VBA
<lang vb>Option Explicit
Private Type Aliquot
Sequence() As Double Classification As String
End Type
Sub Main() Dim result As Aliquot, i As Long, j As Double, temp As String 'display the classification and sequences of the numbers one to ten inclusive
For j = 1 To 10
result = Aliq(j)
temp = vbNullString
For i = 0 To UBound(result.Sequence)
temp = temp & result.Sequence(i) & ", "
Next i
Debug.Print "Aliquot seq of " & j & " : " & result.Classification & " " & Left(temp, Len(temp) - 2)
Next j
'show the classification and sequences of the following integers, in order: Dim a
'15 355 717 786 080 : impossible in VBA ==> out of memory
a = Array(11, 12, 28, 496, 220, 1184, 12496, 1264460, 790, 909, 562, 1064, 1488)
For j = LBound(a) To UBound(a)
result = Aliq(CDbl(a(j)))
temp = vbNullString
For i = 0 To UBound(result.Sequence)
temp = temp & result.Sequence(i) & ", "
Next i
Debug.Print "Aliquot seq of " & a(j) & " : " & result.Classification & " " & Left(temp, Len(temp) - 2)
Next
End Sub
Private Function Aliq(Nb As Double) As Aliquot Dim s() As Double, i As Long, temp, j As Long, cpt As Long
temp = Array("non-terminating", "Terminate", "Perfect", "Amicable", "Sociable", "Aspiring", "Cyclic")
ReDim s(0)
s(0) = Nb
For i = 1 To 15
cpt = cpt + 1
ReDim Preserve s(cpt)
s(i) = SumPDiv(s(i - 1))
If s(i) > 140737488355328# Then Exit For
If s(i) = 0 Then j = 1
If s(1) = s(0) Then j = 2
If s(i) = s(0) And i > 1 And i <> 2 Then j = 4
If s(i) = s(i - 1) And i > 1 Then j = 5
If i >= 2 Then
If s(2) = s(0) Then j = 3
If s(i) = s(i - 2) And i <> 2 Then j = 6
End If
If j > 0 Then Exit For
Next
Aliq.Classification = temp(j)
Aliq.Sequence = s
End Function
Private Function SumPDiv(n As Double) As Double 'returns the sum of the Proper divisors of n Dim j As Long, t As Long
If n > 1 Then
For j = 1 To n \ 2
If n Mod j = 0 Then t = t + j
Next
End If
SumPDiv = t
End Function </lang>
- Output:
Aliquot seq of 1 : Terminate 1, 0 Aliquot seq of 2 : Terminate 2, 1, 0 Aliquot seq of 3 : Terminate 3, 1, 0 Aliquot seq of 4 : Terminate 4, 3, 1, 0 Aliquot seq of 5 : Terminate 5, 1, 0 Aliquot seq of 6 : Perfect 6, 6 Aliquot seq of 7 : Terminate 7, 1, 0 Aliquot seq of 8 : Terminate 8, 7, 1, 0 Aliquot seq of 9 : Terminate 9, 4, 3, 1, 0 Aliquot seq of 10 : Terminate 10, 8, 7, 1, 0 Aliquot seq of 11 : Terminate 11, 1, 0 Aliquot seq of 12 : Terminate 12, 16, 15, 9, 4, 3, 1, 0 Aliquot seq of 28 : Perfect 28, 28 Aliquot seq of 496 : Perfect 496, 496 Aliquot seq of 220 : Amicable 220, 284, 220 Aliquot seq of 1184 : Amicable 1184, 1210, 1184 Aliquot seq of 12496 : Sociable 12496, 14288, 15472, 14536, 14264, 12496 Aliquot seq of 1264460 : Sociable 1264460, 1547860, 1727636, 1305184, 1264460 Aliquot seq of 790 : Aspiring 790, 650, 652, 496, 496 Aliquot seq of 909 : Aspiring 909, 417, 143, 25, 6, 6 Aliquot seq of 562 : Cyclic 562, 284, 220, 284 Aliquot seq of 1064 : Cyclic 1064, 1336, 1184, 1210, 1184 Aliquot seq of 1488 : non-terminating 1488, 2480, 3472, 4464, 8432, 9424, 10416, 21328, 22320, 55056, 95728, 96720, 236592, 459792, 881392, 882384
Wren
<lang ecmascript>import "/fmt" for Conv, Fmt import "/math" for Int, Nums import "/seq" for Lst
class Classification {
construct new(seq, aliquot) {
_seq = seq
_aliquot = aliquot
}
seq { _seq}
aliquot { _aliquot }
}
var THRESHOLD = 2.pow(47)
var classifySequence = Fn.new { |k|
if (k <= 0) Fiber.abort("K must be positive")
var last = k
var seq = [k]
while (true) {
last = Nums.sum(Int.properDivisors(last))
seq.add(last)
var n = seq.count
var aliquot =
(last == 0) ? "Terminating" :
(n == 2 && last == k) ? "Perfect" :
(n == 3 && last == k) ? "Amicable" :
(n >= 4 && last == k) ? "Sociable[%(n-1)]" :
(last == seq[n-2]) ? "Aspiring" :
(n > 3 && seq[1..n-3].contains(last)) ? "Cyclic[%(n-1-Lst.indexOf(seq, last))]" :
(n == 16 || last > THRESHOLD) ? "Non-terminating" : ""
if (aliquot != "") return Classification.new(seq, aliquot)
}
}
System.print("Aliquot classifications - periods for Sociable/Cyclic in square brackets:\n") for (k in 1..10) {
var c = classifySequence.call(k)
System.print("%(Fmt.d(2, k)): %(Fmt.s(-15, c.aliquot)) %(c.seq)")
}
System.print() var a = [11, 12, 28, 496, 220, 1184, 12496, 1264460, 790, 909, 562, 1064, 1488] for (k in a) {
var c = classifySequence.call(k)
System.print("%(Fmt.d(7, k)): %(Fmt.s(-15, c.aliquot)) %(c.seq)")
}
System.print() var k = 15355717786080 var c = classifySequence.call(k) var seq = c.seq.map { |i| Conv.dec(i) }.toList // ensure 15 digit integer is printed in full System.print("%(k): %(Fmt.s(-15, c.aliquot)) %(seq)")</lang>
- Output:
Aliquot classifications - periods for Sociable/Cyclic in square brackets:
1: Terminating [1, 0]
2: Terminating [2, 1, 0]
3: Terminating [3, 1, 0]
4: Terminating [4, 3, 1, 0]
5: Terminating [5, 1, 0]
6: Perfect [6, 6]
7: Terminating [7, 1, 0]
8: Terminating [8, 7, 1, 0]
9: Terminating [9, 4, 3, 1, 0]
10: Terminating [10, 8, 7, 1, 0]
11: Terminating [11, 1, 0]
12: Terminating [12, 16, 15, 9, 4, 3, 1, 0]
28: Perfect [28, 28]
496: Perfect [496, 496]
220: Amicable [220, 284, 220]
1184: Amicable [1184, 1210, 1184]
12496: Sociable[5] [12496, 14288, 15472, 14536, 14264, 12496]
1264460: Sociable[4] [1264460, 1547860, 1727636, 1305184, 1264460]
790: Aspiring [790, 650, 652, 496, 496]
909: Aspiring [909, 417, 143, 25, 6, 6]
562: Cyclic[2] [562, 284, 220, 284]
1064: Cyclic[2] [1064, 1336, 1184, 1210, 1184]
1488: Non-terminating [1488, 2480, 3472, 4464, 8432, 9424, 10416, 21328, 22320, 55056, 95728, 96720, 236592, 459792, 881392, 882384]
15355717786080: Non-terminating [15355717786080, 44534663601120, 144940087464480]
zkl
<lang zkl>fcn properDivs(n){ [1.. (n + 1)/2 + 1].filter('wrap(x){ n%x==0 and n!=x }) } fcn aliquot(k){ //-->Walker
Walker(fcn(rk){ k:=rk.value; if(k)rk.set(properDivs(k).sum()); k }.fp(Ref(k)))
}(10).walk(15).println();</lang> Or, refactoring to remove saving the intermediate divisors (and adding white space): <lang zkl>fcn aliquot(k){ //-->Walker
Walker(fcn(rk){
k:=rk.value;
rk.set((1).reduce((k + 1)/2, fcn(s,n,k){
s + (k%n==0 and k!=n and n) // s + False == s + 0
},0,k));
k
}.fp(Ref(k)))
}(10).walk(15).println();</lang> <lang zkl>fcn classify(k){
const MAX=(2).pow(47); // 140737488355328
ak,aks:=aliquot(k), ak.walk(16);
_,a2,a3:=aks;
if(a2==k) return("perfect");
if(a3==k) return("amicable");
aspiring:='wrap(){
foreach n in (aks.len()-1){ if(aks[n]==aks[n+1]) return(True) }
False
};
cyclic:='wrap(){
foreach n in (aks.len()-1){ if(aks[n+1,*].holds(aks[n])) return(aks[n]) }
False
};
(if(aks.filter1('==(0))!=False) "terminating"
else if(n:=aks[1,*].filter1n('==(k))) "sociable of length " + (n+1)
else if(aks.filter1('>(MAX))) "non-terminating"
else if(aspiring()) "aspiring"
else if((c:=cyclic())!=False) "cyclic on " + c
else "non-terminating" )
+ " " + aks.filter();
}</lang> <lang zkl>[1..10].pump(fcn(k){ "%6d is %s".fmt(k,classify(k)).println() }); T(11,12,28,496,220,1184,12496,1264460,790,909,562,1064,1488)
.pump(fcn(k){ "%6d is %s".fmt(k,classify(k)).println() });</lang>
- Output:
L(10,8,7,1,0,0,0,0,0,0,0,0,0,0,0)
1 is terminating L(1)
2 is terminating L(2,1)
3 is terminating L(3,1)
4 is terminating L(4,3,1)
5 is terminating L(5,1)
6 is perfect
7 is terminating L(7,1)
8 is terminating L(8,7,1)
9 is terminating L(9,4,3,1)
10 is terminating L(10,8,7,1)
11 is terminating L(11,1)
12 is terminating L(12,16,15,9,4,3,1)
28 is perfect
496 is perfect
220 is amicable
1184 is amicable
12496 is sociable of length 5 L(12496,14288,15472,14536,14264,12496,14288,15472,14536,14264,12496,14288,15472,14536,14264,12496)
1264460 is sociable of length 4 L(1264460,1547860,1727636,1305184,1264460,1547860,1727636,1305184,1264460,1547860,1727636,1305184,1264460,1547860,1727636,1305184)
790 is aspiring L(790,650,652,496,496,496,496,496,496,496,496,496,496,496,496,496)
909 is aspiring L(909,417,143,25,6,6,6,6,6,6,6,6,6,6,6,6)
562 is cyclic on 284 L(562,284,220,284,220,284,220,284,220,284,220,284,220,284,220,284)
1064 is cyclic on 1184 L(1064,1336,1184,1210,1184,1210,1184,1210,1184,1210,1184,1210,1184,1210,1184,1210)
1488 is non-terminating L(1488,2480,3472,4464,8432,9424,10416,21328,22320,55056,95728,96720,236592,459792,881392,882384)
The loop to calculate 15355717786080 takes forever (literally)
ZX Spectrum Basic
This program is correct. However, a bug in the ROM of the ZX Spectrum makes the number 909 of an erroneous result. However, the same program running on Sam BASIC (a superset of Sinclair BASIC that ran on the computer Sam Coupé) provides the correct results. <lang zxbasic>10 PRINT "Number classification sequence" 20 INPUT "Enter a number (0 to end): ";k: IF k>0 THEN GO SUB 2000: PRINT k;" ";s$: GO TO 20 40 STOP 1000 REM sumprop 1010 IF oldk=1 THEN LET newk=0: RETURN 1020 LET sum=1 1030 LET root=SQR oldk 1040 FOR i=2 TO root-0.1 1050 IF oldk/i=INT (oldk/i) THEN LET sum=sum+i+oldk/i 1060 NEXT i 1070 IF oldk/root=INT (oldk/root) THEN LET sum=sum+root 1080 LET newk=sum 1090 RETURN 2000 REM class 2010 LET oldk=k: LET s$=" " 2020 GO SUB 1000 2030 LET oldk=newk 2040 LET s$=s$+" "+STR$ newk 2050 IF newk=0 THEN LET s$="terminating"+s$: RETURN 2060 IF newk=k THEN LET s$="perfect"+s$: RETURN 2070 GO SUB 1000 2080 LET oldk=newk 2090 LET s$=s$+" "+STR$ newk 2100 IF newk=0 THEN LET s$="terminating"+s$: RETURN 2110 IF newk=k THEN LET s$="amicable"+s$: RETURN 2120 FOR t=4 TO 16 2130 GO SUB 1000 2140 LET s$=s$+" "+STR$ newk 2150 IF newk=0 THEN LET s$="terminating"+s$: RETURN 2160 IF newk=k THEN LET s$="sociable (period "+STR$ (t-1)+")"+s$: RETURN 2170 IF newk=oldk THEN LET s$="aspiring"+s$: RETURN 2180 LET b$=" "+STR$ newk+" ": LET ls=LEN s$: LET lb=LEN b$: LET ls=ls-lb 2190 FOR i=1 TO ls 2200 IF s$(i TO i+lb-1)=b$ THEN LET s$="cyclic (at "+STR$ newk+") "+s$: LET i=ls 2210 NEXT i 2220 IF LEN s$<>(ls+lb) THEN RETURN 2300 IF newk>140737488355328 THEN LET s$="non-terminating (term > 140737488355328)"+s$: RETURN 2310 LET oldk=newk 2320 NEXT t 2330 LET s$="non-terminating (after 16 terms)"+s$ 2340 RETURN</lang>
- Output:
Number classification sequence 1 terminating 0 2 terminating 1 0 3 terminating 1 0 4 terminating 3 1 0 5 terminating 1 0 6 perfect 6 7 terminating 1 0 8 terminating 7 1 0 9 terminating 4 3 1 0 10 terminating 8 7 1 0 11 terminating 1 0 12 terminating 16 15 9 4 3 1 0 28 perfect 28 496 perfect 496 220 amicable 284 220 1184 amicable 1210 1184 12496 sociable (period 5) 14288 15472 14536 14264 12496 1264460 sociable (period 4) 1547860 1727636 1305184 1264460 790 aspiring 650 652 496 496 909 aspiring 417 143 25 6 6 562 cyclic (at 284) 284 220 284 1064 cyclic (at 1184) 1336 1184 1210 1184 1488 non-terminating (after 16 terms) 2480 3472 4464 8432 9424 10416 21328 22320 55056 95728 96720 236592 459792 881392 882384
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