Abundant, deficient and perfect number classifications: Difference between revisions

=={{header|Bracmat}}==

Two solutions are given. The first solution first decomposes the current number into a multiset of prime factors and then constructs the proper divisors. The second solution finds proper divisors by checking all candidates from 1 up to the square root of the given number. The first solution is a few times faster, because establishing the prime factors of a small enough number (less than 2^32 or less than 2^64, depending on the bitness of Bracmat) is fast.

<lang bracmat>( clk$:?t0

& ( multiples

= prime multiplicity

. !arg:(?prime.?multiplicity)

& !multiplicity:0

& 1

| !prime^!multiplicity*(.!multiplicity)

+ multiples$(!prime.-1+!multiplicity)

)

& ( P

= primeFactors prime exp poly S

. !arg^1/67:?primeFactors

& ( !primeFactors:?^1/67&0

| 1:?poly

& whl

' ( !primeFactors:%?prime^?exp*?primeFactors

& !poly*multiples$(!prime.67*!exp):?poly

)

& -1+!poly+1:?poly

& 1:?S

& ( !poly

: ?

+ (#%@?s*?&!S+!s:?S&~)

+ ?

| 1/2*!S

)

)

)

& 0:?deficient:?perfect:?abundant

& 0:?n

& whl

' ( 1+!n:~>20000:?n

& P$!n

: ( <!n&1+!deficient:?deficient

| !n&1+!perfect:?perfect

| >!n&1+!abundant:?abundant

)

)

& out$(deficient !deficient perfect !perfect abundant !abundant)

& clk$:?t1

& out$(flt$(!t1+-1*!t0,2) sec)

& clk$:?t2

& ( P

= f h S

. 0:?f

& 0:?S

& whl

' ( 1+!f:?f

& !f^2:~>!n

& ( !arg*!f^-1:~/:?g

& !S+!f:?S

& ( !g:~!f&!S+!g:?S

|

)

|

)

)

& 1/2*!S

)

& 0:?deficient:?perfect:?abundant

& 0:?n

& whl

' ( 1+!n:~>20000:?n

& P$!n

: ( <!n&1+!deficient:?deficient

| !n&1+!perfect:?perfect

| >!n&1+!abundant:?abundant

)

)

& out$(deficient !deficient perfect !perfect abundant !abundant)

& clk$:?t3

& out$(flt$(!t3+-1*!t2,2) sec)

);</lang>

Output:

<pre>deficient 15043 perfect 4 abundant 4953

4,27*10E0 sec

deficient 15043 perfect 4 abundant 4953

1,63*10E1 sec</pre>