How to resolve the algorithm Abundant, deficient and perfect number classifications step by step in the Bracmat programming language

Published on 12 May 2024 09:40 PM
#Bracmat

How to resolve the algorithm Abundant, deficient and perfect number classifications step by step in the Bracmat programming language

Table of Contents

Problem Statement

These define three classifications of positive integers based on their   proper divisors. Let   P(n)   be the sum of the proper divisors of   n   where the proper divisors are all positive divisors of   n   other than   n   itself.

6   has proper divisors of   1,   2,   and   3. 1 + 2 + 3 = 6,   so   6   is classed as a perfect number.

Calculate how many of the integers   1   to   20,000   (inclusive) are in each of the three classes. Show the results here.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Abundant, deficient and perfect number classifications step by step in the Bracmat programming language

Source code in the bracmat programming language

( 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+!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+!perfect:?perfect
        | >!n&1+!abundant:?abundant
        )
    )
& out$(deficient !deficient perfect !perfect abundant !abundant)
& clk$:?t3
& out$(flt$(!t3+-1*!t2,2) sec)
);

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