How to resolve the algorithm Abundant odd numbers step by step in the AutoHotkey programming language

Published on 12 May 2024 09:40 PM
#Autohotkey

How to resolve the algorithm Abundant odd numbers step by step in the AutoHotkey programming language

Table of Contents

Problem Statement

An Abundant number is a number n for which the   sum of divisors   σ(n) > 2n, or,   equivalently,   the   sum of proper divisors   (or aliquot sum)       s(n) > n.

12   is abundant, it has the proper divisors     1,2,3,4 & 6     which sum to   16   ( > 12 or n);        or alternately,   has the sigma sum of   1,2,3,4,6 & 12   which sum to   28   ( > 24 or 2n).

Abundant numbers are common, though even abundant numbers seem to be much more common than odd abundant numbers. To make things more interesting, this task is specifically about finding   odd abundant numbers.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Abundant odd numbers step by step in the AutoHotkey programming language

Source code in the autohotkey programming language

Abundant(num){
	sum := 0, str := ""
	for n, bool in proper_divisors(num)
		sum += n,  str .= (str?"+":"") n
	return sum > num ? str " = " sum : 0
}
proper_divisors(n) {
	Array := []
	if n = 1
		return Array
	Array[1] := true
	x := Floor(Sqrt(n))
	loop, % x+1
		if !Mod(n, i:=A_Index+1) && (floor(n/i) < n)
			Array[floor(n/i)] := true
	Loop % n/x
		if !Mod(n, i:=A_Index+1) && (i < n)
			Array[i] := true
	return Array
}


output := "First 25 abundant odd numbers:`n"
while (count<1000)
{
	oddNum := 2*A_Index-1
	if (str := Abundant(oddNum))
	{
		count++
		if (count<=25)
			output .= oddNum " " str "`n"
		if (count = 1000)
			output .= "`nOne thousandth abundant odd number:`n" oddNum " " str "`n"
	}
}
count := 0
while !count
{
	num := 2*A_Index -1 + 1000000000
	if (str := Abundant(num))
	{
		count++
		output .= "`nFirst abundant odd number greater than one billion:`n" num " " str "`n"
	}
}
MsgBox % output
return

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