Numbers which are the cube roots of the product of their proper divisors: Difference between revisions

</pre>

Alternatively and a bit quicker, inspired by the C++ entry and the OEIS comment that (apart from 1) n must have exactly 8 divisors:

<syntaxhighlight lang="ecmascript">import "./fmt" for Fmt

var divisorCount = Fn.new { |n|

var i = 1

var k = (n%2 == 0) ? 1 : 2

var count = 0

while (i <= n.sqrt) {

if (n%i == 0) {

count = count + 1

var j = (n/i).floor

if (j != i) count = count + 1

}

i = i + k

}

return count

}

var numbers50 = []

var count = 0

var n = 1

System.print("First 50 numbers which are the cube roots of the products of their proper divisors:")

while (true) {

var dc = divisorCount.call(n)

if (n == 1 || dc == 8) {

count = count + 1

if (count <= 50) {

numbers50.add(n)

if (count == 50) Fmt.tprint("$3d", numbers50, 10)

} else if (count == 500) {

Fmt.print("\n500th : $,d", n)

} else if (count == 5000) {

Fmt.print("5,000th : $,d", n)

} else if (count == 50000) {

Fmt.print("50,000th: $,d", n)

break

}

}

n = n + 1

}</syntaxhighlight>

{{out}}

<pre>

Same as first version.