Practical numbers

Built-in:

say is_practical(2**128 + 1)   #=> false
say is_practical(2**128 + 4)   #=> true

Slow implementation (as the task requires):

func is_practical(n) {

    var set = Set()

    n.divisors.grep { _ < n }.subsets {|*a|
        set << a.sum
    }

    1..n-1 -> all { set.has(_) }
}

var from = 1
var upto = 333

var list = (from..upto).grep { is_practical(_) }

say "There are #{list.len} practical numbers in the range #{from}..#{upto}."
say "#{list.first(10).join(', ')} ... #{list.last(10).join(', ')}\n"

for n in ([666, 6666, 66666]) {
    say "#{'%5s' % n } is practical? #{is_practical(n)}"
}

Efficient algorithm:

func is_practical(n) {

    n.is_odd && return (n == 1)
    n.is_pos || return false

    var p = 1
    var f = n.factor_exp

    f.each_cons(2, {|a,b|
        p *= sigma(a.head**a.tail)
        b.head > (1 + p) && return false
    })

    return true
}

Output:

There are 77 practical numbers in the range 1..333.
1, 2, 4, 6, 8, 12, 16, 18, 20, 24 ... 288, 294, 300, 304, 306, 308, 312, 320, 324, 330

  666 is practical? true
 6666 is practical? true
66666 is practical? false