2020-02-17 23:21:07 -08:00
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using Primes
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divisors(n) = foldl((a, (p, e)) -> vcat((a * [p^i for i in 0:e]')...), factor(n), init=[1])
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2015-11-18 06:14:39 +00:00
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2020-02-17 23:21:07 -08:00
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isperfect(n) = sum(1 // d for d in divisors(n)) == 2
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lo, hi = 2, 2^19
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println("Perfect numbers between ", lo, " and ", hi, ": ", collect(filter(isperfect, lo:hi)))
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