| definiendum | $ \mathrm{ld}:\mathbb N^+\to\{1\}\cup\mathrm{Prime\ number} $ |
| definiendum | $ \mathrm{ld}(n):=\mathrm{min}\left(\mathrm{divisors}(n)\right) $ |
divides :: Integral a => a -> a -> Bool divides d n = rem n d == 0
ld :: Integral a => a -> a ld n = ldf 2 n ldf :: Integral a => a -> a -> a ldf k n | divides k n = k | k^2 > n = n | otherwise = ldf (k+1) n
using Set of divisors function:
divides :: Integral a => a -> a -> Bool divides d n = rem n d == 0
If $n$ isn't a prime, then $n$ divided by the least divisor is some number bigger than $\mathrm{ld}(n)$ and hence
| $\mathrm{ld}(n)^2\le n$ |
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