Mertens function
Built-in:
say mertens(123456789) #=> 1170
say mertens(1234567890) #=> 9163
Algorithm for computing M(n) in sublinear time:
func mertens(n) is cached {
var lookup_size = (2 * n.iroot(3)**2)
var mertens_lookup = [0]
for k in (1..lookup_size) {
mertens_lookup[k] = (mertens_lookup[k-1] + k.moebius)
}
static cache = Hash()
func (n) {
if (n <= lookup_size) {
return mertens_lookup[n]
}
if (cache.has(n)) {
return cache{n}
}
var M = 1
var s = n.isqrt
for k in (2 .. floor(n/(s+1))) {
M -= __FUNC__(floor(n/k))
}
for k in (1..s) {
M -= (mertens_lookup[k] * (floor(n/k) - floor(n/(k+1))))
}
cache{n} = M
}(n)
}
Task:
with (200) {|n|
say "Mertens function in the range 1..#{n}:"
(1..n).map { mertens(_) }.slices(20).each {|line|
say line.map{ "%2s" % _ }.join(' ')
}
}
with (1000) {|n|
say "\nIn the range 1..#{n}, there are:"
say (1..n->count_by { mertens(_)==0 }, " zeros")
say (1..n->count_by { mertens(_)==0 && mertens(_-1)!=0 }, " zero crossings")
}
Output:
Mertens function in the range 1..200:
1 0 -1 -1 -2 -1 -2 -2 -2 -1 -2 -2 -3 -2 -1 -1 -2 -2 -3 -3
-2 -1 -2 -2 -2 -1 -1 -1 -2 -3 -4 -4 -3 -2 -1 -1 -2 -1 0 0
-1 -2 -3 -3 -3 -2 -3 -3 -3 -3 -2 -2 -3 -3 -2 -2 -1 0 -1 -1
-2 -1 -1 -1 0 -1 -2 -2 -1 -2 -3 -3 -4 -3 -3 -3 -2 -3 -4 -4
-4 -3 -4 -4 -3 -2 -1 -1 -2 -2 -1 -1 0 1 2 2 1 1 1 1
0 -1 -2 -2 -3 -2 -3 -3 -4 -5 -4 -4 -5 -6 -5 -5 -5 -4 -3 -3
-3 -2 -1 -1 -1 -1 -2 -2 -1 -2 -3 -3 -2 -1 -1 -1 -2 -3 -4 -4
-3 -2 -1 -1 0 1 1 1 0 0 -1 -1 -1 -2 -1 -1 -2 -1 0 0
1 1 0 0 -1 0 -1 -1 -1 -2 -2 -2 -3 -4 -4 -4 -3 -2 -3 -3
-4 -5 -4 -4 -3 -4 -3 -3 -3 -4 -5 -5 -6 -5 -6 -6 -7 -7 -8 -8
In the range 1..1000, there are:
92 zeros
59 zero crossings
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