Mertens function
Mertens function is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
The Mertens function M(x) is the count of square-free integers up to x that have an even number of prime factors, minus the count of those that have an odd number.
It is an extension of the Möbius function. Given the Möbius function μ(n), the Mertens function M(x) is the sum of the Möbius numbers from n == 1 through n == x.
- Task
- Write a routine (function, procedure, whatever) to find the Mertens number for any positive integer x.
- Use that routine to find and display here, on this page, at least the first 99 terms in a grid layout. (Not just one long line or column of numbers.)
- Use that routine to find and display here, on this page, the number of times the Mertens function sequence is equal to zero in the range M(1) through M(1000).
- Use that routine to find and display here, on this page, the number of times the Mertens function sequence crosses zero in the range M(1) through M(1000). (Crossing defined as this term equal to zero but preceding term not.)
- See also
This is not code golf. The stackexchange link is provided as an algorithm reference, not as a guide.
- Related tasks
Go
<lang go>package main
import "fmt"
func mertens(to int) ([]int, int, int) {
if to < 1 {
to = 1
}
merts := make([]int, to+1)
primes := []int{2}
var sum, zeros, crosses int
for i := 1; i <= to; i++ {
j := i
cp := 0 // counts prime factors
spf := false // true if there is a square prime factor
for _, p := range primes {
if p > j {
break
}
if j%p == 0 {
j /= p
cp++
}
if j%p == 0 {
spf = true
break
}
}
if cp == 0 && i > 2 {
cp = 1
primes = append(primes, i)
}
if !spf {
if cp%2 == 0 {
sum++
} else {
sum--
}
}
merts[i] = sum
if sum == 0 {
zeros++
if i > 1 && merts[i-1] != 0 {
crosses++
}
}
}
return merts, zeros, crosses
}
func main() {
merts, zeros, crosses := mertens(1000)
fmt.Println("Mertens sequence - First 199 terms:")
for i := 0; i < 200; i++ {
if i == 0 {
fmt.Print(" ")
continue
}
if i%20 == 0 {
fmt.Println()
}
fmt.Printf(" % d", merts[i])
}
fmt.Println("\n\nEquals zero", zeros, "times between 1 and 1000")
fmt.Println("\nCrosses zero", crosses, "times between 1 and 1000")
}</lang>
- Output:
Mertens sequence - First 199 terms:
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
Equals zero 92 times between 1 and 1000
Crosses zero 59 times between 1 and 1000
Perl 6
Mertens number is not defined for n == 0. Perl 6 arrays are indexed from 0 so store a blank value at position zero to keep x and M(x) aligned.
<lang perl6>use Prime::Factor;
sub μ (Int \n) {
my @p = prime-factors(n); +@p == +Bag(@p).keys ?? +@p %% 2 ?? 1 !! -1 !! 0
}
my @mertens = lazy [\+] flat ' ', 1, (2..*).map: -> \n { μ(n) };
put "Mertens sequence - First 199 terms:\n",
@mertens[^200]».fmt('%3s').rotor(20).join("\n"),
"\n\nEquals zero ", +@mertens[1..1000].grep( !* ),
' times between 1 and 1000', "\n\nCrosses zero ",
+@mertens[1..1000].kv.grep( {!$^v and @mertens[$^k - 1]} ),
' times between 1 and 1000';</lang>
- Output:
Mertens sequence - First 199 terms:
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
Equals zero 92 times between 1 and 1000
Crosses zero 67 times between 1 and 1000