Mertens’ Conjecture
January 24, 2020
We begin with a function to compute Moebius(n). Our factors function uses a 2,3,5-wheel; you can use something more complex if you are ambitious:
(define (factors n)
(let ((wheel (vector 1 2 2 4 2 4 2 4 6 2 6)))
(let loop ((n n) (f 2) (w 0) (fs (list)))
(if (< n (* f f)) (reverse (cons n fs))
(if (zero? (modulo n f))
(loop (/ n f) f w (cons f fs))
(loop n (+ f (vector-ref wheel w))
(if (= w 10) 3 (+ w 1)) fs))))))
(define (moebius n)
; (-1)^k if n has k factors,
; or 0 if any factors duplicated
(if (< n 1) (error 'moebius "must be positive")
(if (= n 1) 1
(let loop ((m 1) (f 0) (fs (factors n)))
(if (null? fs) m
(if (= f (car fs)) 0
(loop (- m) (car fs) (cdr fs))))))))
Then the Mertens function is simple:
(define (mertens n)
(do ((k 1 (+ k 1))
(m 0 (+ m (moebius k))))
((< n k) m)))
Here are some of the OEIS sequences:
(define (a008683 n) ; moebius function (map moebius (range 1 (+ n 1)))) > (a008683 25) (1 -1 -1 0 -1 1 -1 0 0 1 -1 0 -1 1 1 0 -1 0 -1 0 1 1 -1 0 0)
(define (a002321 n) ; mertens function (map mertens (range 1 (+ n 1)))) > (a002321 20) (1 0 -1 -1 -2 -1 -2 -2 -2 -1 -2 -2 -3 -2 -1 -1 -2 -2 -3 -3)
(define (a028442 n)
; numbers k such that mertens(k) == 0
(let loop ((k 1) (M 0) (ks (list)))
(if (< n k) (reverse ks)
(let* ((m (moebius k)) (M (+ M m)))
(if (zero? M)
(loop (+ k 1) M (cons k ks))
(loop (+ k 1) M ks))))))
> (a028442 200)
(2 39 40 58 65 93 101 145 149 150 159 160 163 164 166)
(define (a100306 n)
; numbers k such that moebius(k) == mertens(k)
(let loop ((k 1) (M 0) (ks (list)))
(if (< n k) (reverse ks)
(let* ((m (moebius k)) (M (+ M m)))
(if (= m M)
(loop (+ k 1) M (cons k ks))
(loop (+ k 1) M ks))))))
> (a100306 200)
(1 3 40 41 59 66 94 102 146 150 151 160 161 164 165 167)
You can run the program at https://ideone.com/eYYkdG.
Programming Praxis 
I’d have thought using some sort of sieve would be better for this.
For example:
import sympy import itertools def mobius(N): primes = sympy.primerange(2,N) a = [0]+[1]*(N-1) for p in primes: for i in range(p,N,p): a[i] = -a[i] for i in range(p*p,N,p*p): a[i] = 0 return a def mertens(N): return itertools.accumulate(mobius(N)) print(list(mobius(20))) # [0, 1, -1, -1, 0, -1, 1, -1, 0, 0, 1, -1, 0, -1, 1, 1, 0, -1, 0, -1] print(list(mertens(20))) # [0, 1, 0, -1, -1, -2, -1, -2, -2, -2, -1, -2, -2, -3, -2, -1, -1, -2, -2, -3]Here’s a solution in Python, seemingly requiring excessive computation relative to @matthew’s sieve approach above.
import itertools def moebius(n): last_factor = -1 parity = 1 x = 2 while x * x <= n: if n % x == 0: if x == last_factor: return 0 parity *= -1 last_factor = x n //= x else: x += 1 + (x & 1) return 0 if n == last_factor else -parity def merten(): return itertools.accumulate((moebius(x) for x in itertools.count(2)), initial=1) print(list(itertools.islice((x for x in merten()), 20)))Output:
Similar to finding primes with the sieve of Eratosthenes compared with checking all numbers with trial division.