Mertens function

From Rosetta Code
Task
Mertens function
You are encouraged to solve this task according to the task description, using any language you may know.

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



11l

Translation of: Python
F mertens(count)
   ‘Generate Mertens numbers’
   V m = [-1, 1]
   L(n) 2 .. count
      m.append(1)
      L(k) 2 .. n
         m[n] -= m[n I/ k]
   R m

V ms = mertens(1000)

print(‘The first 99 Mertens numbers are:’)
print(‘  ’, end' ‘ ’)
V col = 1
L(n) ms[1.<100]
   print(‘#2’.format(n), end' ‘ ’)
   col++
   I col == 10
      print()
      col = 0

V zeroes = sum(ms.map(x -> Int(x == 0)))
V crosses = sum(zip(ms, ms[1..]).map((a, b) -> Int(a != 0 & b == 0)))
print(‘M(N) equals zero #. times.’.format(zeroes))
print(‘M(N) crosses zero #. times.’.format(crosses))
Output:
The first 99 Mertens numbers are:
    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 
M(N) equals zero 92 times.
M(N) crosses zero 59 times.

360 Assembly

*        Mertens function - 01/05/2023
MERTENS  CSECT
         USING  MERTENS,R13        base register
         B      72(R15)            skip savearea
         DC     17F'0'             savearea
         SAVE   (14,12)            save previous context
         ST     R13,4(R15)         link backward
         ST     R15,8(R13)         link forward
         LR     R13,R15            set addressability         
         LA     R0,1               1
         STH    R0,MM              m(1)=1
         LA     R6,2               i=2 
       DO WHILE=(CH,R6,LE,=AL2(NN)) do i=2 to n
         LR     R1,R6                i
         SLA    R1,1                 *2 (H)
         LA     R0,1                 1
         STH    R0,MM-2(R1)          m(i)=1
         LA     R7,2                 j=2
       DO WHILE=(CR,R7,LE,R6)        do j=2 to i
         LR     R4,R6                  i
         SRDA   R4,32                  ~
         LR     R1,R7                  j
         DR     R4,R1                  i/j
         LR     R8,R5                  d=i/j
         LR     R4,R6                  i
         SLA    R4,1                   *2 (H)
         LH     R2,MM-2(R4)            m(i)
         LR     R1,R8                  d
         SLA    R1,1                   *2 (H)
         LH     R3,MM-2(R1)            m(d)
         SR     R2,R3                  m(i)-m(d)
         STH    R2,MM-2(R4)            m(i)=m(i)-m(d)
         LA     R7,1(R7)               j++
       ENDDO    ,                    enddo j
         LA     R6,1(R6)             i++ 
       ENDDO    ,                  enddo i
         XPRNT  =C'the first 99 Mertens numbers are:',34  print buffer
         LA     R9,PG              @buffer=pg
         MVC    PG,=CL80' '        clean buffer
         MVC    0(3,R9),=CL3' '    output '   '
         LA     R9,3(R9)           @buffer+=3
         LA     R7,9               j=9
         LA     R6,1               i=1 
       DO WHILE=(CH,R6,LE,=AL2(99)) do i=1 to 99
         LR     R1,R6                i
         SLA    R1,1                 *2 (H)
         LH     R2,MM-2(R1)          m(i)
         XDECO  R2,XDEC              edit m(i)
         MVC    0(3,R9),XDEC+9       output m(i)
         LA     R9,3(R9)             @buffer+=3
         BCTR   R7,0                 j=j-1
       IF   LTR,R7,Z,R7 THEN         if j=0 then do;
         LA     R7,10                  j=10
         XPRNT  PG,L'PG                print buffer
         LA     R9,PG                  @buffer=pg
       ENDIF    ,                    endif
         LA     R6,1(R6)             i++ 
       ENDDO    ,                  enddo i
         SR     R10,R10            zero=0
         SR     R11,R11            cross=0
         LA     R6,1               i=2 
       DO WHILE=(CH,R6,LE,=AL2(NN)) do i=2 to n
         LR     R1,R6                i
         SLA    R1,1                 *2 (H)
         LH     R2,MM-2(R1)          m(i)
       IF   LTR,R2,Z,R2 THEN         if m(i)=0 then
         LA     R10,1(R10)             zero=zero+1
         LR     R1,R6                  i
         BCTR   R1,0                   i-1
         SLA    R1,1                   *2 (H)
         LH     R2,MM-2(R1)            m(i-1)
       IF   LTR,R2,NZ,R2 THEN          if m(i-1)^=0 then
         LA     R11,1(R11)               cross=cross+1
       ENDIF    ,                      endif
       ENDIF    ,                    endif
         LA     R6,1(R6)             i++ 
       ENDDO    ,                  enddo i
         MVC    PG,=CL80' '        clean buffer
         MVC    PG(13),=C'm(i) is zero '
         XDECO  R10,XDEC           edit zero
         MVC    PG+13(2),XDEC+10   output zero
         MVC    PG+15(7),=C' times.'
         XPRNT  PG,L'PG            print buffer
         MVC    PGI,=H'0'
         MVC    PG,=CL80' '        clean buffer
         MVC    PG(18),=C'm(i) crosses zero '
         XDECO  R11,XDEC           edit cross
         MVC    PG+18(2),XDEC+10   output cross
         MVC    PG+20(7),=C' times.'
         XPRNT  PG,L'PG            print buffer
         L      R13,4(0,R13)       restore previous savearea pointer
         RETURN (14,12),RC=0       restore registers from calling save
NN       EQU    1000               n
PG       DS     CL80               buffer
PGI      DC     H'0'               buffer index
XDEC     DS     CL12               temp for xdeci xdeco
MM       DS     (NN)H              m
         REGEQU
         END    MERTENS
Output:
the first 99 Mertens numbers are:
     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
m(i) is zero 92 times.
m(i) crosses zero 59 times.

8080 Assembly

MAX:	equ	1000	; Amount of numbers to generate
	org	100h
	;;;	Generate Mertens numbers
	lxi	b,1	; Start at place 1; BC = current Mertens number	
	lxi	h,MM	; First one is 1 
	dad	b
	mvi	m,1
outer:	inx 	b	; Next Mertens number
	lxi	h,MM
	dad	b
	mvi	m,1	; Initialize at 1
	lxi	d,2	; DE = inner loop counter ('k'), starts at 2
	;;;	Now we need to find BC/DE, but there is no hardware divide
	;;;	We also need to be somewhat clever so it doesn't take forever
inner:	push	d	; Keep both loop counters safe on the stack
	push 	b
	xchg		; Divisor in HL
	mov	d,b	; Dividend in DE
	mov	e,c 
	lxi	b,100h	; B = counter, C = zero
double:	dad	h	; Double divisor
	inr	b	; Increment counter
	call	cdehl	; Dividend <= divisor?
	jnc	double	; If so, keep doubling	
	mov	a,b	; Keep counter
	mov	b,c	; BC = 0
	push 	b	; Push result variable on stakc (initial 0)
	mov	b,a	; Restore counter
	xchg		; HL = dividend, DE = doubled divisor
subtr:	mov	a,l	; Try HL -= DE
	sub	e
	mov	l,a
	mov	a,h
	sbb	d
	mov	h,a	
	xthl		; Get result accumulator from stack 
	cmc		; Flip borrow
	mov	a,l	; Rotate into result
	ral
	mov	l,a
	mov	a,h
	ral
	mov	h,a
	mov	a,l	; Retrieve flag
	rar
	xthl		; Retrieve rest of divisor
	jc	$+4	; If borrow,
	dad	d	; Add dividend back into divisor	
	xra     a       ; DE >> 1
	ora     d
	rar
	mov     d,a
	mov     a,e
	rar
	mov     e,a
	dcr     b       ; Are we there yet?
	jnz     subtr   ; If not, try another subtraction
	pop	h  	; HL = quotient
	;;;	Division is done, do lookup and subraction
	lxi	d,MM	; Look up M[outer/inner]
	dad	d
	mov	e,m	; E = M[BC/DE]
	pop	b	; Restore BC (n)
	lxi	h,MM
	dad	b
	mov	a,m 	; A = M[BC]
	sub	e	; A = M[BC] - M[BC/DE]
	mov	m,a	; M[BC] = A
	pop	d	; Restore DE (k)
	;;;	Update loops
	inx	d	; k++	
	call	cbcde	; DE <= BC? 
	jnc	inner
	lxi	h,MAX
	call	chlbc	; BC <= MAX?
	jnc	outer
	;;;	Print table
	lxi	d,frst99
	call	puts
	lxi	h,MM+1	; Start of Merten numbers
	mvi	c,9	; Column counter
table:	mov	a,m	; Get Merten number
	ana	a	; Set flags
	mvi	b,' '	; Space
	jp 	prtab	; If positive, print space-number-space
	mvi	b,'-'	; Otherwise, print minus sign
	cma		; And negate the number (make positive)
	inr	a
prtab:	adi	'0'	; Make ASCII digit
	mov	d,a	; Keep number
	mov	a,b	; Print space or minus sign
	call	putc
	mov	a,d	; Restore number
	call	putc	; Print number
	mvi	a,' '	; Print space
	call	putc
	dcr	c	; Decrement column counter
	jnz	tnext
	lxi	d,nl	; End of columns - print newline
	call	puts
	mvi	c,10	; Column counter
tnext:	inx	h	; Table done?
	mov	a,l
	cpi	100
	jnz	table	; If not, keep going
	;;;	Find zeroes and crossings
	lxi	b,0	; B=zeroes, C=crossings 
	lxi	d,MAX	; Counter
	lxi	h,MM+1
count:	mov	a,m	; Get number
	ana	a 	; Zero?
	jnz	cnext 
	inr	b	; If so, add zero
	dcx	h	; Previous number also zero?
	mov	a,m
	inx	h
	ana	a
	jz	cnext
	inr	c	; If not, add crossiong
cnext:	inx	h
	dcx	d
	mov	a,d
	ora	e
	jnz	count
	lxi	d,zero	; Print zeroes
	call	puts
	mov	a,b
	call	puta
	lxi	d,cross ; Print crossings
	call	puts
	mov	a,c
	call	puta
	lxi	d,tms
	jmp	puts
	;;;	Print character in A using CP/M, keeping registers
putc:	push	b
	push	d
	push	h
	mov	e,a
	mvi	c,2
	call	5
	jmp	resrgs
	;;;	Print number in A, keeping registers
puta:	push	b
	push	d
	push 	h
	lxi	h,num
putad:	mvi	c,-1	
putal:	inr 	c
	sui	10
	jnc	putal
	adi	10+'0'
	dcx	h
	mov	m,a
	mov	a,c
	ana	a
	jnz	putad
	xchg
	mvi	c,9
	call	5
	jmp	resrgs
	;;;	Print string in DE using CP/M, keeping registers
puts:	push	b
	push	d
	push	h
	mvi	c,9
	call	5
resrgs:	pop	h
	pop	d
	pop 	b
	ret
cdehl:	mov	a,d
	cmp 	h
	rnz
	mov	a,e
	cmp	l
	ret
cbcde:	mov	a,b
	cmp	d
	rnz
	mov	a,c
	cmp	e
	ret 
chlbc:	mov	a,h
	cmp	b
	rnz
	mov	a,l
	cmp	c
	ret
	;;;	Strings
	db	'***'
num:	db	'$'
frst99:	db	'First 99 Mertens numbers:',13,10,'   $'
nl:	db	13,10,'$'	
zero:	db	'M(N) is zero $'
cross:	db	' times.',13,10,'M(N) crosses zero $'
tms:	db	' times.$'
	;;; Numbers are stored page-aligned after program
MM:	equ	($/256)*256+256
Output:
First 99 Mertens numbers:
    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
M(N) is zero 92 times.
M(N) crosses zero 59 times.

8086 Assembly

MAX:	equ	1000		; Amount of Mertens numbers to generate
puts:	equ	9		; MS-DOS syscall to print a string
putch:	equ	2		; MS-DOS syscall to print a character
	cpu	8086
	org	100h
section	.text
	;;;	Generate Mertens numbers 
	mov	bx,M		; BX = pointer to start of Mertens numbers
	mov	si,1		; Current Mertens number
	mov	[si+bx],byte 1	; First Mertens number is 1
outer:	inc	si		; Next Mertens number
	mov	[si+bx],byte 1	; Starts out at 1...
	mov	cx,2		; CX = from 2 to current number,
inner:	mov	ax,si		; Divide current number,
	xor	dx,dx
	div	cx		; By CX
	mov	di,ax
	mov	al,[di+bx]	; Get value at that location
	sub	[si+bx],al 	; Subtract from current number
	inc	cx
	cmp	cx,si
	jbe	inner
	cmp	si,MAX
	jbe	outer
	;;;	Print the table
	mov	dx,frst99	; First string
	call	outstr
	mov	si,1		; Start at index 1
	mov	dh,9		; Column count
table:	mov	cl,[si+bx]	; Get item
	test	cl,cl
	mov	dl,' '
	jns	.print		; Positive?
	mov	dl,'-'		; Otherwise, it is negative,
	neg	cl		; print ' ' and negate
.print:	call	putc		; Print space or minus
	add	cl,'0'		; Add ASCII 0
	mov	dl,cl
	call	putc		; Print number
	mov 	dl,' '
	call	putc		; Print space
	dec	dh		; One less column left
	jnz	.next
	mov	dx,nl		; Print newline
	call	outstr
	mov	dh,10
.next:	inc	si		; Done yet?
	cmp	si,100
	jb	table		; If not, print next item from table
	;;;	Calculate zeroes and crossings
	xor	cx,cx		; CL = zeroes, CH = crossings
	mov	si,1
	mov	al,[si+bx]	; AL = current item
zc:	inc	si
	mov	ah,al		; AH = previous item
	mov	al,[si+bx]
	test	al,al		; Zero?
	jnz	.next
	inc	cx		; Then increment zero counter
	test	ah,ah		; Previous one also zero?
	jz 	.next
	inc	ch		; Then increment crossing counter
.next:	cmp	si,MAX		; Done yet?
	jbe	zc
	;;;	Print zeroes and crossings
	mov	dx,zero
	call	outstr
	mov	al,cl
	call	putal
	mov	dx,cross
	call 	outstr
	mov	al,ch
	call	putal
	mov	dx,tms
	jmp	outstr 
putc:	mov	ah,putch	; Print character
	int	21h
	ret
	;;;	Print AL in decimal format
putal:	mov	di,num
.loop:	aam			; Extract digit
	add	al,'0'		; Store digit
	dec	di
	mov	[di],al
	mov	al,ah		; Rest of number
	test	al,al		; Done?
	jnz	.loop		; If not, get more digits
	mov	dx,di		; Otherwise, print string
outstr:	mov	ah,puts
	int	21h
	ret
section .data
	db	'***'		; Number output placeholder
num:	db	'$'
frst99:	db	'First 99 Mertens numbers:',13,10,'   $'
nl:	db	13,10,'$'	
zero:	db	'M(N) is zero $'
cross:	db	' times.',13,10,'M(N) crosses zero $'
tms:	db	' times.$'
section .bss
mm:	resb	MAX	; Mertens numbers
M:	equ	mm-1	; 1-based indexing
Output:
First 99 Mertens numbers:
    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
M(N) is zero 92 times.
M(N) crosses zero 59 times.

Action!

Calculations on a real Atari 8-bit computer take quite long time. It is recommended to use an emulator capable with increasing speed of Atari CPU.

INCLUDE "D2:PRINTF.ACT" ;from the Action! Tool Kit

PROC MertensNumbers(INT ARRAY m INT count)
  INT n,k

  m(1)=1
  FOR n=2 TO count
  DO
    m(n)=1
    FOR k=2 TO n
    DO
      m(n)==-m(n/k)
    OD
  OD
RETURN

PROC PrintMertens(INT ARRAY m INT count)
  CHAR ARRAY s(6)
  INT i,col

  PrintF("First %I Mertens numbers:%E   ",count)
  col=1
  FOR i=1 TO count
  DO
    StrI(m(i),s)
    PrintF("%3S",s)
    col==+1
    IF col=10 THEN
      col=0 PutE()
    FI
  OD
RETURN

PROC Main()
  DEFINE MAX="1001"
  INT ARRAY m(MAX)
  INT i,zeroCnt=[0],crossCnt=[0],prev=[0]

  Put(125) PutE() ;clear the screen
  PrintF("Calculation of Mertens numbers,%E  please wait...")
  MertensNumbers(m,MAX)

  Put(125) PutE() ;clear the screen
  PrintMertens(m,99)

  FOR i=1 TO MAX
  DO
    IF m(i)=0 THEN
      zeroCnt==+1
      IF prev THEN
        crossCnt==+1
      FI
    FI
    prev=m(i)
  OD
  PrintF("%EM(n) is zero %I times for 1<=n<=%I.%E",zeroCnt,MAX-1)
  PrintF("%EM(n) crosses zero %I times for 1<=n<=%I.%E",crossCnt,MAX-1)
RETURN
Output:

Screenshot from Atari 8-bit computer

First 99 Mertens numbers:
     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

M(n) is zero 92 times for 1<=n<=1000.

M(n) crosses zero 59 times for 1<=n<=1000.

ALGOL 68

Translation of: ALGOL W

...which is...

Translation of: Fortran
BEGIN # compute values of the Mertens function                      #
    # Generate Mertens numbers                                      #       
    [ 1 : 1000 ]INT m;
    m[ 1 ] := 1;
    FOR n FROM 2 TO UPB m DO
        m[ n ] := 1;
        FOR k FROM 2 TO n DO m[ n ] -:= m[ n OVER k ] OD
    OD;
    # Print table                                                   #
    print( ( "The first 99 Mertens numbers are:", newline ) );
    print( ( "   " ) );
    INT k := 9;
    FOR n TO 99 DO
        print( ( whole( m[ n ], -3 ) ) );
        IF ( k -:= 1 ) = 0 THEN
            k := 10;
            print( ( newline ) )
        FI
    OD;
    # Calculate zeroes and crossings                                #
    INT zero  := 0;
    INT cross := 0;
    FOR n FROM 2 TO UPB m DO
        IF m[ n ] = 0 THEN
            zero +:= 1;
            IF m[ n - 1 ] /= 0 THEN cross +:= 1 FI
        FI
    OD;
    print( ( newline ) );
    print( ( "M(N) is zero      ", whole( zero,  -4 ), " times.", newline ) );
    print( ( "M(N) crosses zero ", whole( cross, -4 ), " times.", newline ) )
END
Output:
The first 99 Mertens numbers are:
     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

M(N) is zero        92 times.
M(N) crosses zero   59 times.

ALGOL W

Translation of: Fortran
begin % compute values of the Mertens function                      %
    integer array M ( 1 :: 1000 );
    integer k, zero, cross;
    % Generate Mertens numbers                                      %       
    M( 1 ) := 1;
    for n := 2 until 1000 do begin
        M( n ) := 1;
        for k := 2 until n do M( n ) := M( n ) - M( n div k )
    end for_n ;
    % Print table                                                   %
    write( "The first 99 Mertens numbers are:" );
    write( "   " );
    k := 9;
    for n := 1 until 99 do begin
        writeon( i_w := 3, s_w := 0, M( n ) );
        k := k - 1;
        if k = 0 then begin
            k := 10;
            write()
        end if_k_eq_0
    end for_n ;
    % Calculate zeroes and crossings                                %
    zero  := 0;
    cross := 0;
    for n :=2 until 1000 do begin
        if M( n ) = 0 then begin
            zero := zero + 1;
            if M( n - 1 ) not = 0 then cross := cross + 1
        end if_M_n_eq_0
    end for_n ;
    write( i_w := 2, s_w := 0, "M(N) is zero ", zero, " times." );
    write( i_w := 2, s_w := 0, "M(N) crosses zero ", cross, " times." )
end.
Output:
The first 99 Mertens numbers are:
     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

M(N) is zero 92 times.
M(N) crosses zero 59 times.

APL

Works with: Dyalog APL
mertens{
    step   {,-⍨/1,[n÷1↓⍳n1+≢]}
    m1000  step999⊢,1
    zero   m1000+.=0
    cross  +/(~∧1⌽⊢)m10000
    'First 99 Mertens numbers:'
    10 10'∘',m1000
    'M(N) is zero ',(zero),' times.'
    'M(N) crosses zero ',(cross),' times.'
}
Output:
First 99 Mertens numbers:
 ∘  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
M(N) is zero 92 times.
M(N) crosses zero 59 times.

Arturo

mobius: function [n][
    if n=0 -> return ""
    if n=1 -> return 1
    f: factors.prime n

    if f <> unique f -> return 0
    if? odd? size f -> return neg 1
    else -> return 1
]

mertens: function [z][sum map 1..z => mobius]

print "The first 99 Mertens numbers are:"
loop split.every:20 [""]++map 1..99 => mertens 'a [
    print map a 'item -> pad to :string item 2
]

print ""

mertens1000: map 1..1000 => mertens
print ["Times M(x) is zero between 1 and 1000:" size select mertens1000 => zero?]

crossed: new 0
fold mertens1000 [a,b][if and? zero? b not? zero? a -> inc 'crossed, b]
print ["Times M(x) crosses zero between 1 and 1000:" crossed]
Output:
The first 99 Mertens numbers are:
    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 

Times M(x) is zero between 1 and 1000: 92 
Times M(x) crosses zero between 1 and 1000: 59

AutoHotkey

result := "first 100 terms:`n"
loop 100
    result .= SubStr("  " Mertens(A_Index), -1) . (Mod(A_Index, 10) ? "  " : "`n")

eqZero := crZero := 0, preced:=1
loop 1000
{
    if !(x := Mertens(A_Index))
        eqZero++, crZero += preced<>0 ? 1 : 0
    preced := x
}
result .= "`nfirst 1000 terms:"
MsgBox, 262144, , % result .= "`nequal to zero : " eqZero "`ncrosses zero : " crZero
return

Mertens(n){
    loop % n
        result += Möbius(A_Index)
    return result
}


Möbius(n){
    if n=1
        return 1
    x := prime_factors(n)
    c := x.Count()
    sq := []
    for i, v in x
        if sq[v]
            return 0
        else
            sq[v] := 1
    return (c/2 = floor(c/2)) ? 1 : -1
}

prime_factors(n) {
    if (n <= 3)
        return [n]
    ans := [], done := false
    while !done {
        if !Mod(n, 2)
            ans.push(2), n /= 2
        else if !Mod(n, 3)
            ans.push(3), n /= 3
        else if (n = 1)
            return ans
        else {
            sr := sqrt(n), done := true, i := 6
            while (i <= sr+6) {
                if !Mod(n, i-1) { ; is n divisible by i-1?
                    ans.push(i-1), n /= i-1, done := false
                    break
                }
                if !Mod(n, i+1) { ; is n divisible by i+1?
                    ans.push(i+1), n /= i+1, done := false
                    break
                }
                i += 6
    }}}
    ans.push(Format("{:d}", n))
    return ans
}
Output:
first 100 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

first 1000 terms:
equal to zero : 92
crosses zero : 59

BASIC

Note that if you actually try this on a real 8-bit micro, this will take literal hours to run. Interpreted BASIC is not fast.

For comparison, the 8080 and 8086 assembly versions above run in around 4 minutes and 15 seconds respectively, on real hardware. This took nearly 4 hours on the 8080 (using MBASIC) and a little over 1 hour on the 8086 (using GWBASIC).

10 DEFINT C,Z,N,K,M: DIM M(1000)
20 M(1)=1
30 FOR N=2 TO 1000
40 M(N)=1
50 FOR K=2 TO N: M(N) = M(N)-M(INT(N/K)): NEXT
60 NEXT
70 PRINT "First 99 Mertens numbers:"
80 PRINT "   ";
90 FOR N=1 TO 99
100 PRINT USING "###";M(N);
110 IF N MOD 10 = 9 THEN PRINT
120 NEXT
130 C=0: Z=0 
140 FOR N=1 TO 1000
150 IF M(N)=0 THEN Z=Z+1: IF M(N-1)<>0 THEN C=C+1
160 NEXT
170 PRINT "M(N) is zero";Z;"times."
180 PRINT "M(N) crosses zero";C;"times."
190 END
Output:
First 99 Mertens numbers:
     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
M(N) is zero 92 times.
M(N) crosses zero 59 times.

BASIC256

arraybase 1
dim M(1000)
M[1] = 1
for n = 2 to 1000
	M[n] = 1
	for k = 2 to n
		M[n] = M[n] - M[int(n/k)]
	next k
next n
print "First 99 Mertens numbers:"
print "   ";
for n = 1 to 99
	print rjust(string(M[n]),3);
	if n mod 10 = 9 then print
next n
numCruza = 0
numEsCero = 0
for n = 1 to 1000
	if M[n] = 0 then
		numEsCero += 1
		if M[n-1] <> 0 then numCruza += 1
	end if
next n
print
print "M(n) is zero "; numEsCero; " times."
print "M(n) crosses zero "; numCruza; " times."
Output:
Same as BASIC entry.

Run BASIC

Works with: Just BASIC
Works with: Liberty BASIC
dim M(1000)
M(1) = 1
for n = 2 to 1000
    M(n) = 1
    for k = 2 to n
        M(n) = M(n)-M(int(n/k))
    next k
next n
print "First 99 Mertens numbers:"
print "   ";
for n = 1 to 99
    print using("###", M(n));
    if n mod 10 = 9 then print
next n
numCruza = 0
numEsCero = 0
for n = 1 to 1000
    if M(n) = 0 then
        numEsCero = numEsCero +1
        if M(n-1) <> 0 then numCruza = numCruza +1
    end if
next n
print
print "M(n) is zero "; numEsCero; " times."
print "M(n) crosses zero "; numCruza; " times."
Output:
Same as BASIC entry.

True BASIC

DIM m(1000)
LET m(1) = 1
FOR n = 2 TO 1000
    LET m(n) = 1
    FOR k = 2 TO n
        LET m(n) = m(n)-m(INT(n/k))
    NEXT k
NEXT n
PRINT "First 99 Mertens numbers:"
PRINT "   ";
FOR n = 1 TO 99
    PRINT " "; 
    PRINT USING "##": m(n);
    !IF REMAINDER(ROUND(n),10) = 9 THEN PRINT
    IF MOD(n,10) = 9 THEN PRINT
NEXT n
LET numcruza = 0
LET numeszero = 0
FOR n = 1 TO 1000
    IF m(n) = 0 THEN
       LET numeszero = numeszero+1
       IF m(n-1) <> 0 THEN LET numcruza = numcruza+1
    END IF
NEXT n
PRINT
PRINT "M(n) is zero"; numeszero; "times."
PRINT "M(n) crosses zero"; numcruza; "times."
END
Output:
Same as FreeBASIC entry.

XBasic

Works with: Windows XBasic
PROGRAM	"Mertens"
VERSION	"0.0000"

DECLARE FUNCTION  Entry ()

FUNCTION  Entry ()
DIM M[1000]
M[1] = 1
FOR n = 2 TO 1000
	M[n] = 1
	FOR k = 2 TO n
		M[n] = M[n] - M[INT(n/k)]
	NEXT k
NEXT n
PRINT "First 99 Mertens numbers:"
PRINT "   ";
FOR n = 1 TO 99
	PRINT FORMAT$("###", M[n]);
	IF n MOD 10 = 9 THEN PRINT
NEXT n
numCruza = 0
numEsCero = 0
FOR n = 1 TO 1000
	IF M[n] = 0 THEN
		INC numEsCero
		IF M[n-1] <> 0 THEN INC numCruza
	END IF
NEXT n
PRINT
PRINT "M(n) is zero"; numEsCero; " times."
PRINT "M(n) crosses zero"; numCruza; " times."

END FUNCTION
END PROGRAM
Output:
Same as FreeBASIC entry.

Yabasic

dim M(1000)
M(1) = 1
for n = 2 to 1000
	M(n) = 1
	for k = 2 to n
		M(n) = M(n) - M(int(n/k))
	next k
next n
print "First 99 Mertens numbers:"
print "    ";
for n = 1 to 99
	print M(n) using("###");
	if mod(n, 10) = 9  print
next n
numCruza = 0
numEsCero = 0
for n = 1 to 1000
	if M(n) = 0 then
		numEsCero = numEsCero + 1
		if M(n-1) <> 0  numCruza = numCruza + 1
	end if
next n
print
print "M(n) is zero ", numEsCero, " times."
print "M(n) crosses zero ", numCruza, " times."
Output:
Same as FreeBASIC entry.
==Bash==
#!/bin/bash
MAX=1000

m[1]=1
for n in `seq 2 $MAX`
do
    m[n]=1
    for k in `seq 2 $n`
    do
        m[n]=$((m[n]-m[n/k]))
    done
done

echo 'The first 99 Mertens numbers are:'
echo -n '   '
for n in `seq 1 99`
do
    printf '%2d ' ${m[n]}
    test $((n%10)) -eq 9 && echo
done

zero=0
cross=0
for n in `seq 1 $MAX`
do
    if [ ${m[n]} -eq 0 ]
    then
        ((zero++))
        test ${m[n-1]} -ne 0 && ((cross++))
    fi
done

echo "M(N) is zero $zero times."
echo "M(N) crosses zero $cross times."
Output:
The first 99 Mertens numbers are:
    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
M(N) is zero 92 times.
M(N) crosses zero 59 times.

BCPL

get "libhdr"

manifest $( limit = 1000 $)

let mertens(v, max) be
$(  v!1 := 1
    for n = 2 to max do
    $(  v!n := 1
        for k = 2 to n do
            v!n := v!n - v!(n/k)
    $)
$)

let start() be
$(  let m = vec limit
    let eqz, crossz = 0, 0
    
    writes("The first 99 Mertens numbers are:*N")
    mertens(m, limit)
    for y=0 to 90 by 10 do
    $(  for x=0 to 9 do
            test x+y=0
                then writes("   ")
                else writed(m!(x+y),3)
        wrch('*N')
    $)
    
    for x=2 to limit do
        if m!x=0 then 
        $(  eqz := eqz + 1
            unless m!(x-1)=0 do crossz := crossz + 1
        $)
        
    writef("M(N) is zero %N times.*N", eqz)
    writef("M(N) crosses zero %N times.*N", crossz)
$)
Output:
The first 99 Mertens numbers are:
     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
M(N) is zero 92 times.
M(N) crosses zero 59 times.

C

#include <stdio.h>
#include <stdlib.h>

int* mertens_numbers(int max) {
    int* m = malloc((max + 1) * sizeof(int));
    if (m == NULL)
        return m;
    m[1] = 1;
    for (int n = 2; n <= max; ++n) {
        m[n] = 1;
        for (int k = 2; k <= n; ++k)
            m[n] -= m[n/k];
    }
    return m;
}

int main() {
    const int max = 1000;
    int* mertens = mertens_numbers(max);
    if (mertens == NULL) {
        fprintf(stderr, "Out of memory\n");
        return 1;
    }
    printf("First 199 Mertens numbers:\n");
    const int count = 200;
    for (int i = 0, column = 0; i < count; ++i) {
        if (column > 0)
            printf(" ");
        if (i == 0)
            printf("  ");
        else
            printf("%2d", mertens[i]);
        ++column;
        if (column == 20) {
            printf("\n");
            column = 0;
        }
    }
    int zero = 0, cross = 0, previous = 0;
    for (int i = 1; i <= max; ++i) {
        int m = mertens[i];
        if (m == 0) {
            ++zero;
            if (previous != 0)
                ++cross;
        }
        previous = m;
    }
    free(mertens);
    printf("M(n) is zero %d times for 1 <= n <= %d.\n", zero, max);
    printf("M(n) crosses zero %d times for 1 <= n <= %d.\n", cross, max);
    return 0;
}
Output:
First 199 Mertens numbers:
    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
M(n) is zero 92 times for 1 <= n <= 1000.
M(n) crosses zero 59 times for 1 <= n <= 1000.

C++

#include <iomanip>
#include <iostream>
#include <vector>

std::vector<int> mertens_numbers(int max) {
    std::vector<int> m(max + 1, 1);
    for (int n = 2; n <= max; ++n) {
        for (int k = 2; k <= n; ++k)
            m[n] -= m[n / k];
    }
    return m;
}

int main() {
    const int max = 1000;
    auto m(mertens_numbers(max));
    std::cout << "First 199 Mertens numbers:\n";
    for (int i = 0, column = 0; i < 200; ++i) {
        if (column > 0)
            std::cout << ' ';
        if (i == 0)
            std::cout << "  ";
        else
            std::cout << std::setw(2) << m[i];
        ++column;
        if (column == 20) {
            std::cout << '\n';
            column = 0;
        }
    }
    int zero = 0, cross = 0, previous = 0;
    for (int i = 1; i <= max; ++i) {
        if (m[i] == 0) {
            ++zero;
            if (previous != 0)
                ++cross;
        }
        previous = m[i];
    }
    std::cout << "M(n) is zero " << zero << " times for 1 <= n <= 1000.\n";
    std::cout << "M(n) crosses zero " << cross << " times for 1 <= n <= 1000.\n";
    return 0;
}
Output:
First 199 Mertens numbers:
    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
M(n) is zero 92 times for 1 <= n <= 1000.
M(n) crosses zero 59 times for 1 <= n <= 1000.

CLU

% Generate Mertens numbers up to a given limit
mertens = proc (limit: int) returns (array[int])
    M: array[int] := array[int]$fill(1,limit,0)
    M[1] := 1
    for n: int in int$from_to(2,limit) do
        M[n] := 1
        for k: int in int$from_to(2,n) do
            M[n] := M[n] - M[n/k]
        end
    end
    return (M)
end mertens

start_up = proc ()
    max = 1000

    po: stream := stream$primary_output()
    M: array[int] := mertens(max)

    stream$putl(po, "The first 99 Mertens numbers are:")
    for y: int in int$from_to_by(0,90,10) do
        for x: int in int$from_to(0,9) do
            stream$putright(po, int$unparse(M[x+y]), 3)
            except when bounds:
                stream$putright(po, "", 3)
            end
        end
        stream$putl(po, "")
    end

    eqz: int := 0
    crossz: int := 0
    for i: int in int$from_to(2,max) do
        if M[i]=0 then
            eqz := eqz + 1
            if M[i-1]~=0 then crossz := crossz + 1 end
        end
    end

    stream$putl(po, "M(N) is zero " || int$unparse(eqz) || " times.")
    stream$putl(po, "M(N) crosses zero " || int$unparse(crossz) || " times.")
end start_up
Output:
The first 99 Mertens numbers are:
     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
M(N) is zero 92 times.
M(N) crosses zero 59 times.

COBOL

       IDENTIFICATION DIVISION.
       PROGRAM-ID.  MERTENS.

       DATA DIVISION.
       WORKING-STORAGE SECTION.
       01 VARIABLES.
          03 M             PIC S99 OCCURS 1000 TIMES.
          03 N             PIC 9(4).
          03 K             PIC 9(4).
          03 V             PIC 9(4).
          03 IS-ZERO       PIC 99 VALUE 0.
          03 CROSS-ZERO    PIC 99 VALUE 0.

       01 OUTPUT-FORMAT.
          03 OUT-ITEM. 
             05 OUT-NUM    PIC -9.
             05 FILLER     PIC X VALUE SPACE.
          03 OUT-LINE      PIC X(30) VALUE SPACES.
          03 OUT-PTR       PIC 99 VALUE 4.

       PROCEDURE DIVISION.
       BEGIN.
           PERFORM GENERATE-MERTENS.
           PERFORM WRITE-TABLE.
           PERFORM COUNT-ZEROES.
           STOP RUN.

       GENERATE-MERTENS.
           MOVE 1 TO M(1).
           PERFORM MERTENS-OUTER-LOOP VARYING N FROM 2 BY 1
               UNTIL N IS GREATER THAN 1000.
    
       MERTENS-OUTER-LOOP.
           MOVE 1 TO M(N).
           PERFORM MERTENS-INNER-LOOP VARYING K FROM 2 BY 1
               UNTIL K IS GREATER THAN N.

       MERTENS-INNER-LOOP.
           DIVIDE N BY K GIVING V.
           SUBTRACT M(V) FROM M(N).

       WRITE-TABLE.
           DISPLAY "The first 99 Mertens numbers are: "
           PERFORM WRITE-ITEM VARYING N FROM 1 BY 1
               UNTIL N IS GREATER THAN 99.
  
       WRITE-ITEM.
           MOVE M(N) TO OUT-NUM.
           STRING OUT-ITEM DELIMITED BY SIZE INTO OUT-LINE
               WITH POINTER OUT-PTR.
           IF OUT-PTR IS EQUAL TO 31,
               DISPLAY OUT-LINE,
               MOVE 1 TO OUT-PTR.

       COUNT-ZEROES.
           PERFORM TEST-N-ZERO VARYING N FROM 2 BY 1
               UNTIL N IS GREATER THAN 1000.
           DISPLAY "M(N) is zero " IS-ZERO " times.".
           DISPLAY "M(N) crosses zero " CROSS-ZERO " times.".

       TEST-N-ZERO.
           IF M(N) IS EQUAL TO ZERO,
               ADD 1 TO IS-ZERO,
               SUBTRACT 1 FROM N GIVING K,
               IF M(K) IS NOT EQUAL TO ZERO,
                   ADD 1 TO CROSS-ZERO.
Output:
The first 99 Mertens numbers are: 
    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 
M(N) is zero 92 times.
M(N) crosses zero 59 times.

Cowgol

include "cowgol.coh";

const MAX := 1000;

# Table output
sub printtab(n: int8) is
    if n<0 then 
        print_char('-');
        n := -n;
    else 
        print_char(' ');
    end if;
    print_char(n as uint8 + '0');
    print_char(' ');
end sub;
    
# Generate Merten numbers
var M: int8[MAX+1];
M[0] := 0;
M[1] := 1;

var n: @indexof M := 2;
while n < @sizeof M loop
    M[n] := 1;
    var k: @indexof M := 2;
    while k <= n loop
        M[n] := M[n] - M[n/k];
        k := k + 1;
    end loop;
    n := n + 1;
end loop;

# Find zeroes and crossings
var zero: uint8 := 0;
var cross: uint8 := 0;
n := 1;
while n < @sizeof M loop
    if M[n] == 0 then
        zero := zero + 1;
        if M[n-1] != 0 then
            cross := cross + 1;
        end if;
    end if;
    n := n + 1;
end loop;

# Print table
print("The first 99 Mertens numbers are:\n");
print("   ");
n := 1;
var col: uint8 := 9;
while n < 100 loop
    printtab(M[n]);
    col := col - 1;
    if col == 0 then    
        print_nl();
        col := 10;
    end if;
    n := n + 1;
end loop;

print("M(n) is zero "); print_i8(zero); print(" times\n");
print("M(n) crosses zero "); print_i8(cross); print(" times\n");
Output:
The first 99 Mertens numbers are:
    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
M(n) is zero 92 times
M(n) crosses zero 59 times

Delphi

Translation of: Go
program Mertens_function;

{$APPTYPE CONSOLE}

uses
  System.SysUtils;

type
  TMertens = record
    merts: TArray<Integer>;
    zeros, crosses: Integer;
    class function Mertens(_to: Integer): TMertens; static;
  end;

{ TMertens }

class function TMertens.Mertens(_to: Integer): TMertens;
var
  sum, zeros, crosses: Integer;
begin
  if _to < 1 then
    _to := 1;

  sum := 0;
  zeros := 0;
  crosses := 0;

  SetLength(Result.merts, _to + 1);
  var primes := [2];
  for var i := 1 to _to do
  begin
    var j := i;
    var cp := 0;
    var spf := false;
    for var p in primes do
    begin
      if p > j then
        Break;
      if j mod p = 0 then
      begin
        j := j div p;
        inc(cp);
      end;

      if j mod p = 0 then
      begin
        spf := true;
        Break;
      end;
    end;
    if (cp = 0) and (i > 2) then
    begin
      cp := 1;
      SetLength(primes, Length(primes) + 1);
      primes[High(primes)] := i;
    end;

    if not spf then
    begin
      if cp mod 2 = 0 then
        inc(sum)
      else
        dec(sum);
    end;

    Result.merts[i] := sum;
    if sum = 0 then
    begin
      inc(zeros);
      if (i > 1) and (Result.merts[i - 1] <> 0) then
        inc(crosses);
    end;
  end;
  Result.zeros := zeros;
  Result.crosses := crosses;
end;

begin
  var m := TMertens.mertens(1000);
  writeln('Mertens sequence - First 199 terms:');
  for var i := 0 to 199 do
  begin
    if i = 0 then
    begin
      write('    ');
      Continue;
    end;
    if i mod 20 = 0 then
      writeln;
    write(format(' %3d', [m.merts[i]]));
  end;
  writeln(#10#10'Equals zero ', m.zeros, ' times between 1 and 1000');
  writeln(#10'Crosses zero ', m.crosses, ' times between 1 and 1000');
  {$IFNDEF UNIX} readln; {$ENDIF}
end.

Draco

proc nonrec mertens([*] short m) void:
    word n,k;
    m[1] := 1;
    for n from 2 upto dim(m,1)-1 do
        m[n] := 1;
        for k from 2 upto n do
            m[n] := m[n] - m[n/k]
        od
    od
corp

proc nonrec main() void:
    [1001] short m;
    word x, y, eqz, crossz;
    
    mertens(m);
    writeln("The first 99 Mertens numbers are:");
    for y from 0 by 10 upto 90 do
        for x from 0 upto 9 do
            if x+y=0
                then write("   ")
                else write(m[x+y]:3)
            fi
        od;
        writeln()
    od;
    
    eqz := 0;
    crossz := 0;
    for x from 2 upto dim(m,1)-1 do
        if m[x]=0 then
            eqz := eqz + 1;
            if m[x-1]~=0 then crossz := crossz + 1 fi
        fi
    od;
    writeln("M(N) is zero ",eqz," times.");
    writeln("M(N) crosses zero ",crossz," times.")
corp
Output:
The first 99 Mertens numbers are:
     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
M(N) is zero 92 times.
M(N) crosses zero 59 times.

Dyalect

Translation of: Swift
func mertensNumbers(max) {
    let mertens = Array.Empty(max + 1, 1)
    for n in 2..max {
        for k in 2..n {
            mertens[n] -= mertens[n / k]
        }
    }
    mertens
}
 
let max = 1000
let mertens = mertensNumbers(max)
 
let count = 200
let columns = 20
print("First \(count - 1) Mertens numbers:")
 
for i in 0..<count {
    if i % columns > 0 {
        print(" ", terminator: "")
    }
    print(i == 0 ? "   " : mertens[i].ToString().PadLeft(2, ' ') + " ", terminator: "")
    if (i + 1) % columns == 0 {
        print()
    }
}
 
var (zero, cross, previous) = (0, 0, 0)
for i in 1..max {
    let m = mertens[i]
    if m == 0 {
        zero += 1
        if previous != 0 {
            cross += 1
        }
    }
    previous = m
}
 
print("M(n) is zero \(zero) times for 1 <= n <= \(max).")
print("M(n) crosses zero \(cross) times for 1 <= n <= \(max).")
Output:
First 199 Mertens numbers:
     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 
M(n) is zero 92 times for 1 <= n <= 1000.
M(n) crosses zero 59 times for 1 <= n <= 1000.

EasyLang

Translation of: FutureBasic
len mertens[] 1000
mertens[1] = 1
for n = 2 to 1000
   mertens[n] = 1
   for k = 2 to n
      mertens[n] -= mertens[n div k]
   .
.
print "First 99 Mertens numbers:"
write "   "
numfmt 0 2
for n = 1 to 99
   write mertens[n] & " "
   if n mod 10 = 9
      print ""
   .
.
for n = 1 to 1000
   if mertens[n] = 0
      zeros += 1
      if mertens[n - 1] <> 0
         crosses += 1
      .
   .
.
print ""
print "In the first 1000 terms of the Mertens sequence there are:"
print zeros & " zeros"
print crosses & " zero crosses"
Output:
First 99 Mertens numbers:
    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 

In the first 1000 terms of the Mertens sequence there are:
92 zeros
59 zero crosses

F#

This task uses Möbius_function (F#)

// Mertens function. Nigel Galloway: January 31st., 2021
let mertens=mobius|>Seq.scan((+)) 0|>Seq.tail
mertens|>Seq.take 500|>Seq.chunkBySize 25|>Seq.iter(fun n->Array.iter(printf "%3d") n;printfn "\n####")
let n=mertens|>Seq.take 1000|>Seq.mapi(fun n g->(n+1,g))|>Seq.groupBy snd|>Map.ofSeq
n|>Map.iter(fun n g->printf "%3d->" n; g|>Seq.iter(fun(n,_)->printf "%3d " n); printfn "\n####")
printfn "%d Zeroes\n####" (Seq.length (snd n.[0]))
printfn "Crosses zero %d times" (mertens|>Seq.take 1000|>Seq.pairwise|>Seq.sumBy(fun(n,g)->if n<>0 && g=0 then 1 else 0)))
Output:
  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
 -7 -6 -5 -5 -4 -3 -3 -3 -2 -1 -2 -2 -1  0  1  1  2  3  4  4  5  4  3  3  3
  4  3  3  2  1  0  0 -1 -1  0  0  1  0 -1 -1 -2 -2 -2 -2 -2 -3 -2 -2 -1 -1
 -2 -2 -1  0 -1 -1 -2 -3 -2 -2 -2 -1 -2 -2 -1 -2 -1 -1 -2 -2 -3 -3 -4 -3 -3
 -3 -4 -3 -3 -3 -4 -5 -6 -6 -7 -8 -7 -7 -7 -8 -7 -7 -8 -8 -7 -7 -7 -6 -5 -5
 -4 -3 -2 -2 -1 -1 -2 -2 -1 -2 -3 -3 -4 -3 -3 -3 -4 -5 -4 -4 -3 -4 -3 -3 -3
 -2 -1 -1  0  1  0  0  0  1  2  2  1  1  2  2  3  3  3  3  2  3  2  2  1  1
  1  1  0 -1  0  0 -1  0 -1 -1 -1  0  0  0  1  0 -1 -1 -1 -2 -1 -1 -2 -3 -3
 -3 -2 -2 -3 -3 -2 -1 -2 -2 -3 -2 -2 -2 -3 -2 -1 -1  0  1  2  2  1  2  1  1
  0 -1  0  0  0 -1  0  0 -1 -2 -1 -1  0  0  1  1  2  1  0  0 -1  0  0  0  0
 -1  0  0 -1 -2 -3 -3 -4 -5 -6 -6 -5 -6 -7 -7 -7 -8 -9 -9 -8 -7 -6 -6 -7 -7
 -6 -6 -5 -4 -5 -5 -6 -5 -5 -5 -6 -5 -6 -6 -7 -6 -7 -7 -6 -7 -6 -6 -5 -6 -6
 -6 -6 -5 -6 -6 -5 -4 -5 -5 -4 -4 -5 -5 -4 -4 -5 -5 -4 -5 -5 -5 -4 -5 -6 -6
####
-12->665 666 678 683 684
-11->661 663 664 667 668 670 673 677 679 680 682 685 686
-10->659 660 662 669 671 672 674 675 676 681 687 688
 -9->443 444 654 658 689 691 692 693
 -8->199 200 286 290 293 294 442 445 653 655 656 657 690 694
 -7->197 198 201 285 287 288 289 291 292 295 296 297 439 440 441 446 449 450 465 467 468 470 647 648 651 652 695 696
 -6->114 193 195 196 202 283 284 298 435 436 438 447 448 451 452 457 461 463 464 466 469 471 472 474 475 476 477 479 480 499 500 509 646 649 650 697
 -5->110 113 115 116 117 182 191 192 194 203 204 282 299 300 318 434 437 453 455 456 458 459 460 462 473 478 481 483 484 487 488 491 492 494 495 496 498 501 503 504 506 507 508 510 619 620 621 645 698 701 702 705 710 711 712 743 744 762
 -4-> 31  32  73  79  80  81  83  84 109 111 112 118 139 140 174 175 176 181 183 184 186 190 205 273 277 281 301 313 317 319 320 322 433 454 482 485 486 489 490 493 497 502 505 511 512 513 618 622 643 644 699 700 703 704 706 709 713 715 716 742 745 761 763 764 765 830 834
 -3-> 13  19  20  30  33  43  44  45  47  48  49  50  53  54  71  72  74  75  76  78  82  85 105 107 108 119 120 121 131 132 138 141 173 177 179 180 185 187 188 189 206 207 208 246 258 271 272 274 275 276 278 279 280 302 311 312 314 315 316 321 323 324 325 374 375 376 379 380 385 389 431 432 514 523 524 525 617 623 624 625 627 628 631 632 642 707 708 714 717 719 720 730 733 741 746 747 748 751 752 754 757 759 760 766 769 777 829 831 832 833 835 836 837 839 840 841 861 863 864
 -2->  5   7   8   9  11  12  14  17  18  21  23  24  25  29  34  37  42  46  51  52  55  56  61  67  68  70  77  86  89  90 103 104 106 122 127 128 130 133 137 142 154 157 170 171 172 178 209 211 212 241 242 243 244 245 247 248 251 252 257 259 260 261 263 264 266 269 270 303 304 307 308 310 326 370 373 377 378 381 383 384 386 387 388 390 410 430 515 516 518 521 522 526 530 531 532 534 610 613 615 616 626 629 630 633 641 718 721 722 727 728 729 731 732 734 735 736 739 740 749 750 753 755 756 758 767 768 770 773 774 775 776 778 787 788 790 827 828 838 842 857 859 860 862 865
 -1->  3   4   6  10  15  16  22  26  27  28  35  36  38  41  57  59  60  62  63  64  66  69  87  88  91  92 102 123 124 125 126 129 134 135 136 143 144 151 152 153 155 156 158 165 167 168 169 210 213 233 234 239 240 249 250 253 255 256 262 265 267 268 305 306 309 327 328 354 357 359 360 361 367 368 369 371 372 382 391 392 402 406 409 411 412 421 426 429 517 519 520 527 528 529 533 535 536 609 611 612 614 634 638 639 640 723 724 725 726 737 738 771 772 779 780 782 783 784 786 789 791 792 797 826 843 844 845 846 847 848 854 855 856 858 866 867 868 885 887 888 890 891 892 894 897 907 908 909 911 912
  0->  2  39  40  58  65  93 101 145 149 150 159 160 163 164 166 214 231 232 235 236 238 254 329 331 332 333 353 355 356 358 362 363 364 366 393 401 403 404 405 407 408 413 414 419 420 422 423 424 425 427 428 537 541 607 608 635 636 637 781 785 793 795 796 798 811 812 814 823 824 825 849 850 853 869 877 883 884 886 889 893 895 896 898 903 904 906 910 913 915 916 919 920
  1->  1  94  97  98  99 100 146 147 148 161 162 215 216 230 237 330 334 337 338 349 350 351 352 365 394 397 399 400 415 416 418 538 539 540 542 606 794 799 800 801 806 809 810 813 815 816 822 851 852 870 874 875 876 878 881 882 899 900 902 905 914 917 918 921 947 948 971 972 978 987 988 991 992 994 997
  2-> 95  96 217 229 335 336 339 340 345 347 348 395 396 398 417 543 544 602 603 604 605 802 805 807 808 817 821 871 872 873 879 880 901 922 942 946 949 950 953 954 957 970 973 977 979 980 981 983 984 986 989 990 993 995 996 998 999 1000
  3->218 223 224 225 227 228 341 342 343 344 346 545 547 548 549 550 601 803 804 818 819 820 923 924 925 929 938 941 943 944 945 951 952 955 956 958 962 963 964 969 974 975 976 982 985
  4->219 220 222 226 546 551 552 557 558 561 563 564 577 578 599 600 926 927 928 930 931 932 937 939 940 959 960 961 965 967 968
  5->221 553 555 556 559 560 562 565 569 571 572 574 575 576 579 580 582 595 596 598 933 935 936 966
  6->554 566 567 568 570 573 581 583 584 585 587 588 590 593 594 597 934
  7->586 589 591 592
####
92 Zeroes
####
Crosses zero 59 times

Factor

Works with: Factor version 0.99 2020-01-23
USING: formatting grouping io kernel math math.extras
math.ranges math.statistics prettyprint sequences ;

! Take the cumulative sum of the mobius sequence to avoid
! summing lower terms over and over.
: mertens-upto ( n -- seq ) [1,b] [ mobius ] map cum-sum ;

"The first 199 terms of the Mertens sequence:" print
199 mertens-upto " " prefix 20 group
[ [ "%3s" printf ] each nl ] each nl

"In the first 1,000 terms of the Mertens sequence there are:"
print 1000 mertens-upto
[ [ zero? ] count bl pprint bl "zeros." print ]
[
    2 <clumps> [ first2 [ 0 = not ] [ zero? ] bi* and ] count bl
    pprint bl "zero crossings." print
] bi
Output:
The first 199 terms of the Mertens sequence:
     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

In the first 1,000 terms of the Mertens sequence there are:
 92 zeros.
 59 zero crossings.

Forth

: AMOUNT 1000 ;

variable mertens AMOUNT cells allot
: M 1- cells mertens + ; \ 1-indexed array 

: make-mertens
  1 1 M !
  2 begin dup AMOUNT <= while
    1 over M !
    2 begin over over >= while
      over over / M @ 
      2 pick M @ swap -
      2 pick M !  
    1+ repeat
    drop
  1+ repeat
  drop
;
  
: print-row
  begin dup while 
    swap dup M @ 3 .r 1+
    swap 1-
  repeat
  drop
;

: print-table ."    "
  1 9 print-row cr
  begin dup 100 < while 10 print-row cr repeat
  drop
;

: find-zero-cross
  0 0
  1 begin dup AMOUNT <= while
    dup M @ 0= if
      swap 1+ swap
      dup 1- M @ 0<> if rot 1+ -rot then
    then
    1+
  repeat
  drop
;

make-mertens
." The first 99 Mertens numbers are:" cr print-table
find-zero-cross
." M(N) is zero " . ." times." cr
." M(N) crosses zero " . ." times." cr
bye
Output:
The first 99 Mertens numbers are:
     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
M(N) is zero 92 times.
M(N) crosses zero 59 times.

Fortran

      program Mertens
      implicit none
      integer M(1000), n, k, zero, cross
       
C     Generate Mertens numbers       
      M(1) = 1
      do 10 n=2, 1000
          M(n) = 1
          do 10 k=2, n
              M(n) = M(n) - M(n/k)
 10   continue 

C     Print table
      write (*,"('The first 99 Mertens numbers are:')")
      write (*,"('   ')",advance='no')
      k = 9
      do 20 n=1, 99
          write (*,'(I3)',advance='no') M(n)
          k = k-1
          if (k .EQ. 0) then
              k=10
              write (*,*)
          end if
 20   continue
 
C     Calculate zeroes and crossings
      zero = 0
      cross = 0
      do 30 n=2, 1000
          if (M(n) .EQ. 0) then
              zero = zero + 1
              if (M(n-1) .NE. 0) cross = cross+1
          end if    
 30   continue 
 
 40   format("M(N) is zero ",I2," times.")
      write (*,40) zero
 50   format("M(N) crosses zero ",I2," times.")
      write (*,50) cross
      end program
Output:
The first 99 Mertens numbers are:
     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
M(N) is zero 92 times.
M(N) crosses zero 59 times.

FreeBASIC

function padto( i as ubyte, j as integer ) as string
    return wspace(i-len(str(j)))+str(j)
end function

dim as integer M( 1 to 1000 ), n, col, k, psum
dim as integer num_zeroes = 0, num_cross = 0
dim as string outstr

M(1) = 1
for n = 2 to 1000
    psum = 0
    for k = 2 to n
        psum += M(int(n/k))
    next k
    M(n) = 1 - psum
    if M(n) = 0 then
        num_zeroes += 1
        if M(n-1)<>0 then
            num_cross += 1
        end if
    end if
next n

print using "There are ### zeroes in the range 1 to 1000."; num_zeroes
print using "There are ### crossings in the range 1 to 1000."; num_cross
print "The first 100 Mertens numbers are: "

for n=1 to 100
    outstr += padto(3, M(n))+"  "
    if n mod 10 = 0 then
        print outstr
        outstr = ""
    end if
next n
Output:
There are  92 zeroes in the range 1 to 1000.
There are  59 crossings in the range 1 to 1000.
The first 100 Mertens numbers are: 
  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

FutureBasic

void local fn MertensFunction
  long mertens(1000), n, k, crossesTotal = 0, zerosTotal = 0
  
  mertens(1) = 1
  
  for n = 2 to 1000
    mertens(n) = 1
    for k = 2 to n
      mertens(n) = mertens(n) - mertens(n/k)
    next
  next
  
  printf @"First 99 Mertens numbers:\n    \b"
  
  for n = 1 to 99
    printf @"%3ld \b", mertens(n)
    if ( n mod 10 == 9 ) then print
  next
  
  for n = 1 to 1000
    if ( mertens(n) == 0 )
      zerosTotal++
      if mertens(n-1) != 0 then crossesTotal++
    end if
  next
  
  print
  printf @"mertens(n) array is zero %ld times.", zerosTotal
  printf @"mertens(n) array crosses zero %ld times.", crossesTotal
end fn

fn MertensFunction

HandleEvents
Output:
First 99 Mertens numbers:
      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 

mertens(n) array is zero 92 times.
mertens(n) array crosses zero 59 times.

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")
}
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

Haskell

import           Data.List.Split          (chunksOf)
import qualified Data.MemoCombinators  as Memo
import           Math.NumberTheory.Primes (unPrime, factorise)
import           Text.Printf              (printf)

moebius :: Integer -> Int
moebius = product . fmap m . factorise
  where
    m (p, e)
      | unPrime p == 0 = 0
      | e == 1 = -1
      | otherwise = 0

mertens :: Integer -> Int
mertens = Memo.integral (\n -> sum $ fmap moebius [1..n])

countZeros :: [Integer] -> Int
countZeros = length . filter ((==0) . mertens)

crossesZero :: [Integer] -> Int
crossesZero = length . go . fmap mertens
  where
    go (x:y:xs) 
      | y == 0 && x /= 0 = y : go (y:xs)
      | otherwise        = go (y:xs)
    go _ = []

main :: IO ()
main = do
  printf "The first 99 terms for M(1..99):\n\n   "
  mapM_ (printf "%3d" . mertens) [1..9] >> printf "\n"
  mapM_ (\row -> mapM_ (printf "%3d" . mertens) row >> printf "\n") $ chunksOf 10 [10..99]
  printf "\nM(n) is zero %d times for 1 <= n <= 1000.\n" $ countZeros [1..1000]
  printf "M(n) crosses zero %d times for 1 <= n <= 1000.\n" $ crossesZero [1..1000]
Output:
The first 99 terms for M(1..99):

     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

M(n) is zero 92 times for 1 <= n <= 1000.
M(n) crosses zero 59 times for 1 <= n <= 1000.

J

mu    =: 0:`(1 - 2 * 2|#@{.)@.(1: = */@{:)@(2&p:)"0
M     =: +/@([: mu 1:+i.)

m1000 =: (M"0) 1+i.1000
zero  =: +/ m1000 = 0
cross =: +/ (-.*.1:|]) m1000 ~: 0 

echo 'The first 99 Merten numbers are'
echo 10 10$ __, 99{.m1000
echo 'M(N) is zero ',(":zero),' times.'
echo 'M(N) crosses zero ',(":cross),' times.'
exit''
Output:
The first 99 Merten numbers are
__  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
M(N) is zero 92 times.
M(N) crosses zero 0 times.

Java

public class MertensFunction {

    public static void main(String[] args) {
        System.out.printf("First 199 terms of the merten function are as follows:%n    ");
        for ( int n = 1 ; n < 200 ; n++ ) {
            System.out.printf("%2d  ", mertenFunction(n));
            if ( (n+1) % 20 == 0 ) {
                System.out.printf("%n");
            }
        }
        
        for ( int exponent = 3 ; exponent<= 8 ; exponent++ ) {
            int zeroCount = 0;
            int zeroCrossingCount = 0;
            int positiveCount = 0;
            int negativeCount = 0;
            int mSum = 0;
            int mMin = Integer.MAX_VALUE;
            int mMinIndex = 0;
            int mMax = Integer.MIN_VALUE;
            int mMaxIndex = 0;
            int nMax = (int) Math.pow(10, exponent);
            for ( int n = 1 ; n <= nMax ; n++ ) {
                int m = mertenFunction(n);
                mSum += m;
                if ( m < mMin ) {
                    mMin = m;
                    mMinIndex = n;
                }
                if ( m > mMax ) {
                    mMax = m;
                    mMaxIndex = n;
                }
                if ( m > 0 ) {
                    positiveCount++;
                }
                if ( m < 0 ) {
                    negativeCount++;
                }
                if ( m == 0 ) {
                    zeroCount++;
                }
                if ( m == 0 && mertenFunction(n - 1) != 0 ) {
                    zeroCrossingCount++;
                }
            }
            System.out.printf("%nFor M(x) with x from 1 to %,d%n", nMax);        
            System.out.printf("The maximum of M(x) is M(%,d) = %,d.%n", mMaxIndex, mMax);
            System.out.printf("The minimum of M(x) is M(%,d) = %,d.%n", mMinIndex, mMin);
            System.out.printf("The sum of M(x) is %,d.%n", mSum);
            System.out.printf("The count of positive M(x) is %,d, count of negative M(x) is %,d.%n", positiveCount, negativeCount);
            System.out.printf("M(x) has %,d zeroes in the interval.%n", zeroCount);
            System.out.printf("M(x) has %,d crossings in the interval.%n", zeroCrossingCount);
        }
    }
    
    private static int MU_MAX = 100_000_000;
    private static int[] MU = null;
    private static int[] MERTEN = null;
        
    //  Compute mobius and merten function via sieve
    private static int mertenFunction(int n) {
        if ( MERTEN != null ) {
            return MERTEN[n];
        }
        
        //  Populate array
        MU = new int[MU_MAX+1];
        MERTEN = new int[MU_MAX+1];
        MERTEN[1] = 1;
        int sqrt = (int) Math.sqrt(MU_MAX);
        for ( int i = 0 ; i < MU_MAX ; i++ ) {
            MU[i] = 1;
        }
        
        for ( int i = 2 ; i <= sqrt ; i++ ) {
            if ( MU[i] == 1 ) {
                //  for each factor found, swap + and -
                for ( int j = i ; j <= MU_MAX ; j += i ) {
                    MU[j] *= -i;
                }
                //  square factor = 0
                for ( int j = i*i ; j <= MU_MAX ; j += i*i ) {
                    MU[j] = 0;
                }
            }
        }
        
        int sum = 1;
        for ( int i = 2 ; i <= MU_MAX ; i++ ) {
            if ( MU[i] == i ) {
                MU[i] = 1;
            }
            else if ( MU[i] == -i ) {
                MU[i] = -1;
            }
            else if ( MU[i] < 0 ) {
                MU[i] = 1;               
            }
            else if ( MU[i] > 0 ) {
                MU[i] = -1;
            }
            sum += MU[i];
            MERTEN[i] = sum;
        }
        return MERTEN[n];
    }

}
Output:
First 199 terms of the merten function are as follows:
     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  

For M(x) with x from 1 to 1,000
The maximum of M(x) is M(586) = 7.
The minimum of M(x) is M(665) = -12.
The sum of M(x) is -1,572.
The count of positive M(x) is 254, count of negative M(x) is 654.
M(x) has 92 zeroes in the interval.
M(x) has 59 crossings in the interval.

For M(x) with x from 1 to 10,000
The maximum of M(x) is M(8,511) = 35.
The minimum of M(x) is M(9,861) = -43.
The sum of M(x) is -20,409.
The count of positive M(x) is 3,965, count of negative M(x) is 5,629.
M(x) has 406 zeroes in the interval.
M(x) has 256 crossings in the interval.

For M(x) with x from 1 to 100,000
The maximum of M(x) is M(48,433) = 96.
The minimum of M(x) is M(96,014) = -132.
The sum of M(x) is -516,879.
The count of positive M(x) is 47,830, count of negative M(x) is 50,621.
M(x) has 1,549 zeroes in the interval.
M(x) has 949 crossings in the interval.

For M(x) with x from 1 to 1,000,000
The maximum of M(x) is M(992,998) = 311.
The minimum of M(x) is M(926,265) = -368.
The sum of M(x) is -14,244,200.
The count of positive M(x) is 472,963, count of negative M(x) is 521,676.
M(x) has 5,361 zeroes in the interval.
M(x) has 3,269 crossings in the interval.

For M(x) with x from 1 to 10,000,000
The maximum of M(x) is M(9,993,034) = 1,143.
The minimum of M(x) is M(7,109,110) = -1,078.
The sum of M(x) is -194,680,528.
The count of positive M(x) is 4,938,188, count of negative M(x) is 5,049,266.
M(x) has 12,546 zeroes in the interval.
M(x) has 7,646 crossings in the interval.

For M(x) with x from 1 to 100,000,000
The maximum of M(x) is M(92,418,127) = 3,290.
The minimum of M(x) is M(76,015,339) = -3,448.
The sum of M(x) is -608,757,258.
The count of positive M(x) is 54,659,906, count of negative M(x) is 45,298,186.
M(x) has 41,908 zeroes in the interval.
M(x) has 25,525 crossings in the interval.

jq

Adapted from C

Works with: jq

Works with gojq, the Go implementation of jq

This entry will use the strategy exemplified in the entry for C and some others, that is, it will begin by defining a function that constructs an array of a specified number of Merten numbers, and use that function to solve the other tasks, which, however, will be solved independently for the sake of modularity and to illustrate efficient approaches to the problems considered separately. It would be trivial but uninteresting to merge the answers for efficiency.

Preliminaries

def sum(s): reduce s as $x (null; . + $x);

def nwise($n):
  def n: if length <= $n then . else .[0:$n] , (.[$n:] | n) end;
  n;

def lpad($len): tostring | ($len - length) as $l | (" " * $l)[:$l] + .;

# input: an array
# output: number of crossings at $value
def count_crossings($value):
  . as $a
  | reduce range(0; length) as $i ({};
    if $a[$i] == $value
      then if $i == 0 or .prev != $value then .count += 1 else . end
      else .
      end
      | .prev = $a[$i] )
  | .count;

Mertens Numbers

# Input: $max >= 1
# Output: an array of size $max with $max mertenNumbers beginning with 1
def mertensNumbers:
  . as $max
  | reduce range(2; $max + 1) as $n ( [1];
      .[$n-1]=1
      | reduce range(2; $n+1) as $k (.;
          .[$n-1] -= .[($n / $k) | floor - 1] ));

The Tasks

# Task 0:
def mertens_number:
  mertensNumbers[.-1];

def task1:
  "The first \(.) Mertens numbers are:",
   (mertensNumbers | nwise(10) | map(lpad(2)) | join(" ") );

def task2:
  . as $n
  | sum(mertensNumbers[] | select(.==0) | 1)
  | "M(n) is zero \(.) times for 1 <= n <= \($n)\n";

def task3:
  . as $n
  | mertensNumbers
  | count_crossings(0)
  | "M(n) crosses zero \(.) times for 1 <= n <= \($n).\n" ;

(99|task1),
"",
(1000 | (task2, task3))
Output:
The first 99 Mertens numbers are:
 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

M(n) is zero 92 times for 1 <= n <= 1000

M(n) crosses zero 59 times for 1 <= n <= 1000.

Julia

The OEIS A002321 reference suggests the Mertens function has a negative bias, which it does below 1 million, but this bias seems to switch to a positive bias by 1 billion. There may simply be large swings in the bias overall, which get larger and longer as the sequence continues.

using Primes, Formatting

function moebius(n::Integer)
    @assert n > 0
    m(p, e) = p == 0 ? 0 : e == 1 ? -1 : 0
    return reduce(*, m(p, e) for (p, e) in factor(n) if p   0; init=1)
end
μ(n) = moebius(n)

mertens(x) = sum(n -> μ(n), 1:x)
M(x) = mertens(x)

print("First 99 terms of the Mertens function for positive integers:\n   ")
for n in 1:99
    print(lpad(M(n), 3), n % 10 == 9 ? "\n" : "")
end

function maximinM(N)
    z, cros, lastM, maxi, maxM, mini, minM, sumM, pos, neg = 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
    for i in 1:N
        m = μ(i) + lastM
        if m == 0 && lastM != 0
            cros += 1
        end
        sumM += m
        lastM = m
        if m > maxM
            maxi = i
            maxM = m
        elseif m < minM
            mini = i
            minM = m
        end
        if m > 0
            pos += 1
        elseif m < 0
            neg += 1
        else
            z += 1
        end
    end
    println("\nFor M(x) with x from 1 to $(format(N, commas=true)):")
    println("The maximum of M(x) is M($(format(maxi, commas=true)) = $maxM.")
    println("The minimum of M(x) is M($(format(mini, commas=true))) = $minM.")
    println("The sum of M(x) is $(format(sumM, commas=true)).")
    println("The count of positive M(x) is $(format(pos, commas=true)), count of negative M(x) is $(format(neg, commas=true)).")
    println("M(x) has $(format(z, commas=true)) zeroes in the interval.")
    println("M(x) has $(format(cros, commas=true)) crossings in the interval.")
    diff = pos - neg
    if diff > 0
        println("Positive M(x) exceed negative ones by $(format(diff, commas=true)).")
    else
        println("Negative M(x) exceed positive ones by $(format(-diff, commas=true)).")
    end
end

foreach(maximinM, (1000, 1_000_000, 1_000_000_000))
Output:
First 99 terms of the Mertens function for positive integers:
     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

For M(x) with x from 1 to 1,000:
The maximum of M(x) is M(586 = 7.
The minimum of M(x) is M(665) = -12.
The sum of M(x) is -1,572.
The count of positive M(x) is 254, count of negative M(x) is 654.
M(x) has 92 zeroes in the interval.
M(x) has 59 crossings in the interval.
Negative M(x) exceed positive ones by 400.

For M(x) with x from 1 to 1,000,000:
The maximum of M(x) is M(992,998 = 311.
The minimum of M(x) is M(926,265) = -368.
The sum of M(x) is -14,244,200.
The count of positive M(x) is 472,963, count of negative M(x) is 521,676.
M(x) has 5,361 zeroes in the interval.
M(x) has 3,269 crossings in the interval.
Negative M(x) exceed positive ones by 48,713.

For M(x) with x from 1 to 1,000,000,000:
The maximum of M(x) is M(903,087,703 = 10246.
The minimum of M(x) is M(456,877,618) = -8565.
The sum of M(x) is 510,495,361,261.
The count of positive M(x) is 510,200,302, count of negative M(x) is 489,658,577.
M(x) has 141,121 zeroes in the interval.
M(x) has 85,652 crossings in the interval.
Positive M(x) exceed negative ones by 20,541,725.

MAD

            NORMAL MODE IS INTEGER
            DIMENSION M(1000)
                  
            M(1) = 1
            THROUGH GENMRT, FOR N=2, 1, N.G.1000
            M(N) = 1
            THROUGH GENMRT, FOR K=2, 1, K.G.N
GENMRT      M(N) = M(N) - M(N/K)
            
            PRINT COMMENT $ FIRST 99 MERTEN NUMBERS ARE$
            VECTOR VALUES F9 = $S3,9(I2,S1)*$
            VECTOR VALUES F10 = $10(I2,S1)*$
            
            PRINT FORMAT F9, M(1), M(2), M(3), M(4), M(5), M(6),
          0                  M(7), M(8), M(9)
          
            THROUGH SHOW, FOR N=10, 10, N.GE.100
SHOW        PRINT FORMAT F10, M(N), M(N+1), M(N+2), M(N+3), M(N+4),
          0   M(N+5), M(N+6), M(N+7), M(N+8), M(N+9), M(N+10)
          
            ZERO = 0
            CROSS = 0
            THROUGH ZC, FOR N=1, 1, N.G.1000
            WHENEVER M(N).E.0, ZERO = ZERO + 1
ZC          WHENEVER M(N).E.0 .AND. M(N-1).NE.0, CROSS = CROSS + 1

            VECTOR VALUES FZ = $13HM(N) IS ZERO ,I2,S1,5HTIMES*$
            PRINT FORMAT FZ, ZERO
            
            VECTOR VALUES FC = $18HM(N) CROSSES ZERO ,I2,S1,5HTIMES*$
            PRINT FORMAT FC, CROSS
            
            END OF PROGRAM
Output:
FIRST 99 MERTEN NUMBERS ARE
    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
M(N) IS ZERO 92 TIMES
M(N) CROSSES ZERO 59 TIMES

Mathematica/Wolfram Language

ClearAll[Mertens]
Mertens[n_] := Total[MoebiusMu[Range[n]]]
Grid[Partition[Mertens /@ Range[99], UpTo[10]]]
Count[Mertens /@ Range[1000], 0]
SequenceCount[Mertens /@ Range[1000], {Except[0], 0}]
Output:
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	
92
59

Modula-2

MODULE Mertens;
FROM InOut IMPORT WriteString, WriteInt, WriteCard, WriteLn;

CONST Max = 1000;
VAR n, k, x, y, zero, cross: CARDINAL;
    M: ARRAY [1..Max] OF INTEGER;

BEGIN
    M[1] := 1;
    FOR n := 2 TO Max DO
        M[n] := 1;
        FOR k := 2 TO n DO
            M[n] := M[n] - M[n DIV k];
        END;
    END;

    WriteString("The first 99 Mertens numbers are:");
    WriteLn();
    FOR y := 0 TO 90 BY 10 DO
        FOR x := 0 TO 9 DO
            IF x+y=0 THEN WriteString("   ");
            ELSE WriteInt(M[x+y], 3);
            END;
        END;
        WriteLn();
    END;

    zero := 0;
    cross := 0;
    FOR n := 2 TO Max DO
        IF M[n] = 0 THEN
            zero := zero + 1;
            IF M[n-1] # 0 THEN
                cross := cross + 1;
            END;
        END;
    END;

    WriteString("M(n) is zero "); 
    WriteCard(zero,0); 
    WriteString(" times.");
    WriteLn();
    WriteString("M(n) crosses zero ");
    WriteCard(cross,0);
    WriteString(" times.");
    WriteLn();
END Mertens.
Output:
The first 99 Mertens numbers are:
     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
M(n) is zero 92 times.
M(n) crosses zero 59 times.

Nim

Translation of: C
import sequtils, strformat

func mertensNumbers(max: int): seq[int] =
  result = repeat(1, max + 1)
  for n in 2..max:
    for k in 2..n:
      dec result[n], result[n div k]

const Max = 1000
let mertens = mertensNumbers(Max)

echo "First 199 Mertens numbers:"
const Count = 200
var column = 0
for i in 0..<Count:
  if column > 0: stdout.write ' '
  stdout.write if i == 0: "  " else: &"{mertens[i]:>2}"
  inc column
  if column == 20:
    stdout.write '\n'
    column = 0

var zero, cross, previous = 0
for i in 1..Max:
  let m = mertens[i]
  if m == 0:
    inc zero
    if previous != 0:
      inc cross
  previous = m

echo ""
echo &"M(n) is zero {zero} times for 1 ⩽ n ⩽ {Max}."
echo &"M(n) crosses zero {cross} times for 1 ⩽ n ⩽ {Max}."
Output:
    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

M(n) is zero 92 times for 1 ⩽ n ⩽ 1000.
M(n) crosses zero 59 times for 1 ⩽ n ⩽ 1000.

Pascal

Works with: Free Pascal

Nearly the same as Square-free_integers#Pascal Instead here marking all multiples, starting at factor 2, of a prime by incrementing the factor count.
runtime ~log(n)*n

program Merten;
{$IFDEF FPC}
  {$MODE DELPHI}
  {$Optimization ON,ALL}
{$ELSE}
   {$APPTYPE CONSOLE}
{$ENDIF}
uses
  sysutils;
const
  BigLimit = 10*1000*1000*1000;//1e10

type
  tSieveElement = Int8;
  tpSieve = pInt8;
  tMoebVal = array[-1..1] of Int64;
var
  MertensValues : array[-40000..50500] of NativeInt;
  primes : array of byte;
  sieve : array of tSieveElement;

procedure CompactPrimes;
//searching for needed primes
//last primes are marked with -1
var
  pSieve : tpSieve;
  i,lmt,dp:NativeInt;
Begin
  setlength(Primes,74500);//suffices for primes to calc square upto 1e12
  //extract difference of primes
  i := 2;
  lmt := 0;
  dp := 2;
  pSieve :=@sieve[0];
  repeat
    IF pSieve[i]= 0 then
    Begin
      //mark for Moebius
      pSieve[i]:= -1;
      primes[lmt] := dp;
      dp := 0;
      inc(lmt);
    end;
    inc(dp);
    inc(i);
  until i*i >BigLimit;
  setlength(Primes,lmt+1);

  repeat
    IF pSieve[i]= 0 then
      //mark for Moebius
      pSieve[i]:= -1;
    inc(i);
  until i >BigLimit;
end;

procedure SieveSquares;
//mark all powers >=2 of prime  => all powers = 2 is sufficient
var
  pSieve : tpSieve;
  i,sq,k,prime : NativeInt;
Begin
  pSieve := @sieve[0];
  prime := 0;
  For i := 0 to High(primes) do
  Begin
    prime := prime+primes[i];
    sq := prime*prime;
    k := sq;
    if sq > BigLimit then
      break;
    repeat
      pSieve[k] := 0;
      inc(k,sq);
    until k> BigLimit;
  end;
end;

procedure initPrimes;
var
  pSieve : tpSieve;
  fakt,
  sieveprime : NativeUint;
begin
  pSieve := @sieve[0];
  sieveprime := 2;
  repeat
    if pSieve[sieveprime]=0 then
    begin
      fakt := sieveprime+sieveprime;
      while fakt <=BigLimit do
      Begin
        //count divisors
        inc(pSieve[fakt]);
        inc(fakt,sieveprime);
      end;
    end;
    inc(sieveprime);
  until sieveprime>BigLimit DIV 2;
  //Möbius of 1
  pSieve[1] := 1;

  //convert to Moebius
  For fakt := 2 to BigLimit do
  Begin
    sieveprime := pSieve[fakt];
    IF sieveprime<>0 then
      pSieve[fakt] := 1-(2*(sieveprime AND 1)) ;
  end;
  CompactPrimes;
  SieveSquares;
end;

procedure OutMerten10(Lmt,ZeroCross:NativeInt;Const MoebVal:tMoebVal);
var
  i,j: NativeInt;
Begin
  Writeln(lmt:11,MoebVal[-1]:11,MoebVal[1]:11,MoebVal[-1]+MoebVal[1]:11,
  MoebVal[-1]-MoebVal[1]:7,MoebVal[0]:11);
  i:= low(MertensValues);
  while MertensValues[i] = 0 do
    inc(i);
  j:= High(MertensValues);
  while MertensValues[j] = 0 do
    dec(j);
  write('Merten min ',i:6,' max ',j:6,' zero''s ',MertensValues[0]:8);
  writeln(' zeroCross ',ZeroCross);
  writeln;
end;

procedure Count_x10;
var
  MoebCount: tMoebVal;
  pSieve : tpSieve;
  i,lmt,Merten,Moebius,LastMert,ZeroCross: NativeInt;
begin
  writeln('[1 to limit]');
  Writeln('Limit        Moeb. odd   Moeb.even  sqr-free Merten     Zero''s');

  pSieve := @sieve[0];
  For i := -1 to 1 do
    MoebCount[i]:=0;
  ZeroCross := 0;
  LastMert :=1;
  Merten :=0;
  lmt := 10;
  i := 1;
  repeat
    while i <= lmt do
    Begin
      Moebius := pSieve[i];
      inc(MoebCount[Moebius]);
      inc(Merten,Moebius);
      inc(MertensValues[Merten]);//MoebCount[1]-MoebCount[-1]]);
      inc(ZeroCross,ORD( (Merten = 0) AND (LastMert <> 0)));
      LastMert := Merten;
      inc(i);
    end;
    OutMerten10(Lmt,ZeroCross,MoebCount);

    IF lmt >= BigLimit then
      BREAK;
    lmt := lmt*10;
    IF lmt >BigLimit then
      lmt := BigLimit;
  until false;
  writeln;
end;

procedure OutMerten(lmt:NativeInt);
var
  i,k,m : NativeInt;
Begin
  iF lmt> BigLimit then
    lmt := BigLimit;
  writeln('Mertens numbers from 1 to ',lmt);
  k := 9;
  write('':3);
  m := 0;
  For i := 1 to lmt do
  Begin
    inc(m,sieve[i]);
    write(m:3);
    dec(k);
    IF k = 0 then
    Begin
      writeln;
      k := 10;
    end;
  end;
  writeln;
end;

procedure OutMoebius(lmt:NativeInt);
var
  i,k : NativeInt;
Begin
  iF lmt> BigLimit then
    lmt := BigLimit;
  writeln('Möbius numbers from 1 to ',lmt);
  k := 19;
  write('':3);
  For i := 1 to lmt do
  Begin
    write(sieve[i]:3);
    dec(k);
    IF k = 0 then
    Begin
      writeln;
      k := 20;
    end;
  end;
  writeln;
end;

Begin
  setlength(sieve,BigLimit+1);
  InitPrimes;
  SieveSquares;
  Count_x10;
  OutMoebius(199);
  OutMerten(99);
  setlength(primes,0);
  setlength(sieve,0);
end.
Output:
[1 to limit]
Limit        Moeb. odd   Moeb.even  sqr-free Merten     Zero's
         10          4          3          7      1          3
Merten min     -2 max      1 zero's        1 zeroCross 1

        100         30         31         61     -1         39
Merten min     -4 max      2 zero's        6 zeroCross 5

       1000        303        305        608     -2        392
Merten min    -12 max      7 zero's       92 zeroCross 59

      10000       3053       3030       6083     23       3917
Merten min    -43 max     35 zero's      406 zeroCross 256

     100000      30421      30373      60794     48      39206
Merten min   -132 max     96 zero's     1549 zeroCross 949

    1000000     303857     304069     607926   -212     392074
Merten min   -368 max    311 zero's     5361 zeroCross 3269

   10000000    3039127    3040164    6079291  -1037    3920709
Merten min  -1078 max   1143 zero's    12546 zeroCross 7646

  100000000   30395383   30397311   60792694  -1928   39207306
Merten min  -3448 max   3290 zero's    41908 zeroCross 25525

 1000000000  303963673  303963451  607927124    222  392072876
Merten min  -8565 max  10246 zero's   141121 zeroCross 85652

10000000000 3039652332 3039618610 6079270942  33722 3920729058
Merten min -35517 max  50286 zero's   431822 zeroCross 262605


Möbius numbers from 1 to 199
     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  1  0  0 -1 -1 -1  0  1  1  1  0 -1  1  1
  0 -1 -1 -1  0  0  1 -1  0  0  0  1  0 -1  0  1  0  1  1 -1
  0 -1  1  0  0  1 -1 -1  0  1 -1 -1  0 -1  1  0  0  1 -1 -1
  0  0  1 -1  0  1  1  1  0 -1  0  1  0  1  1  1  0 -1  0  0
  0 -1 -1 -1  0 -1  1 -1  0 -1 -1  1  0 -1 -1  1  0  0  1  1
  0  0  1  1  0  0  0 -1  0  1 -1 -1  0  1  1  0  0 -1 -1 -1
  0  1  1  1  0  1  1  0  0 -1  0 -1  0  0 -1  1  0 -1  1  1
  0  1  0 -1  0 -1  1 -1  0  0 -1  0  0 -1 -1  0  0  1  1 -1
  0 -1 -1  1  0  1 -1  1  0  0 -1 -1  0 -1  1 -1  0 -1  0 -1

Mertens numbers from 1 to 99
     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

real    3m54,249s = 234s //BigLimit = 100*1000*1000;takes 2.017s

Perl

use utf8;
use strict;
use warnings;
use feature 'say';
use List::Util 'uniq';

sub prime_factors {
    my ($n, $d, @factors) = (shift, 1);
    while ($n > 1 and $d++) {
        $n /= $d, push @factors, $d until $n % $d;
    }
    @factors
}

sub μ {
    my @p = prime_factors(shift);
    @p == uniq(@p) ? 0 == @p%2 ? 1 : -1 : 0
}

sub progressive_sum {
    my @sum = shift @_;
    push @sum, $sum[-1] + $_ for @_;
    @sum
}

my($upto, $show, @möebius) = (1000, 199, ());
push @möebius, μ($_) for 1..$upto;
my @mertens = progressive_sum @möebius;

say "Mertens sequence - First $show terms:\n" .
    (' 'x4 . sprintf "@{['%4d' x $show]}", @mertens[0..$show-1]) =~ s/((.){80})/$1\n/gr .
    sprintf("\nEquals zero %3d times between 1 and $upto", scalar grep { ! $_ } @mertens) .
    sprintf "\nCrosses zero%3d times between 1 and $upto", scalar grep { ! $mertens[$_-1] and $mertens[$_] } 1 .. @mertens;
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

Phix

with javascript_semantics
requires("1.0.2") -- (skip arg added to join_by)
sequence mcache = {1}

function Mertens(integer n)
    for m=length(mcache)+1 to n do
        integer mm = 1
        for k=2 to m do
            mm -= mcache[floor(m/k)]
        end for
        mcache &= mm
    end for
    return mcache[n]
end function
constant first = 99, perline = 10
--constant first = 199, perline = 20    -- matches C/Go/etc
--constant first = 143, perline = 12    -- matches wp
sequence s = {"  ."}&apply(tagset(first),Mertens)
printf(1,"First %d Mertens numbers:\n",first)
puts(1,join_by(s,1,perline," ",fmt:="%3d",skip:=1))
 
integer prev = 1, zeroes = 0, crosses = 0
for n=2 to 1000 do
    integer m = Mertens(n)
    if m=0 then
        zeroes += 1
        crosses += prev!=0 
    end if
    prev = m
end for
printf(1,"\nMertens[1..1000] equals zero %d times and crosses zero %d times\n",{zeroes,crosses})
Output:
First 99 Mertens numbers:
  .   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

Mertens[1..1000] equals zero 92 times and crosses zero 59 times

PL/I

mertens: procedure options(main);
    %replace MAX by 1000;
    
    declare M(1:MAX) fixed binary(5);
    declare (n, k) fixed binary(10);
    declare (isZero, crossZero) fixed binary(8);
    
    M(1) = 1;
    do n = 2 to MAX;
        M(n) = 1;
        do k = 2 to n;
            M(n) = M(n) - M(divide(n,k,10));
        end;
    end;
    
    put skip list('The first 99 Mertens numbers are:');
    put skip list('   ');
    do n = 1 to 99;
        put edit(M(n)) (F(3));
        if mod(n,10) = 9 then put skip;
    end;
    
    isZero = 0;
    crossZero = 0;
    do n = 2 to MAX;
        if M(n) = 0 then do;
            isZero = isZero + 1;
            if M(n-1) ^= 0 then
                crossZero = crossZero + 1;
        end;
    end;
    
    put skip list('Zeroes:   ',isZero); 
    put skip list('Crossings:',crossZero);
end mertens;
Output:
The first 99 Mertens numbers are:
     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

Zeroes:         92
Crossings:      59

Prolog

Works with: SWI Prolog
:- dynamic mertens_number_cache/2.

mertens_number(1, 1):- !.
mertens_number(N, M):-
    mertens_number_cache(N, M),
    !.
mertens_number(N, M):-
    N >= 2,
    mertens_number(N, 2, M, 0),
    assertz(mertens_number_cache(N, M)).

mertens_number(N, N, M, M):- !.
mertens_number(N, K, M, S):-
    N1 is N // K,
    mertens_number(N1, M1),
    K1 is K + 1,
    S1 is S - M1,
    mertens_number(N, K1, M, S1).

print_mertens_numbers(Count):-
    print_mertens_numbers(Count, 0).

print_mertens_numbers(Count, Count):-!.
print_mertens_numbers(Count, N):-
    (N == 0 ->
        write('   ')
        ;
        mertens_number(N, M),
        writef('%3r', [M])
    ),
    N1 is N + 1,
    Column is N1 mod 20,
    (N > 0, Column == 0 ->
        nl
        ;
        true
    ),
    print_mertens_numbers(Count, N1).

count_zeros(From, To, Z, C):-
    count_zeros(From, To, Z, C, 0, 0, 0).

count_zeros(From, To, Z, C, Z, C, _):-
    From > To,
    !.
count_zeros(From, To, Z, C, Z1, C1, P):-
    mertens_number(From, M),
    (M == 0 -> Z2 is Z1 + 1 ; Z2 = Z1),
    (M == 0, P \= 0 -> C2 is C1 + 1 ; C2 = C1),
    Next is From + 1,
    count_zeros(Next, To, Z, C, Z2, C2, M).

main:-
    writeln('First 199 Mertens numbers:'),
    print_mertens_numbers(200),
    count_zeros(1, 1000, Z, C),
    writef('M(n) is zero %t times for 1 <= n <= 1000.\n', [Z]),
    writef('M(n) crosses zero %t times for 1 <= n <= 1000.\n', [C]).
Output:
First 199 Mertens numbers:
     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
M(n) is zero 92 times for 1 <= n <= 1000.
M(n) crosses zero 59 times for 1 <= n <= 1000.

PureBasic

Dim M.i(1000)

M(1)=1
For n=2 To 1000
  psum=0
  For k=2 To n : psum+M(Int(n/k)) : Next : M(n)=1-psum
  If M(n)=0 : z+1 : If M(n-1)<>0 : c+1 : EndIf : EndIf
Next

OpenConsole("")
PrintN("First 99 Mertens numbers:") : Print(Space(4))
For n=1 To 99 : Print(RSet(Str(M(n)),4)) : If n%10=9 : PrintN("") : EndIf : Next
PrintN("M(N) is zero "+Str(z)+" times.") : PrintN("M(N) crosses zero "+Str(c)+" times.")
Input()
Output:
First 99 Mertens numbers:
       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
M(N) is zero 92 times.
M(N) crosses zero 59 times.

Python

def mertens(count):
    """Generate Mertens numbers"""
    m = [None, 1]
    for n in range(2, count+1):
        m.append(1)
        for k in range(2, n+1):
            m[n] -= m[n//k]
    return m
    

ms = mertens(1000)

print("The first 99 Mertens numbers are:")
print("  ", end=' ')
col = 1
for n in ms[1:100]:
    print("{:2d}".format(n), end=' ')
    col += 1
    if col == 10:
        print()
        col = 0
        
zeroes = sum(x==0 for x in ms)
crosses = sum(a!=0 and b==0 for a,b in zip(ms, ms[1:]))
print("M(N) equals zero {} times.".format(zeroes))
print("M(N) crosses zero {} times.".format(crosses))
Output:
The first 99 Mertens numbers are:
    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
M(N) equals zero 92 times.
M(N) crosses zero 59 times.

Quackery

mobius is defined at Möbius function#Quackery.

   [ ' [ 0 ]
     swap 1+ times
       [ dup -1 peek
         i^ 1+ mobius
         + join ]
     behead drop ]        is mertens   ( n --> [ )

  [ say "   "
    99 times
      [ dup i^ peek
        dup dup
        -1 > if sp
        abs 10 < if sp
        echo
        i^ 1+ 10 mod
        9 = if cr ]
     drop ]               is grid      ( [ -->   )

  [ 0 swap
    witheach
     [ 0 = + ] ]          is zeroes    ( [ --> n )

  [ 0 0
    rot witheach
      [ dup 0 =
        rot 0 !=
        and
        rot + swap ]
    drop ]                is crossings ( [ --> n )

  1000 mertens
  say "First 99 terms:"
  cr
  dup grid
  cr
  dup zeroes echo say " zeroes and "
  crossings  echo say " crossings"
Output:
First 99 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

92 zeroes and 59 crossings

Raku

(formerly Perl 6)

Works with: Rakudo version 2019.11

Mertens number is not defined for n == 0. Raku arrays are indexed from 0 so store a blank value at position zero to keep x and M(x) aligned.

use Prime::Factor;

sub μ (Int \n) {
    return 0 if n %% 4 or n %% 9 or n %% 25 or n %% 49 or n %% 121;
    my @p = prime-factors(n);
    +@p == +@p.unique ?? +@p %% 2 ?? 1 !! -1 !! 0
}

my @mertens = lazy [\+] flat '', 1, (2..*).hyper.map: -> \n { μ(n) };

put "Mertens sequence - First 199 terms:\n",
    @mertens[^200]».fmt('%3s').batch(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]} ),
    " times between 1 and 1000\n\nFirst Mertens equal to:";

for 10, 20, 30100 -> $threshold {
    printf "%4d: M(%d)\n", -$threshold, @mertens.first: * == -$threshold, :k;
    printf "%4d: M(%d)\n",  $threshold, @mertens.first: * ==  $threshold, :k;
}
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

First Mertens equal to:
 -10: M(659)
  10: M(1393)
 -20: M(2791)
  20: M(3277)
 -30: M(9717)
  30: M(8503)
 -40: M(9831)
  40: M(11770)
 -50: M(24018)
  50: M(19119)
 -60: M(24105)
  60: M(31841)
 -70: M(24170)
  70: M(31962)
 -80: M(42789)
  80: M(48202)
 -90: M(59026)
  90: M(48405)
-100: M(59426)
 100: M(114717)

REXX

The   Mertens   function is named after Franz Mertens.

Programming note:   This REXX version supports the specifying of the low and high values to be generated,
as well as the "group" size for the grid   (it can be specified as   1   which will show a vertical list).

A null value will be shown as a bullet (•) when showing the Möbius and/or Mertens value of for zero   (which can be changed easily).

The above "feature" was added to make the grid to be aligned with other solutions.

/*REXX pgm computes & shows a value grid of the Mertens function for a range of integers*/
parse arg LO HI grp eqZ xZ .                     /*obtain optional arguments from the CL*/
if  LO=='' |  LO==","  then  LO=    0            /*Not specified?  Then use the default.*/
if  HI=='' |  HI==","  then  HI=  199            /* "      "         "   "   "     "    */
if grp=='' | grp==","  then grp=   20            /* "      "         "   "   "     "    */
if eqZ=='' | eqZ==","  then eqZ= 1000            /* "      "         "   "   "     "    */
if  xZ=='' |  xZ==","  then  xZ= 1000            /* "      "         "   "   "     "    */
call genP                                        /*generate primes up to max  √  HIHI   */
               call Franz LO, HI
if eqZ>0  then call Franz 1, -eqZ
if  xZ>0  then call Franz -1, xZ
exit 0                                           /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
Franz: parse arg a 1 oa,b 1 ob;        @Mertens= ' The Mertens sequence from '
a= abs(a);   b= abs(b);    grid= oa>=0 & ob>=0   /*semaphore used to show a grid title. */
if grid  then say center(@Mertens LO " ──► " HI" ", max(50, grp*3), '═')    /*show title*/
         else say
zeros= 0                                         /*# of  0's found for Mertens function.*/
Xzero= 0                                         /*number of times that zero was crossed*/
$=;                 prev=                        /*$  holds output grid of GRP numbers. */
   do j=a  to b;     _= Mertens(j)               /*process some numbers from  LO ──► HI.*/
   if _==0  then zeros= zeros + 1                /*Is Zero?  Then bump the zeros counter*/
   if _==0  then if prev\==0 then Xzero= Xzero+1 /*prev ¬=0?   "   "    "  Xzero    "   */
   prev= _
   if grid  then $= $ right(_, 2)                /*build grid if A & B are non─negative.*/
   if words($)==grp  then do;  say substr($, 2);  $=    /*show grid if fully populated, */
                          end                           /*  and nullify it for more #s. */
   end   /*j*/                                   /*for small grids, using wordCnt is OK.*/

if $\==''  then say substr($, 2)                 /*handle any residual numbers not shown*/
if oa<0  then say @Mertens   a    " to "    b   ' has crossed zero '    Xzero    " times."
if ob<0  then say @Mertens   a    " to "    b   ' has '                 zeros    " zeros."
return
/*──────────────────────────────────────────────────────────────────────────────────────*/
Mertens: procedure expose @. !!. M.;  parse arg n; if M.n\==.  then return M.n
         if n<1  then return '∙'; m= 0           /*handle special cases of non─positive#*/
              do k=1  for n;   m= m + mobius(k)  /*sum the  MU's  up to  N.             */
              end   /*k*/                        /* [↑] mobius function uses memoization*/
         M.n= m;          return m               /*return the sum of all the  MU's.     */
/*──────────────────────────────────────────────────────────────────────────────────────*/
mobius: procedure expose @. !!.;  parse arg x 1 ox  /*get integer to be tested for  mu  */
        if !!.x\==.  then return !!.x            /*X computed before?  Return that value*/
        if x<1  then return '∙';      mu= 0      /*handle special case of non-positive #*/
             do k=1;  p= @.k; if p>x  then leave       /* (P)    > X?   Then we're done.*/
             if p*p>x    then do; mu= mu+1; leave; end /* (P**2) > X?   Bump # and leave*/
             if x//p==0  then do; mu= mu+1       /*X divisible by P?   Bump  mu  number.*/
                                  x= x % p       /*                    Divide by prime. */
                                  if x//p==0  then return 0  /*X÷by P?  Then return zero*/
                              end
             end   /*k*/                         /*MU (below) is almost always small, <9*/
        !!.ox=  -1 ** mu;         return !!.ox   /*raise -1 to the mu power, memoize it.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
genP: @.1=2; @.2=3; @.3=5; @.4=7; @.5=11; @.6=13 /*initialize some low primes; # primes.*/
      !!.=.;  M.=!!.;      #= 6;  sq.#= @.6**2   /*     "     2 arrays for memoization. */
       do j=@.#+4  by 2  to max(HI, eqZ, xZ); parse var j '' -1 _   /*odd Ps from now on*/
       if _==5 then iterate; if j//3==0 then iterate; if j//7==0  then iterate /*÷ 5 3 7*/
          do k=7  while sq.k<=j                  /*divide by some generated odd primes. */
          if j//@.k==0  then iterate j           /*Is J divisible by  P?  Then not prime*/
          end   /*k*/                            /* [↓]  a prime  (J)  has been found.  */
       #= #+1;          @.#=j;  sq.j= j*j        /*bump P count;  P──►@.;  compute  J**2*/
       end      /*j*/;                    return /*calculate the squares of some primes.*/
output   when using the default inputs:

Output note:   note the use of a bullet (•) to signify that a "null" is being shown (for the 0th entry).

══════════ The Mertens sequence from  0  ──►  199 ══════════
 ∙  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

 The Mertens sequence from  1  to  1000  has  92  zeros.

 The Mertens sequence from  1  to  1000  has crossed zero  59  times.

Ruby

require 'prime'

def μ(n)
  return 1 if self == 1
  pd = n.prime_division
  return 0 unless pd.map(&:last).all?(1)
  pd.size.even? ? 1 : -1
end

def M(n)
  (1..n).sum{|n| μ(n)}
end

(["  "] + (1..199).map{|n|"%2s" % M(n)}).each_slice(20){|line| puts line.join(" ") }

ar = (1..1000).map{|n| M(n)}
puts "\nThe Mertens function is zero #{ar.count(0)} times in the range (1..1000);"
puts "it crosses zero #{ar.each_cons(2).count{|m1, m2| m1 != 0 && m2 == 0}} times."
Output:
    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

The Mertens function is zero 92 times in the range (1..1000);
it crosses zero 59 times.

SETL

program mertens;
    m := [1] * 1000;
    loop for n in [1..#m] do
        m(n) -:= 0 +/[m(n div k) : k in [2..n]];
    end loop;

    print("The first 99 Mertens numbers:");
    putchar("   ");
    loop for n in [1..99] do
        putchar(lpad(str m(n), 3));
        if n mod 10=9 then print; end if;
    end loop;

    zero := #[n : n in [1..#m] | m(n) = 0];
    cross := #[n : n in [1..#m] | m(n) = 0 and m(n-1) /= 0];

    print("M(N) is zero " + str zero + " times.");
    print("M(N) crosses zero " + str cross + " times.");
end program;
Output:
The first 99 Mertens numbers:
     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
M(N) is zero 92 times.
M(N) crosses zero 59 times.

Sidef

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

Swift

Translation of: C
import Foundation

func mertensNumbers(max: Int) -> [Int] {
    var mertens = Array(repeating: 1, count: max + 1)
    for n in 2...max {
        for k in 2...n {
            mertens[n] -= mertens[n / k]
        }
    }
    return mertens
}

let max = 1000
let mertens = mertensNumbers(max: max)

let count = 200
let columns = 20
print("First \(count - 1) Mertens numbers:")
for i in 0..<count {
    if i % columns > 0 {
        print(" ", terminator: "")
    }
    print(i == 0 ? "  " : String(format: "%2d", mertens[i]), terminator: "")
    if (i + 1) % columns == 0 {
        print()
    }
}

var zero = 0, cross = 0, previous = 0
for i in 1...max {
    let m = mertens[i]
    if m == 0 {
        zero += 1
        if previous != 0 {
            cross += 1
        }
    }
    previous = m
}
print("M(n) is zero \(zero) times for 1 <= n <= \(max).")
print("M(n) crosses zero \(cross) times for 1 <= n <= \(max).")
Output:
First 199 Mertens numbers:
    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
M(n) is zero 92 times for 1 <= n <= 1000.
M(n) crosses zero 59 times for 1 <= n <= 1000.

V (Vlang)

Translation of: go
fn mertens(t int) ([]int, int, int) {
    mut to:=t
    if to < 1 {
        to = 1
    }
    mut merts := []int{len:to+1}
    mut primes := [2]
    mut sum := 0
    mut zeros := 0
    mut crosses := 0
    for i := 1; i <= to; i++ {
        mut j := i
        mut cp := 0      // counts prime factors
        mut spf := false // true if there is a square prime factor
        for p in 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 << 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
}
 
fn main() {
    merts, zeros, crosses := mertens(1000)
    println("Mertens sequence - First 199 terms:")
    for i := 0; i < 200; i++ {
        if i == 0 {
            print("    ")
            continue
        }
        if i%20 == 0 {
            println('')
        }
        print("  ${merts[i]:2}")
    }
    println("\n\nEquals zero $zeros times between 1 and 1000")
    println("\nCrosses zero $crosses times between 1 and 1000")
}
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

Wren

Library: Wren-fmt
Library: Wren-math
import "./fmt" for Fmt
import "./math" for Int

var isSquareFree = Fn.new { |n|
    var i = 2
    while (i * i <= n) {
        if (n%(i*i) == 0) return false
        i = (i > 2) ? i + 2 : i + 1
    }
    return true
}

var mu = Fn.new { |n|
    if (n < 1) Fiber.abort("Argument must be a positive integer")
    if (n == 1) return 1
    var sqFree = isSquareFree.call(n)
    var factors = Int.primeFactors(n)
    if (sqFree && factors.count % 2 == 0) return 1
    if (sqFree) return -1
    return 0
}

var M = Fn.new { |x| (1..x).reduce { |sum, n| sum + mu.call(n) } }

System.print("The first 199 Mertens numbers are:")
for (i in 0..9) {
    for (j in 0..19) {
        if (i == 0 && j == 0) {
            System.write("    ")
        } else {
            System.write("%(Fmt.dm(3, M.call(i*20 + j))) ")
        }
    }
    System.print()
}

// use the recurrence relationship for the last 2 parts rather than calling M directly
var count = 0
var mertens = M.call(1)
for (i in 2..1000) {
    mertens = mertens + mu.call(i)
    if (mertens == 0) count = count + 1
}
System.print("\nThe Mertens function is zero %(count) times in the range [1, 1000].")

count = 0
var prev = M.call(1)
for (i in 2..1000) {
    var next = prev + mu.call(i)
    if (next == 0 && prev != 0) count = count + 1
    prev = next
}
System.print("\nThe Mertens function crosses zero %(count) times in the range [1, 1000].")
Output:
The first 199 Mertens numbers are:
      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 

The Mertens function is zero 92 times in the range [1, 1000].

The Mertens function crosses zero 59 times in the range [1, 1000].

XPL0

Translation of: ALGOL W
integer M ( 1+1000 );
integer K, Zero, Cross, N;
begin   \compute values of the Mertens function
\Generate Mertens numbers
M( 1 ) := 1;
for N := 2 to 1000 do begin
    M( N ) := 1;
    for K := 2 to N do M( N ) := M( N ) - M( N / K )
    end;
\Print table
Text(0, "The first 99 Mertens numbers are:^m^j");
Text(0, "   " );
K := 9;
for N := 1 to 99 do begin
    Format(3, 0);
    RlOut(0, float(M(N)));
    K := K - 1;
    if K = 0 then begin
        K := 10;
        CrLf(0);
        end
    end;
\Calculate zeroes and crossings
Zero  := 0;
Cross := 0;
for N :=2 to 1000 do begin
    if M( N ) = 0 then begin
        Zero := Zero + 1;
        if M( N - 1 ) # 0 then Cross := Cross + 1
        end
    end;
Text(0, "M(N) is zero ");  IntOut(0, Zero);  Text(0, " times.^m^j" );
Text(0, "M(N) crosses zero ");  IntOut(0, Cross);  Text(0, " times.^m^j" );
end
Output:
The first 99 Mertens numbers are:
     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
M(N) is zero 92 times.
M(N) crosses zero 59 times.

zkl

fcn mertensW(n){
   [1..].tweak(fcn(n,pm){
      pm.incN(mobius(n));
      pm.value
   }.fp1(Ref(0)))
}
fcn mobius(n){
   pf:=primeFactors(n);
   sq:=pf.filter1('wrap(f){ (n % (f*f))==0 });  // False if square free
   if(sq==False){ if(pf.len().isEven) 1 else -1 }
   else 0
}
fcn primeFactors(n){  // Return a list of prime factors of n
   acc:=fcn(n,k,acc,maxD){  // k is 2,3,5,7,9,... not optimum
      if(n==1 or k>maxD) acc.close();
      else{
	 q,r:=n.divr(k);   // divr-->(quotient,remainder)
	 if(r==0) return(self.fcn(q,k,acc.write(k),q.toFloat().sqrt()));
	 return(self.fcn(n,k+1+k.isOdd,acc,maxD))  # both are tail recursion
      }
   }(n,2,Sink(List),n.toFloat().sqrt());
   m:=acc.reduce('*,1);      // mulitply factors
   if(n!=m) acc.append(n/m); // opps, missed last factor
   else acc;
}
mertensW().walk(199)
.pump(Console.println, T(Void.Read,19,False),
	fcn{ vm.arglist.pump(String,"%3d".fmt) });

println("\nIn the first 1,000 terms of the Mertens sequence there are:");
otm:=mertensW().pump(1_000,List);
otm.reduce(fcn(s,m){ s + (m==0) },0) : println(_," zeros");
otm.reduce(fcn(p,m,rs){ rs.incN(m==0 and p!=0); m }.fp2( s:=Ref(0) ));
println(s.value," zero crossings");
Output:
  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

In the first 1,000 terms of the Mertens sequence there are:
92 zeros
59 zero crossings