Meissel–Mertens constant

Meissel–Mertens constant
You are encouraged to solve this task according to the task description, using any language you may know.

Task

Calculate Meissel–Mertens constant up to a precision your language can handle.

Motivation

Analogous to Euler's constant, which is important in determining the sum of reciprocal natural numbers, Meissel-Mertens' constant is important in calculating the sum of reciprocal primes.

Example

We consider the finite sum of reciprocal natural numbers:

1 + 1/2 + 1/3 + 1/4 + 1/5 ... 1/n

this sum can be well approximated with:

log(n) + E

where E denotes Euler's constant: 0.57721...

log(n) denotes the natural logarithm of n.

Now consider the finite sum of reciprocal primes:

1/2 + 1/3 + 1/5 + 1/7 + 1/11 ... 1/p

this sum can be well approximated with:

log( log(p) ) + M

where M denotes Meissel-Mertens constant: 0.26149...

See

•   Details in the Wikipedia article:  Meissel–Mertens constant

ALGOL 68

Sieving odd primes only, not using extended precision.
Note that to run this with Algol 68G, a large heap size must be specified, e.g.: by adding -heap 256M to the command line.

BEGIN # compute an approximation to the Meissel-Mertens constant             #
    # construct a sieve of odd primes                                        #
    [ 0 : 10 000 000 ]BOOL primes;
    BEGIN
        FOR i TO UPB primes DO primes[ i ] := TRUE  OD;
        INT ip := 1;
        FOR i WHILE i + ( ip +:= 2 ) <= UPB primes DO
            IF primes[ i ] THEN
                FOR s FROM i + ip BY ip TO UPB primes DO primes[ s ] := FALSE OD
            FI
        OD
    END;
    # sum the reciprocals of the primes                                      #
    INT       p count := 1;
    INT       last p  := 0;
    LONG REAL sum     := long ln( 0.5 ) + 0.5;
    INT       p       := 1;
    INT       p10     := 10;
    # Euler's constant from the wikipedia, truncated for LONG REAL           #
    LONG REAL eulers constant = 0.5772156649015328606 # 0651209008240243104215933593992 #;
    FOR i TO UPB primes DO
        p +:= 2;
        IF primes[ i ] THEN
            LONG REAL rp = 1 / LENG p;
            sum     +:= long ln( 1 - rp ) + rp;
            p count +:= 1;
            last p   := p;
            IF p count = p10 THEN
                print( ( "after ", whole( p count, -8 ), " primes, the approximation is: "
                       , fixed( sum + eulers constant, -14, 12 )
                       , ", last prime considered: ", whole( last p, 0 )
                       , newline
                       )
                     );
                p10 := IF p10 < 1 000 000 THEN p10 * 10 ELSE p10 + 1 000 000 FI
            FI
        FI   
    OD;
    print( ( "after ", whole( p count, -8 ), " primes, the approximation is: "
           , fixed( sum + eulers constant, -14, 12 )
           , ", last prime considered: ", whole( last p, 0 )
           , newline
           )
         )
END

Output:

after       10 primes, the approximation is: 0.265160104017, last prime considered: 29
after      100 primes, the approximation is: 0.261624626173, last prime considered: 541
after     1000 primes, the approximation is: 0.261503563617, last prime considered: 7919
after    10000 primes, the approximation is: 0.261497594498, last prime considered: 104729
after   100000 primes, the approximation is: 0.261497238454, last prime considered: 1299709
after  1000000 primes, the approximation is: 0.261497214692, last prime considered: 15485863
after  1270607 primes, the approximation is: 0.261497214255, last prime considered: 19999999

BASIC

BASIC256

Translation of: FreeBASIC

Euler = 0.5772156649
m = 0
for x = 2 to 1e6    # more prime numbers do not add more precision
	if isPrime(x) then m += log(1-(1/x)) + (1/x)
next x
print "MM = "; Euler + m
print Euler
end

function isPrime(v)
	if v < 2 then return False
	if v mod 2 = 0 then return v = 2
	if v mod 3 = 0 then return v = 3
	d = 5
	while d * d <= v
		if v mod d = 0 then return False else d += 2
	end while
	return True
end function

Output:

MM = 0.26149724673

Run BASIC

Translation of: FreeBASIC

function isPrime(n)
if n < 2       then isPrime = 0 : goto [exit]
if n = 2       then isPrime = 1 : goto [exit]
if n mod 2 = 0 then isPrime = 0 : goto [exit]
isPrime = 1
for i = 3 to int(n^.5) step 2
 if n mod i = 0 then isPrime = 0 : goto [exit]
next i
[exit]
end function

e = 0.5772156

for x = 2 to 100000    ' more prime numbers do not add more precision
    if isPrime(x) then m = m + log(1-(1/x)) + (1/x)
next x
print "MM = "; e + m

Output:

MM = 0.261274

Yabasic

Translation of: FreeBASIC

e = 0.5772156

for x = 2 to 1e6    // more prime numbers do not add more precision
	if isPrime(x)  m = m + log(1-(1/x)) + (1/x)
next x
print "MM = ", e + m
end

sub isPrime(v)
    if v < 2  return False
    if mod(v, 2) = 0  return v = 2
    if mod(v, 3) = 0  return v = 3
    d = 5
    while d * d <= v
        if mod(v, d) = 0 then return False else d = d + 2 : fi
    wend
    return True
end sub

Output:

MM = 0.261497

FreeBASIC

On my CPU i5 it takes about 3.5 minutes

'#include "isprime.bas"

Const As Double Euler = 0.57721566490153286

Dim As Double m = 0
For x As Ulongint = 2 To 1e8
    If isPrime(x) Then m += Log(1-(1/x)) + (1/x)
Next x
Print "MM ="; Euler + m
Sleep

Output:

MM = 0.2614972131104154

Go

Translation of: Wren

Library: Go-rcu

package main

import (
    "fmt"
    "math"
    "rcu"
)

func contains(a []int, f int) bool {
    for _, e := range a {
        if e == f {
            return true
        }
    }
    return false
}

func main() {
    const euler = 0.57721566490153286
    primes := rcu.Primes(1 << 31)
    pc := len(primes)
    sum := 0.0
    fmt.Println("Primes added         M")
    fmt.Println("------------  --------------")
    for i, p := range primes {
        rp := 1.0 / float64(p)
        sum += math.Log(1.0-rp) + rp
        c := i + 1
        if (c%1e7) == 0 || c == pc {
            fmt.Printf("%11s   %0.12f\n", rcu.Commatize(c), sum+euler)
        }
    }
}

Output:

Primes added         M
------------  --------------
 10,000,000   0.261497212987
 20,000,000   0.261497212912
 30,000,000   0.261497212889
 40,000,000   0.261497212878
 50,000,000   0.261497212871
 60,000,000   0.261497212867
 70,000,000   0.261497212864
 80,000,000   0.261497212862
 90,000,000   0.261497212861
100,000,000   0.261497212859
105,097,565   0.261497212858

J

   Euler=: 0.57721566490153286
   MM=: {{ Euler + +/ (+ ^.@-.)@% p:i. y }}
   0j12 ": MM 1e7
0.261497212987

Julia

Off by one in the 11th digit after 10^8 primes.

using Base.MathConstants  # sets constant γ = 0.5772156649015...
using Primes

""" Approximate the Meissel-Mertons constant. """
function meissel_mertens(iterations = 100_000_000)
    return mapreduce(p ->(d = 1/p; log(1 - d) + d), +, primes(prime(iterations))) + γ
end

@show meissel_mertens(100_000_000) # meissel_mertens(100000000) = 0.2614972128591237

PARI/GP

Summation method

{
MM(t)=
  my(s=0);
  forprime(p = 2, t,
    s += log(1.-1./p)+1./p
  );
  Euler+s
};

Output:

Running 10^9 summations to get 9 valid digits.

? \p10
   realprecision = 19 significant digits (10 digits displayed)
? MM(1e9)
?
%1 = 0.2614972129
?
? ##
  ***   last result: cpu time 1min, 18,085 ms, real time 1min, 18,094 ms.
?

Analytic method

The Analytic method requires high precision calculation of Riemann zeta function.

{
Meissel_Mertens(d)=
  default(realprecision, d);
  my(prec = default(realprecision), z = 0, y = 0, q);
  forprime(p = 2 , 7, 
    z += log(1.-1./p)+1./p
  );
  for(k = 2, prec,
    q = 1;
    forprime(p = 2, 7, 
      q *= 1.-p^-k
    );
    y += moebius(k)*log(zeta(k)*q)/k
  );
  Euler+z+y
};

Output:

1000 valid digits.

? Meissel_Mertens(1001)
%1 = 0.26149721284764278375542683860869585905156664826119920619206421392492451089736820971414263143424665105161772887648602199778339032427004442454348740197238640666194955709392581712774774211985258807266272064144464232590023543105177232173925663229980314763831623758149059290382284758265972363422015971458785446941586825460538918007031787714156680620570605257601785334398970354507934530971953511716888598019955346947142883673537117910619342522616975101911159537244599605203558051780574237201332999961769676911386909654186249097435916294862238555389898241954857937738258646582212506260380084370067541379219020626760709633535981989783010762417792511961619355361391684002933280522289185167238258837930443067100391254985761418536020400457460311825670423438456551983202200477824746954606715454777572171338072595463648319687279859427306787306509669454587505942593547068846408425666008833035029366514525328713339609172639368543886291288200447611698748441593459920236225093315001729474600911978170842383659092665509
? ##
  ***   last result: cpu time 283 ms, real time 284 ms.

Phix

Converges very slowly, I started running out of memory (from generating so many primes) before it showed any signs of getting stuck.

with javascript_semantics -- (but perhaps a bit too slow)
constant mmc = 0.2614972128476427837554268386086958590516,
       smmc = "0.2614972128476427837554268386086958590516"
atom t = 0.57721566490153286, dpa = 1, p10=1
integer pn = 1, adp = 0
string st = "0", fmt = "%.0f"
printf(1,"Primes added         M\n")
printf(1,"------------  --------------\n")
while adp<=10 do
    atom rp = 1/get_prime(pn)
    t += log(1-rp) + rp
    if (t-mmc)<dpa then
        string ft = sprintf(fmt,trunc(t*p10)/p10) -- (as below)
        if ft=st then
            --
            -- We have to artifically calculate a "truncated t", aka tt, 
            -- to prevent say 0.2..299[>5..] being automatically rounded 
            -- by printf() to 0.2..300, otherwise it just "looks wrong".
            --
            atom tt = trunc(t*1e12)/1e12
            printf(1,"%,11d   %0.12f (accurate to %d d.p.)\n", {pn, tt, adp})
            adp += 1
            fmt = sprintf("%%.%df",adp)
            st = smmc[1..2+adp]
            dpa /= 10
            p10 *= 10
        end if
    end if
    pn += 1
end while
printf(1,"(actual value 0.26149721284764278375542683860869)\n")

Output:

(Couldn't be bothered with making it an inner loop to upgrade the "accurate to 4dp" to 5dp)

Primes added         M
------------  --------------
          1   0.384068484341 (accurate to 0 d.p.)
          3   0.288793158252 (accurate to 1 d.p.)
          6   0.269978901636 (accurate to 2 d.p.)
         38   0.261990332075 (accurate to 3 d.p.)
      1,940   0.261499998883 (accurate to 4 d.p.)
      1,941   0.261499997116 (accurate to 5 d.p.)
      5,471   0.261497999951 (accurate to 6 d.p.)
     34,891   0.261497299999 (accurate to 7 d.p.)
    303,447   0.261497219999 (accurate to 8 d.p.)
  9,246,426   0.261497212999 (accurate to 9 d.p.)
 24,304,615   0.261497212899 (accurate to 10 d.p.)
(actual value 0.26149721284764278375542683860869)

analytical/gmp

Translation of: PARI/GP

without js -- no mpfr_zeta[_ui]() in mpfr.js, as yet, or mpfr_log() or mpfr_const_euler(), for that matter
include mpfr.e
function moebius(integer n)
    if n=1 then return 1 end if
    sequence f = prime_factors(n,true)
    for i=2 to length(f) do
        if f[i] = f[i-1] then return 0 end if
    end for
    return iff(odd(length(f))?-1:+1)
end function

function Meissel_Mertens(integer d)
    mpfr_set_default_precision(-d) -- (d decimal places)
    mpfr {res,rp,z,y,q} = mpfr_inits(5)
    for p in {2,3,5,7} do
        -- z += log(1-1/p)+1/p
        mpfr_set_si(rp,1)
        mpfr_div_si(rp,rp,p)
        mpfr_add(z,z,rp)
        mpfr_si_sub(rp,1,rp)
        mpfr_log(rp,rp)
        mpfr_add(z,z,rp)
    end for
    for k=2 to d do -- (see note)
        integer m = moebius(k)
        if m then
            mpfr_set_si(q,1)
            for p in {2,3,5,7} do
                -- q *= 1-power(p,-k)
                mpfr_set_si(rp,p)
                mpfr_pow_si(rp,rp,-k)
                mpfr_si_sub(rp,1,rp)
                mpfr_mul(q,q,rp)
            end for
            -- y += moebius(k)*log(zeta(k)*q)/k
            mpfr_zeta_ui(rp,k)
            mpfr_mul(rp,rp,q)
            mpfr_log(rp,rp)
            mpfr_div_si(rp,rp,k)
            mpfr_mul_si(rp,rp,m)
            mpfr_add(y,y,rp)
        end if
    end for
    -- res := EULER+z+y
    mpfr_const_euler(res)
    mpfr_add(z,z,y)
    mpfr_add(res,res,z)
    return res
end function

mpfr m = Meissel_Mertens(1001)
?shorten(mpfr_get_str(m,10,1001))

Output:

Agrees with PARI/GP except for the very last digit being 8 instead of 9. Note that playing with precision and/or iterations (which seems to differ quite wildly) gave utterly incorrect results... not that I actually comprehend what the algorithm is doing. The 1,003 includes 2 from the "0.", fairly obviously.

"0.261497212847642783...70842383659092665508 (1,003 digits)"

Wren

Summation method

Library: Wren-math

Library: Wren-fmt

It will be seen that this is converging to the correct answer though we'd need to add a lot more primes to obtain a valid 11th digit.

import "./math" for Int
import "./fmt" for Fmt

var euler = 0.57721566490153286
var primes = Int.primeSieve(2.pow(31))
var pc = primes.count
var sum = 0
var c = 0
System.print("Primes added         M")
System.print("------------  --------------")
for (p in primes) {
    var rp = 1/p
    sum = (1-rp).log + rp + sum
    c = c + 1
    if ((c % 1e7) == 0 || c == pc) Fmt.print("$,11d   $0.12f", c, sum + euler)
}

Output:

Primes added         M
------------  --------------
 10,000,000   0.261497212987
 20,000,000   0.261497212912
 30,000,000   0.261497212889
 40,000,000   0.261497212878
 50,000,000   0.261497212871
 60,000,000   0.261497212867
 70,000,000   0.261497212864
 80,000,000   0.261497212862
 90,000,000   0.261497212861
100,000,000   0.261497212859
105,097,565   0.261497212858

Analytic method

Translation of: PARI/GP

Library: Wren-gmp

This agrees with the Phix entry that the 1,000th digit after the decimal point is '8' after rounding (according to OEIS it's actually '7' but the next one is '7' also).

import "./gmp" for Mpf
import "./math" for Int
import "./fmt" for Fmt

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 meisselMertens = Fn.new { |d|
    Mpf.defaultPrec = d
    var z = Mpf.zero
    var y = Mpf.zero
    var r = Mpf.new()
    var q = Mpf.new()
    var t = Mpf.new()
    var m = Mpf.new()
    for (p in [2, 3, 5, 7]) {
        r.setUi(p).inv
        t.uiSub(1, r).log.add(r)
        z.add(t, z)
    }
    for (k in 2..d) {
        q.setUi(1)
        for (p in [2, 3, 5, 7]) {
            r.setUi(p).inv
            t.uiSub(1, r.pow(k))
            q.mul(t)
        }
        m.setSi(mu.call(k))
        t.zetaUi(k).mul(q).log.mul(m).div(k)
        y.add(t, y)
    }
    return Mpf.euler.add(z).add(y)
}

Fmt.print("$20a", meisselMertens.call(3300).toString(1001))

Output:

0.261497212847642783...70842383659092665508