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
Go
Unlike the Wren example, slowly converging towards the correct answer.
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
0j12 ": Euler + +/(+ ^.@-.)@% p: i. 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
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
Wren's only native number type is a 64 bit float (15 or 16 digits accuracy) and it appears from the following results that, no matter how many primes we use, we're not going to be able to improve on 8 valid digits for M with this particular number type.
However, it's curious that the other examples, which are using (or appear to be using) a 64 bit float type, are able to get much closer to the correct answer. It's clearly something that needs to be investigated.
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 = sum + (1-rp).log + rp
c = c + 1
if ((c % 1e7) == 0 || c == pc) Fmt.print("$,11d $0.12f", c, sum + euler)
}
- Output:
Primes added M ------------ -------------- 10,000,000 0.261497213008 20,000,000 0.261497213008 30,000,000 0.261497213009 40,000,000 0.261497213008 50,000,000 0.261497213009 60,000,000 0.261497213009 70,000,000 0.261497213009 80,000,000 0.261497213009 90,000,000 0.261497213009 100,000,000 0.261497213009 105,097,565 0.261497213009