11
$\begingroup$

Short version

If we define $$ f(n) = \prod_{2 < p \le n} \left( 1 - \frac{1}{p-1}\right) $$ where the product is over prime numbers $p$, then is it true that asymptotically $$ f(n) \sim \frac{c}{\log n} $$ and if so for what value of $c$? Or if not, can we get an asymptotic formula for $f(n)$?


Longer version

Define $f(n)$ as above; for example $$ \begin{align} f(3) &= \left(1 - \frac{1}{2}\right) &&= \frac{1}{2} \\ f(5) &= \left(1 - \frac{1}{2}\right) \left(1 - \frac{1}{4}\right) &&= \frac{3}{8} \\ f(7) &= \left(1 - \frac{1}{2}\right) \left(1 - \frac{1}{4}\right) \left(1 - \frac{1}{6}\right) &&= \frac{5}{16} \\ f(11) &= \left(1 - \frac{1}{2}\right) \left(1 - \frac{1}{4}\right) \left(1 - \frac{1}{6}\right) \left(1 - \frac{1}{10}\right) &&= \frac{9}{32} \\ \end{align} $$ and so on. In code (Python):

from fractions import Fraction
import math
def isprime(n):
    return n > 1 and all(n % d != 0 for d in range(2, min(n, int(math.sqrt(n))+5)))
n = 2
f = {}
cf = Fraction(1, 1)
while True:
    n += 1
    if not isprime(n): continue
    cf *= (1 - Fraction(1, n - 1))
    f[n] = cf
    print(n, f[n] * math.log(n))

If we let this run for up to $n = 100000$, we see output like:

(99881, 0.7410660117923158)
(99901, 0.7410714826009325)
(99907, 0.7410679310376648)
(99923, 0.7410708229998296)
(99929, 0.7410672721476895)
(99961, 0.7410804687613776)
(99971, 0.7410794950190189)
(99989, 0.7410836723109864)
(99991, 0.7410775482554816)

so $f(n) \log n$ does seem to approach a value around $0.74$.

Note that the third theorem of Mertens says that the similar product $$ \prod_{p \le n}\left(1 - \frac{1}{p}\right) \sim \frac{e^{-\gamma}}{\log n} $$ where $\gamma \approx 0.577$ is Euler's constant. This is my reason for trying to see whether $$ f(n) = \prod_{2 < p \le n} \left( 1 - \frac{1}{p-1}\right) \sim \frac{c}{\log n} $$ as well, for some other $c$.


What I've tried

(Obviously not successfully, so it may be best to ignore everything that follows.)

Taking logs, we can write $$ \log f(n) = \sum_{2 < p \le n} \log\left(1 - \frac{1}{p-1}\right). \tag{1}\label{one} $$ We can try to relate this to the theorem of Mertens that $$ \sum_{p \le x} \log\left(1 - \frac{1}{p}\right) = -\log\log x - \gamma + o(1) $$ or (peeling off the $p=2$ term) $$ \sum_{2 < p \le n} \log\left(1 - \frac{1}{p}\right) = -\log\log n - \gamma + \log 2 + o(1) \tag{2}\label{two} $$ To try to relate $\eqref{one}$ to $\eqref{two}$, we can write $\log\left(1 - \frac{1}{p-1}\right)$ in terms of $\log\left(1 - \frac{1}{p}\right)$: for $p>2$ we have $$ \begin{align} -\log\left(1 - \frac{1}{p-1}\right) &= \frac{1}{p} + \frac{3}{2p^2} + \frac{7}{3p^3} + \frac{15}{4p^4} + \frac{31}{5p^5} + \frac{63}{6p^6} + \frac{127}{7p^7} + \dots \\ &= -\log\left(1 - \frac1p\right) + \left(\frac{2}{2p^2} + \frac{6}{3p^3} + \frac{14}{4p^4} + \frac{30}{5p^5} + \frac{62}{6p^6} + \cdots \right) \end{align} $$ where the second term is $\log\left(\frac{(p - 1)^2}{p(p-2)}\right)$. So, summing the above over $2 < p \le n$, $$ \begin{align} -\log f(n) &= -\sum_{2 < p \le n}\log\left(1 - \frac{1}{p-1}\right) \\ &= -\sum_{2 < p \le n}{\log\left(1 - \frac1p\right)} + \sum_{2 < p \le n}\left(\frac{2}{2p^2} + \frac{6}{3p^3} + \frac{14}{4p^4} + \frac{30}{5p^5} + \frac{62}{6p^6} + \cdots \right) \tag{3}\label{three}\\ &\approx \log\log n + \gamma - \log 2 + \frac{2}{2}\left(P(2)-\frac{1}{2^2}\right) + \frac{6}{3}\left(P(3)-\frac{1}{2^3}\right) + \frac{14}{4}\left(P(4)-\frac{1}{2^4}\right) + \dots \end{align} $$ where $P(k) = \sum_{p} \frac{1}{p^k}$ denotes the prime zeta function. On the face of it, this seems like it may give an expression of the form $-\log f(n) = \log\log n + c + o(1)$ for some constant $c$, and therefore $\log f(n) = -c - \log\log n + o(1)$ or $$ f(n) \sim \frac{e^{-c}}{\log n} $$ which is what we wanted. The problem with this is that, in addition to the $\approx$ on the last line of $\eqref{three}$ being sloppy, it appears that in fact the subtracted term $\left(\frac{2}{2\cdot2^2} + \frac{6}{3\cdot2^3} + \frac{14}{4\cdot2^4} + \frac{30}{5\cdot2^5} + \frac{62}{6\cdot2^6} + \cdots \right)$ diverges! So it's not clear whether $\eqref{three}$ is meaningful in any way (and even if it were, whether this is a “proper” way to express the constant $c$).

Update: On actually trying the final expression of $\eqref{three}$, it seems to match the numerical data. The following Sage program (using mpmath.primezeta, the equivalent of PrimeZetaP in Mathematica):

import mpmath
mpmath.mp.dps = 25 # Set precision to 25 decimal digits
ans = mpmath.euler - mpmath.log(2)
for k in range(2, 100):
    ans += (2**k - 2) * (mpmath.primezeta(k) - 1/2**k) / k
print(ans)
print(mpmath.exp(-ans))

prints (compare the second output with the output from an earlier program above):

0.2993387828283008984224987
0.7413082243919210826540034

This is quite persuasive, so the main thing that's missing is a more rigorous proof of $\eqref{three}$ (I guess we need to say something about the rate of convergence, to justify the “$\approx$”), and (if it exists) a more concise expression for the constant (something that isn't itself an infinite sum). Or of course, a completely different alternative solution.

$\endgroup$
  • $\begingroup$ What happens if you use the expansion $$\sum_{2<p \le n} \log\left(1 - \frac{1}{p}\right) =\log2+\sum_{p \le n} \log\left(1 - \frac{1}{p}\right) = -\log\log n - \gamma+\log2 + o(1)\ ?$$ $\endgroup$ – Did May 21 '18 at 9:24
  • $\begingroup$ For each $k>1$, $$\sum_{p>2}\frac1{p^k}=P(k)-\frac1{2^k}$$ $\endgroup$ – Did May 21 '18 at 9:33
  • $\begingroup$ @Did Thanks for your comments; I've mentioned them in the question — the problem is with the second term, where the $p=2$ term that is to be pulled out diverges (as far as I can tell). The funny thing is that with some combination of signs I was able to get something close to what seems a numerically correct value, but I was not able to justify the sign error… maybe it's just very late at night for me. :-) Thanks for any help… $\endgroup$ – ShreevatsaR May 21 '18 at 9:39
5
+300
$\begingroup$

We can write that $$ \bbox[lightyellow] { \prod\limits_{2\, \le \,p\, \le \,n} {\left( {1 - {1 \over p}} \right)} \le \prod\limits_{2\, \le \,p\, \le \,n - 1} {\left( {1 - {1 \over p}} \right)} < \prod\limits_{3\, \le \,p\, \le \,n} {\left( {1 - {1 \over {p - 1}}} \right)} < \prod\limits_{3\, \le \,p\, \le \,n} {\left( {1 - {1 \over p}} \right)} = 2\prod\limits_{2\, \le \,p\, \le \,n} {\left( {1 - {1 \over p}} \right)} }$$

Since all the products are non-increasing in $n$, and since for the third Merten's Theorem as you cited, it is $$ \prod\limits_{\,\left( {2\, \le } \right)\,p\, \le \,n} {\left( {1 - {1 \over p}} \right)} \sim {{e^{\, - \gamma } } \over {\ln n}} $$ then we can tell that $$ \bbox[lightyellow] { \prod\limits_{3\, \le \,p\, \le \,n} {\left( {1 - {1 \over {p - 1}}} \right)} \;{\sim}\;{{c\;e^{\, - \gamma } } \over {\ln n}}\quad \left| {\;1 < c < 2} \right. }$$

Concerning the actual value of $c$, the ratio $$ {{\prod\limits_{3\, \le \,p\, \le \,n} {\left( {1 - {1 \over {p - 1}}} \right)} } \over {\prod\limits_{3\, \le \,p\, \le \,n} {\left( {1 - {1 \over p}} \right)} }} = \prod\limits_{3\, \le \,p\, \le \,n} {\left( {{{p\left( {p - 2} \right)} \over {\left( {p - 1} \right)^2 }}} \right)} = \prod\limits_{3\, \le \,p\, \le \,n} {\left( {1 - {1 \over {\left( {p - 1} \right)^2 }}} \right)} = \prod\limits_{3\, \le \,p\, \le \,n} {\left( {1 - {1 \over {\left( {p - 1} \right)}}} \right)} \prod\limits_{3\, \le \,p\, \le \,n} {\left( {1 + {1 \over {\left( {p - 1} \right)}}} \right)} $$ tells us that $$ \bbox[lightyellow] { c = 2\,\prod\limits_{3\, \le \,p\,} {\left( {1 - {1 \over {\left( {p - 1} \right)^2 }}} \right)} = 1.3203236... }$$ that is

$ 2$ times the Twin Primes Constant.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.