Euler's totient function can be formulated involving a product of the form $\prod\left(1-\frac{1}{p}\right)$. In particular, if every prime is included in the product, the product can be stated in a closed form $$\prod_{i=1}^n\left(1-\frac{1}{p_i}\right)=\frac{\phi(p_n\#)}{p_n\#}$$

With regard to a sieving method related to twin primes, I have found that by eliminating two numbers, a distance of $k$ on each side of each prime of the form $6k\pm 1$, candidates for twin primes can be identified. Thus, for each such prime considered, the sieve removes a fraction $\frac{2}{p}$ numbers, leaving $\left(1-\frac{2}{p}\right)$. Since each prime is coprime to every other prime, sieving over multiple primes would leave a fraction of numbers equal to $$\prod_{i=3}^n\left(1-\frac{2}{p_i}\right)$$

Note that since the primes being considered are of the form $6k\pm 1$, the index runs from $3$. I have thought about this a long time, but I can come up with no closed form for this product, and in particular no form related to the totient function.

My first question is: Can anyone provide a closed form for the product? Is a (hypothetical) closed form related in any way to the totient function?

Since there is no limit to the number of primes, the product $\prod\left(1-\frac{1}{p}\right)$, which in effect sieves primes from natural numbers, cannot exhaust every number, no matter how many primes are included in the product. The product I am interested in, $\prod\left(1-\frac{2}{p}\right)$, sieves numbers at a greater rate. A quick calculation reveals that the product with respect to the first $200$ primes of the form $6k\pm 1$ eliminate over $95\%$ of numbers, but the product decreases ever more slowly as the number of primes included in the product increases. Other than manual calculation, I do not know how to evaluate the product further.

My second question is: Does the limit of my product approach $0$ as the number of primes included in the product tends toward $\infty$?


1 Answer 1


The closest form we have is the one due to the generalization of given by Merten's Third Theorem.

$$\prod_{m<p\le x}(1-\frac{m}{p})\sim \frac{c(m)}{(\ln(x))^m}$$

where $m$ is a positive integer and $c(m)$ is a real number that depends on $m$.

For you case $m = 2$, you have the complete answer here. So clearly as $x \to \infty$ the product approaches $0$.


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