# Closed form for totient related product

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

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

$$\prod_{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 $$0$$.