Return to Answer

To answer the main question "Is this close enough to be of use in any practical application?" I do not believe so. However, one cannot say "no" with certainty, so it seems unlikely that you will get a concrete answer.

I wanted to add a derivation of a more precise formula of what you gave above. In particular, notice that that $\log\zeta(s)\approx P(s)$, so the $\zeta(s)$ and $\exp(-P(s))$ terms should cancel out in some way. Judging by your previous questions, this may not be new to you, and you likely used this same derivation to arrive at the formula given in your question.

Let $\mathcal{R}=\{{n:\ \Omega(n)=2\}}$ and consider the logarithm of the product above. This equals $$\sum_{n\in \mathcal{R}} -\log(1-n^{-s})=\sum_{n\in \mathcal{R}}\sum_{k=1}^{\infty} \frac{n^{-ks}}{k}.$$ Setting $P_2(s)=\sum_{n\in \mathcal{R}} n^{-s}$, the left hand side above equals then equals $\sum_{k=1}^{\infty} P_2(ks)/k.$ Now, $$P(s)^2+P(2s)=\sum_{p}\sum_{q}(pq)^{-s}+\sum_{p}p^{-2s}$$ $$=2\sum_{p}\sum_{q\geq p} p^{-s}q^{-s},$$ and so $$\sum_{k=1}^{\infty} P_2(ks)/k=\frac{1}{2}\sum_{k=1}^\infty \frac{P(ks)^2+P(2ks)}{k}.$$ Now, $\log\zeta(s)=\sum_{k=1}^\infty P(ks)/k$, so we find that

$$\prod_{n:\Omega(n)=2}\frac{1}{1-n^{-s}}=\sqrt{\zeta(2s)}\exp\left(\frac{1}{2}\sum_{k=1}^\infty P(ks)^2/k\right),$$

Now, in fact, $$\sum_{k=1}^\infty P(ks)^2/k=\sum_{q}\sum_{p}\sum_{k=1}^\infty\frac{q^{-ks}p^{-ks}}{k}$$ $$=\sum_{p}\sum_{q} -\log\left(1-(qp)^{-s}\right),$$ and hence

$$\prod_{n:\Omega(n)=2}\left(1-n^{-s}\right)^{-2}=\zeta(2s)\prod_{p}\prod_{q}\left(1-(pq)^{-s}\right)^{-1}$$ which I will rewrite as

$$\prod_{n:\Omega(n)=2}\left(1-n^{-s}\right)^{-2}=\zeta(2s)\mathcal{P}$$

where $\mathcal{P}$ is precisely the double product appearing in your previous question Zeta function double product.

To answer the main question "Is this close enough to be of use in any practical application?" I do not believe so. However, one cannot say "no" with certainty, so it seems unlikely that you will get a concrete answer.

I wanted to add a derivation of a more precise formula of what you gave above. In particular, notice that that $\log\zeta(s)\approx P(s)$, so the $\zeta(s)$ and $\exp(-P(s))$ terms should cancel out in some way. Judging by your previous questions, this may not be new to you, and you likely used this same derivation to arrive at the formula given in your question.

Let $\mathcal{R}=\{{n:\ \Omega(n)=2\}}$ and consider the logarithm of the product above. This equals $$\sum_{n\in \mathcal{R}} -\log(1-n^{-s})=\sum_{n\in \mathcal{R}}\sum_{k=1}^{\infty} \frac{n^{-ks}}{k}.$$ Setting $P_2(s)=\sum_{n\in \mathcal{R}} n^{-s}$, the left hand side above equals then equals $\sum_{k=1}^{\infty} P_2(ks)/k.$ Now, $$P(s)^2+P(2s)=\sum_{p}\sum_{q}(pq)^{-s}+\sum_{p}p^{-2s}$$ $$=2\sum_{p}\sum_{q\geq p} p^{-s}q^{-s},$$ and so $$\sum_{k=1}^{\infty} P_2(ks)/k=\frac{1}{2}\sum_{k=1}^\infty \frac{P(ks)^2+P(2ks)}{k}.$$ Now, $\log\zeta(s)=\sum_{k=1}^\infty P(ks)/k$, so we find that

$$\prod_{n:\Omega(n)=2}\frac{1}{1-n^{-s}}=\sqrt{\zeta(2s)}\exp\left(\frac{1}{2}\sum_{k=1}^\infty P(ks)^2/k\right),$$

Now, in fact, $$\sum_{k=1}^\infty P(ks)^2/k=\sum_{q}\sum_{p}\sum_{k=1}^\infty\frac{q^{-ks}p^{-ks}}{k}$$ $$=\sum_{p}\sum_{q} -\log\left(1-(qp)^{-s}\right),$$ and hence

$$\prod_{n:\Omega(n)=2}\left(1-n^{-s}\right)^{-2}=\zeta(2s)\prod_{p}\prod_{q}\left(1-(pq)^{-s}\right)^{-1}$$ which I will rewrite as

$$\prod_{n:\Omega(n)=2}\left(1-n^{-s}\right)^{-2}=\zeta(2s)\mathcal{P}$$

where $\mathcal{P}$ is precisely the double product appearing in your previous question Zeta function double product.

To answer the main question "Is this close enough to be of use in any practical application?" I do not believe so. However, one cannot say "no" with certainty, so it seems unlikely that you will get a concrete answer.

I wanted to add a derivation of a more precise formula of what you gave above. In particular, notice that that $\log\zeta(s)\approx P(s)$, so the $\zeta(s)$ and $\exp(-P(s)$ terms should cancel out in some way. Judging by your previous questions, this may not be new to you, and you likely used this same derivation to arrive at the formula given in your question.

Let $\mathcal{R}=\{{n:\ \Omega(n)=2\}}$ and consider the logarithm of the product above. This equals $$\sum_{n\in \mathcal{R}} -\log(1-n^{-s})=\sum_{n\in \mathcal{R}}\sum_{k=1}^{\infty} \frac{n^{-ks}}{k}.$$ Setting $P_2(s)=\sum_{n\in \mathcal{R}} n^{-s}$, the left hand side above equals then equals $\sum_{k=1}^{\infty} P_2(ks)/k.$ Now, $$P(s)^2+P(2s)=\sum_{p}\sum_{q}(pq)^{-s}+\sum_{p}p^{-2s}$$ $$=2\sum_{p}\sum_{q\geq p} p^{-s}q^{-s},$$ and so $$\sum_{k=1}^{\infty} P_2(ks)/k=\frac{1}{2}\sum_{k=1}^\infty \frac{P(ks)^2+P(2ks)}{k}.$$ Now, $\log\zeta(s)=\sum_{k=1}^\infty P(ks)/k$, so we find that

$$\prod_{n:\Omega(n)=2}\frac{1}{1-n^{-s}}=\sqrt{\zeta(2s)}\exp\left(\frac{1}{2}\sum_{k=1}^\infty P(ks)^2/k\right),$$

Now, in fact, $$\sum_{k=1}^\infty P(ks)^2/k=\sum_{q}\sum_{p}\sum_{k=1}^\infty\frac{q^{-ks}p^{-ks}}{k}$$ $$=\sum_{p}\sum_{q} -\log\left(1-(qp)^{-s}\right),$$ and hence

$$\prod_{n:\Omega(n)=2}\left(1-n^{-s}\right)^{-2}=\zeta(2s)\prod_{p}\prod_{q}\left(1-(pq)^{-s}\right)^{-1}$$ which I will rewrite as

$$\prod_{n:\Omega(n)=2}\frac{1}{1-n^{-s}}=\sqrt{\zeta(2s)}\mathcal{P}$$

where $\mathcal{P}$ is precisely the double product appearing in your previous question Zeta function double product.

To answer the main question "Is this close enough to be of use in any practical application?" I do not believe so. However, one cannot say "no" with certainty, so it seems unlikely that you will get a concrete answer.

I wanted to add a derivation of a more precise formula of what you gave above. In particular, notice that that $\log\zeta(s)\approx P(s)$, so the $\zeta(s)$ and $\exp(-P(s))$ terms should cancel out in some way. Judging by your previous questions, this may not be new to you, and you likely used this same derivation to arrive at the formula given in your question.

Let $\mathcal{R}=\{{n:\ \Omega(n)=2\}}$ and consider the logarithm of the product above. This equals $$\sum_{n\in \mathcal{R}} -\log(1-n^{-s})=\sum_{n\in \mathcal{R}}\sum_{k=1}^{\infty} \frac{n^{-ks}}{k}.$$ Setting $P_2(s)=\sum_{n\in \mathcal{R}} n^{-s}$, the left hand side above equals then equals $\sum_{k=1}^{\infty} P_2(ks)/k.$ Now, $$P(s)^2+P(2s)=\sum_{p}\sum_{q}(pq)^{-s}+\sum_{p}p^{-2s}$$ $$=2\sum_{p}\sum_{q\geq p} p^{-s}q^{-s},$$ and so $$\sum_{k=1}^{\infty} P_2(ks)/k=\frac{1}{2}\sum_{k=1}^\infty \frac{P(ks)^2+P(2ks)}{k}.$$ Now, $\log\zeta(s)=\sum_{k=1}^\infty P(ks)/k$, so we find that

$$\prod_{n:\Omega(n)=2}\frac{1}{1-n^{-s}}=\sqrt{\zeta(2s)}\exp\left(\frac{1}{2}\sum_{k=1}^\infty P(ks)^2/k\right),$$

Now, in fact, $$\sum_{k=1}^\infty P(ks)^2/k=\sum_{q}\sum_{p}\sum_{k=1}^\infty\frac{q^{-ks}p^{-ks}}{k}$$ $$=\sum_{p}\sum_{q} -\log\left(1-(qp)^{-s}\right),$$ and hence

$$\prod_{n:\Omega(n)=2}\left(1-n^{-s}\right)^{-2}=\zeta(2s)\prod_{p}\prod_{q}\left(1-(pq)^{-s}\right)^{-1}$$ which I will rewrite as

$$\prod_{n:\Omega(n)=2}\left(1-n^{-s}\right)^{-2}=\zeta(2s)\mathcal{P}$$

where $\mathcal{P}$ is precisely the double product appearing in your previous question Zeta function double product.

I wanted to provide a clean derivation of a more precise formula of what you gave above. In particular, notice that that $\log\zeta(s)\approx P(s)$, so the $\zeta(s)$ and $\exp(-P(s)$ terms should cancel out in some way.

Let $\mathcal{R}=\{{n:\ \Omega(n)=2\}}$ and consider the logarithm of the product above. This equals $$\sum_{n\in \mathcal{R}} -\log(1-n^{-s})=\sum_{n\in \mathcal{R}}\sum_{k=1}^{\infty} \frac{n^{-ks}}{k}.$$ Setting $P_2(s)=\sum_{n\in \mathcal{R}} n^{-s}$, the left hand side above equals then equals $\sum_{k=1}^{\infty} P_2(ks)/k.$ Now, $$P(s)^2+P(2s)=\sum_{p}\sum_{q}(pq)^{-s}+\sum_{p}p^{-2s}$$ $$=2\sum_{p}\sum_{q\geq p} p^{-s}q^{-s},$$ and so $$\sum_{k=1}^{\infty} P_2(ks)/k=\frac{1}{2}\sum_{k=1}^\infty \frac{P(ks)^2+P(2ks)}{k}.$$ Now, $\log\zeta(s)=\sum_{k=1}^\infty P(ks)/k$, so we find that

$$\prod_{n:\Omega(n)=2}\frac{1}{1-n^{-s}}=\zeta(2s)\exp\left(\frac{1}{2}\sum_{k=1}^\infty P(ks)^2/k\right),$$

Now, in fact, $$\sum_{k=1}^\infty P(ks)^2/k=\sum_{q}\sum_{p}\sum_{k=1}^\infty\frac{q^{-ks}p^{-ks}}{k}$$ $$=\sum_{p}\sum_{q} -\log\left(1-(qp)^{-s}\right),$$ and hence

$$\prod_{n:\Omega(n)=2}\frac{1}{1-n^{-s}}=\zeta(2s)\sqrt{\prod_{p}\prod_{q}\left(1-(pq)^{-s}\right)^{-1}}$$ which I will rewrite as

$$\prod_{n:\Omega(n)=2}\frac{1}{1-n^{-s}}=\zeta(2s)\mathcal{P}$$

where $\mathcal{P}$ is precisely the double product appearing in your previous question Zeta function double product.

To answer the main question "Is this close enough to be of use in any practical application?" my response is a pessimistic one, however this too vague of a question to give a concrete answer.

To answer the main question "Is this close enough to be of use in any practical application?" I do not believe so. However, one cannot say "no" with certainty, so it seems unlikely that you will get a concrete answer.

I wanted to add a derivation of a more precise formula of what you gave above. In particular, notice that that $\log\zeta(s)\approx P(s)$, so the $\zeta(s)$ and $\exp(-P(s)$ terms should cancel out in some way. Judging by your previous questions, this may not be new to you, and you likely used this same derivation to arrive at the formula given in your question.

Let $\mathcal{R}=\{{n:\ \Omega(n)=2\}}$ and consider the logarithm of the product above. This equals $$\sum_{n\in \mathcal{R}} -\log(1-n^{-s})=\sum_{n\in \mathcal{R}}\sum_{k=1}^{\infty} \frac{n^{-ks}}{k}.$$ Setting $P_2(s)=\sum_{n\in \mathcal{R}} n^{-s}$, the left hand side above equals then equals $\sum_{k=1}^{\infty} P_2(ks)/k.$ Now, $$P(s)^2+P(2s)=\sum_{p}\sum_{q}(pq)^{-s}+\sum_{p}p^{-2s}$$ $$=2\sum_{p}\sum_{q\geq p} p^{-s}q^{-s},$$ and so $$\sum_{k=1}^{\infty} P_2(ks)/k=\frac{1}{2}\sum_{k=1}^\infty \frac{P(ks)^2+P(2ks)}{k}.$$ Now, $\log\zeta(s)=\sum_{k=1}^\infty P(ks)/k$, so we find that

$$\prod_{n:\Omega(n)=2}\frac{1}{1-n^{-s}}=\sqrt{\zeta(2s)}\exp\left(\frac{1}{2}\sum_{k=1}^\infty P(ks)^2/k\right),$$

Now, in fact, $$\sum_{k=1}^\infty P(ks)^2/k=\sum_{q}\sum_{p}\sum_{k=1}^\infty\frac{q^{-ks}p^{-ks}}{k}$$ $$=\sum_{p}\sum_{q} -\log\left(1-(qp)^{-s}\right),$$ and hence

$$\prod_{n:\Omega(n)=2}\left(1-n^{-s}\right)^{-2}=\zeta(2s)\prod_{p}\prod_{q}\left(1-(pq)^{-s}\right)^{-1}$$ which I will rewrite as

$$\prod_{n:\Omega(n)=2}\frac{1}{1-n^{-s}}=\sqrt{\zeta(2s)}\mathcal{P}$$

where $\mathcal{P}$ is precisely the double product appearing in your previous question Zeta function double product.